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2 Way to Success ⓬ MATHS PUBLIC EXAM SPECIAL GUIDE Prepared by Mr. K. Dinesh M.Sc., M.Phil., P.G.D.C.A., Ph.D., For subject related clarifications Mail us: & Call us : (doubts) (orders) Visit us: You can download free study materials from our website

3 Government Question Pattern Section Qn. Given To be Question Type No Questions written Marks Section A 40 Choose the best answer ( mark questions) Section B Mark questions Mark Compulsory question ( either or form) 6 Section C Mark questions Mark Compulsory question ( either or form) Blue Print Chap. One marks Six marks Ten marks Chapter No No. of Qn Marks No. of Qn Marks No. of Qn Marks Application of Matrices and Determinants Vector Algebra Complex Numbers Analytical Geometry Differential Calculus: Applications I Differential Calculus: Applications II Integral Calculus : Applications Differential Equations Discrete Mathematics Probability Distributions Total For Single Copy by VPP : Send your address through SMS to For Bulk Orders: Please contact your district Co-ordinators from the list given in the last page of this book or Contact following phone Numbers , , Note : At most care is taken to fulfill your requirements without mistakes. If you find any mistakes, kindly inform us by to ways00@gmail.com. Corrections will be updated then and there in our website and that will be carried out in our further edition.

4 Dear students Greetings You have Way to Success Easy way Maths guide now in your hands. This guide is not written like other usual guides. There is a huge difference between this guide and other guides. This guide will help the students understand the concepts clearly and make them to score very high marks. This guide gives you the exact answers for textbook questions, previous years Govt. Examination questions and additional questions. Tips to slow learners: Believe in yourself that you can get good marks if you do practice and hard work. Study the following chapters should well practiced One mark 7 Book back questions (Volume I questions, Volume II 50 questions) Six Marks.Applications of Matrices and Determinants,. Complex Numbers 9. Discrete Mathematics Ten Marks. Vector Algebra,. Complex numbers, 4. Analytical Geometry, 6. Differential Calculus Applications II, 8. Differential Equations (Applications Question 0 only) 9. Discrete Mathematics Try to follow the above tips and score more in MATHS. Wish you all the best. CONTENT Chapter Page Number Mark Book Back Questions (All Chapters) 4 6 Mark Application of Matrices and Determinants 5 6 Mark Complex Numbers 6 Mark Discrete Mathematics 9 0 Mark Vector Algebra 45 0 Mark Complex Numbers 54 0 Mark Analytical Geometry 58 0 Mark Differential Calculus: Applications II 76 0 Mark Differential Equations 80 0 Mark Discrete Mathematics 8 - Marked sums are creative questions asked in the Public exams

5 th Maths Public Exam Special Guide Way to Success Section A ( Marks). APPLICATIONS OF MATRICES AND DETERMINANTS. The rank of the matrix 4 is () () () (4) 4 (OCT-). The rank of the diagonal matrix (MAR-0, JUN-6) ()0 () () (4)5. If A 0, then the rank of AA T is (OCT-06,MAR-08,JUN-09,MAR-, OCT-5) () () () (4)0 4. If A, then the rank of AA T is (MAR-09, JUN-,OCT-4,6) () ()0 () (4) λ 0 5. If the rank of the matrix 0 λ is, then λ is (JUN-08,OCT-09,JUN-,OCT-5) 0 λ () () () (4) any real number 6. If A is a scalar matrix with scalar k 0, of order, then A is (OCT-07,MAR-08,JUN-08,OCT-08,MAR-0,MAR-4,JUN-4,OCT-4) () k I () k () I k (4)kI 7. If the matrix k has an inverse then the values of k 4 5 (OCT-06,OCT-09,MAR-,OCT-,MAR-5) () k is any real number () k 4 () k 4 (4) k 4 8. If A, then (adj A)A (MAR-07, JUN-07,OCT-08,OCT-0,OCT-,JUN-5, OCT-5, MAR-6) 4 () () 0 0 () If A is a square matrix of order n, then adj A is (MAR-06,JUN-06,MAR-,JUN-4) () A () A n () A n (4) A 0. The inverse of the matrix () is () () (4) (4) (JUN-,JUN-,JUN-6) If A is a matrix of order, then det (ka) (OCT-06, OCT-07,JUN-09,JUN-0,JUN-, MAR-6, 7) ()k det(a) () k det(a) () k det(a) (4) det(a) ways00@gmail.com

6 th Maths Public Exam Special Guide Way to Success. If I is the unit matrix of order n, where k 0 is a constant, then adj(ki ) (OCT-0,MAR-,JUN-,MAR-,OCT-,JUN-5,JUN-6) () k n (adj I) () k(adj I) () k (adj I) (4) k n (adj I). If A and B are any two matrices such that ABO and A is non-singular, then (MAR-07,OCT-07,MAR-08,MAR-09,MAR-,MAR-,OCT-4,MAR-5) ()B0 ()B is singular () B is non-singular (4)BA 4. If A , then A is (JUN-07,JUN-09,JUN-0, OCT-6, MAR-7) () () 0 60 () (4) Inverse of is (MAR-06, OCT-07, OCT-08,OCT-09,OCT-,OCT-, MAR-4,JUN-4) () 5 5 () 5 () 5 (4) 5 6. In a system of linear non-homogeneous equation with three unknowns, if 0, x 0, y 0 and z 0, then the system has (JUN-06,JUN-07,MAR-0,MAR-,JUN-) () unique solution () two solutions () infinitely many solutions (4 ) no solutions 7. The system of equations ax + y + z 0; x + by + z 0; x + y + cz 0 has a non-trivial solution then + + a b c (MAR-07,MAR-09,JUN-0,OCT-,MAR-4,JUN-5,MAR-6) () () () (4)0 8. If ae x + be y c; pe x + qe y d and a b p q, c b d q, a c p d then the value of (x, y) is (JUN-08,OCT-0,MAR-,OCT-6,MAR-7) (), () log, log () log, log (4) log, log 9. If the equation x + y + z l, x y + z m, x + y z n, such that l + m + n 0, then the system has (MAR-06,JUN-06,JUN-,OCT-,JUN-,OCT-,MAR-5) () a non- zero unique solution () trivial solutions () infinitely many solution (4) no solution. VECTOR ALGEBRA. If a is a non-zero vector and m is a non-zero scalar then ma is a unit vector if (MAR-07,OCT-,MAR-4,MAR-6) () m ± () a m () a m (4) a. If a and b are two unit vectors and θ is the angle between them, then (a + b) is a unit vector if (OCT-06,OCT-07,MAR-08,OCT-09,JUN-0,MAR-,MAR-,JUN-5, OCT-5) () θ π () θ π 4 () θ π (4) θ π. If a and b include an angle 0 and their magnitude are and then a.b is equal to (JUN-07,JUN-09,JUN-4) () () () (4) ways00@gmail.com

7 th Maths Public Exam Special Guide Way to Success 4. If u a b c + b c a + c a b, then (JUN-06,MAR-07,JUN-08,OCT-08,MAR-,MAR-5) () u is a unit vector () u a+b + c () u 0 (4)u 0 5. If a+b + c0, a, b 4, c 5 then the angle between a and b is (JUN-,MAR-,MAR-4, OCT-6) () π 6 () π () 5π (4) π 6. The vectors i+j+4k and ai + bj + ck are perpendicular when (JUN-07,JUN-,JUN-4) () a,b,c 4 ()a4,b4,c5 () a4,b4,c 5 (4)a-,b,c4 7. The area of the parallelogram having a diagonal i+j k and a side i j+4k is (JUN-08,OCT-09,MAR-0,JUN-,OCT-,OCT-4,MAR-6) ()0 ()6 0 () 0 (4) 0 8. If a + b a b then. (MAR-06,JUN-06,MAR-07,JUN-09,MAR-5) () a is parallel to b () a is perpendicular to b () a b (4) a and b are unit vectors 9. If p, q and p + q are vectors of magnitude λ, then the magnitude of p q is (OCT-08,MAR-09,OCT-) () λ () λ () λ (4) 0. If a b c + b c a + c a b x y then (JUN-,JUN-,OCT-5) () x 0 () y 0 () x and y are parallel (4) x 0 or y 0 or x and y are parallel. If PR i + j + k, QS i + j + k then the area of the quadrilateral PQRS is (OCT-06,OCT-07,OCT-0,MAR-,MAR-5, MAR-7) ()5 ()0 (). The projection of OP on a unit vector OQ equals thrice the area of parallelogram OPRQ. Then POQ is ( JUN-06,MAR-09,JUN-0, MAR-, JUN-6) ()tan ()cos 0 5 () sin 0 (4) (4) sin. If the projection of a on b and projection of b on a are equal then the angle between a + band a b is. (JUN-07,OCT-09,MAR-,OCT-,JUN-4) () π () π () π 4 (4) π 4. If a b c a b c for non-coplanar vectors a, b, c then (MAR-09,OCT-) () a parallel to b () b parallel to c ()c parallel to a (4) a+b + c0 5. If a line makes 45, 60 with positive direction of axes x and y then the angle it makes with z axis is (JUN-,MAR-4,OCT-4, MAR-7) () 0 () 90 () 45 (4) If a b, b c, c a 64 then a, b, c is (MAR-06,MAR-08,OCT-08,MAR-, OCT-6) () () 8 () 8 (4)0 7. If a + b, b + c, c + a 8 then a, b, c is (JUN-07,MAR-0,MAR-,OCT-,MAR-6) () 4 () 6 () (4)-4 ways00@gmail.com

8 th Maths Public Exam Special Guide Way to Success 8. The value i + j, j + k, k + i is equal to (MAR-09,MAR-5, OCT-5, MAR- 7) ()0 () () (4)4 9. The shortest distance of the point (,0,) from the plane r. i j + 4k 6 is (MAR-07,MAR-08,JUN-09,OCT-09,MAR-,JUN-5) () 6 () 6 () (4) 6 0. The vector a b (c d) is (OCT-,OCT-5) () perpendicular to a, b, c and d () parallel to the vectors a b and (c d) () parallel to the line of intersection of the plane containing a and b and the plane containing c and d (4) perpendicular to the line of intersection of the plane containing a and b and the plane containing c and d. If a, b, c are a right handed triad of mutually perpendicular vectors of magnitude a, b, c then the value of a, b, c is (JUN-08,JUN-, JUN-6) ()a b c () 0 () abc (4) abc. If a, b, c are non-coplanar and a b, b c, c a a + b, b + c, c + a then a, b, c is (OCT-,OCT-4) () () () (4)0. r si + tj is the equation of (JUN-08,OCT-0) () a straight line joining the points i and j () xoy plane () yoz plane (4) zox plane 4. If the magnitude of moment about the point j + k of a force i + aj k acting through the point i + j is 8 then the value of a is (OCT-0,JUN-,OCT-,6) () () () (4)4 5. The equation of the line parallel to x y+ z 5 and passing through the point (,,5) in vector 5 form is (MAR-7) () r i + 5j + k + t(i + j + 5k) () r i + j + 5k + t(i + 5j + k) () r i + 5j + k + t(i + j + 5k) (4) r i + j + 5k + t i + 5j + k 6. The point of intersection of the line r i k + t(i + j + 7k) and the plane r. i + j k 8 is (MAR-07,MAR-08, JUN-6) ()(8,6,) ()(-8,-6,-) ()(4,,) (4)(-4,-,-) 7. The equation of the plane passing through the point (,, ) and the line of intersection of the planes r. i + j k 0 and r. j + k 0 (MAR-0,OCT-0,MAR-,JUN-5) ()x + 4y z 0 () x + 9y + z 0 () x + y z (4) x y + z 0 8. The work done by the force F i + j + k acting on a particle, if the particle is displaced from A(,,) to the point B(4,4,4) is (MAR-06, MAR-,JUN-) () units () units ()4 units (4)7 units ways00@gmail.com

9 th Maths Public Exam Special Guide Way to Success 9. If a i j + k and b i + j + k then a unit vector perpendicular to a and b is (OCT-06, JUN-6) i+j +k () i j +k () 0. The point of intersection of the lines x 6 6 i+j +k () y+4 z 4 x+ and y+ z+ is (4) i j k (OCT-07,JUN-09,MAR-,MAR-,OCT-,MAR-4,JUN-4, OCT-6, MAR-7) ()(0,0,-4) ()(,0,0) ()(0,,0) (4)(,,0). The point of intersection of the lines r i + j + k + t( i + j + k) and r i + j + 5k + s(i + j + k) is (OCT-06, JUN-0, JUN-,JUN-,MAR-5,MAR-6) () (,,) ()(,,) ()(,,) (4)(,,). The shortest distance between the lines x y z x and y 4 z 5 is (MAR-06,OCT-,OCT-4) () () 6. The shortest distance between the parallel lines x y z 5 x and y z is 4 4 ( OCT-07, JUN-, OCT-,MAR-6) () () () (4)0 4. The following two lines are x x 5 is. (JUN-06,MAR-0) () () parallel ()intersecting ()skew (4) perpendicular 5. The centre and radius of the sphere given by x + y + z 6x + 8y 0z + 0 is ( OCT-08,JUN-0,OCT-,JUN-,MAR-4,JUN-4) () (,4, 5),49 ()( 6,8, 0), ()(, 4,5),7 (4)(6, 8,0), 7 (4) 6. The value of 00 +i + i. COMPLEX NUMBERS 00 is () ()0 () (4). The modulus and amplitude of the complex number e iπ 4 are respectively ()e 9, π () e 9, π (JUN-0,JUN-,MAR-6) (JUN-07,MAR-08,JUN-08,OCT-09,MAR-0,MAR-5,OCT-5) () e 6, π (4) e 9, π 4 4. If m 5 + i(n + 4) is the complex conjugate of m + + i(n ) then (n, m) are (MAR-07,OCT-0,JUN-6, MAR-7) (), 8 (), 8 (), 8 (4), 8 4. If x + y then the value of +x+iy +x iy is (JUN-09,MAR-,MAR-,JUN-4, OCT-6) () x iy () x () iy (4) x + iy 5. The modulus of the complex number + i is (JUN-) () () () 7 (4)7 ways00@gmail.com

10 th Maths Public Exam Special Guide Way to Success 6. If A + ib (a + ib )(a + ib )(a + ib ) then A + B is (JUN-5) ()a + b + a + b + a + b ()(a + a + a ) + (b + b + b ) ()( a + b )( a + b )( a + b ) (4)( a + a + a )( b + b + b ) 7. If a + i and z i then the points on the Argand diagram representing az, az, and az are (OCT-06,JUN-08,OCT-4) () Vertices of a right angled triangle ()Vertices of an equilateral triangle () Vertices of an isosceles triangle (4)Collinear 8. The points z, z, z, z 4 in the complex plane are the vertices of a parallelogram taken in order if and only if (OCT-5,MAR-6) () z + z 4 z + z () z + z z + z 4 () z + z z + z 4 (4) z z z z 4 9. If z represents a complex number then arg z + arg(z ) is (OCT-08,MAR-09,MAR-, OCT-6) () π 4 () π ()0 (4) π 4 0. If the amplitude of a complex number is π then the number is (OCT-0,OCT-,OCT-,MAR-4) () purely imaginary () purely real ()0 (4) neither real nor imaginary. If the point represented by the complex number iz is rotated about the origin through the angle π in the counter clockwise direction then the complex number representing the new position is (JUN-,OCT-,MAR-,OCT-5) () iz () iz () z (4) z. The polar form of the complex number i 5 is (MAR-06,OCT-06,MAR-07,OCT-07,JUN-09,JUN-5) () cos π + i sin π ()cos π + i sin π () cos π i sin π (4) cos π i sin π. If P represents the variable complex number z and if z z then the locus of P is (JUN-06,MAR-0,MAR-,JUN-,OCT-4,MAR-7) () the straight line x 4 () the straight line z () the straight line y 4 (4) the circle x + y 4x e iθ +e iθ (JUN-07,MAR-,JUN-) () cos θ + i sin θ () cos θ i sin θ () sin θ i cos θ (4) sin θ + i cos θ 5. If z n cos nπ + i sin nπ then z z. z 6 is (JUN-08,OCT-,JUN-4,JUN-6) () () () i (4) i 6. If z lies in the third quadrant then z lies in the (OCT-,MAR-4,MAR-6) () first quadrant () second quadrant ( )third quadrant (4)fourth quadrant 7. If x cos θ + i sin θ the value of x n + n is (MAR-08,OCT-08,MAR-09,MAR-5) x () cos n θ () i sin nθ () sin nθ (4) i cos n θ 8. If a cos α i sin α, b cos β i sin β, c cos γ i sin γ then (a c b ) abc is ()cos (α β + γ) + i sin (α β + γ) () i sin (α β + γ) () cos(α β + γ) (4) cos(α β + γ) (MAR-4,JUN-4) ways00@gmail.com

11 th Maths Public Exam Special Guide Way to Success 9. z 4 + 5i, z + i then z z is () i () + i () (OCT-06,OCT-07,OCT-,JUN-,JUN-6) i (4) + i 0. The value of i + i + i + i 4 + i 5 is (MAR-06,JUN-06,JUN-07,JUN-09) () i () i () (4). The conjugate of i + i 4 + i 5 + i 6 is () () ()0 (4) i. If i + is one root of the equation ax bx + c 0, then the other root is () i () i () + i (4) i + i (MAR-08,OCT-09,MAR-0,MAR-). The quadratic equation whose roots are ±i 7 is (OCT-09,JUN-0,MAR-) () x () x 7 0 () x + x (4) x x The equation having 4 i and 4+i as roots is (MAR-07) ()x + 8x () x + 8x 5 0 () x 8x (4) x 8x If i +i is a root of the equation ax + bx + 0, where a, b are real then a, b is () (,) ()(, ) () (0,) (4)(,0) (OCT-07,JUN-) 6. If i+ is a root of x 6x + k 0 then the value of k is (OCT-08,MAR-) ()5 () 5 () 0 (4)0 7. If ω is a cube root of unity then the value of ( ω + ω ) 4 + ( + ω ω ) 4 is (MAR-06,JUN-06,OCT-0,JUN-,OCT-,OCT-4,JUN-5) ()0 () () 6 (4) 8. If ω is the n th root of unity then (JUN-0,JUN-,MAR-5) () + ω + ω 4 + ω + ω + ω 5 + () ω n 0 () ω n (4)ω ω n 9. If ω is the cube root of unity then the value of ω ω ω 4 ( ω 8 ) is (MAR-09,OCT-,OCT-, OCT-6, MAR-7) () 9 () 9 ()6 (4) 4. ANALYTICAL GEOMETRY. The axis of the parabola y y + 8x 0 is (OCT-08,MAR-, OCT-,JUN-, OCT-6) () y () x () x (4) y. 6x y x y 44 0 represents (JUN-08,OCT-0) () an ellipse ()a circle () a parabola (4) a hyperbola. The line 4x + y c is a tangent to the parabola y 6x then c is (JUN-09,MAR-0,JUN-,OCT-,MAR-5) () () ()4 (4) 4 4. The point of intersection of the tangents at t t and t t to the parabola y 8x is (MAR-08,OCT-09,JUN-4) () (6t, 8t) () (8t, 6t ) () (t, 4t) (4) (4t, t ) ways00@gmail.com

12 th Maths Public Exam Special Guide Way to Success 5. The length of the latus rectum of the parabola y 4x + 4y is (MAR-07,MAR-09,MAR-4,JUN-5,JUN-6) ()8 ()6 ()4 (4) 6. The directrix of the parabola y x + 4 is (JUN-0) ()x 5 () x 5 () x 7 (4) x The length of the latus rectum of the parabola whose vertex is (, ) and the directrix x 4 is (OCT-07,MAR-6) () () 4 () 6 (4) 8 8. The focus of the parabola x 6y is (OCT-5) () (4,0) () (0,4) ()(-4,0) (4)(0,-4) 9. The vertex of the parabola x 8y is (MAR-) () 8, 0 () 8, 0 () 0, 8 (4) 0, 8 0. The line x + y touches the parabola y 8x at the point (MAR-06,MAR-) ()(0,-) ()(,4) () 6, 9 (4) 9, 6. The tangents at the end of any focal chord to the parabola y x intersect on the line (JUN-07,OCT-) () x 0 () x + 0 () y + 0 (4) y 0. The angle between the two tangents drawn from the point (-4,4) to y 6x is (JUN-08) ()45 () 0 () 60 (4) 90. The eccentricity of the conic 9x + 5y 54x 40y is (MAR-07,OCT-,6, MAR-7) () () () 4 9 (4) 5 4. The length of the semi-major and the length of semi minor axis of the ellipse x + y are (JUN-0,MAR-5) () 6, (),4 (),6 (4), 5. The distance between the foci of the ellipse 9x + 5y 80 is (JUN-09,JUN-,MAR-,JUN-,OCT-,MAR-4) () 4 () 6 () 8 (4) 6. If the length of major and semi-minor axes of an ellipse are 8, and their corresponding equations are y 6 0 and x then the equations of the ellipse is () x+4 + y 6 x+4 y 6 () + () x+4 y 6 (4) x+4 y The straight line x y + c 0 is a tangent to the ellipse 4x + 8y if c is (OCT-08,OCT-0,MAR-,OCT-) ()± () ±6 ()6 (4)±4 8. The sum of the distance of any point on the ellipse 4x + 9y 6 from 5, 0, ( 5, 0) is () 4 () 8 () 6 (4) 8 (MAR-06,OCT-09) ways00@gmail.com - -

13 th Maths Public Exam Special Guide Way to Success 9. The radius of the director circle of the conic 9x + 6y 44 is (OCT-06,JUN-08,MAR-,OCT-4,MAR-6) () 7 () 4 () (4) 5 0. The locus of foot of perpendicular from the focus to a tangent of the curve 6x + 5y 400 is (MAR-09,JUN-6) () x + y 4 () x + y 5 () x + y 6 (4) x + y 9. The eccentricity of the hyperbola y 4x 4x + 48y 7 0 is (OCT-09,MAR-0,OCT-5) () 4 () () (4)6. The eccentricity of the hyperbola whose latus rectum is equal to half of its conjugate axis is (JUN-06,JUN-07,MAR-5) () () 5. The difference between the focal distances of any point on the hyperbola x y a b is 4 and the eccentricity is. Then the equation of the hyperbola is (MAR-07,JUN-) () () x y x () y x () y (4) x y The directrices of the hyperbola x 4(y ) 6 are (MAR-06,OCT-09,OCT-4, MAR-6) ()y ± 8 5 () x ± 8 5 () y ± 5 8 (4) 5 (4) x ± The line 5x y + 4k 0 is a tangent to 4x y 6 then k is (OCT-06,JUN-5) () 4 9 () () 9 4 (4) The equation of the chord of contact of tangents from (,) to the hyperbola x 6 y () 9x 8y 7 0 () 9x + 8y ()8x 9y 7 0 (4) 8x + 9y is (JUN-) 7. The angle between the asymptotes to the hyperbola x y is (MAR-08,MAR-,OCT-) 6 9 () π tan 4 () π tan 4 () tan 4 (4) tan 4 8. The asymptotes of the hyperbola 6y 5x are (MAR-7) ()y ± 6 x () y ± x () y ± x (4) y ± x The product of the perpendiculars drawn from the point (8,0) on the hyperbola x y to its 64 6 asymptotes is (JUN-0,JUN-4,JUN-6) () () () 6 5 (4) The locus of the point of intersection of perpendicular tangents to the hyperbola x 6 y () x + y 5 () x + y 4 () x + y (4) x + y 7 is 9 (OCT-08,JUN-). The eccentricity of the hyperbola with asymptotes x + y 5 0,x y is (OCT-06,OCT-07) () () () (4) ways00@gmail.com - -

14 th Maths Public Exam Special Guide Way to Success. Length of the semi-transverse axis of the rectangular hyperbola xy 8 is (OCT-0,OCT-,MAR-4) () () 4 () 6 (4) 8. The asymptotes of the rectangular hyperbola xy c are (MAR-) ()x c, y c () x 0, y c () x c, y 0 (4) x 0, y 0 4. The co-ordinate of the vertices of the rectangular hyperbola xy 6 are (MAR-07,OCT-) () (4,4),(-4,-4) () (,8),(-,-8) ()(4,0),(-4,0) (4) (8,0),(-8,0) 5. One of the foci of the rectangular hyperbola xy 8 is (MAR-09, MAR-7) ()(6,6) ()(,) ()(4,4) (4)(5,5) 6. The length of the latus rectum of the rectangular hyperbola xy is (MAR-08,MAR-0,JUN-,OCT-4,OCT-5, OCT-6) ()8 () ()8 (4)6 7. The area of the triangle formed by the tangent at any point on the rectangular hyperbola xy 7 and its asymptotes is (MAR-,JUN-5) ()6 ()8 ()7 (4)44 8. The normal to the rectangular hyperbola xy 9 at 6, meets the curve again at (JUN-,JUN-4) (), 4 () 4, (), 4 (4) 4, DIFFERENTIAL CALCULUS APPLICATIONS I. The gradient of the curve y x + x+5 at x is (OCT-6) () 0 ()7 () 6 (4). The rate of change of area A of a circle of radius r is (OCT-08) ()πr () πr dr dt ()πr dr dt (4) π dr dt. The velocity v of a particle moving along a straight line when at a distance x from the origin is given by a + bv x where a and b are constants. Then the acceleration is (MAR-09) () b x () a x () x b (4) x a 4. A spherical snowball is melting in such a way that its volume is decreasing at a rate of cm min. The rate at which the diameter is decreasing when the diameter is 0 cms is (MAR-0,MAR-,JUN-4) () 50π cm min () 50π cm min () 75π cm min (4) cm min 75π 5. The slope of the tangent to the curve y x + sin x at x 0 is (MAR-07,JUN-08,JUN-09,MAR-,JUN-4) () () () (4)- 6. The slope of the normal to the curve y x at the point whose x coordinate is is (MAR-06,OCT-06,JUN-07,OCT-09,JUN-,MAR-5) () () 4 () (4) ways00@gmail.com - -

15 th Maths Public Exam Special Guide Way to Success 7. The point on the curve y x 6x 4 at which the tangent is parallel to the x axis is (OCT-0,JUN-5) () 5, 7 () 5, 7 () 5, 7 (4), 7 8. The equation of the tangent to the curve y x at the point (, ) is (MAR-08) 5 5 ()5y + x () 5y x ()x 5y (4) x + y 9. The equation of the normal to the curve θ t at the point (, ) is (OCT-,MAR-4) ()θ 7t 80 () 5θ 7t 80 ()θ 7t + 80 (4) θ t 0. The angle between the curves x + y x and y is (JUN-07,MAR-09,OCT-,6) () π 4 () π () π 6 (4) π. The angle between the curves y e mx and y e mx for m > is (OCT-,MAR-6) m ()tan m m () tan m m () tan +m. The parametric equations of the curve x + y a are ()x a sin θ ; y a cos θ () x a cos θ ; y a sin θ ()x a sin θ ; y a cos θ (4) x a cos θ ; y a sin θ m (4) tan m +. If the normal to the curve x + y a makes an angle θ with the x axis then the slope of the normal is (MAR-,MAR-) () cot θ () tan θ () tan θ (4) cot θ 4. If the length of the diagonal of a square is increasing at the rate of 0. cm sec. what is the rate of increase of its area when the side is 5 cm? (). 5 cm sec () cm sec () cm sec (4) 0.5 cm sec (OCT-,JUN-6) 5. What is the surface area of a sphere when the volume is increasing at the same rate as its radius? (JUN-06,MAR-0) () () () 4π (4) 4π π 6. For what values of x is the rate of increase of x x + x+8 is twice the rate of increase of x (JUN-08,JUN-) (), (), (), (4), 7. The radius of a cylinder is increasing at the rate of cm sec and its altitude is decreasing at the rate of cm sec. The rate of change of volume when the radius is cm and the altitude is 5cm is (JUN-07) ()π () π () 4π (4) 5π 8. If y 6x x and x increases at the rate of 5 units per second, the rate of change of slope when x is (MAR-) () - 90 units/sec ()90 units sec ()80 units sec (4)-80 units sec 9. If the volume of an expanding cube is increasing at the rate of 4 cm sec then the rate of change of surface area when the volume of the cube is 8 cubic cm is () 8 cm sec () 6 cm sec () cm sec (4) 4 cm sec ways00@gmail.com

16 th Maths Public Exam Special Guide Way to Success 0. The gradient of the tangent to the curve y 8 + 4x x at the point where the curve cuts the y-axis is ()8 () 4 ()0 (4)-4. The angle between the parabolas y x and x y at the origin is (JUN-06,JUN-0,MAR-4,OCT-4) () tan 4 () tan 4 () π (4) π 4. For the curve x e t cos t; y e t sin t the tangent line is parallel to the x-axis when t is equal to (JUN-) () π 4 () π 4 ()0 (4) π. If a normal makes an angle θ with positive x-axis then the slope of the curve at the point where the normal is drawn is (OCT-07) () cot θ () tan θ () tan θ (4) cot θ 4. The value of a so that the curves y e x and y a e x intersect orthogonally is (OCT-0) () () () 5. If s t 4t + 7, the velocity when the acceleration is zero is... (OCT-06,MAR-07,OCT-09,JUN-5) () 6 m/sec () m/sec () 6 m/sec (4) m/sec 6. If the velocity of a particle moving along a straight line is directly proportional to the square of its distance from a fixed point on the line. Then its acceleration is proportional to (OCT-, JUN-6) ()s () s () s (4) s 4 7. The Rolle s constant for the function y x on [,] is () (4) () 0 () (4) 8. The value of c of Lagranges Mean Value Theorem for the function f x x + x ; a 0, b is (MAR-09,OCT-,OCT-4) () () ()0 (4) 9. The value of c in Rolle s Theorem for the function f x cos x on [π, π] is (MAR-06,08,, 7) ()0 () π () π (4) π 0. The value of c of Lagranges Mean Value Theorem for f x x when a and b 4 is (JUN-0,JUN-, OCT-5, OCT-6) () 9 4 () () (4) 4 (OCT-07,OCT-08) x. lim x e x is () () 0 () (4). lim x 0 a x b x c x d x () () 0 ()log ab cd (4) log(a/b) log(c/d) (MAR-07,OCT-09) ways00@gmail.com

17 th Maths Public Exam Special Guide Way to Success. If f a ; f a ; g a ; g a then the value of lim x a g x f a g a f(x) ()5 () 5 () (4) x a is (JUN-08,MAR-6) 4. Which of the following function is increasing in (0, ) (OCT-06,OCT-,OCT-5) ()e x () x () x (4)x 5. The function f x x 5x + 4 is increasing in (MAR-,JUN-) ()(, ) ()(,4) ()(4, ) (4) everywhere 6. The function f x x is decreasing in (JUN-09,MAR-5) ()(, ) ()(, 0) ()(0, ) (4) (, ) 7. The function y tan x x is (JUN-4) () an increasing function in 0, π () increasing in 0, π and decreasing in π, π 4 4 (4) decreasing in 0, π and increasing in π, π () a decreasing function in 0, π 8. In a given semi circle of diameter 4 cm a rectangle is to be inscribed. The maximum area of the rectangle is (MAR-06,OCT-4) () ()4 ()8 (4)6 9. The least possible perimeter of a rectangle of area 00m is (OCT-07,OCT-08,OCT-0,OCT-,JUN-5) ()0 ()0 ()40 (4) If f x x 4x + 5 on [0,] then the absolute maximum value is (MAR-,OCT-5, MAR-7) () () ()4 (4)5 4. The curve y e x is (MAR-0,MAR-) () concave upward for x > 0 () concave downward for x > 0 () everywhere concave upward (4) everywhere concave downward 4. Which of the following curves is concave down? (MAR-08,JUN-09,OCT-) ()y x () y x ()y e x (4) y x + x 4. The point of inflexion of the curve y x 4 is at (JUN-,MAR-5) ()x 0 () x ()x (4) nowhere 44. The curve y ax + bx + cx + d has a point of inflexion at x then (MAR-4) ()a + b 0 () a + b 0 ()a + b 0 (4) a + b 6. DIFFERENTIAL CALCULUS APPLICATIONS II. If u x y then u is equal to (OCT-08,OCT-0,OCT-5, OCT-6) x ()yx y ()u log x () u log y (4) xy x. If u sin x 4 +y 4 x +y and f sin u then f is a homogeneous function of degree ()0 () () (4) 4 ways00@gmail.com

18 th Maths Public Exam Special Guide Way to Success. If u u u x +y, then x + y is equal to (MAR-08,JUN-09,JUN-0,JUN-) x y () u () u () u (4) u 4. The curve y x x ( + x) has (OCT-09,MAR-, MAR-7) () an asymptote parallel to x-axis () an asymptote parallel to y-axis () asymptotes parallel to both axes (4) no asymptotes 5. If x r cos θ, y r sin θ, then r is equal to (JUN-09,MAR-,MAR-4) x ()sec θ ()sin θ ()cos θ (4) cosec θ 6. Identify the true statements in the following: (MAR-,JUN-6) (i) If a curve is symmetrical about the origin, then it is symmetrical about both axes (ii) If a curve is symmetrical about both the axes, then it is symmetrical about the origin (iii) A curve f x, y 0 is symmetrical about the line y x if f x, y f(y, x) (iv) For the curve f x, y 0, if f x, y f( y, x),then it is symmetrical about the origin () (ii),(iii) ()(i),(iv) ()(i),(iii) (4)(ii),(iv) 7. If u log x +y then x u u + y is (JUN-06,MAR-07,OCT-07,MAR-0,OCT-,JUN-5) xy x y ()0 () u () u (4)u 8. The percentage error in the th root of the number 8 is approximately times the percentage error in 8 (MAR-06,OCT-06,MAR-,OCT-4) () 8 () () (4)8 9. The curve a y x (a x ) has (OCT-07,OCT-09,MAR-0,JUN-,OCT-,JUN-5) () only one loop between x 0 and x a () two loops between x 0 and x a () two loops between x a and x a (4) no loop 0. An asymptote to the curve y a + x x (a x) is (JUN-06,07,08,, OCT-,, 6) () x a () x a/ () x a/ (4) x 0. In which region the curve y a + x x (a x) does not lie? (MAR-09,JUN-0,,4,6) () x > 0 () 0 < x < a () x a and x > a (4) a < x < a. If u y sin x, then u is equal to (JUN-07,JUN-08) x y () cos x ()cos y ()sin x (4)0. If u f y x then x u u + y is equal to x y (OCT-08,MAR-09,JUN-) () 0 () ()u (4) u 4. The curve 9y x (4 x ) is symmetrical about (MAR-06,OCT-06,OCT-0,MAR-5,OCT-5) () y-axis () x- axis () y x (4) both the axes 5. The curve ay x (a x) cuts the y-axis at (MAR-08,MAR-5,MAR-6) () x a, x 0 ()x 0, x a () x 0, x a (4) x 0 ways00@gmail.com

19 th Maths Public Exam Special Guide Way to Success. The value of () π. The value of π/ 0 π/ 0 7. INTEGRAL CALCULUS AND ITS APPLICATIONS cos 5/ x cos 5/ x+sin 5/ x () π 4 sin x cos x +sin x cos x dx is dx is ()0 (4) π (MAR-,JUN-,JUN-5) (JUN-0, MAR-7) () π () 0 () π (4) π 4. The value of x( x) 4 dx is (MAR-06,MAR-09,JUN-0,MAR-,OCT-,JUN-4,MAR-5,OCT-5) 0 () 4. The value of π/ π/ () 0 sin x +cos x dx is () 4 (4) 0 (JUN-07,OCT-07,0,6) ()0 () ()log (4)log 4 π 5. The value of sin 4 x dx is (OCT-06,OCT-09,JUN-,MAR-4) 0 () π 6 π/4 0 () 6 ()0 (4) π 8 6. The value of cos x dx is (MAR-07, 08, 0, 4, 7, JUN-09, OCT-08,, 4) () () 7. The value of π sin x cos xdx is 0 ()0 (4) π ()π () π () π (4) The area bounded by the line y x, the x-axis, the ordinates x, x is (JUN-08,MAR-,OCT-, JUN-6) (JUN-07,OCT-08,MAR-09,JUN-,OCT-5) () () 5 () (4) 7 9. The area of the region bounded by the graph y sin x and y cos x between x 0 and x π 4 is () + () () (4) + (JUN-06,OCT-07,MAR-0,OCT-,JUN-4,MAR-6) 0. The area between the ellipse x + y a b and its auxillary circle is (MAR-06,JUN-06,MAR-07,JUN-09) () πb(a b) () πa(a b) () πa(a b) (4) πb(a b). The area bounded by the parabola y x and its latus rectum is (MAR-08,MAR-,OCT-,JUN-5) () 4 () () 6. The volume of the solid obtained by revolving x 9 + y 6 (4) 8 about the minor axis is ()48π ()64π () π (4)8π (JUN-07,JUN-,MAR-,OCT-, OCT-6). The volume, when the curve y + x from x 0 to x 4 is rotated about x- axis is ()00π () 00 9 π ()00 π (4)00 (MAR-06,OCT-08,OCT-09,MAR-,JUN-) ways00@gmail.com

20 th Maths Public Exam Special Guide Way to Success 4. The volume generated when the region bounded by y x, y, x 0 is rotated about y-axis is () π 4 () π () π (JUN-08,OCT-0,JUN-,JUN-4,OCT-4) 5. Volume of solid obtained by revolving the area of the ellipse x + y a b about major and minor axes are in the ratio (JUN-06,MAR-09,JUN-0,JUN-,MAR-,OCT-,MAR-5, MAR-7) (4) π ()b : a ()a : b () a: b (4) b: a 6. The volume generated by rotating the triangle with vertices at (0,0),(,0) and (,) about x-axis is (OCT-06, OCT-07,JUN-09,OCT-,MAR-4,MAR-6) ()8π ()π () 6π (4)9π 7. The length of the arc of the curve x + y 4 is (MAR-08,JUN-08,MAR-,OCT-4,MAR-5,JUN-6) ()48 ()4 () (4)96 8. The surface area of the solid of revolution of the region bounded by y x, x 0 and x about x-axis is (OCT-06,MAR-07,OCT-0,OCT-,MAR-,JUN-5,OCT-5,MAR-6) ()8 5π () 5π () 5π (4)4 5π 9. The curved surface area of a sphere of radius 5, intercepted between two parallel planes of distance and 4 from the centre is (OCT-09,MAR-0,JUN-,JUN-6, OCT-6) ()0π ()40π ()0π (4)0π 8. DIFFERENTIAL EQUATIONS. The integrating factor of dy + y dx x e4x is () log x () x () e x (4) x. If cos x is an integrating factor of the differential equation dy + Py Q then P () cot x () cot x () tan x (4) tan x dx (JUN-09,JUN-) (JUN-07,OCT-08,MAR-09,OCT-5, OCT-6). The integrating factor of dx + xdy e y sec ydy is (OCT-0, MAR-,OCT-4, JUN-6) () e x () e x () e y (4) e y 4. Integrating factor of dy dx + x log x. y x is (OCT-06,MAR-07,JUN-07,OCT-09,MAR-,JUN-5) () e x () log x () (4) e x x 5. Solution of dx + mx 0, where m < 0 is (MAR-08,0,, 4, 7 JUN-09,0, OCT-) dy ()x ce my () x ce my () x my + c (4) x c 6. y cx c is the general solution of the differential equation (OCT-08,MAR-4) () y xy + y 0 () y 0 () y c (4) y + xy + y 0 dx 7. The differential equation + 5y / x (MAR-06, JUN-08,MAR-0, MAR-7) dy () of order and degree ()of order and degree () of order and degree 6 (4) of order and degree ways00@gmail.com

21 th Maths Public Exam Special Guide Way to Success 8. The differential equation of all non-vertical lines in a plane is () dy 0 () d y dy dx dx 0 () m (4) d y m dx dx 9. The differential equation of all circles with centre at the origin is (MAR-09,,JUN-07,5,6) ()xdy + y dx 0 () xdy ydx 0 () xdx + ydy 0 (4) xdx ydy 0 0. The integrating factor of the differential equation dy + py Q is (MAR-06) dx () pdx () Q dx () e Q dx (4) e. The complementary function of (D + )y e x is (JUN-09,JUN-,MAR-5) () (Ax + B)e x () A cos x + B sin x ()(Ax + B)e x (4) (Ax + B)e x. A particular integral of (D 4D + 4)y e x is (OCT-06,JUN-08,OCT-08,OCT-09,JUN-5) () x ex () xe x () xe x (4) x e x. The differential equation of the family of lines y mx is (OCT-07) () dy m () ydx xdy 0 y dx ()d dx 0 (4) ydx + xdy 0 p dx 4. The degree of the differential equation + dy dx / d y dx (MAR-07,OCT-,MAR-,MAR-5,OCT-5,MAR-6) () () () (4)6 + dy dx / 5. The degree of differential equation c d y where c is a constant is (JUN-) dx () () ()- (4) 6. The amount present in a radio active element disintegrates at a rate proportional to its amount. The differential equation corresponding to the above statement is (k is negative) (OCT-07,JUN-,JUN-) () dp k () dp kt dp () kp dp (4) kt dt p dt dt dt 7. The differential equation satisfied by all the straight lines in xy plane is (MAR-6, OCT-6) () dy a constant () d y dy dx dx 0 () y + 0 (4) d y + y 0 dx dx 8. If y ke λx then its differential equation is (JUN-06,,4, OCT-0,6, MAR-7) () dy λy dy () ky dy () + ky 0 dy (4) dx dx dx dx eλx 9. The differential equation obtained by eliminating a and b from y ae x + be x is (OCT-06,MAR-09,JUN-0,OCT-,OCT-4) () d y dx () d y dx 9y 0 () d y dx dx (4) d y dx 0. The differential equation formed by eliminating A and B from the relation y e x (A cos x + B sin x) is (OCT-09,OCT-,OCT-) ()y + y 0 () y y 0 () y y + y 0 (4) y y y 0. If dy x y then dx x+y ()xy + y + x c () x + y x + y c () x + y xy c (4) x y xy c (MAR-08,MAR-,OCT-,JUN-, JUN-6) ways00@gmail.com

22 th Maths Public Exam Special Guide Way to Success. If f x x and f then f x is (OCT-07, MAR-0,JUN-5,OCT-5) () (x x + ) () (x x + ) () (x x + ) (4) x( x + ). On putting y vx, then homogeneous differential equation x dy + y x + y dx 0 becomes (JUN-06, MAR-, JUN-4,OCT-4) ()xdv + v + v dx 0 ()vdx + x + x dv 0 ()v dx x + x dv 0 (4)vdv + x + x dx 0 4. The integrating factor of the differential equation dy y tan x cos x is (MAR-08,JUN-08,JUN-0,OCT-,MAR-,OCT-,OCT-,MAR-4,JUN-4, MAR-7) ()sec x ()cos x ()e tan x (4)cot x 5. The P.I of (D + D 4)y e x is (MAR-06,MAR-07,JUN-,MAR-) () 6xe x () xe x () xe x (4)x /e x 6. The particular integral of the differential equation f D y e ax where f D D a g D, g(a) 0 is (JUN-06,OCT-0,JUN-,MAR-6) () me ax () e ax g(a) dx () g(a)e ax (4) xeax g(a) 9. DISCRETE MATHEMATICS. Which of the following are statements? (MAR-,JUN-,MAR-6 ) (i) May God bless you. (ii)rose is a flower (iii)milk is white (iv) is a prime number ()(i),(ii),(iii) ()(i),(ii),(iv) ()(i),(iii),(iv) (4)(ii),(iii),(iv). If a compound statement is made up of three simple statements, then the number of rows in the truth table is (OCT-09,OCT-,MAR-, MAR-5 ) () 8 () 6 () 4 (4). If p is T and q is F, then which of the following have the truth value T? (OCT-07,MAR-0,OCT-,MAR-4,MAR-0,JUN-6, MAR-7 ) (i) p q (ii) p q (iii) p q (iv) p q ()(i),(ii),(iii) ()(i),(ii),(iv) ()(i),(iii),(iv) (4)(ii),(iii),(iv) 4. The number of rows in the truth table of ~[ p (~q)] is (JUN-06,JUN-08,OCT-0,OCT-,OCT-6 ) () () 4 () 6 (4) 8 5. The conditional statement p q is equivalent to (MAR-06,MAR-09,MAR-,OCT-,JUN-4,OCT-4,JUN-5,OCT-5) () p q () p ~q () ~p q (4) p q 6. Which of the following is a tautology? (JUN-07,MAR-08,MAR-09,JUN-09,JUN-0,MAR-,MAR-,OCT-) () p q () p q () p ~p (4) p ~p 7. Which of the following is a contradiction? (MAR-06,OCT-06,OCT-08,MAR-4,MAR-6) () p q () p q () p ~p (4) p ~p 8. p q is equivalent to (MAR-07,JUN-,MAR-5) ()p q () q p () p q (q p) (4) p q (q p) ways00@gmail.com - -

23 th Maths Public Exam Special Guide Way to Success 9. Which of the following is not a binary operation on R (OCT-07,JUN-08, 09,MAR-0,, 5) ()a b ab ()a b a b () a b ab (4) a b a + b 0. A monoid becomes a group if it also satisfies the (MAR-08,JUN- ) () closure axiom () associative axiom ()identity axiom (4)inverse axiom. Which of the following is not a group? (OCT-08,JUN-0,JUN-6 ) ()(Z n, + n ) () (Z, +) () (Z,. ) (4) (R, +). In the set of integers with operation * defined by a b a + b ab, the value of *(4 *5) is (JUN-06,OCT- ) ()5 ()5 ()0 (4)5. The order of [7] in (Z 9, + 9 ) is (OCT-06,4, MAR-07, 09,,7, JUN-08,0,4) ()9 ()6 () (4) 4. In the multiplicative group of cube root of unity, the order of ω is (JUN-07,OCT-08,JUN-5,OCT-5) ()4 () () (4) 5. The value of + ([5]+ [6]) is (MAR-06,JUN-09,JUN-) ()[0] ()[] ()[] (4)[] 6. In the set of real numbers R, an operation * is defined by a b a + b. Then the value of (*4)*5 is (JUN-07,OCT-0,JUN-5, JUN-6) ()5 ()5 ()5 (4)50 7. Which of the following is correct? (OCT-06,JUN-,OCT-4) () An element of a group can have more than one inverse () If every element of a group is its own inverse, then the group is abelian () The set of all real matrices forms a group under matrix multiplication (4) a b a b for all a, b G 8. The order i in the multiplicative group of 4 th roots of unity is (JUN-,,OCT-5,6, MAR-6) ()4 () () (4) 9. In the multiplicative group of nth roots of unity, the inverse of ω k is k < n (JUN-06,MAR-08,OCT-09,OCT-,MAR-,OCT-,JUN-4, MAR-7) () ω /k () ω () ω n k (4) ω n/k 0. In the set of integers under the operation * defined by a b a + b, the identity element is (MAR-07,MAR-0,OCT-0,JUN-,MAR-,MAR-4, OCT-6) ()0 () () a (4) b. If f x k x, 0 < x < 0, elsewhere () 0. PROBABILITY DISTRIBUTIONS is a probability density function then the value of k is (JUN-08, 09,OCT-, 6) () 6 () 9 (4). If f x A π 6+x, < x < is a p.d.f. of a continuous random variable X, then the value of A is (MAR-06,MAR-07,OCT-08,OCT-09,MAR-,OCT-,MAR-4,JUN-4,JUN-5) ()6 () 8 () 4 (4) ways00@gmail.com - -

24 th Maths Public Exam Special Guide Way to Success. A random variable X has the following probability distribution X P(Xx) /4 a a 4a 5a /4 Then P( x 4) is () 0 () 7 () 4 (JUN-0,JUN-, MAR-7) 4. A random variable X has the following probability mass function as follows: X λ λ λ P(X x) 6 4 Then the value of λ is (JUN-6 ) () () () (4)4 5. X is a discrete random variable which takes the values 0,, and P X 0 44, P X then the value of P X is (MAR-09,OCT-0,JUN- ) () () 4 69 () 69 (4) 69 (4) A random variable X has the following p.d.f. (MAR-6 ) X P(Xx) 0 k k k k k k 7k + k The value of k is () 8 () 0 () 0 (4) or 7. Given E X + c 8 and E X c, then the value of c is (OCT-06,OCT-07,JUN-09,MAR-,OCT-,MAR-5,OCT-5, MAR-6 ) () () 4 () 4 (4) 8. X is a random variable taking the values, 4, with probabilities, 4 and 5 0. Then E X is (OCT-08, OCT-5, JUN-6) () 5 () 7 () 6 (4) 9. Variance of the random variable X is 4. Its mean is. Then E(X ) is (JUN-07,MAR-09,JUN-,OCT-5 ) () () 4 () 6 (4) 8 0. μ 0, μ 76 for a discrete random variable X. Then the mean of the random variable X is (OCT-09,MAR-0,JUN-,MAR-4 ) ()6 ()5 () (4). Var(4X + ) is (MAR-06, JUN-06,MAR-08,JUN-08,JUN-5 ) ()7 ()6 Var(X) ()9 (4)0. In 5 throws of a die, getting or is a success. The mean number of successes is (OCT-06,JUN-,MAR-) () 5 () 5. The mean of a binomial distribution is 5 and its standard deviation is. Then the value of n and p are (OCT-,OCT-,OCT-5, MAR-7) () 4 5, 5 () 5, () 5 9 () (4) 9 5 5, 5 (4) 5, 5 69 ways00@gmail.com - -

25 th Maths Public Exam Special Guide Way to Success 4. If the mean and standard deviation of a binomial distribution are and respectively. Then the value of its parameter p is (OCT-0,MAR-,MAR-4,OCT-4 ) () () 5. In 6 throws of a die getting an even number is considered a success. Then the variance of the successes is (JUN-07,MAR-0,MAR-,MAR-6) () 4 () 6 () (4) A box contains 6 red and 4 white balls. If balls are drawn at random, the probability of getting white balls without replacement, is (JUN-0,OCT-,OCT-4, JUN-6) () 0 () 8 5 () () 4 5 (4) 4 (4) 0 7. If cards are drawn from a well shuffled pack of 5 cards, the probability that they are of the same colours without replacement, is (JUN-4,MAR-5 ) () () () 5 5 (4) If in a Poisson distribution P X 0 k then the variance is (MAR-07,09,OCT-07,OCT-08,,JUN-,4) ()log k ()log k ()e λ (4) k 9. If a random variable X follows Poisson distribution such that E X 0 then the variance of the distribution is (JUN-08,MAR-, JUN-6, OCT-6) ()6 ()5 ()0 (4)5 0. The distribution function F(X) of a random variable X is (MAR-08, JUN-,OCT-4 ) () a decreasing function ()a non-decreasing function () a constant function (4) increasing first and the decreasing. For a Poisson distribution with parameter λ 0.5 the value of the nd moment about the origin is (OCT-09,OCT-,OCT-,MAR-,MAR-5 ) ()0.5 ()0.5 ()0.065 (4)0.05. In a Poisson distribution if P X P X then the value of its parameter λ is (MAR-06, JUN-06,MAR-08,JUN-09,MAR-,JUN-5) ()6 () () (4)0. If f(x) is a p.d.f of a normal distribution with mean μ then f(x)dx is (OCT-06,MAR-07,JUN-07,OCT-0 ) () ()0.5 ()0 (4)0.5 (x 00 ) 4. The random variable X follows normal distribution f x ce 5 Then the value of c is (JUN-06,MAR-0,JUN-,MAR-, MAR-7) () π () ()5 π (4) π 5. If f(x) is a p.d.f of a normal variate X and X~N(μ, σ μ ) then f(x)dx is () undefined () (). 5 (4).5 5 π (OCT-6) 6. The marks secured by 400 students in a Mathematics test were normally distributed with mean 65,.If 0 students got more marks above 85, the number of students securing marks between 45 and 65 is (OCT-07,JUN-0,JUN-, MAR-6 ) ()0 ()0 ()80 (4)60 ways00@gmail.com

26 th Maths Public Exam Special Guide Way to Success Section B (6 Marks). APPLICATIONS OF MATRICES AND DETERMINANTS. For any non-singular matrix A, show that A T T A (OCT-07) We know that AA I A A Taking transpose on both sides of AA I, We have AA T I T By reversal law for transpose we get A T A T I... Similarly, by taking transposes on both sides A A I, we have A T A T I. From () & () A T A T A T A T I A T is the inverse of A T A T T A. State and prove reversal law for inverses of matrices. ( JUN-,OCT-4) If A, B are any two non-singular matrices of the same order, then AB is also non-singular and (AB) B A i.e. the inverse of a product is the product of the inverses taken in the reverse order. Proof: Since A and B are non-singular. A 0, and B 0 We know that AB A B A 0, B 0 A B 0 AB 0 Hence AB is also non-singular. So AB is invertible (AB)( B A ) A(BB ) A A I A A A I Similarly we can show that ( B A ) AB I (AB)( B A ) ( B A ) AB I B A is the inverse of AB. (AB) B A. Verify that A T A T for the matrix A ( MAR-0 ) 5 6 A 5 6 A adj A 6 5 A A A T 7 adj A A T A T A T adj AT AT adj A T 6 5 A T From () and () A T T A 4. Find the inverse of the matrix 4 (OCT-07) A, then A A is a non-singular matrix. Hence it is invertible. The matrix formed by the cofactors is A ij 4 adj A 4 A A adj A 4 5. Find the inverse of the matrix A 0 (OCT-09,OCT-,MAR-7) A 0 A 0 0 A is non-singular and hence A exists 6 A ij 5 8 adj A A A adj A If A and B, then the verify that (AB) B A ( JUN-09,JUN-0) A 0 B So, A and B are invertible ways00@gmail.com

27 th Maths Public Exam Special Guide Way to Success AB 0 AB 0 AB is invertible adj AB (AB) AB adj AB.. A adj A A adj A A B adj B B adj B 0 B 0 B A 0. From () and () we have (AB) B A 5 7. If A and B, then verify that 7 (AB) B A ( JUN-06,JUN-,JUN-,JUN-5) A 5 ; B 7 AB To find A A adj A 7 5 A adj A A A 7 5 To find B B 0 adj B B adj B B B Now, B A () To find (AB) AB adj AB 4 8 (AB) AB adj AB (AB) 4 8 () From () and () (AB) B A 5 8. If A and B, then verify that 7 (AB) T B T A T (JUN-4) A 5 and B 7 AB 8 4 (AB) T 8 4 B T, AT 5 7 B T A T From (),() we get (AB) T B T A T 9. Find the adjoint of the matrix A 5 and verify the result. A adj A adj A A A. I (MAR-07,MAR-09,MAR-) A 5 A adj A 5 A adj A A. I.. adj A A A. I... From () and () A adj A adj A A A. I 0. Show that the adjoint of A itself. A (MAR-08,MAR-,MAR-6, OCT-6) is A ways00@gmail.com

28 th Maths Public Exam Special Guide Way to Success 4 4 A ij adj A (A ij ) T A. For A 4 4, show that A A ( MAR-06,MAR-4) A A A ij adj A A (adj A) 4 4 A A Hence A A.. Solve by matrix inversion method x + y, x + y 8 (JUN-08,OCT-08,OCT-0,OCT-) The given system of equations can be written in the form of A X B x y 8 Here, A 0 Since A is non-singular, A exists A The solution is X A B x y x, y 8. Solve by matrix inversion method of the following system of linear equations x y 7 and x y ( JUN-07,MAR-,MAR-5) x y 7 x y x y 7 It is the form A X B, where A ; X x y ; B 7 Solution is given by X A B To find A A A ij adj A Hence X A B A adj A A A 4. Find the rank of the matrix A ~ ~ ~ ~ x, y C C ( MAR-5) R R 4R R 5 R R R + R The last equivalent matrix is in the echelon form. The number of non-zero rows in this matrix is two. ρ A 5. Find the rank of the matrix A ~ ~ ~ (OCT-06,MAR-07, OCT-6) R R R R R R R R R R R The last equivalent matrix is in the echelon form. It has three non-zero rows. ρ A ways00@gmail.com

29 th Maths Public Exam Special Guide Way to Success 6. Find the rank of the matrix A ~ ~ ~ (OCT-0) R R R R + R R R 7R This equivalent matrix is in the echelon form. Since the number of non-zero rows is, ρ A 7. Find the rank of the matrix A ~ ~ ~ R R (JUN -08 ) R R R R R R R R R This equivalent matrix is in the echelon form. Since the number of non-zero rows is, ρ A 8. Find the rank of the matrix A ~ ~ ~ ( JUN-,OCT-, MAR-7) R R R R R R 5R + R The last equivalent matrix is in the echelon form. It has three non-zero rows. ρ A 9. Find the rank of the matrix A ~ ~ (OCT-08, OCT-5) R R R R R R R R R This matrix is in the echelon form. It has two nonzero rows. ρ A 0. Find the rank of the matrix A ~ (JUN-07) R R R R R R This equivalent matrix is in the echelon form. Since the number of non-zero rows of the matrix in this echelon form is, ρ A. Find the rank of A ~ ~ (MAR-06,JUN-0) R R + R R R + R R R R The last equivalent matrix is in the echelon form. It has two non-zero rows. ρ A. Find the rank of the matrix 4 A (JUN-) ways00@gmail.com

30 th Maths Public Exam Special Guide Way to Success ~ ~ ~ R R R R + R R R R The last equivalent matrix has non zero rows ρ A Rank of matrix is.. Solve the following system of linear equations by determinant method. (OCT-5) (i) x y (ii) x + y + z 0 y x 7 x + y z 0 x + y + z 0 (i) x y y x 7 0 x 7 Since 0 and x 0 the system is inconsistent. It has no solution. (ii) x + y + z 0 x + y z 0 x + y + z 0 0, the system has unique solution. The above system of homogeneous equation has only trivial solution. x, y, z (0,0,0) 4. Solve the following system of linear equations by determinant method x + y 8, 4x + 6y 6 ( JUN-06, MAR-) x + y 8.. 4x + 6y x y Since 0, and x y 0 and atleast one of the coefficients a ij of 0 the system is consistent and has infinitely many solutions. All minor are zero and atleast minor is non zero. The system is reduced to a single equation. We assign arbitrary value to x (or y) and solve for y(or x) Suppose we assign x t, from equation () We get y (8 t) The solution set is x, y t, (8 t) t R 5. Solve the following system of linear equations by determinant method x + y + z 4; x + y + 4z 8; x + y + 6z 0 y ( JUN-, MAR-6) 4 0, x , z x y z 0. Also all minors of 0, but not all the minors of x, y, z are zero. Therefore the system is inconsistent. It has no solution. 6. Solve the following system of linear equations by determinant method x + y + z 4; x + y + 6z 7; x + y + z 0 (MAR-) x (7 0) Since 0, x 0 The system is inconsistent. It has no solution. 7. Solve x + y + z 0; x y z 5; x + z 6 by determinant method. (MAR-4) x and x 0 The system of equations are inconsistent and no solution. ways00@gmail.com

31 th Maths Public Exam Special Guide Way to Success 8. Solve by determinant method x y ; x 4y 4 ( JUN-4) x y , x 0, y 0 The system of equations are consistent and having many solution Assign y k x k x + k The solution is ( + k, k),k R 9. Solve the following system of linear equation by determinant method x + y 4, 4x + 8y 6 ( JUN-5) x + y 4. 4x + 8y x y Since 0, and x y 0 and atleast one of the coefficients a ij of 0 the system is consistent and has infinitely many solutions. All minor are zero and atleast minor is non zero. The system is reduced to a single equation. We assign arbitrary value of x (or y) and solve for y(or x) Suppose we assign x t, from equation () y 4 t We get y (4 t) The solution set is x, y t, (4 t) t R 0. Solve the following non-homogeneous equations of three unknowns using determinants x + y + z 5, x y + z, x + y + z 4 (MAR-08,MAR-09,MAR-,OCT-) 0 5 x Since, 0 and x 0 ( atleast one of the values of x, y, z non zero ) the system is inconsistent. It has no solution.. Solve the following non-homogeneous system of linear equations by determinant method 4x + 5y 9, 8x + 0y 8. (OCT-06,OCT-09) x y Since 0, and x y 0 and atleast one of the coefficient a, a, a, a is non-zero the system is consistent and has infinitely many solutions. Let y k Then x 9 5k The solution set is 4 9 5k 4, k where k R. Solve the following system of linear equation by determinant method x y 7, 4x 6y 4 (JUN-09) x y Since x y 0 and atleast one of a, a, a, a is non zero, it has infinitely many solutions. The above system is reduced to a single equation x y 7. To solve this equation x y 7, assign y k x k 7 x 7 + k x 7 + k The solution is x, y 7+k, k, k R. Find the values of α for which αx + y + z 0, 4x + y + 8z 0 4x + y + 4z 0 have (i) Only trivial solution (ii) trivial and non-trivial solutions By using determinant method (JUN-6) αx + y + z 0 4x + y + 8z 0 4x + y + 4z 0 α α (8 ) α ( 4) 4α + 6 4α + 4 ways00@gmail.com

32 th Maths Public Exam Special Guide Way to Success (i) Only trivial solution α 0 The above system of homogeneous equation has only trivial solution. x, y, z (0,0,0) The system of equation has only trivial solution (ii) trivial and non-trivial solutions α 0 It has infinitely many solution. Also atleast one minors of 0, the system is reduced to equations. Assigning arbitrary value to one of the unknowns, say z k x + y k 4x + y 8k x k 9k + 8k k 8k y k 8k + k 4k 4 8k By Cramer s Rule x k, y 4k Solution is x, y, z (k, 4k, k) 4. Examine the consistence of the system of equations: x + y + z 7, x + y + z 8, y + z 6 by rank method (OCT-07,MAR-0, JUN-6) x + y + z 7 x + y + z 8 y + z 6 The matrix form of the system of equations is 0 x y z A X B The augmented matrix is 7 [A, B] ~ ~ R R R R R R It is in the echelon form. ρ A, B, and ρ A ρ A, B ρ A The given system is inconsistent and has no solution. 5. Solve x y + z ; x + y z 7; x + y by rank method. (OCT-4) [A, B] 7 0 ~ ~ R R R R R R R R R The last equivalent matrix is in the echelon form. It has three non-zero rows. ρ A, ρ A, B ρ A, B ρ A The system of equations are inconsistent and no solution. 6. Using rank method, examine the consistency of the system x 4y + 7z 4, x + 8y z, 7x 8y + 6z 5. (OCT-,OCT-) The system of equations can be put in the matrix form as x y z A X B The augmented matrix is [A, B] ~ ~ R R R R R 7R R R R It is in the echelon form. ρ A, B and ρ A ρ A, B ρ A The system is inconsistent and no solution. COMPLEX NUMBERS 4 5. Find the real and imaginary part of +i (OCT-08) i i + i +i +i i Real part. Imaginary part ways00@gmail.com - -

33 th Maths Public Exam Special Guide Way to Success. Find the least positive integer n such that +i n (MAR-6) i +i +i +i +i i i +i +i i i i n i n The least positive integer n is 4. Find the real values of x and y for which the following equation is satisfied. i x + + i y i (JUN-) x + y + i( x + y) i Equating real and imaginary parts x + y, x + y Solving we get x, y 4. Find the real value of x and y for which x + x x + 4 i y( + i) (OCT-0,JUN-4) x + x x + 4 i y( + i) x + x + 8 y + i(y x 4) Equating real and imaginary parts x + x + 8 y x + x + 8 4y.. y x 4 0 Using () in () we get x + x + 8 4(x x) x + 9x x + 7 x x 7 or x 8 When x 7, y When 8, y x 7, y and x 8, y 4 5. State and prove Triangle inequality (OCT-09,JUN-0,MAR-,MAR-4) The modules of sum of two complex numbers is always less than or equal to the sum of their moduli. z + z z + z Let z and z be two complex numbers We know that z + z (z + z )(z + z ) z zz z + z (z + z ) z z + z z + z z + z z z z + z z + z z + z z z + z + Re (z z ) z + z + z z (Re z z ) z + z + z z z + z z + z z + z Thus taking positive square root we get z + z z + z 6. For any two complex numbers z, z (i) z z z z (ii)arg z. z arg z + arg z (OCT-07,JUN-08,JUN-) Let z r (cos θ + i sin θ ) and z r (cos θ + i sin θ ) z. z r r (cos θ + i sin θ )(cos θ + i sin θ ) r r [ cos θ. cos θ sin θ sin θ + i sin θ. cos θ + cos θ sin θ ] r r [cos(θ + θ ) + i sin ( θ + θ ) z z r r z. z and arg z. z θ + θ arg z + arg z 7. For any two complex numbers z and z r r (i) z z z z,(z 0) (ii) arg z arg z z arg z (JUN-4) Let z r (cos θ + i sin θ ) and z r (cos θ + i sin θ ) z r, arg z θ and z r, arg z θ z r (cos θ + i sin θ ) z r (cos θ + i sin θ ) r cos θ +i sin θ cos θ i sin θ r cos θ +i sin θ cos θ i sin θ cos θ. cos θ + sin θ sin θ + i sin θ. cos θ cos θ sin θ cos θ + i sin θ r [cos(θ r θ ) + i sin ( θ θ )] z r z and z r z arg z z θ θ arg z arg z 8. If a + ib a + ib (a n + ib n ) A + ib Prove that (i) a + b a + b a n + b n (A + B ) ways00@gmail.com - -

34 th Maths Public Exam Special Guide Way to Success (ii) tan b a + tan b a + + tan b n a n kπ + tan B, k Z (OCT-) A Given a + ib a + ib (a n + ib n ) A + ib a + ib a + ib (a n + ib n A + ib a + ib a + ib (a n + ib n A + ib a + b a + b a n + b n A + B Squaring on both sides a + b a + b a n + b n A + B Also, arg[ a + ib a + ib (a n + ib n )] arg (A + ib) arg a + ib + arg a + ib + + arg(a n + ib n ) arg (A + ib) tan b a + tan b a By taking the general value tan b a + tan b a + + tan b n a n tan B A + + tan b n a n kπ + tan B A, k Z 9. Show that the points representing the complex numbers 7 + 9i, + 7i, ( + i) form a right angled triangle on the Argand diagram. ( JUN-07,MAR-0,OCT-4, MAR-7) Let A, B and C represent the complex numbers i, + 7i and + i in the Argand diagram respectively. AB 7 + 9i ( + 7i) 0 + i BC + 7i ( + i) 6 + 4i CA + i (7 + 9i) 4 6i AB BC + CA BCA 90 Hence ABC is a right angled isosceles triangle. 0. Prove that the triangle formed by the points representing the complex numbers 0 + 8i, + 4i, ( + i) on the Argand plane is right angled. (OCT-0) Let A, B and C represent the points 0,8,,4, (,) respectively AB 0 + 8i ( + 4i) + 4i BC + 4i ( + i) 9 7i CA + i (0 + 8i) + i AB + BC CA 970 Hence the given points represents a right angled triangle on the Argand plane. Prove that the points representing the complex numbers 7 + 5i, 5 + i, (4 + 7i) and + 4i form a parallelogram (OCT-6) Let A, B, C and D represent the points, 4, 5,, (7, 5) and (4, 7) respectively Mid point of AC +7, 4+5, +7 9, 9 Mid point of BD 5+4 9, 9 Mid points of AC and BD are same. The given points form a parallelogram. ways00@gmail.com - -

35 th Maths Public Exam Special Guide Way to Success. Find the square roots of 8 6i Let x + iy 8 6i (MAR-06, OCT-06,JUN-5) Squaring on both sides x y + xyi 8 6i Equating real and imaginary parts, x y 8.() xy 6.. We know that x + y x y + (xy) 8 + ( 6) 00 x + y ()+() x,, x ± When x, y When x, y The roots are i, + i. Find the square roots of 7 + 4i Let x + iy 7 + 4i (MAR-07,JUN-09,MAR-5) Squaring on both sides x y + xyi 7 + 4i Equating real and imaginary parts, x y 7. xy 4.. We know that x + y x y + (xy) 7 + (4) 65 x + y x 9, y 6 x ±, y ±4 Since xy is positive, x and y have the same sign. x, y 4 or x, y i + 4i or ( 4i) 4. P represents the variable z. Find the locus of P if Re z+ z i 0 (OCT-) Let z x + iy, z x iy z + x iy + x + iy z i x iy i x i(y + ) Re z + z i z + z i 0 x+ x+y(y+) x + y+ 0 x + x + y + y 0 x + y + x + y 0 x + iy x + i(y + ) x i(y + ) x + i(y + ) x+ x+i x+ y+ iyx +y(y+) x + y+ 5. P represents the variable z. Find the locus of P if Re z+ z i 0 (MAR-08) Let z x + iy z + x + iy + x + + iy z i x + iy i x + i(y ) z+ z i x+ +iy x i(y ) x+i(y ) x i(y ) x+ x i x+ y +iyx +y(y ) x + y Re z+ 0 z i x+ x+y(y ) x + y 0 x + x + y y 0 x + x + y y 0 x + y + x y 0 6. P represents the variable complex number z. Find the locus of P. If Z 5i Z + 5i (MAR-5) Let z x + iy x + iy 5i x + iy + 5i x + i(y 5) x + i(y + 5) x + (y 5) x + (y + 5) Squaring on both sides, x + (y 5) x + (y + 5) (y 5) (y + 5) y 0y + 5 y + 0y + 5 0y 0 y 0 The locus of P is y 0 7. P represents the variable complex number z. Find the locus of P if z 5 z + (JUN-) z x + iy (x + iy) 5 x + iy + x + iy 5 x + + iy x 5 + iy x + + iy ways00@gmail.com

36 th Maths Public Exam Special Guide Way to Success x 5 + (y) x + + y Squaring on both sides and then expand 9x + 5 0x + 9y 9[ x + + x + y ] 9x + 5 0x + 9y 9x x + 9y 0 8x + 0x x 6 48x 6 x The locus in a straight lines which is parallel to y- axis. 8. P represents the variable complex number Z. Find the locus of P, if Z i Z + i (MAR-09) Let z x + iy x + iy i x + iy + i x + i(y ) x + i(y + ) x + (y ) x + (y + ) Squaring on both sides, x + (y ) x + (y + ) (y ) (y + ) y 6y + 9 y + 6y + 9 y 0 y 0 9. P represents the variable complex number Z. Find the locus of P, if Z 4i Z + 4i (OCT-5) Let z x + iy x + iy 4i x + iy + 4i x + i(y 4) x + i(y + 4) x + (y 4) x + (y + 4) Squaring on both sides, x + (y 4) x + (y + 4) (y 4) (y + 4) y 8y + 6 y + 8y + 6 6y 0 y 0 0. P represents the variable complex number z. Find the locus of P if z z (MAR-06) Let z x + iy (x + iy) x + iy x + iy x + iy x + (y) x + (y) Squaring on both sides and expand 4x + 4x + 4y x + 4 4x + y x + y x + y The locus is a circle, where centre is (0,0) and radius unit.. For any polynomial equation P x 0 with real coefficients, imaginary(complex) roots occur in conjugate pairs. (OCT-) Let P x a n x n + a n x n + + a x + a 0 0 be a polynomial equation of degree n with real coefficients. Let z be a root of P x 0. We show that z is also a root of P x 0. Since z is a root of P x 0 P z a n z n + a n z n + + a z + a 0 0 Taking the conjugate on both sides P z a n z n + a n z n + + a z + a 0 0 Using the idea that the conjugate of the sum of the complex numbers is equal to the sum of their conjugates, a n z n + a n z n + + a z + a 0 0 a n z n + a n z n + + a z + a 0 0 Since z n z n and a 0, a, a a n are real numbers, each of them is its own conjugate and hence we get a n z n + a n z n + + a z + a 0 0 Which is same as P z 0 This means z is also a root of P x 0 Hence the result.. Solve the equation x 4 4x + 8x if one roots is + i (MAR-6, JUN-6) Since + i is a root, i is also a root. Sum of the roots 4 Products of the roots ( + i)( i) The corresponding factor is x (sum of the roots)x+product of the roots x 4x + 7 x 4 4x + 8x + 5 x 4x + 7 (x + px + 5) Equating x term, we get 8 7p 0 p 4 The other factor is x + 4x + 5 x ± i Thus the roots are ± i, ± i ways00@gmail.com

37 th Maths Public Exam Special Guide Way to Success. Solve the equation x 4x + 6x 4 0 if ( i) is a root. ( JUN-07 ) One root is i The other root is + i Sum of the roots + i + i Product of the roots + i i +. The quadratic equation having the above roots x x Sum of the roots + Product of the roots 0 x x + 0 x 4x + 6x 4 x x + (x ) x 0 { + i, i, } 4. Solve the equation x 4 8x + 4x x if + i is a root (MAR-09,MAR-) Since + i is a root, i is also a root. Sum of the roots 6 Product of the roots 0 The corresponding factor is x (sum of the roots)x+product of the roots x 6x + 0 x 4 8x + 4x x + 0 x 6x + 0 (x + px + ) Equating x term, we get 0p p The other factor is x x + 0 x ± i Thus the roots are ± i, ± i 5. Solve the equation x 4 4x + x 4x if one roots is + i (JUN-09,MAR-) Since + i is a root, i is also a root. Sum of the roots Products of the roots 5 The corresponding factor is x (sum of the roots)x+product of the roots x x + 5 x 4 4x + x 4x + 0 x x + 5 (x + px + ) Equating x term, we get 5p 4 4 p The other factor is x x + 0 x ± i Thus the roots are ± i, ± i 6. Solve x , if + i is one of the roots. (JUN-06) If one of the root is + i The other root is i Sum of the roots + i + i Product of the roots + i i +. The quadratic equation having the above roots x x Sum of the roots + Product of the roots 0 x x + 0 x x + 0 x + 0x + 4 x x + x + px + 0 Equating coefficient of x on both sides 0 p 4 p 4 p Therefore, the other quadratic equation is x + x + 0 x 7. Simplify: ± 4 8 ± 4 ±i ± i { + i, i, + i, i} cos θ+i sin θ 4 (OCT-06,,6) sin θ+i cos θ 5 cos θ+i sin θ 4 cos θ+i sin θ 4 sin θ+i cos θ 5 cos π θ +i sin π θ 5 cos 4θ 5 π θ +i sin 4θ 5 π θ cos 9θ 5π +i sin 9θ 5π cos 5π 9θ i sin 5π 9θ cos π 9θ i sin π 9θ sin 9θ i cos 9θ 8. If n is a positive integer, prove that +sin θ+i cos θ n π cos n θ + i sin n π θ +sin θ i cos θ (MAR-) Let z sin θ + i cos θ z sin θ i cos θ + sin θ + i cos θ + z + sin θ i cos θ + z n z (sin θ + i cos θ) n n n cos π θ + i sin π n θ cos n π θ + i sin n π θ ways00@gmail.com

38 th Maths Public Exam Special Guide Way to Success 9. Simplify : cos θ i sin θ 7 cos θ+i sin θ 5 cos 4θ+i sin 4θ cos 5θ i sin 5θ 6 cos θ i sin θ 7 cos θ+i sin θ 5 cos 4θ+i sin 4θ cos 5θ i sin 5θ 6 ( JUN-) (cos θ+i sin θ) 4 (cos θ+i sin θ) 5 (cos θ+i sin θ) 48 (cos θ+i sin θ) 0 (cos θ + i sin θ) (cos θ + i sin θ) 07 cos 07θ i sin 07 θ 0. If n is a positive integer, then prove that ( + cos θ + i sin θ) n + ( + cos θ i sin θ) n n+ cos n θ ( + cos θ + i sin θ) n cos nθ cos θ + i sin θ cos θ cos θ cos θ + i sin θ ( JUN-,MAR-4) n cos n θ cos n θ + i sin n θ.. Replace i by i in () we get ( + cos θ i sin θ) n n cos n θ cos n θ i sin n θ. ()+() + cos θ + i sin θ n + + cos θ i sin θ n n cos n θ n n cos nθ n+ cos n θ cos nθ. If n is a positive integer, prove that ( + i) n + ( i) n n+ cos nπ 4 (OCT-07,MAR-08,MAR-0,OCT-5) + i r(cos θ + i sin θ) Equating real and imaginary parts r cos θ, r sin θ r + Also, cos θ, sin θ θ π 4 + i cos π 4 + i sin π 4 + i n n cos π 4 + i sin π 4 n cos nπ 4 + i sin nπ 4.. Replace i by i in () we get, i n n cos nπ 4 i sin nπ 4 n Adding () and () we get + i n + i n n cos nπ 4 n + cos nπ 4. If n N, prove that ( + i ) n + ( i ) n n+ cos nπ (JUN-08) + i r(cos θ + i sin θ) Equating real and imaginary parts r cos θ,r sin θ r + ( ) Also cos θ, sin θ θ π + i cos π + i sin π + i n n cos nπ + i sin nπ. Replace i by i in () we get, i n n cos nπ i sin nπ.. Adding () and () we get + i n + i n n cos nπ n+ cos nπ. If x + cos θ prove that x (i) x n + xn cos nθ, (ii) xn xn i sin nθ (JUN-0) x + cos θ x x cos θ x + 0 x cos θ± 4 cos θ 4 cos θ± cos θ x cos θ ± sin θ cos θ ± i sin θ Take x cos θ + i sin θ x n cos θ + i sin θ n x n cos nθ + i sin nθ.. x n cos nθ i sin nθ.. ()+ () x n + x n cos nθ () () x n x n i sin nθ ways00@gmail.com

39 th Maths Public Exam Special Guide Way to Success 4. If x cos α + i sin α, y cos β + i sin β Prove that x m y n + x m y n cos(mα + nβ) (MAR-07,OCT-) x m y n (cos α + i sin α) m (cos β + i sin β) n cos mα + i sin mα (cos nβ + i sin nβ) cos(mα + nβ) + i sin(mα + nβ) x m y n cos(mα + nβ) i sin(mα + nβ) x m y n + cos mα + nβ + i sin mα + nβ + x m y n cos mα + nβ i sin(mα + nβ) x m y n + x m y n cos(mα + nβ) 5. cos α + cos β + cos γ 0 sin α + sin β + sin γ then prove that cos α + cos β + cos γ cos( α + β + γ) and sin α + sin β + sin γ sin ( α + β + γ) (MAR-,OCT-) a cos α + i sin α b cos β + i sin β c cos γ + i sin γ a + b + c cos α + cos β + cos γ +i sin α + sin β + sin γ 0 + i0 0 We know that if a + b + c 0 then a + b + c abc cos α + i sin α + cos β + i sin β + cos γ + i sin γ (cos α + i sin α)(cos β + i sin β)(cos γ + i sin γ) cos α + i sin α + cos β + i sin β + (cos γ + i sin γ) [cos( α + β + γ) + i sin ( α + β + γ) ] cos α + cos β + cosγ + i(sin α +sin β + i sin γ) [cos( α + β + γ) + i sin ( α + β + γ) ] Equating real and imaginary parts, we get cos α + cos β + cos γ cos( α + β + γ) and sin α + sin β + sin γ sin ( α + β + γ) 6. cos α + cos β + cos γ 0 sin α + sin β + sin γ then prove that cos α + cos β + cos γ 0 and sin α + sin β + sin γ 0 (JUN-06,JUN-) a cos α + i sin α cos α i sin α a b cos β + i sin β cos β i sin β b c cos γ + i sin γ cos γ i sin γ c Here + + (cos α + cos β + cos γ) a b c (sin α + sin β + sin γ) 0 i 0 0 But a + b + c + ab + bc + ca (a + b + c) a + b + c + abc c a b a + b + c + abc 0 0 a + b + c 0 cos α + i sin α + cos β + i sin β + cos γ + i sin γ 0 cos α + i sin α + cos β + i sin β + cos γ + i sin γ 0 cos α + cos β + cos γ + i (sin α + sin β + sin γ) 0 Equating real and imaginary parts, we get cos α + cos β + cos γ 0, sin α + sin β + sin γ 0 7. Find the all values of i (MAR-, MAR-6) i cos π + i sin π i cos π + i sin π cos kπ + π + i sin kπ + π cos 4k + π + i sin 4k + π cos 4k + π 6 + i sin 4k + π, k 0,, 6 The values are cis π 6, cis 5π 6, cis 9π 6 8. If ω, prove that +i i (OCT-08,JUN-,MAR-) If ω is a cube root of unity, then ω +i ω i +i 5 + i 5 ω 5 + ω 5 ω 5 + ω 0 ω + ω 9. If x a + b, y aω + bω, z aω + bω Show that (i) xyz a + b (ii) x + y + z (a + b ) where ω is the complex cube root of unity (JUN-5) (i) xyz a + b aω + bω aω + bω ways00@gmail.com

40 th Maths Public Exam Special Guide Way to Success a + b a + abω + abω + b a + b a + ab(ω + ω) + b a + b a ab + b a + b (ii) x + y + z a + b + aω + bω + aω + bω a + ω + ω + b + ω + ω + c + ω + ω a 0 + b 0 + c(0) 0 If x + y + z 0 then x + y + z xyz x + y + z (a + b ) [by (i)] 40. Solve x (OCT-08,OCT-09,OCT-4) x x 4 4 4( ) x 4 4( ) 4 ( ) 4(cos π + i sin π) 4 (cos(kπ + π) + i sin(kπ + π)) 4 cos k + π 4 + i sin k + π 4, The values are k 0,,, cis π 4, cis π 4, cis 5π 4, cis 7π 4 4. If α and β are complex conjugates to each other and α + i then find α + β αβ(mar-7) Given, α and β are complex conjugates. α + i, β i α + i + i + i i i β i + i i + i + i αβ + i i i + α + β αβ i + + i 9. DISCRETE MATHEMATICS. (a) Construct the truth table for (p q) (~q) (b) Construct the truth table for p (~p) (JUN-06) The truth table for (p q) (~q) p q ~q p q (p q) (~q) T T F T F T F T T T F T F T F F F T F F The truth table for p (~p) p ~p p ~p T F F F T F The last column contains only F p (~p) is a contradiction. Construct the truth table for the statement (p q) [~(p q)] (JUN-4, MAR-5) Truth table for (p q) [~(p q)] p q p q ~(p q) (p q) [~(p q)] T T T F T T F F T T F T F T T F F F T T. Construct the truth table for ~( ~p ~q ) (OCT-06) The truth table of ~( ~p ~q ) p q ~p ~q (~p) (~q) ~( ~p ~q ) T T F F F T T F F T F T F T T F F T F F T T T F 4. Construct the truth table for (p q) (~r) (JUN-07, OCT-, MAR-,7) p q r p q ~ r (p q) (~r) T T T T F T T T F T T T T F T F F F T F F F T T F T T F F F F T F F T T F F T F F F F F F F T T ways00@gmail.com

41 th Maths Public Exam Special Guide Way to Success 5. Construct the truth table for (p q) r (MAR-08) p q r p q (p q) r T T T T T T T F T F T F T T T T F F T F F T T T T F T F T F F F T F F F F F F F 6. Construct the truth table for (p q) r (OCT-08, 5) p q r p q (p q) r T T T T T T T F T T T F T F T T F F F F F T T F T F T F F F F F T F T F F F F F 7. Show that ((~p) (~q)) p is a tautology.(oct-09) Truth table for ((~p) (~q)) p p q ~p ~q (~p) (~q) ((~p) (~q)) p T T F F F T T F F T T T F T T F T T F F T T T T The last column contains only T. ((~p) (~q)) p is a tautology 8. Verify whether (p ~q ) ((~p) q) is a tautology or contradiction (OCT-09, JUN-) p q ~p ~q p (~q) (~p) q (p (~q)) ((~p) q) T T F F F T T T F F T T F T F T T F F T T F F T T F T T (p ~q ) ((~p) q) is a tautology 9. Verify whether the statement q p ~q is a tautology or contradiction (MAR-7) p q ~q p (~q) q p ~q T T F T T T F T T T F T F F T F F T T T q p ~q is a tautology 0. Show that ((~q) p) q is a contradiction (MAR-, 6) Truth table for ((~q) p) q p q ~q (~q) p ((~q) p) q T T F F F T F T T F F T F F F F F T F F The last column contains only F. ((~q) p) q is a contradiction.. Show that p q (p q) is a tautology. (JUN-06,JUN-,OCT-,OCT-4) p q p p p q (p q) q q T T T T T T F T F T F T T F T F F F F T p q (p q) is a tautology.. Use the truth table to determine whether the statement ( ~p q) (p ( q)) (OCT-07) p q ~p ~q ~p q p (~q) ( ~p q) (p ( q)) T T F F T F T T F F T F T T F T T F T F T F F T T T F T The last column contains only T. The given statement is a tautology.. Use the truth table to establish which of the following statement is tautology or contradiction (p (~p)) ((~q) p) (JUN-08,0, MAR-) Truth table of (p (~p)) ((~q) p) p q ~p ~q p (~q) (~q) p (p (~p)) ((~q) p) T T F F F F F T F F T F T F F T T F F F F F F T T F F F (p (~p)) ((~q) p) is a contradiction. ways00@gmail.com

42 th Maths Public Exam Special Guide Way to Success 4. Show that p q (p q) (q p) (OCT-06, JUN-09, JUN-5) Truth table for p q p q p q T T T T F F F T F F F T Truth table for (p q) (q p) p q p q q p (p q) (q p) T T T T T T F F T F F T T F F F F T T T Both the tables have identical last columns p q (p q) (q p) 5. Show that ~ p q (~p) (~q) (MAR-06) Truth table for ~ p q p q p q ~ p q T T T F T F T F F T T F F F F T Truth table for (~p) (~q) p q ~p ~q (~p) (~q) T T F F F T F F T F F T T F F F F T T T The last columns are identical ~ p q (~p) (~q) 6. Show that p q (~p) q (MAR-) Truth table of p q p q p q T T T T F F F T T F F T Truth table of (~p) q p q ~p (~p) q T T F T T F F F F T T T F F T T The last columns in the truth tables of p q and (~p) q are identical p q (~p) q 7. Show that ~ p q ~p (~q) (OCT-0, 6) Truth table for ~ p q p q p q ~ p q T T T F T F F T F T F T F F F T Truth table for ~p (~q) p q ~p ~q (~p) (~q) T T F F F T F F T T F T T F T F F T T T Both the truth tables have identical last columns ~ p q ~p (~q) 8. Show that p q and q p are not equivalent. (MAR-09,JUN-6) Truth table for p q p q p q T T T T F F F T T F F T Truth table for q p p q q p T T T T F T F T F F F T The columns corresponding to p q and q p are not identical. p q is not equivalent to q p 9. Show that p q ~p q ( q) p (MAR-07,0,4, JUN-07-, OCT-08,,) Truth table for p q p q p q T T T T F F F T F F F T ways00@gmail.com

43 th Maths Public Exam Special Guide Way to Success The truth table of ~p q ( q) p p q ~p ~q ~p q ( q) p ~p q ( q) p T T F F T T T T F F T F T F F T T F T F F F F T T T T T Both the truth tables have last columns as identical. p q ~p q ( q) p 0. Show that the cube roots of unity forms a finite abelian group under multiplication. (OCT-4) G {, ω, ω }. The Cayley s table is. ω ω ω ω ω ω ω ω ω ω From the table, we see that (i) all the entries in the table are members of G. So the closure property is true. (ii) Multiplication is always associative. (iii) the identity element is and it satisfies the identity axiom. (iv) The inverse of is The inverse of ω is ω The inverse of ω is ω and it satisfies the inverse axiom also (G,. ) is a group. (v) the commutative property is also true. (G,. ) is an abelian group. (vi) Since G is a finite set, (G,. ) is a finite abelian group.. Prove that the set of all 4 th roots of unity forms an abelian group under multiplication. (JUN-) We know that the fourth roots of unity are, i,, i Let G {, i,, i }. The Cayley s table is. i i i i i i i i i i i i From the table, (i) the closure axiom is true. (ii) multiplication is always associative in C and hence in G (iii) the identity element is G and it satisfies the identity axiom (iv) the inverse of is ; the inverse of i is i the inverse of is ; the inverse of i is i. Further it satisfies the inverse axiom. Hence (G,. ) is a group. (v) From the table, the commutative property is also true. (G,. ) is an abelian group.. Prove that (Z, +) is an infinite abelian group. (OCT-08) (i) Closure axiom: we know that sum of two integers is again an integer. (ii) Associative axiom: Addition is always associative in Z a, b, c Z, a + b + c a + (b + c) (iii)identity axiom: The identity element O Z and it satisfies 0 + a a + 0 a, a Z Identity axiom is true. (iv) Inverse axiom: For every a Z, there exist an element a Z such that a + a a + a 0 Inverse axiom is true (Z, +) is a group (v) a, b Z, a + b b + a addition is commutative (Z, +) is an abelian group (vi) Since Z is an infinite set (Z, +) is an abelian group.. Show that the set of all non-zero complex numbers is an abelian group under multiplication of complex numbers (JUN-6) (i) Closure axiom: Let G C {0} Product of two non-zero complex numbers is again a nonzero complex number Closure axiom is true (ii) Associative axiom: Multiplication is always associative Associative property is true (iii) Identity axiom: + i0 G, is the identity element and ways00@gmail.com

44 th Maths Public Exam Special Guide Way to Success. z z. z z G Identity axiom is true. (iv) Inverse axiom: Let z x + iy G. Here z 0 x and y are not both zero x + y 0 z x+iy x iy x+iy x iy x iy x +y x y x +y + i G x +y Further z. z z. z z has the inverse z G Thus inverse axiom is satisfied G,. is a group (v) Commutative property: z z a + ib c + id ac bd + i(ad + bc) ca db + i da + cb z z it satisfies the commutative property G is an abelian group under the usual multiplication of complex numbers. 4. Show that the set of all non- singular matrices forms a non- abelian infinite group under matrix multiplication (Where the entries belongs to R) (OCT-07) Let G be the set of all non-singular matrices, where the entries belongs to R. (i) Closure axiom: Since product of two non-singular matrices is again non-singular and the order is, the closure axiom is satisfied. A, B G AB G (ii) Associative axiom: Matrix multiplication is always associative and hence associative axiom is true. A BC AB C, A, B, C G (iii) Identity axiom: The identity element is I 0 G and it satisfies the identity property. (iv) Inverse axiom: 0 The inverse of A G, exists, i.e. A exists and is of order and AA A A I. Thus the inverse axiom is satisfied. Hence the set of all non-singular matrices forms a group under matrix multiplication. Further, matrix multiplication is noncommutative (in general) and the set contain infinitely many elements. The group is an infinite non- abelian group. 5. Show that the set of four matrices ,,, form an abelian group, under multiplication of matrices. (MAR-,OCT-,MAR-5) I 0 0 B 0, A 0 0,, C and let G {I, A, B, C} By computing the products of these matrices, taken in pairs, we can form the multiplication table as given below. I A B C I I A B C A A I C B B B C I A C C B A I (i) All the entries in the multiplication tables are members of G. So, G is closed under. Closure axiom is true. (ii) Matrix multiplication is always associative (iii) Since the row headed by I coincides with the top row and the column headed by I coincides with the extreme left column, I is the identity element in G (iv) I. I I I is the inverse of I A. A I A is the inverse of A B. B I B is the inverse of B C. C I C is the inverse of C From the table it is clear that. is commutative G is an abelian group under matrix multiplication. ways00@gmail.com

45 th Maths Public Exam Special Guide Way to Success 6. With usual symbols prove that (Z 5 0,. 5 ) is a group (JUN-0) G Z 5 0 {,,, [4]} The Cayley s table is. 5 [] [] [] [4] [] [] [] [] [4] [] [] [4] [] [] [] [] [] [4] [] [4] [4] [] [] [] From the table (i) all the elements of the composition table are the elements of G The closure axiom is true. (ii) multiplication modulo 5 is always associative. (iii) the identity element [] G and satisfies the identity axiom (iv) the inverse of [] is [], [] is [], [] is [] and [4] is [4] and it satisfies the inverse axiom. The given set forms a group under multiplication modulo Find the order of each element of the group (Z 7 0,. 7 ) (MAR-) The given group is (,,, 4, 5, 6,. 7 ) O ; O O 6 ; O 4 O 5 6 ; O 6 8. Find the order of all the elements of the group (z 6, + 6 ) (JUN-08) z 6 { 0,,,, 4, [5]} O 0 ; O 6 O ; O O 4 ; O State and prove cancellation laws on groups. (MAR-06,MAR-08,OCT-0, MAR-4) Let G be a group. Then for all a, b, c G (i) a b a c b c (left cancellation law) (ii) b a c a b c (right cancellation law) (i) a b a c a a b a a c a a b (a a) c e b e c b c (ii)b a c a b a a c a a b a a c (a a ) b e c e b c 0. State and prove reversal law on inverses of a group (OR) Let G is a group a, b G. Then a b b a (MAR-07,0, JUN-09,,4,5, OCT-, 5) It is enough to prove b a is the inverse of (a b) To prove (i) a b b a e (ii) b a a b e (i) a b b a a b b a a e a a a e (ii) b a a b b a a b b e b b b e b a is the inverse of a b a b b a. The identity element of a group is unique (JUN-,OCT-6) Let G be a group. If possible let e and e be identity elements in G. Treating e as an identity element we have e e e. Treating e as an identity element, we have e e e From () and (), e e Identity element of a group is unique.. The inverse of each element of a group is unique. (JUN-, OCT-6) Let G be a group and let a G If possible, let a and a be two inverses of a. Treating a as an inverse of a we have a a a a e Treating a as an inverse of a we have a a a a e Now a a e a a a (a a) a e a a Inverse of an element is unique. ways00@gmail.com

46 th Maths Public Exam Special Guide Way to Success. If every element of a group is its own inverse then prove that the group is abelian (MAR-6) Let G be a group Let a, b G Given a a and b b Let x G, x G Let x ab Given x x ab ab b a ba Commutative axiom satisfied. Hence G is an abelian group. Section C (0 Marks). Vector Algebra. Altitudes of a triangle are concurrent Prove by vector method. ( OCT-06,JUN-08,OCT-,MAR-5) Let ABC be a triangle and let AD, BE be its two altitudes intersecting at O. In order to prove that the altitudes are concurrent it is sufficient to prove that CO is perpendicular to AB. Taking O as the origin, let the position vectors of A, B, C be a, b, c respectively. Then OA a, OB b, OC c Now AD BC OA BC OA. BC 0 a. c b 0 a. c a. b 0 BE CA OB CA OB. CA 0 b. a c 0 b. a b. c 0. Adding () and (), we get, a. c a. b + b. a b. c 0 a. c b. c 0 ways00@gmail.com

47 th Maths Public Exam Special Guide Way to Success (a b). c 0 BA. OC 0 OC AB Hence, the three altitudes are concurrent.. If a i + j k, b i + 5k, c j k Verify that a b c a. c b a. b c b c ( MAR-07,OCT-08,OCT-09, JUN-6) i j k i 6j k a b c i j k 5 6 i + 9j + k a. c i + j k. j k a. c b 6 i + 5k i + 0k a. b i + j k. i + 5k a. b c 9 j k 9j + 7k. Verify a b c d a b d c a b c d, for a i + j + k, b i + k, c i + j + k, d i + j + k. ( MAR-09,OCT-,OCT-,MAR-6) a b c d a b c d a b c a b d a b d c a b c d From (),() i j k 0 i + j k i j k i j + k i j k 5i j 4k. 0 0 i + j + k (i + j + k) 4i j k i j k 5i j 4k.. a b c d a b d c a b c d a. c b a. b c i + 0k + 9j 7k a. c b a. b c i + 9j + k Hence, a b c a. c b a. b c ways00@gmail.com

48 th Maths Public Exam Special Guide Way to Success 4. Prove that cos A B cos A cos B + sin A sin B (JUN-,JUN-) 5. Prove that cos A + B cos A cos B sin A sin B ( MAR-06,08,4,7, JUN-, OCT-4) 6. Prove that sin A + B sin A cos B + cos A sin B (OCT-08,MAR-,MAR-,JUN-4) 7. Prove that sin A B sin A cos B cos A sin B ( JUN-07, MAR-, OCT-07,0, 5,6) Take the points P and Q on the unit circle with centre at the origin O. Assume that OP and OQ make angles A and B with x-axis respectively. POQ POx QOx A B POQ POx + QOx A + B POQ POx + QOx A + B POQ POx QOx A B 4 Clearly the coordinates of P and Q are (cos A, sin A) and (cos B, sin B) Clearly the coordinates of P and Q are (cos A, sin A), (cos B, sin B) Clearly the coordinates of P and Q are cos A, sin A, (cos B, sin B) Clearly the coordinates of P and Q are (cos A, sin A), (cos B, sin B) 5 Take the unit vectors i and j along x and y axes. 6 OP OM + MP cos Ai +sin A j OP OM + MP cos Ai +sin A j OP OM + MP cos Ai +sin A j OP OM + MP cos Ai +sin A j OQ OL + LQ cos Bi +sin B j OQ ON + NQ cos Bi sin B j OQ ON + NQ cos Bi sin B j OQ OL + LQ cos Bi +sin B j 7 By value, OP. OQ cos Ai +sin A j. cos Bi +sin B j cos A cos B + sin A sin B. By value, OP. OQ cos Ai +sin A j. cos Bi sin B j cos A cos B sin A sin B. By value, OQ OP OQ. OP sin A + B k sin(a + B) k By value, OQ OP OQ. OP sin A B k sin(a B) k 8 By definition, By definition, By definition, By definition, OP. OQ OP. OQ cos(a B) cos (A B).. OP. OQ OP. OQ cos(a + B) cos (A + B).. OQ OP i j k cos B sin B 0 cos A sin A 0 OQ OP i j k cos B sin B 0 cos A sin A 0 k(sin A cos B + cos A sin B) k(sin A cos B cos A sin B)...() 9 From () and () cos A B cos A cos B + sin A sin B From () and () cos A + B cos A cos B sin A sin B From () and () sin A + B sin A cos B + cos A sin B From () and () sin A B sin A cos B cos A sin B ways00@gmail.com

49 th Maths Public Exam Special Guide Way to Success 8. Show that the lines x y z+ and 0 x 4 y z+ intersect and hence find the point 0 of intersection. ( JUN-07,JUN-09,JUN-5 ) The condition for intersecting is x x y y z z l m n 0 l m n Compare with x x x x l y y m z z n y y z z l m n and We get, x x 4 l l y y 0 m m 0 z z n 0 n The determinant becomes ( 0) (0 + ) The lines are intersecting lines Point of intersection: Take x y z+ λ 0 x λ y λ z+ λ 0 x λ y λ z + 0 x λ + y λ + z Any point on the line is of the form λ +, λ +, Take x 4 y 0 z+ μ x 4 μ y 0 μ z+ μ x 4 μ y 0 z + μ x μ + 4 y 0 z μ Any point on the line is of the form μ + 4,0,μ Since they are intersecting, for some λ, μ λ +, λ +, μ + 4,0,μ λ + μ + 4 λ + 0 λ μ μ 0 The point of intersection is (4,0, ) 9. Show that the lines x x y y+ z and z intersect and find the point of intersection. ( JUN-06,JUN-0,JUN- ) The condition for intersecting is x x y y z z l m n 0 l m n Compare with x x x x l y y m z z n We get, y y z z l m n and x x l l y y m m z 0 z n n The determinant becomes ( + ) The lines are intersecting lines Point of intersection: Take x x y+ z x λ y + λ z λ x λ + y λ z λ Any point on the line is of the form λ +, λ, λ Take x y z+ μ x μ y μ z+ μ x μ y μ z + μ x μ + y μ + z μ Any point on the line is of the form μ +, μ +, μ Since they are intersecting, for some λ, μ λ +, λ, λ μ +, μ +, μ λ + μ +. λ μ +. ways00@gmail.com

50 th Maths Public Exam Special Guide Way to Success λ μ.. Solve () and () μ + 0 μ μ Substituting in () λ + + λ + λ 0 The point of intersection (,0) 0. Find the vector and Cartesian equations of the plane through the point (,, ) and parallel to the lines x Vector form: y z 4 x and y+ z (MAR-0,OCT-,JUN-6, OCT-6) The required plane passes through A(,, ) and parallel to u i + j 4k and v i j + k. a i j k The required equation is r a + su + tv r i j k + s i + j 4k Cartesian form: +t(i j + k) The equation of the plane is x x y y z z l m n 0 l m n x l l y m m z n 4 n x y + z x 4 y x 8 y z z + 0 8x + 6 4y 4 z 9 0 8x 4y z x + 4y + z This is the required equation in Cartesian form.. Find the vector and Cartesian equations of the plane passing through the points (,, ) and (,, ) and perpendicular to the plane x + y + z 5 ( MAR-07,MAR-09,JUN-0,OCT-4) Vector form: The vector equation of the plane passing through two given points and parallel to a vector is r ( s)a + sb + tv Here a i + j + k, b i j + k, v i + j + k r s i + j + k + s i j + k Cartesian form: The equation of the plane is x x y y z z x x y y z z l m n x x l y y m z z n +t(i + j + k) x + y z 0 0 x y z x + 4 y (4) + z 6 0 4x 4 4y z 6 0 4x 4y + 6z 6 0 4x + 4y 6z x + y z + 0. Find the vector and Cartesian equations of the plane passing through the points,,,, 4,,and (7, 0, 6) Vector form: (OCT-09) Vector equation of the plane passing through three given non collinear points is r ( s t)a + sb + tc ways00@gmail.com

51 th Maths Public Exam Special Guide Way to Success Here a i + j k, b i + 4j + k, c 7i + 6k r s t i + j k +s(i + 4j + k) + t(7i + 6k) Cartesian form: The equation of the plane is x x y y z z x x y y z z 0 x x y y z z x x x 7 y y 4 y 0 z z z 6 x y z x y z + ( 0) 0 x 0 y 8 + z + ( ) 0 0x y 6 z 0 0x + 8y z x + y z 7 0. Find the vector and Cartesian equation of the plane containing the line x y Vector form: z and x+ y u i + j + k and v i + j + k a i + j + k z+ (MAR-4) The required equation is r a + su + tv r i + j + k + s i + j + k Cartesian form: +t(i + j + k) The equation of the plane is x x y y z z l m n 0 l m n x l l y m m z n n x y z x 6 y 9 + z x y 7 + z x y 4 5z x + 7y 5z 0 x 7y + 5z + 0 This is the required equation in Cartesian form. 4. Find the vector and Cartesian equation of the plane through the point (,, ) and parallel to the lines x+ Vector form: y+ z+ and x u i j + k and v i + j + k a i + j + k y+ z+ (JUN-) The required equation is r a + su + tv r i + j + k + s i j + k Cartesian form: +t(i + j + k) The equation of the plane is x x y y z z l m n 0 l m n x l l y m m z n n x y z 0 x 6 y 4 + z x 8 y + z 5 0 8x + 8 y + + 5z 0 0 8x y + 5z + 0 8x + y 5z 0 ways00@gmail.com

52 th Maths Public Exam Special Guide Way to Success 5. Find the vector and Cartesian equation to the plane through the point (,, ) and perpendicular to the planes x + y + z 5 and x + y + z 8 (JUN-) Vector form: The required equation is r a + su + tv u i + j + k and v i + j + k a i + j + k The required equation is r a + su + tv r i + j + k + s i + j + k +t(i + j + k) Cartesian form: The equation of the plane is x x y y z z l m n 0 l m n x l l y m m z n n x + y z 0 x + 4 y 6 + z 6 0 x + y 4 + z 5 0 x + + 4y 5z x + 4y 5z 0 6. Find the vector and Cartesian equation of the plane passing through the points A(,, ) and B(,, ) and is parallel to the line x y+ z 4 Vector form: The required equation is r ( s)a + sb + tv ( JUN-4 ) Here a i j + k, b i + j k, v i + j + 4k r s i j + k + s i + j k +t(i + j + 4k) Cartesian form: The equation of the plane is x x y y z z x x y y z z 0 l m n x x l y y m z z n 4 x y + z x 6 + y z x 8 y z 4 0 8x 8 4z x 4z x z Find the vector and Cartesian equation of the plane through the points (,, ) and (,, ) perpendicular to the plane x y + 4z 5 0 Vector form: (MAR-06,, OCT-06,07,5, JUN-08, 5) The required equation is r ( s)a + sb + tv Here a i + j + k, b i + j + k, v i j + 4k r s i + j + k + s i + j + k Cartesian form: The equation of the plane is x x y y z z x x y y z z l m n x x l y y m z z n 4 x y z 4 +t(i j + 4k) 0 0 ways00@gmail.com

53 th Maths Public Exam Special Guide Way to Success x 4 4 y z 0 x 0 y 0 + z 5 0 0y + 0 5z y + 5z y + z Find the vector and Cartesian equation of the plane containing the line x y z and passing through (,, ) ( MAR-,MAR-6 ) Vector form: The required equation is r ( s)a + sb + tv Here a i + j k, b i + j + k, v i + j k r s i + j k + s i + j + k Cartesian form: The equation of the plane is x x y y z z x x y y z z l m n x x l y y m z z n +t(i + j k) 0 x + y z + 0 x + 6 y z x + 8 y 0 + z x 8 + 0y 0 + 7z x + 0y + 7z 0 8x 0y 7z Find the vector and Cartesian equation of the plane passing through the points with position vectors i + 4j + k, i j k and 7i + k ( JUN-09, MAR-,OCT-, MAR-7) Vector form: The required equation is r ( s t)a + sb + tc Here a i + 4j + k, b i j k, c 7i + k r s t i + 4j + k Cartesian form: +s(i j k) + t(7i + k) The equation of the plane is x x y y z z x x y y z z 0 x x y y z z x x x 7 y 4 y y 0 z z z x y 4 z x 6 y z (4 + 4) 0 x 6 y 4 + z (8) 0 6x + 8 y z x y + 8z x + y 8z Derive the equation of the plane in the intercept form: (MAR-0, MAR-5) Note: We can derive this in either Cartesian form or Vector form. But both are vector method. Cartesian form: Let a, b and c be the x, y and z intercepts of the plane respectively. The plane passes through the points a, 0,0, 0, b, 0, (0,0, c) Here The equation is x x y y z z x x y y z z 0 x x y y z z ways00@gmail.com

54 th Maths Public Exam Special Guide Way to Success x a x 0 x 0 y 0 y b y 0 z 0 z 0 z c x a y 0 z 0 a b a 0 c 0 x a bc y 0 ac + z 0 (0 + ab) 0 x a bc + yac + zab 0 xbc abc + yac + zab 0 xbc + yac + zab abc xbc abc + yac + zab abc abc abc abc abc x + y + z a b c Vector form: The equation of the plane passing through three given points is r ( s t)a + sb + tc r ( s t)ai + sbj + tck xi + yj + zk ( s t)ai + sbj + tck x x s t a; y sb; z tc x a s t, y b s, z c t + y + z s t + s + t a b c x + y + z a b c. Find the vector and Cartesian equations of the plane passing through the point (,, ) and perpendicular to two planes x + y + 4z and x y + z + 0 (JUN-06,MAR-08) Vector form: u i + j + 4k and v i j + k a i j + k The required equation is r a + su + tv r i j + k + s i + j + 4k Cartesian form: The equation of the plane is +t(i j + k) x x y y z z l m n 0 l m n x l l y m m z n 4 n x + y + z 4 x y + 8 x + 0 y z z 5 0 0x y + 0 5z x + 5y 5z x + y z Find the vector and Cartesian equations of the plane, through the point (,, ) and parallel to the line x+ y+ z 4 and perpendicular to 4 the plane x + y + z 8 (OCT-0) Vector form: u i j 4k and v i + j + k a i + j k The required equation is r a + su + tv r i + j k + s i j 4k Cartesian form: +t(i + j + k) The equation of the plane is x x y y z z l m n 0 l m n x l l y m m z n 4 n x y z x 6 + y z ways00@gmail.com

55 th Maths Public Exam Special Guide Way to Success x 6 y 7 + z + 0 6x 6 7y z x 7y + z Find the vector and Cartesian equations of the plane which contains the line x y z+ x y + z 0 Vector form: and perpendicular to the plane (OCT-) The required equation is r a + su + tv a i + 0j k, u i j + k, v i j + k r i + 0j k + s i j + k Cartesian form: +t(i j + k) The equation of the plane is x x y y z z l m n 0 l m n x l l y 0 m m z n n x y z + x 9 + y z x 7 y 5 + z + 0 7x + 7 5y z 0 7x + 5y + z 6 0. Complex Numbers. P represents the variable complex number z. Find the locus of P, if Im z+ iz+ (MAR-0,JUN-) Let z x + iy z+ (x+iy )+ x+ +iy iz+ i(x+iy )+ y +ix x+ +iy y ix y +ix y ix x+ y +xy +i[y y x(x+)] y +x But, Im z+ iz+ y y x(x+) y +x y y x x [ + y y + x ] y y x x y + 4y x y x + 0 The locus of P is x + y 0. P represents the variable complex number z. Find the locus of P, if Re z (JUN-) z+i Let z x + iy z x+iy x +iy z+i x+iy +i x+i(y+) Re z z+i x +iy x i(y+) x+i(y+) x i(y+) x x +y(y+) + i [ imaginary part] x +(y+) x x +y(y+) x +(y+) x + y x + y x + y + y + x y Locus of P is x + y + 0. P represents the variable complex number z. Find the locus of P, if Im z+i (MAR-4) iz Let z x + iy z+i (x+iy )+i x+iy+i iz i x+iy ix y x+i(y+) y+ ix y+ +ix y+ ix x y+ ix i y+ y+ +x(y+) y+ +x But, Im z+i iz x y+ y+ y+ +x ways00@gmail.com

56 th Maths Public Exam Special Guide Way to Success x y y y y x x + y + y 0 The locus of P is x + y + y 0 4. P represents the variable complex number z. Find the locus of P, if arg z π z+ (MAR-) arg z arg z + π arg x + iy arg x + iy + π arg (x ) + iy arg x + + iy π y tan x tan tan y y x y x+ + y x y x+ x +y tan π y x +y y x+ π π y (x + y ) x + y y 0 is the required locus. 5. P represents the variable complex number z. Find the locus of P, if Re z+ z+i Let z x + iy z + z + i (MAR-6) x + iy + x + iy + i x+ +iy x i(y+) x+i(y+) x i(y+) x + + iy x + i(y + ) x x+ +y(y+) + i [ imaginary part] x +(y+) Re z z + i x x + + y(y + ) x + (y + ) x + y + x + y x + y + y + x y Locus of P is x y 6. If α and β are the roots of x x + 0 and cot θ y +,Show that y+α n y+β n α β sin nθ sin n θ (MAR-06,OCT-,OCT-4) The roots of the equation x x + 0 are ± i Let α + i, β i y + α n cot θ + ( + i) n cot θ + i n sin n θ n cos θ + i sin θ y + α n [cos nθ + i sin nθ] sin n θ Similarly, y + β n [cos nθ i sin nθ] sin n θ i sin nθ y + α n y + β n sin n θ α β + i i i Further, y + α n y + β n i sin nθ sin nθ α β i sin n θ sin n θ 7. If α and β are the roots of x px + p + q 0 and tan θ q y+p y+α n y+β n, Show that n sin nθ q α β sin n θ x px + p + q 0 x p± 4p 4 p +q p ± iq Let α p + iq, β p iq α β qi Given tan θ q y+p q y + p tan θ y + p q cot θ y q cot θ p y + α q cot θ p + p + iq q cot θ + i q cos θ+i sin θ sin θ y + α n n cos θ+i sin θ n q y + α n Similarly y + β n () () gives q n sin n θ q n sin n θ sin n θ y + α n y + β n y+α n y+β n α β (MAR-07,OCT-09,6) [cos nθ + i sin nθ].. [cos nθ i sin nθ] q n q n sin n θ iq sin n θ n sin nθ q sin n θ [i sin nθ] i sin nθ ways00@gmail.com

57 th Maths Public Exam Special Guide Way to Success 8. If α and β are the roots of x x prove that α n β n i n+ sin nπ and deduct α 9 β 9 (OCT-06,OCT-08,MAR-09,MAR-,JUN-5) x x x ± i α + i, β i α n + i n n cos nπ + i sin nπ β n i n n cos nπ i sin nπ α n β n n cos nπ + i sin nπ α n β n n i sin nπ α n β n i n+ sin nπ Put n 9 in the above result α 9 β 9 i 0 sin 9π i 0 sin π 0 n cos nπ i sin nπ 9. If a and b are the roots of x + x then find value of a n + b n. Also deduce the value of a + b (n is an integer)(jun-6) x + x x ± 4 (4) () ± 6 ± 4 ± i ( ± i) x ± i a + i, b i a n + i n n cos 5π 6 + i sin 5π 6 n cos 5nπ 5nπ + i sin 6 6 b n i n n cos 5nπ 5nπ i sin 6 6 a n + b n n cos 5nπ 6 n cos 5nπ 6 + i sin 5nπ 6 + i sin 5nπ 6 + n cos 5nπ 6 + cos 5nπ 6 i sin 5nπ 6 n i sin 5nπ 6 n cos 5nπ 6 a n + b n n+ cos 5nπ 6 Put n in the above result a + b cos 5()π 6 cos 0 π Solve the equation x 9 + x 5 x 4 0 (JUN-06,) x 9 + x 5 x 4 0 x 5 x 4 + x x 5 x x 5 0; x (i) x 5 cos 0 + i sin 0 5 cos k π + i sin kπ 5 cos kπ kπ + i sin 5 5 k 0,,,,4 (ii) x 4 cos π + i sin π 4 cos(k + ) π + i sin(k + )π 4 (k + )π (k + )π cos + i sin 4 4 k 0,,, Thus we have 9 roots.. Solve the equation x 7 + x 4 + x + 0 (JUN-09) x 7 + x 4 + x + 0 x 4 x + + x + 0 x 4 + x + 0 x 4, x (i) x 4 cos π + i sin π 4 cos(k + ) π + i sin(k + )π 4 cos (k+)π 4 + i sin (k+)π 4 k 0,,, (ii) x cos π + i sin π cos(k + ) π + i sin(k + )π cos (k+)π + i sin (k+)π k 0,, ways00@gmail.com

58 th Maths Public Exam Special Guide Way to Success. Solve the equation x 4 x + x x + 0 ( JUN-08,JUN-0,OCT-, MAR-7) The terms of the polynomial are in G.P with r x, a, n 5 x + x x + x 4 a rn x5 + x + r Where x Solve x and remove the root x x x 5 cos π + i sin π 5 cos(k + ) π + i sin(k + )π 5 cos (k+)π + i sin (k+)π 5 5 Where k 0,,,,4 The values are cis π 5, cis π 5, cis π, cis 7π 5, cis 9π 5 Remove the root cis π The roots are cis π 5, cis π 5, cis 7π 5, cis 9π 5. If x + cos θ and y + cos φ show that x y (i) xm y (i) xm y yn n + x yn n x m cos( mθ nφ) m i sin( mθ nφ) (JUN-4) x + cos θ x x cos θ x + 0 x cos θ ± i sin θ Take x cos θ + i sin θ, Similarly, y cos φ + i sin φ x m cos mθ + i sin mθ y n cos nφ + i sin nφ x m y n cos mθ + i sin mθ cos nφ + i sin nφ cos mθ + i sin mθ (cos nφ + i sin( nφ)) x m y n cos( mθ nφ) + i sin( mθ nφ) y n x m cos( mθ nφ) i sin( mθ nφ) () + () () x m y n x m y n y n + cos( mθ nφ) x m y n i sin( mθ nφ) x m 4. If a cos α + i sin α, b cos β + i sin β, c cos γ + i sin γ prove that (i) abc + abc cos( α + β + γ) (ii) a b +c cos( α + β γ) (OCT-0) abc abc (cos α + i sin α) (cos β + i sin β) (cos γ + i sin γ) abc (cos α + i sin α)(cos β + i sin β)(cos γ + i sin γ) abc cos( α + β + γ) + i sin( α + β + γ) abc cos( α + β + γ) i sin( α + β + γ) (ii) a b +c abc ab c ab c ab abc + abc a b abc + c abc ab c cos( α + β + γ) + c ab cos α+i sin α (cos β+i sin β ) (cos γ+i sin γ) cos( α + β γ) + i sin( α + β γ) c cos( α + β γ) i sin( α + β γ) ab + c cos( α + β γ) + ab i sin( α + β γ) + cos( α + β γ) i sin( α + β γ) cos( α + β γ) c a b +c abc cos( α + β γ) 5. Find all the values of + i + i r(cos θ + i sin θ) r cos θ, r sin θ r ( ) + cos θ, sin θ θ π 6 + i cos π + i sin π 6 6 cos π + i sin π (JUN-07) cos kπ + π + i sin kπ + π cos 6k + π 9 + i sin 6k + π 9 where k 0,, The values are cis π 9, cis 7π 9, cis π 9 ways00@gmail.com

59 th Maths Public Exam Special Guide Way to Success 6. Find all the values of i i r(cos θ + i sin θ) r cos θ, r sin θ r ( ) + (OCT-) cos θ, sin θ θ π + π 5π 6 6 i cos 5π i 6 cos 5π + i sin 5π i sin 5π 6 cos kπ 5π 6 + i sin kπ 5π 6 cos k 5 π 9 + i sin k 5 π 9 k 0,, The values are, cis 7π 9, cis 9π cis 5π 9 7. Find all the values of i that the product of the value is. i r(cos θ + i sin θ) r cos θ, r sin θ r (or) cis π and hence prove (OCT-07,MAR-08,MAR-, OCT-5) cos θ, sin θ θ in the 4th quadrant θ π i i cos π 4 cos π + i sin π + i sin π cos π + i sin π 4 cos kπ π + i sin kπ π 4 cos k 4 4 π + i sin k π 4 k 0,,, The values are cis π, cis π, cis π, cis 5π Product of these values is Cis π 4 + π 4 + π 4 + 5π 4 cis 8π 4 cisπ cos π + i sin π 4 8. Find all the values of + i and hence prove that the product of the value is. (MAR-5) + i r(cos θ + i sin θ) r cos θ, r sin θ r cos θ, sin θ θ in the st quadrant θ π + i cos π + i sin π + i 4 cos π + i sin π 4 cos π + i sin π 4 cos kπ + π + i sin kπ + π 4 cos k+ 4 π + i sin k+ π 4 k 0,,, The values are cis π 4, cis π 4, cis 5π 4, cis 7π 4 Product of these values is cis π 4 + π 4 + 5π 4 + 7π 4 cis 6π 4 cis4π cos 4π + i sin 4π 4. ANALYTICAL GEOMETRY. A girder of a railway bridge is in the parabolic form with span 00ft. and the highest point on the arch is 0ft. above the bridge. Find the height of the bridge at 0ft. to the left or right from the midpoint of the bridge. (MAR-09,JUN-6) Consider the parabolic girder as open downwards. x 4ay It passes through (50, 0) a( 0) a 50 4 x y ways00@gmail.com

60 th Maths Public Exam Special Guide Way to Success x 50y Let B(0, y ) be a point on the parabola y y Let AB be the height of the bridge at 0ft to the right from the midpoint AC 0 and BC 5 AB ft The height of the bridge at the required place 9 5 ft. On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4mts when it is 6 mts away from the point of projection. Finally it reaches the ground mts away from the starting point. Find the angle of projection. (MAR-06,JUN-09,JUN-0,JUN-,OCT-,MAR-4, MAR-7) The equation of the parabola is of the form x 4ay (by taking the vertex at the origin). It passes through (6, 4) 6 6a a 9 4 The equation is x 9y.. Find the slope at ( 6, 4) Differentiating () with respect to x, we get x 9 dy dx dy dx 9 x At 6, 4, dy 6 4 dx 9 tan θ 4 θ tan 4 The angle of projection is tan 4. Assume that water issuing from the end of a horizontal pipe, 7.5m above the ground, describes a parabolic path. The vertex of the parabolic path is at the end of the pipe. At a position.5m below the line of the pipe, the flow of water has curved outward m beyond the vertical line through the end of the pipe. How far beyond this vertical line will the water strike the ground? (OCT-09,MAR-,OCT-) As per the given information, we can take the parabola as open downwards x 4ay Let P be the point on the flow path,.5m below the line of the pipe and m beyond the vertical line through the end of the pipe. P(,.5) Thus, 9 4a(.5) a 9 0 The equation of the parabola is x y Let x be the distance between the bottom of the vertical line on the ground from the pipe end and the point on which the water touches the ground. But the height of the pipe from the ground is 7.5 m The point (x, 7.5) lies on the parabola x x The water strikes the ground m beyond the vertical line. 4. A comet is moving in a parabolic orbit around the sun which is at the focus of a parabola. When the comet is 80 million kms from the sun, the line segment from the sun to the comet makes an angle of π radians with the axis of the orbit. Find the i equation of the comet s orbit ii how close does the comet come nearer to the sun?(take the orbit as open rightward) (MAR-08,MAR-,JUN-,MAR-6, OCT-5,6) ways00@gmail.com

61 th Maths Public Exam Special Guide Way to Success Take the parabolic orbit as open rightward and the vertex at the origin. Let P be the position of the comet in which FP 80 million kms. Draw a perpendicular PQ from P to the axis of the parabola. Let FQ x From the triangle FQP, PQ FP. sin π Thus, FQ x FP. cos π VQ a + 40 if VF a P is VQ, PQ (a + 40,40 ) Since P lies on the parabola y 4ax (40 ) 4a(a + 40) a 60 or a 0 a 60 is not acceptable. The equation of the orbit is y 4 0 x y 80x The shortest distance between the Sun and the Comet is VF The shortest distance is 0 million kms. 5. A cable of a suspension bridge hangs in the form of a parabola when the load is uniformly distributed horizontally. The distance between two towers is 500 ft, the points of support of the cable on the towers are 00 ft above the road way and the lowest point on the cable is 70ft above the roadway. Find the vertical distance to the cable (parallel to the roadway) from a pole whose height is ft. (OCT-07, OCT-, JUN-4) Take the lowest point on the cable as the vertex and take it as origin. Let AB and CD be the towers. Since the distance between the two towers is 500 ft VA 750ft AB 00ft A B ft Thus the point B is (750,0) The equation of the parabola is x 4ay Since B is a point on x 4ay (750) 4a(0) 4a The equation is x y Let PQ be the vertical distance to the cable from the pole RQ. RQ, RR 70 R Q 5 Let VR be x. Q is (x, 5) Q is a point on parabola x x 50 0 PQ x 00 0 ft 6. A cable of a suspension bridge is in the form of a parabola whose span is 40mts. The road way is 5mts below the lowest point of the cable. If an extra support is provided across the cable 0 mts above the ground level, find the length of the support if the height of the pillars are 55 mts. (JUN-06,JUN-5) Consider the suspension bridge to be open upwards x 4ay From the given data, the vertex of the bridge lies 5 meter above the roadway. The span of the bridge being 40 metres. The point A(0,50) lies on the parabola a(50) a The equation is x 8y The point Q is (x, 5) lies on the parabola x PQ x 0 mts is the length of the support. ways00@gmail.com

62 th Maths Public Exam Special Guide Way to Success 7. An arch is in the form of a semi-ellipse whose span is 48 feet wide. The height of the arch is 0 feet. How wide is the arch at a height of 0 feet above the base? (OCT-06,OCT-) Take the mid point of the base as the centre C(0,0) Since the base wide is 48 feet, the vertices A and A are (4,0) and ( 4,0) respectively Clearly a 48 and b 0 The corresponding equation is x + y. 4 0 Let x be the distance between the pole whose height is 0m and the centre. Then (x, 0) satisfies the equation () x x Clearly the width of the arch at a height of 0 feet is x 4 Thus the required width of arch is 4 feet. 8. The ceiling in a hallway 0ft wide is in the shape of a semi ellipse and 8ft high at the centre. Find the height of the ceiling 4 feet from either wall if the height of the side walls is ft. (MAR-07,0,7) Let PQR be the height of the ceiling which is 4 feet from the wall. From the diagram PQ ft To find the height QR Since the width is 0ft, take A, A as vertices with A as (0,0) and A as ( 0,0). Take the midpoint of AA as the centre which is (0,0) From the diagram AA a 0 a 0 and b 8 6 x + y 00 6 Let QR be y then R is (6, y ) Since R lies on the ellipse 6 + y y PQ + QR The required height of the ceiling is 6.8 feet. 9. The orbit of the earth around the sun is elliptical in shape with sun at a focus. The semi major axis is of length 9.9 million miles and eccentricity is Find how close the earth gets to sun and the greatest possible distance between the earth and the sun. Semi major axis CA is a 9.9 million miles Given e 0.07 The closet distance of the earth from the sun FA, and farthest distance of the earth from the sun FA CF ae 9.9(0.07) FA CA CF million miles FA CA + CF (0.07) 9.9( ) million miles 0. A ladder of length 5m moves with its ends always touching the vertical wall and the horizontal floor. Determine the equation of the locus of a point P on the ladder, which is 6m form the end of the ladder in contact with the floor. (OCT-07,OCT-08,MAR-,JUN-5) Let AB be the ladder and P(x, y ) be a point on the ladder such that AP 6m ways00@gmail.com

63 th Maths Public Exam Special Guide Way to Success Draw PD perpendicular to x-axis and PC perpendicular to y-axis. Clearly the triangles ADP and PCB are similar PC PB BC DA AP PD x 9 BC DA 6 y DA 6x x 9 BC 9y y 6 OA OD + DA x + x 5x OB OC + BC y + y 5y But OA + OB AB 5x 9 + 5y 4 5 x The locus of (x, y ) is x + y 8 6 Which is an ellipse. Another method: PAO BPC θ In PCB cos θ x 9 In ADP sin θ y 6 cos θ + sin θ x + y 8 6 The locus of (x, y ) is x + y 8 6 Which is an ellipse. + y A kho kho player in a practice session while running realises that the sum of the distances form the two kho-kho poles from him is always 8m. Find the equation of the path traced by him if the distance between the poles is 6m.(MAR-,5) From the given data, the two kho-kho poles be taken as the foci F and F. Let P(x, y) be the position of the player at any instant. F P + F P a 8 a 4 F F ae 6 ae 4e e 4 Also b a e b 7 The equation of the path is x 6 + y 7. A satellite is travelling around the earth in an elliptical orbit having the earth at a focus and of eccentricity. The shortest distance that the satellite gets to the earth is 400kms. Find the longest distance that the satellite gets from the earth. ( JUN-07,JUN-08,JUN-,JUN-4,OCT-4) In the adjacent figure, earth at F.The shortest distance of the satellite is F A 400km. we have to findthe longest distance of the satellite F A CA a, CF ae, F A 400km F A CA CF a ae 400 a( e) 400 a a 800km CA 800 and CF ae km F A F C + CA km. The orbit of the planet mercury around the sun is in elliptical shape with sun at a focus. The semi-major axis is of length 6 million miles and the eccentricity of the orbit is Find (i) how close the mercury gets to sun? (ii) the greatest possible distance between mercury and sun. (OCT-09,JUN-0,OCT-,MAR-6) In the adjacent figure F is the position of sun. ways00@gmail.com

64 th Maths Public Exam Special Guide Way to Success CA 6million miles, e 0.06 The closet and farthest position of the planet mercury are A and A respectively. (i) Closet distance F A F A CA CF a ae a( e) Closet distance million miles (ii)the farthest position F A F A F C + CA ae + a a(e + ) million miles Let y be the height of the arch from 9ft right of the centre. (9, y ) is a point on the equation 9 + y 400 y y y The height of the arch 9ft from the right of the centre is feet. FOUR STANDARD TYPES OF PARABOLAS. 4. The arch of a bridge is in the shape of a semiellipse having a horizontal span of 40ft and 6ft high at the centre. How high is the arch, 9ft from the right or left of the centre. (OCT-0, JUN-, MAR-4) Take the midpoint of the base as the centre C(0,0). Since the base is 40 ft, the vertices A and A are (0,0) and ( 0,0) Clearly a 40 a 0, b 6 The corresponding equation x + y ways00@gmail.com

65 th Maths Public Exam Special Guide Way to Success 5. Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabola and hence draw its graph. y 8x + 6y (JUN-08,OCT-0,4) y 8x + 6y y + 6y 8x 9 y + 6y + 8x 9 (y + ) 8x 9 (y + ) 8x (y + ) 8x Y 8X where X x, Y y + Y 4 X a The type is open rightward Referred to X, Y Axis Y 0 Vertex (0, 0) Focus (a, 0) (, 0) Directrix Latus rectum X a X X a X Referred to x, y X x, Y y + Y 0 y + 0 X 0, Y 0 x 0, y + 0 V(0, ) X, Y 0 x, y + 0 F(, ) X x X x Length 4a 8 8 x x 8y 7 x x + 8y 7 (x ) 8y 7 (x ) 8y 7 + (x ) 8y 6 (x ) 8(y + ) X 8Y where X x, Y y + X 4()Y a The type is open downward Referred to X, Y Axis X 0 Vertex (0, 0) Focus Directrix Latus rectum (0, a) (0, ) Y a Y Y a Y Referred to x, y X x, Y y + X 0 x 0 x X 0, Y 0 x 0, y + 0 V(, ) X 0, Y x 0, y + F(, 4) Y y + y 0 Y y + y 4 Length 4a Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabola and hence draw it graph. x x + 8y ( MAR-5) x x + 8y Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabola and hence draw its graph. x 4x + 4y 0 (JUN-) x 4x + 4y 0 x 4x 4y x 4x + 4y (x ) 4 4y ways00@gmail.com

66 th Maths Public Exam Special Guide Way to Success (x ) 4y + 4 (x ) 4(y ) (x ) 4(y ) X 4Y where X x, Y y X 4()Y a The type is open downward Referred to X, Y Axis X 0 Vertex (0, 0) Focus Directrix Latus rectum (0, a) (0, ) Y a Y Y a Y Referred to x, y X x, Y y X 0 x 0 x X 0, Y 0 x 0, y 0 V(,) X 0, Y x 0, y F(, 0) Y y y Y y y Length 4a Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabola and hence draw its graph. y + 8x 6y + 0 (OCT-06,MAR-07,6) y + 8x 6y + 0 y 6y 8x y 6y + 8x (y ) 9 8x y 8x + 9 8x + 8 8(x ) Y 8X where X x, Y y Y 4()X a The type is open leftward Referred to X, Y Referred to x, y X x, Y y Axis Y 0 Y 0 y 0 Vertex (0, 0) Focus ( a, 0) (, 0) Directrix Latus rectum X a X X a X X 0, Y 0 x 0, y 0 V(,) X, Y 0 x, y 0 F(,) X x x 0 X x x + 0 Length 4a Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabola and hence draw its graph. x 6x y 0 (MAR-0) x 6x y 0 x 6x y + x 6x + y + (x ) 9 y + (x ) y + (x ) (y + ) X Y where X x, Y y + X 4()Y a The type is open upward. Referred to X, Y Axis X 0 Vertex (0, 0) Referred to x, y X x, Y y + X 0 x 0 x X 0, Y 0 x 0, y + 0 V(, ) ways00@gmail.com

67 th Maths Public Exam Special Guide Way to Success Focus Directrix Latus rectum (0, a) (0, ) Y a Y Y a Y X 0, Y x 0, y + F(,) Y y + y Y y + y 0 Length 4a 4a 0. Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabola and hence draw its graph. y 8x y ( JUN-07) y 8x y y y 8x 7 y y + 8x 7 (y ) 8x 7 (y ) 8x 7 + (y ) 8x 6 (y ) 8(x ) Y 8X where X x, Y y Y 4()X a The type is open rightward. Referred to X, Y Referred to x, y X x, Y y Axis Y 0 Y 0 y 0 Vertex (0, 0) Focus (a, 0) (, 0) Directrix Latus rectum X a X X a X X 0, Y 0 x 0, y 0 V(,) X, Y 0 x, y 0 F(4,) X x x 0 X x x 4 Length 4a 8 8. Find the vertex, axis, focus, equation of the latus rectum, equation of directrix and length of the latus rectum of the parabola y + 4y + 4x ( MAR- ) y + 4y 4x 8 y + 4y + 4x 8 (y + ) 4x 8 (y + ) 4x (y + ) 4x 4 (y + ) 4(x + ) Y 4X where X x +, Y y + a The type is open leftward Referred to X, Y Referred to x, y X x +, Y y + Axis Y 0 Y 0 y + 0 Vertex (0, 0) Focus ( a, 0) (, 0) Directrix Latus rectum X a X X a X X 0, Y 0 x + 0, y + 0 V(, ) X, Y 0 x +, y + 0 F(, ) X x + x 0 X x + x + 0 Length 4a 4 4 ways00@gmail.com

68 th Maths Public Exam Special Guide Way to Success TWO TYPES OF ELLIPSE Centre x a + y b x b + y a C(0,0) Vertices A(a, 0), A ( a, 0) Foci F (ae, 0), F ( ae, 0) C(0,0) A(0, a) A (0, a) F (0, ae) F (0, ae). Find the eccentricity, centre, foci, vertices of the following ellipse. 6x + 4y 7x + y 44 0 ( MAR-06,JUN-06,OCT-5 ) 6x 7x + 4y + y (x x) + 4(y + 8y) x x + +4(y + 8y ) x + 4 y x y x + 4 y x 44 x 4 X + Y x + 4 y y y where X x, Y y + 4 The major axis is along Y axis e b a 6, b 4, a 6, b a ae 6 4 Referred to X, Y Centre (0, 0) Vertices Foci (0, ±a) (0, ±6) (0, ±ae) (0, ±4 ) Referred to x, y X x, Y y + 4 X 0, Y 0 x 0, y C(, 4) 0, a (0,6) X 0, Y 6 x 0, y A(,) 0, a (0, 6) X 0, Y 6 x 0, y A (, 0) (0, 4 ) X 0, Y 4 x 0, y x, y F (, ) (0, 4 ) X 0, Y 4 x 0, y x, y 4 4 F (, 4 4 ). Find the eccentricity, centre, foci, vertices of the following ellipse. x + 4y 8x 6y 68 0 (x 8x) + (4y 6y) 68 0 (x 8x ) + 4(y 4y + ) 68 x y 4 68 x y x y 00 x 4 00 x y 00 + y ways00@gmail.com

69 th Maths Public Exam Special Guide Way to Success The major axis is parallel to x- axis Let X x 4, Y y X + Y 00 5 a 00, b 5, a 0, b 5 e ae Referred to X, Y Centre (0,0) Foci Vertices (±ae, 0) (±5, 0) (±a, 0) (±0,0) Referred to x, y X x 4, Y y X 0, Y 0 x 4 0, y 0 C(4,) 5, 0 X 5, Y 0 x 4 5, y 0 x 4 + 5, y F (4 + 5, ) 5, 0 X 5, Y 0 x 4 5, y 0 x 4 5, y F (4 5, ) (0, 0) X 0, Y 0 x 4 0, y 0 x 4, y A(4,) ( 0,0) X 0, Y 0 x 4 0, y 0 x 6, y A ( 6,) 4. Find the eccentricity, centre, foci, vertices of the following ellipse 6x + 9y + x 6y 9 (OCT-, MAR-, JUN-6) 6x + 9y + x 6y 9 6x + x + 9y 6y 9 6(x + x) + 9(y 4y) 9 6 x + x + + 9(y 4y + ) 9 X 6 x y Y 6 6 x y x y x + 44 x y 44 + y where X x +, Y y The major axis is along Y axis e b a 6, b 9, a 4, b a ae Referred to X, Y Centre (0,0) Vertices Foci (0, ±a) (0, ±4) (0, ±ae) (0, ± 7) Referred to x, y X x +, Y y X 0, Y 0 x + 0, y 0 C(,) (0,4) X 0, Y 4 x + 0, y 4 A(,6) (0, 4) X 0, Y 4 x + 0, y 4 A (, ) (0, 7) X 0, Y 7 x + 0, y 7 x, y + 7 F (, + 7) (0, 7) X 0, Y 7 x + 0, y 7 x, y 7 F (, 7) ways00@gmail.com

70 th Maths Public Exam Special Guide Way to Success 5. Find the eccentricity, centre, foci, vertices of the following ellipse 6x + 9y x + 6y 9 0 (OCT -6) 6x + 9y x + 6y 9 6x x + 9y + 6y 9 6 x x + 9(y + 4y) 9 6 x x + + 9(y + 4y + ) 9 X 6 x + 9 y Y 6 6 x y x + 9 y x 44 x y y where X x, Y y + The major axis is along Y axis e b a 6, b 9, a 4, b a Referred to X, Y Centre (0,0) Vertices (0, ±a) (0, ±4) ae Referred to x, y X x, Y y + X 0, Y 0 x 0, y + 0 C(, ) (0,4) X 0, Y 4 x 0, y + 4 A(,) (0, 4) X 0, Y 4 x 0, y + 4 A (, 6) Foci (0, ±ae) (0, ± 7) (0, 7) X 0, Y 7 x 0, y + 7 x, y + 7 F (, + 7) (0, 7) X 0, Y 7 x 0, y + 7 x, y 7 F (, 7) 6. Find the eccentricity, centre, foci, vertices of the following ellipse. 9x + 5y 8x 00y 6 0 (MAR-09) 9x + 5y 8x 00y 6 0 9x 8x + 5y 00y 6 9 x x + 5(y 4y) 6 9 x x + +5(y 4y + ) 6 9 x + 5 y x y x + 5 y 5 9 x + 5 y 5 x + y 5 9 X + Y where X x, Y y 5 9 The major axis is along X axis a 5, b 9, a 5, b e ae ways00@gmail.com

71 th Maths Public Exam Special Guide Way to Success Referred to X, Y Centre (0,0) Foci Vertices (±ae, 0) (±4, 0) (±a, 0) (±5,0) TWO TYPES OF HYPERBOLA Referred to x, y X x, Y y X 0, Y 0 x 0, y 0 C(,) 4, 0 X 4, Y 0 x 4, y 0 x 5, y, F (5,) 4, 0 X 4, Y 0 x 4, y 0 x, y F (,) (5, 0) X 5, Y 0 x 5, y 0 x 6, y A(6, ) ( 5,0) X 5, Y 0 x 5, y 0 x 4, y A ( 4,) x a y b x b y a Centre C(0,0) C(0,0) Vertices A(a, 0), A ( a, 0) Foci F (ae, 0), F ( ae, 0) A(0, a) A (0, a) F (0, ae) F (0, ae) 7. Find the eccentricity, centre, foci and vertices of the hyperbola 9x 6y 8x 64y 99 0 and also trace the curve. (OCT-) 9x 8x 6y 64y 99 9(x x) 6(y + 4y) 99 9(x x + ) 6(y + 4y + ) 99 9{(x ) } 6 y (x ) 6 y (x ) 6 y (x ) 6 y (x ) y X Y where X x, Y y The transverse axis is parallel to X axis a 6, b 9, a 4, b e + b a Referred to X, Y Centre (0,0) Foci Vertices (±ae, 0) (±5, 0) (±a, 0) (±4,0) ae Referred to x, y X x, Y y + X 0, Y 0 x 0, y + 0 C(, ) 5, 0 X 5, Y 0 x 5, y + 0 x 6, y F (6, ) 5, 0 X 5, Y 0 x 5, y + 0 x 4, y F ( 4, ) (4, 0) X 4, Y 0 x 4, y + 0 x 5, y A(5, ) ( 4,0) X 4, Y 0 x 4, y + 0 x, y A (, ) ways00@gmail.com

72 th Maths Public Exam Special Guide Way to Success (0, 5) X 0, Y 5 x + 0, y 5 x, y 5 F (, 4) 8. Find the eccentricity, centre, foci and vertices of the following hyperbola 9x 6y + 6x + y and draw the diagram. ( JUN-5) 9x 6y + 6x + y (x + 4x) 6(y y) 64 9(x + 4x + ) 6(y y + ) 64 9{(x + ) 4} 6 y 64 9(x + ) 6 y 44 6 y 9(x + ) 44 Y X y 44 y 9 9(x + ) 44 x where X x +, Y y The transverse axis is parallel to Y axis a 9, b 6, a, b 4 e + b a 9 9 Referred to X, Y Centre (0,0) Vertices Foci (0, ±a) (0, ±) (0, ±ae) (0, ±5) ae Referred to x, y X x +, Y y X 0, Y 0 x + 0, y 0 C(,) (0,) X 0, Y x + 0, y A(,4) (0, ) X 0, Y x + 0, y A(, ) (0, 5) X 0, Y 5 x + 0, y 5 x, y + 5 F (,6) 9. Find the eccentricity, centre, foci and vertices of the following hyperbola x 4y + 6x + 6y 0 and draw its diagram. (MAR-0,JUN-) x 4y + 6x + 6y 0 x + 6x 4y + 6y (x + 6x + ) 4(y 4y + ) {(x + ) 9} 4 y 4 (x + ) 4 y 4 (x + ) y 4 4 X Y 4 where X x +, Y y The transverse axis is along X axis a 4, b, a, b e + b ae 5 5 a + 4 Referred to X, Y Centre (0,0) Foci (±ae, 0) (± 5, 0) Referred to x, y X x +, Y y X 0, Y 0 x + 0, y 0 C(,) 5, 0 X 5, Y 0 x + 5, y 0 x + 5, y ways00@gmail.com

73 th Maths Public Exam Special Guide Way to Success Vertices (±a, 0) (±,0) F ( + 5, ) 5, 0 X 5, Y 0 x + 5, y 0 x 5, y F ( 5, ) (, 0) X, Y 0 x +, y 0 x, y A(,) (,0) X, Y 0 x +, y 0 x 5, y A ( 5,) 0. Find the eccentricity, centre, foci and vertices of the hyperbola x y + 6x + 6y and also trace the curve. (MAR-08,OCT-08,OCT-09,JUN-0,MAR-,MAR-4) x y + 6x + 6y x + 6x y + 6y x + 6x (y y) 8 (x + 6x + ) (y y + ) 8 {(x + ) 9} y 8 (x + ) y (x + ) y y (x + ) Y X 4 y y 4 (x + ) x + where X x +, Y y The transverse axis is parallel to Y axis a 4, b, a, b e + b ae a 4 4 Referred to X, Y Centre (0,0) Vertices Focus (0, ±a) (0, ±) (0, ±ae) (0, ±4) 6 4 Referred to x, y X x +, Y y X 0, Y 0 x + 0, y 0 C(,) (0,) X 0, Y x + 0, y A(,) (0, ) X 0, Y x + 0, y A(, ) (0, 4) X 0, Y 4 x + 0, y 4 x, y + 4 F (,5) (0, 4) X 0, Y 4 x + 0, y 4 x, y 4 F (, ). Find the eccentricity, centre, foci and vertices of the hyperbola 9x 7y + 6x + 4y and draw the diagram. ( JUN- ) 9x 7y + 6x + 4y x + 6x 7y + 4y (x + 4x) 7(y y) 9 9(x + 4x + ) 7(y y + ) 9 ways00@gmail.com

74 th Maths Public Exam Special Guide Way to Success 9{(x + ) 4} 7 y 9 9(x + ) 7 y (x + ) 7 y 6 7 y 9(x + ) 6 Y X y 9(x+) 6 6 y x where X x +, Y y The transverse axis is along Y axis e + b a 9, b 7, a, b 7 a Referred to X, Y Centre (0,0) Vertices Focus (0, ±a) (0, ±) (0, ±ae) (0, ±4) ae 4 4 Referred to x, y X x +, Y y X 0, Y 0 x + 0, y 0 C(,) (0,) X 0, Y x + 0, y A(,4) (0, ) X 0, Y x + 0, y A(, ) (0, 4) X 0, Y 4 x + 0, y 4 x, y + 4 F (,5) (0, 4) X 0, Y 4 x + 0, y 4 x, y 4 F (, ). Find the eccentricity, centre, foci and vertices of the hyperbola 6x 9y x 6y 64 0 and draw its diagram ( JUN-4) 6x 9y x 6y x x 9y 6y (x x) 9(y + 4y) 64 6(x x + ) 9(y + 4y + ) 64 6{(x ) } 9 y (x ) 9 y (x ) 9 y (x ) 9 y (x ) y X 9 Y 6 where X x, Y y + The transverse axis is along X axis a 9, b 6, a, b 4 e + b ae a 9 9 Referred to X, Y Centre (0,0) Foci Vertices (±ae, 0) (±5, 0) (±a, 0) (±,0) Referred to x, y X x, Y y + X 0, Y 0 x 0, y + 0 C(, ) 5, 0 X 5, Y 0 x 5, y + 0 x 6, y F (6, ) 5, 0 X 5, Y 0 x 5, y + 0 x 4, y F ( 4, ) (, 0) X, Y 0 x, y + 0 x 4, y A(4, ) (,0) X, Y 0 x, y + 0 x, y A (, ) ways00@gmail.com

75 th Maths Public Exam Special Guide Way to Success 5, 0 X 5, Y 0 x 5, y 4 0 x 4, y 4 F ( 4,4) 5, 0 X 5, Y 0. Find the eccentricity, centre, foci and vertices of the hyperbola x 4y 4x + y 7 0 and also trace the curve. (OCT-07) x 4y 4x + y 7 0 x 4x 4y + y 7 (x x) 4(y 8y) 7 (x x + ) 4(y 8y ) 7 {(x ) } 4 y (x ) 4 y (x ) 4 y 4 75 X 75 Y 75 4 (x ) 75 y where X x, Y y 4 The transverse axis is along X axis a 75, b 75 4, Vertices (±a, 0) ± 5, 0 x 5, y 4 0 x 7, y 4 A 7, 4 5, 0 X 5, Y 0 x 5, y 4 0 x, y 4 A, 4 e + b a 75 ae Referred to X, Y Centre (0,0) Foci (±ae, 0) (±5, 0) Referred to x, y X x, Y y 4 X 0, Y 0 x 0, y 4 0 C(,4) 5, 0 X 5, Y 0 x 5, y 4 0 x 6, y 4 F (6,4) 4. Prove that the line 5x + y 9 touches the hyperbola x 9y 9 and find its point of contact. ( JUN-09,6, MAR-,OCT-, 4,6) The condition for y mx + c to be a tangent to a hyperbola x y is a b c a m b 5x + y 9 y 5x + 9 y 5x y x + 4 m 5, c 4 x 9y 9 x y 9 a 9, b ways00@gmail.com

76 th Maths Public Exam Special Guide Way to Success c 9 6 ; a m b c a m b Thus the line 5x + y 9 is a tangent to the hyperbola. It touches the hyperbola The point of contact is a m c b a m c, b c c The point of contact is 5, 4 5. Show that the line x y is a tangent to the ellipse x + y. Find the coordinates of the point of contact ( JUN-,MAR-6) The condition for y mx + c to be a tangent to a ellipse x + y is a b c a m + b x y y x + 4 m, c 4 x + y x 4 a, b 4 c 6 ; a m + b c a m + b x y is a tangent to the ellipse + y The point of contact is a m c a m c, b c 4 4 b c 4 4 The point of contact is, 6. Find the equation of the hyperbola if its asymptotes are parallel to x + y 0 and x y + 8 0, (, 4) is centre of the hyperbola and it passes through (, 0) (MAR-06, 09,5, JUN-06,08,, OCT-5) The asymptotes are parallel to x + y 0 and x y Equations of the two asymptotes are of the form x + y + l 0 and x y + m 0 But asymptotes pass through the centre (,4) of the hyperbola l 0 l m 0 m 6 Equations of the asymptotes are x + y 0 0 and x y Combined equation of the asymptotes is x + y 0 x y The equation of the hyperbola is of the form x + y 0 x y k 0 It passes through (,0) k 0 k 64 The equation of the hyperbola is x + y 0 x y Find the equation of the rectangular hyperbola which has for one of its asymptotes the line x + y 5 0 and passes through the points (6, 0) and (, 0) (OCT-06, 08,0,, MAR-07,08,,7, JUN-07) One of the asymptotes is x + y 5 0 The other asymptotes is of the form x y + k 0 Equation of the R.H. is x + y 5 x y + k + c 0 It passes through (6, 0) k + c 0 k + c.. It also passes through (,0) k + c k + c k + c 0 8k + c 48 Solving () and () k 4 and c 6 Equation of the R.H is x + y 5 x y ways00@gmail.com

77 th Maths Public Exam Special Guide Way to Success 6. DIFFERENTIAL CALCULUS APPLICATIONS-II. Use differentials to find an approximate value for the given number 4 y y f x x Take x, dx x 0.0 dy x dx f x + x y + dy f Again, let (.0).0066 y f x x 4 Take x, dx x 0.0 dy 4 x 4dx f x + x y + dy f (.0) (MAR-05) (.0) + (.0) Using Euler s theorem, prove that x u u + y x y tan u if u sin x y x y (MAR-07, MAR-08, JUN-4) R.H.S. is not homogeneous and hence define f sin u x y f tx, ty tx ty tx ty t x y t x y t (x y) ( x y) x y t x y x y t f f is homogeneous of degree By Euler s theorem x f f + y f x y x (sin u) x (sin u) + y (sin u) y x u u cos u + y cos u (sin u) x y cos u x u u + y x y (sin u) x u u + y x y sin u cos u x u u + y x y tan u. Using Euler s theorem, If u tan x + y x y prove that x u u + y sin u (OCT-09,, 5) x y u is not a homogeneous function. But tan u is a homogeneous function. Define f tan u x + y f tx, ty tx + ty tx ty t x + y t x y t (x + y ) (x y) t x + y x y t f x y f is a homogeneous function of degree By Euler s theorem x (tan u) x x f f + y x y f (tan u) + y (tan u) y x. sec u u + y. x sec u u tan u y sec u x u u + y x y x u u tan u + y x y sec u tan u ways00@gmail.com

78 th Maths Public Exam Special Guide Way to Success sin u cos u cos u sin u cos u cos u sin u cos u sin u 4. Prove x u u + y x x u sin x+y x+ y sin u Let x tx, y ty f tx, ty t x+y x+ y u sin x+y x+ y tx +ty tx + ty x+y x+ y t(x+y) (x+y) ( x+ y ) t x+y cos x+ y (JUN-) f(x, y) t( x+ y ) Degree of f is, by Euler s theorem x x f f + y nf x y if (x+y) ( x+ y ) t f x ( sin u) + y ( sin u) n( sin u) x y u + y u u x u y sin sin x+y x+ y u u u x + y x y x u u + y x y x+y x+ y x+y x+ y x+y x+ y x+y x+ y x+y x+ y u sin x+y x+ y cos x+y x+ y x+y cos x+ y 5. Verify Euler s theorem for f x, y f tx, ty x +y (JUN-06,OCT-0,MAR-4) t x +t y t (x +y ) t (x +y ) t t f x, y f x, y f is a homogenous function of degree and by Euler s theorem x f f + y x y f Verification: f x x x +y x f x x f y x +y y f y y y x +y x +y x x +y y x +y x f f + y x x y x +y y x +y x +y x +y x +y f Hence Euler s theorem is verified. 6. If u tan x y then verify that u x y u y x u tan x y u x + x y u y + x y u x y u y x y y +x y +x y x x +y x y + x y y +x y y y y x +y x y u y + x y x y x y x y y +x y x x x +y (MAR-0) y y +x y (x +y ). x.x x +y x (x +y ) (x +y ) y x +y (x +y ) x y (x +y ).. u x y y x +y (x +y ). y.y (x +y ) x +y y (x +y ) x y (x +y ). ways00@gmail.com

79 th Maths Public Exam Special Guide Way to Success From () and () u x y u y x 7. If u x y y x then verify that u x y u y x u x y y x u y x y x + y y x u x. y y x x y x u x y u y x x u y x (JUN-, MAR-7) x ( ) y x y x x y.. y u x y + x x y y From () and () u x y + y y x. ) u y x 8. Verify u u x where u y x y y x y x u x y y x u x ( )y x + y x y x y x u ( )x (y) x 4y y y x y x u x y u y x x u y x 4y x y x x ( )4y x + 4y y x y x 4y x x y. y u x ( )x y y + (y) x x + y y x x y + 4y x 4y x x y From () and () u x y u y x (OCT - 4) 9. Verify u x y u y x for the function u sin x y (MAR-) u x y u y x u sin x y u cos x x y y u x cos x y y y x u y x y x. sin x y y x y. sin x y y u x y sin x y y x y cos x y y + cos x y y cos x y x + cos x y y y x y sin x y + cos x y From () and () u x y u y x + cos x y ().. y 0. If u sin x cos 4y then verify that u u x y y x u sin x cos 4y u x u y u x y u y x (MAR-06) cos 4y. cos x. cos x cos 4y sin x sin 4y. 4 4sin x sin 4y x y u y u x x From () and () 4sin x sin 4y 4 sin 4y cos x. cos x sin 4y.. (cos x cos 4y) y cos x sin 4y. 4 cos x sin 4y u x y u y x ways00@gmail.com

80 th Maths Public Exam Special Guide Way to Success. Trace the curve y x +. Trace the curve y x. Trace the curve y x Domain Extent Intercepts Origin Symmetry Test Asymptotes ( JUN-09, OCT-, JUN-6, OCT-6) The function is defined for all real values of x and hence the domain is the entire interval (, ) Horizontal extent is < x < Vertical extent is < y < x 0, then y y 0, then x The curve does not pass through (0, 0) The symmetry test shows that the curve does not possess any of the symmetry properties. As x c(for c finite ) y does not tend to ±. Therefore the curve does not admit any asymptote. (OCT-06, JUN-07,OCT-07, JUN-08, OCT-08, JUN-0, MAR-, MAR-) The function is defined for all real values of x and hence the domain is the entire interval (, ) Horizontal extent is < x < Vertical extent is < y < x 0, then y 0 y 0, then x 0 The curve passes through (0, 0) It is symmetrical about the origin The curve does not admit any asymptote. (MAR-06, MAR-09, JUN-, OCT-,JUN-5) When x 0, y is well defined. [0, ) Horizontal extent is 0 x < Vertical extent is < y < x 0, then y 0 y 0, then x 0 The curve passes through the origin By symmetry test, we have, the curve is symmetric x axis only. As x +, y ± The curve does not admit any asymptote. Monotonicity Special points The first derivative test shows that the curve is increasing throughout (, ). Since y 0 for all x The curve is concave downward in (, 0) and concave upward in (0, ) since y 6x < 0 for x < 0 y 6x > 0 for x > 0 y 6x for x 0 yields (0, ) as the inflection point Since y 0 for all x The curve is increasing in (, ) Since y 6x, the curve is concave upward in (0, ) and convex downward (, 0) y 0 for x 0 yields (0, 0) as the point of inflection. For the branch y x of the curve is increasing since dy dx > 0 for x > 0. For the branch y x of the curve is decreasing since dy < 0 for x > 0 dx (0,0) is not a point of inflection. Diagram ways00@gmail.com

81 th Maths Public Exam Special Guide Way to Success 8. Differential Equations. In a certain chemical reaction the rate of conversion of a substance at time t is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour, 60 grams remain and at the end of 4 hours grams. How many grams of the substance was there initially? (MAR-, OCT-5) Let A be the substance at time t da da αa ka A cekt dt dt When t, A 60 ce k 60.. () When t 4, A ce 4k. () c 4 e 4k 60 4 () c 604 c 85.5 (by using log) Initially,when t 0, A c 85.5 gms (app.) Hence initially there was 85.5 gms (approximately) of the substance.. A bank pays interest by continuous compounding, that is by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accures at 8% per year compounded continuously. Calculate the percentage increase in such an account over one year. [ Take e ]. (OCT-07) Let A be the principal at time t da da da αa ka 0.08 A, since k 0.08 dt dt dt A t ce 0.08t Percentage increase in year A A 0 00 c.e 0.08 c 00 8.% Hence percentage increase is 8.% A A(0) A(0) 00. For a postmortem report, a doctor requires to know approximately the time of death of the deceased. He records the first temperature at 0.00 a.m. to be 9. 4 o F. After hours he finds the temperature to be 9. 4 o F. If the room temperature (which is constant) is 7 o F, estimate the time of death. (Assume normal temperature of a human body to be o F). [log e and log.4 e ] (JUN-,JUN-6) Let T be temperature of the body at any time t By Newton s law of cooling dt α T 7 since dt S 7 o F dt dt k T 7 T 7 ce kt T 7 + ce kt At t 0, T ce k c c c.4 [First recorded time 0 a.m. is t 0] /T 7 +.4e kt When t 0, T 9.4 e 0k k 0 log 9.4 e.4 ( ) 0 Let t be the elapsed time after the death. When t t ; T e kt t k log e min [For better approximation the hours converted into minutes] 4 hours 6 minutes before the first recorded temperature. The approximate time of death is 0.00 hrs 4 hours 6 minutes. /Approximate time of death is 5.4 A.M. ways00@gmail.com

82 th Maths Public Exam Special Guide Way to Success 4. The number of bacteria in a yeast culture grows at a rate which is proportional to the number present. If the population of a colony of yeast bacteria triples in hour. Show that the number of bacteria at the end of five hours will be 5 times of the population at initial time. (JUN-06, MAR-09, OCT-) Let A be the number of bacteria at any time t da αa dt da ka dt A ce kt Initially, i.e., When t 0, assume that A A 0 /A 0 ce o c /A A 0 e kt When t, A A 0 A 0 A 0 e k e k When t 5, A A 0 e 5k A 0 (e k ) 5 5. A 0 / The number of bacteria at the end of 5 hours will be 5 times of the number of bacteria at initial time. 5. Radium disappears at a rate proportional to the amount present. If 5% of the original amount disappears in 50 years, how much will remain at the end of 00 years. [Take A 0 as the initial amount]. (MAR-06, 0, JUN-09,OCT-,6) Let A be the amount of radium at time t A A(t) da dt αa da dt ka A ce kt At t 0, A A 0 / A 0 ce 0 c / A A 0 e kt But 5 % of the original amount disappears in 50 years. i.e., when t 50, A.95 A 0 / 0.95 A 0 A 0 e 50k e 50k 0.95 Again, when t 00, A A 0 e 00k A 0 e 50k A A 0 Thus the amount of radium that remains at the end of 00 years is 0.905A 0 6. The sum of Rs. 000 is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal? (log e 0. 69). (MAR-5,JUN-07,08,,OCT-06,0) Let A be the principal time t da dt α A da dt ka da dt 0.04 t, since k 0.04 A c. e.04t When t 0, A ce 0 c 000 / A 000e.04t Find t, when A e.04t t log yrs (app.) 7. A cup of coffee at temperature 00 0 C is placed in a room whose temperature is 5 0 C and it cools to 60 0 C in 5 minutes. Find its temperature after a further interval of 5 minutes. (OCT-09, OCT-, JUN-5, MAR-7) Let T be the temperature of the coffee at any time t By Newtons law of cooling, dt dt dt dt α T S k(t S) T S ce kt T 5 + ce kt, ways00@gmail.com

83 th Maths Public Exam Special Guide Way to Success since S 5 0 C When t 0, T ce k(0) 00 5 c c 85 T e kt When t 5, T e 5k e 5k When t 0, T e 0k The rate at which the population of a city increases at any time is proportional to the population at that time. If there were,0,000 people in the city in 960 and,60,000 in 990 what population may be anticipated in 00 log e ; e.4. 5 (MAR-08, JUN-0, MAR-4, JUN-4) Let A be the population at time t da da αa ka A cekt dt dt Take the year 960 as the initial time t 0 When t 0, A ce 0 c A 0000e kt When the year 990 when t 0, A 60,000 /60, e 0k e 0k 6 Find A, when the year 00 when t 60 A 0000 e 60k 0,000 6 ~97000 The approximate population in 00 is A radioactive substance disintegrates at a rate proportional to its mass. When its mass is 0 mgm, the rate of disintegration is 0.05 mgm per day. How long will it take for the mass to be reduced from 0 mgm to 5 mgm. (log e 0. 69) (MAR-, JUN-,OCT-4) Let A be the amount of mass at time t. ways00@gmail.com da dt α A da dt ka A ce kt When t 0, A 0 c 0 A 0e kt Again da dt ka It is given that when A 0, da disintegration k k / A 0e Find t when A 5 5 0e 0.005t e 0.005t e 0.005t log 0.005t t log ~6 days dt 0.05 of 0. A drug is expected in a patients urine. The urine is monitored continuously using a catheter. A patient is administered 0 mg of drug at time t 0, which is excreted at a Rate of t / mg/h (i) What is the general equation for the amount of drug in the patient at time t > 0? (ii) When will the patient be drug free? (i) Let A be the quantum of drug at any time t The drug is excreted at a rate of t da dt t A t + c When t 0, A 0 c 0 At any time t A 0 t (ii) For drug free, A 0 5 t t 5 t.9 hours. Hence the patient will be drug free in.9 hours or hours 54 min.

84 th Maths Public Exam Special Guide Way to Success 9. DISCRETE MATHEMATICS. Show that (Z, ) is an infinite abelian group where is defined as a b a + b + (OCT-08,JUN-0,MAR-6) (i) Closure axiom: Since a, b and are integers a + b + is also an integer. a b z, a, b z Thus closure axiom is true. (ii) Associative axiom: Let a, b, c z a b c a + b + c a + b + + c + a + b + c + 4 a b c a b + c + a + b + c + + a + b + c + 4 a b c a b c Thus associative axiom is true (iii) Identity axiom: Let e be the identity element. By the definition of e, a e a By the definition of, a e a + e + a + e + a e Z. Thus identity axiom is true. (iv) Inverse axiom: Let a G and a be the inverse element of a By the definition of a, a a e By the definition of, a a a + a + a + a + a a 4 Clearly a 4 Z. Inverse axiom is true (Z, ) is a group. (v) Commutative property: Let a, b G a b a + b + b + a + b a is commutative in G and hence (Z, ) is an abelian group. Z is infinite set. The group is an infinite abelian group. Show that the set G of all matrices of the form x x, where x R {0}, is a group under x x matrix multiplication. (JUN-, MAR-5) Let G x x / x R 0 we shall show x x that G is a group under matrix multiplication. (i) Closure axiom: A x x y y G, B x x y y G AB xy xy G, ( x 0, y 0 xy 0) xy xy G is closed under matrix multiplication. (ii) Matrix multiplication is always associative. (iii) Let E e e G be such that AE A for e e every A G AE A x x e e x x x x e e x x xe xe x x xe xe x x xe x e ( x 0) / / Thus E G is such that AE A, / / for every A G We can similarly show that EA A for every A G Eis the identity element in G and hence identity axiom is true. (iv) Suppose A y y y y G is such that A A E xy xy / / Then we have xy xy / / xy y 4x A 4x 4x 4x 4x G is such that A A E Similarly we can show that AA E A is the inverse of A G is a group under matrix multiplication.. Let G be the set of all rational numbers except and be defined on G by a b a + b ab for all a, b G. Show that (G, ) is an infinite abelian group. (JUN-08,5, MAR-) Let G Q { } Let a, b G. Then a and b are rational numbers and a, b. i) Closure axiom: a b a + b ab is a rational number. But to prove a b G. We have to prove that a b ways00@gmail.com

85 th Maths Public Exam Special Guide Way to Success On the contrary, assume that a b then a + b ab b ab a b a a b ( a a 0 This is impossible because b. / Our assumption is wrong / a b and hence a b G / Closure axiom is true. ii) Associative axiom: a b c a b + c bc a + b + c bc a b + c bc a + b + c bc ab ac + abc a b c (a + b ab) c a + b ab + c a + b + ab c a + b + c ab ac bc + abc / a b c a b c, a, b, c G / Associative axiom is true. iii) Identity axiom: Let e be the identity element. By definition of, a e a + e ae By definition of e, a e a a + e ae a e a 0 e 0 since a e 0 G / Identity axiom is satisfied. iv) Inverse axiom: Let a be the inverse of a G By the definition of inverse, a a e 0 By the definition of, a a a + a aa a a a a + a aa 0 a a a G, since a / inverse axiom is satisfied. / (G,*) is a group. v) Commutative axiom: For any a, b G a b a + b ab b + a ba b a / * is commutative in G and hence (G, ) is an abelian group. Since G is infinite, (G, ) is an infinite abelian group. 4. Prove that the set of four functions f, f, f, f 4 on the set of non zero complex numbers C 0 defined by f z z, f z z, f z, f z 4 z z C {0} forms an z abelian group with respect to the composition of functions. (OCT-06,09,5) Let G {f, f, f, f 4 } f f z f f z f (z) f f f, f f f, f f f, f 4 f f 4 Again f f z f f z f z z z f (z) f f f Similarly f f f 4, f f 4 f f f z f f z f z z f 4 z f f f 4 Similarly f f f, f f 4 f (f 4 f ) z f 4 f z f 4 z z z f (z) f 4 f f Similarly f 4 f f, f 4 f 4 f Using these results we have the composition tables as given below: f f f f 4 f f f f f 4 f f f f 4 f f f f 4 f f f 4 f 4 f f f From the table (i) All the entries of the composition table are the elements of G Closure axiom is true. (ii) Composition of functions is in general associative. Associative axiom is true. (iii) Clearly f is the identity element of G and satisfies the identity axiom (iv) From the table: Inverse of f is f ; Inverse of f is f Inverse of f is f ; Inverse of f 4 is f 4 Inverse axiom is satisfied. (G, ) is a group (v) From the table the commutative property is also true (G, ) is an abelian group ways00@gmail.com

86 th Maths Public Exam Special Guide Way to Success 5. Show that (Z n, + n ) forms group. (JUN-, MAR-4) Let Z n { 0,,, [n ]} be the set of all congruence classes modulo n. And let l, m Z n 0 l, m < n (i) Closure axiom: By definition l + m if l + m < n l + n m r if l + m n where l + m q. n + r 0 r < n In both the cases, l + m Z n and [r] Z n Closure axiom is true (ii) Addtion modulo n is always associative in the set of congruence classes modulo n (iii) The identity element [0] Z n and it satisfies the identity axiom. (iv) The inverse of [l] Z n is [n l] Clearly [n l] Z n and l + n n l [0] n l + n l [0] The inverse axiom is also true. Hence (Z n, + n ) is a group. 6. Show that (Z 7 0,. 7 ) forms a group. (MAR-0, OCT-6) Let G {,, [6]} The Cayley s table is. 7 [] [] [] [4] [5] [6] [] [] [] [] [4] [5] [6] [] [] [4] [6] [] [] [5] [] [] [6] [] [5] [] [4] [4] [4] [] [5] [] [6] [] [5] [5] [] [] [6] [4] [] [6] [6] [5] [4] [] [] [] From the table: (i) all the elements of the composition table are the elements of G The closure axiom is true (ii) multiplication modulo 7 is always associative (iii) the identity element is [] G and satisfies the identity axiom (iv) the inverse of [] is []; [] is [4]; [] is [5]; [4] is []; [5] is []; [6] is [6] and it satisfies the inverse axiom. the given set forms a group under multiplication modulo Show that the nth roots of unity form an abelian group of finite order with usual multiplication. (MAR-) We know that, ω, ω ω n are the nth roots of unity, where ω cis π n Let G {, ω, ω ω n } (i) Closure axiom: Let ω l, ω m G, 0 l, m (n ) To prove ω l ω m ω l+m G Case (i) l + m < n If l + m < n then clearly ω l+m G Case (ii) l + m n By division algoritham l + m q. n + r where 0 r < n, q is a positive integer. ω l+m ω qn +r ω n q. ω r q. ω r ω r G 0 r < n Closure property is true. (ii) Associative axiom: Multiplication is always associative in the set of complex numbers and hence in G ω l. ω p. ω m ω l. ω p+m l+ p+m ω ω l+p +m ω l. ω p. ω m ω l, ω p, ω m G (iii) Identity axiom: The identity element G and it satisfies. ω l ω l. ω l ω l G (iv) Inverse axiom: For any ω l G, ω n l G and ω l. ω n l ω n l. ω l ω n Thus inverse axiom is true G,. is a group. (v) Commutative axiom: ω l. ω m ω l+m ω m +l ω m. ω l ω l, ω m G G,. is an abelian group. Since G contains n elements, G,. is a finite abelian group of order n. 8. Show that the set,, 4, 5, 9 forms an abelian group under multiplication modulo. (MAR-07, JUN-09) Let G,, 4, 5, 9 The cayley s table is ways00@gmail.com

87 th Maths Public Exam Special Guide Way to Success From the table i) All the elements of the composition table are the elements of G. / The closure axiom is true. ii) Multiplication modulo is always associative. iii) The identity element G and satisfies the identity axiom. iv) the inverse of is [] the inverse of [] is [4] the inverse of [4] is [] the inverse of [5] is [9] the inverse of [9] is 5 / The given set forms a group under multiplication modulo. v) From the table the commutative property is also true. / The group is abelian. 9. Show that the set G of all positive rationals forms a group under the composition defined by a b ab for all a, b G (MAR-06, JUN-06, OCT-07,0,,4) (i) Closure axiom: Let a, b G. Since a, b are positive rationals, ab is positive rational and hence ab positive rational. ab G, a b G Closure axiom is true. (ii) Associative axiom: Let a, b, c G a b c ab ab c c ab c 9 a b c a bc a b c a b c The associative axiom is true. bc a abc 9 is also (iii) Identity axiom: Let e be the identity element. By the definition of e, a e a By the definition of, a e ae ae a e e G Thus identity axiom is true (iv) Inverse axiom: Let a G and a be the inverse element of a By the definition of a, a a e By the definition of, a a aa aa a a 9 a 9 a G Inverse axiom is true. (G, ) is a group. 0. Show that 0 0, ω 0 0 ω, ω 0 0 ω, 0 0, 0 ω ω 0, 0 ω ω 0 where ω, ω form a group with respect to matrix multiplication. (JUN-,MAR-, 7) Let I 0 0 C 0, A ω 0 0 ω, B ω 0 0 ω,, D 0 ω 0 ω 0 Let G {I, A, B, C, D, E} By computing the products of these matrices, taken in pairs, we can form the multiplication table Cayley s Table as given below: I A B C D E I I A B C D E A A B I E C D B B I A D E C C C D E I A B D D E C B I A E E C D A B I, E 0 ω ω 0 i) All the entries in the multiplication table are members of G. So G is closed under multiplication of matrices Closure axiom is true. ii) Since matrix multiplication is associative in general, we see that. is associative. iii) From the table, it is clear that, I is the identity element in G. ways00@gmail.com

88 th Maths Public Exam Special Guide Way to Success iv) I. I I I is the inverse of I A. B B. A I A and B are inverses of each other. C. C I C is the inverse of C D. D I D is inverse of D. E. E I E is inverse of E / G is a group with respect to matrix multiplication.. Show that the set M of complex numbers z with the condition z forms a group with respect to the operation of multiplication of complex numbers. (OCT-,JUN-4) M z C/ z i) Closure axiom: Let z, z M z z z z. z, z M / Closure axiom is true ii) Associative axiom: The multiplication of complex numbers is always associative. Associative axiom is true z. z. z z. z. z, z, z, z M iii) Identity axiom: C such that z.. z z is the identity element. / Identity axiom is true iv) Inverse axiom: Let z M. Clearly z Now z z z M and z. z z. z / z is the inverse of z M / Inverse axiom is true. / M is a group with respect to the multiplication of complex numbers.. Show that the set G of all rational numbers except forms an abelian group with respect to the operation * given by a b a + b + ab for all a, b G. (JUN-07, MAR-09,JUN-6) Let G Q { } Leta, b G. Then a and b are rational numbers and a, b. i) Closure axiom: a b a + b + ab is a rational number. But to prove a b G. We have to prove that a b On the contrary, assume that a b then a + b + ab b + ab a b + a + a b ( a + a 0 This is impossible because b. / Our assumption is wrong / a b and hence a b G / Closure axiom is true. ii) Associative axiom: a b c a b + c + bc a + b + c + bc + a b + c + bc a + b + c + bc + ab + ac + abc) a b c (a + b + ab) c a + b + ab + c + a + b + ab c a + b + c + ab + ac + bc + abc / a b c a b c, a, b, c G / Associative axiom is true. iii) Identity axiom: Let e be the identity element. By definition of, a e a + e + ae By definition of e, a e a a + e + ae a e + a 0 e 0 since a e 0 G / Identity axiom is satisfied. iv) Inverse axiom: Let a be the inverse of a G By the definition of inverse, a a e 0 By the definition of, a a a + a + aa a + a + aa 0 a + a a a a G, since a +a / inverse axiom is satisfied. / (G,*) is a group. v) Commutative axiom: For any a, b G a b a + b + ab b + a + ba b a / * is commutative in G and hence (G, ) is an abelian group.. Show that the set of all matrices of the form a o, a R 0 forms an abelian group o o under matrix multiplication. (MAR-08) ways00@gmail.com

89 th Maths Public Exam Special Guide Way to Success Let G be the set of all matrices of the form a o where a R 0 o o i) Closure axiom: Let A a o b o G, B o o o o G AB ab o o o G( a 0, b 0 ab 0) G is closed under matrix multiplication. ii) Associative axiom: Matrix multiplication is always associative. iii) Identity axiom: Let E e o o o G be such that AE A for every A G AE A a o e o a o o o o o o o ae o a o o o o o ae a e Thus E o o o G is such that AE A for every A G We can similarly show that EA A for every A G E is the identity element in G and hence the identity axiom is true. (iv) Inverse axiom: Suppose A x o o o G is such that A A E Then we have x o a o o o o o o o o xa o o o o o o xa x a A /a o o o G is such that A A E Similarly we can show that AA E A is the inverse of A and hence the inverse axiom is true. G is a group under matrix multiplication. (v) Commutative axiom: A, B G AB ab o ba o o o o o BA The group is an abelian group under matrix multiplication. 4. Show that the set G { n / n G} is an abelian group under multiplication. (OCT-) G { n / n G} (i) Closure axiom: Let x r, y s G where r, s Z xy r. s r+s G ( r, s Z r + s Z) The closure axiom is true. (ii) Associative axiom: x r, y s, z t G where r, s, t Z r+s +t x. y. z r. s t r+s. t r+(s+t) r s. t x. (y. z) Associative axiom is true. (iii) Identity axiom: For every x r G, there exists 0 G Such that x. r. r x Similarly. x. r r x is the identity, element and hence the identity axiom is true. (iv) Inverse axiom: For every x r G there exists x r G such that x. x r. r r+( r) 0 Similarly x. x r. r r +r 0 Inverse of r is r G Inverse axiom is true and therefore G is a group. (v) Commutative axiom: x r, y s G x. y r. s r+s s+r s r y. x (Since addition is commutative in Z ) (G,. ) is an abelian group For complete material contact : Way to Success + MATHS Registered Office : Way to Success Publications, Plot No.4, Ramamoorthi Nagar, Old Karur Road, Melachinthamani, Trichy 60 00, Tamil Nadu , You can also purchase our Products online in ways00@gmail.com

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