Page 1 MATHEMATICS

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1 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., Page + MATHEMATICS

2 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., SIX MARKS Page MARCH 6 JUNE 6 OCTOBER 6 P represents the variable comple number z. 4 Solve 4, if +I is one of the roots. It is Solve ( D D ) ysin 5 Find the locus of P if z z. given that +I is a root Let z iy ( iy) iy ( ) i ( y) ( ) iy ( ) ( y) ( ) y ( ) ( y) ( ) y 4 4 4y 4 4 y y y Which is a circle whose centre is origin and radius is unit. i is also a root Sum of the roots = + i+-i = Product of the roots = (+i) (+i) = ( i) ( ) The corresponding factor is (sum of roots) + product of roots p 4 ( ) ( ) Equating term we get O p 4 p = 4 p = Other factor is To find the roots of b b 4ac a a =, b=, c = () 4 () () () 4 ( ) i i Other roots are - + i and - i Characteristic equation is a =, b = -, c = p () ( ) ( ) 4() () p p 4 4 p p p i, Complementary function ye [ ACos B Sin] ye [ ACos B Sin ] PI Sin Sin D D D Sin Sin 4 D D D Sin D Sin D [ Cos sin ] D D 5 General solution y = C.F + P.I + P.I y e [ A Cos BSin ] [ Cos Sin ]

3 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., MARCH 7 JUNE 7 OCTOBER 7 Find k,, the normal distribution Solve the equation if iis a root dy whose probability distribution function The given equation has three roots. Page ( y) d is given by 4 f ( ) k e 4 f ( ) k e ( 4 ) ke ke ke We have ( ) ( ) 4 4 ( ) 4 4 ke...() f ( ) e... () From () and (), k Ans : k It is given that -i is a root Its Conjugate +i also is a root Sum of these roots = -i + +i = Product of these roots = (-i)(+i) = -i = -(-) = + = The factor corresponding to these roots is (sum) + product i.e, i.e, The corresponding factor is () () 4 The other factor is ) ( ) The root is = The roots are +i, -i and. dy Let Z = + y d ( y) () () Differentiate w.r, t dz dy d d dy dz d d () dz Put (), () in () () z d dz dz dz z d z d z d z z dz z d ( z ) dz z d dz d z dz dz d C z ztan z c ytan ( y) c ytan ( y) c

4 MARCH 8 JUNE 8 OCTOBER 8 P represents the variable Z Find the order of all the elements t Show that a b, b c, c a the group (Z6,+6) z Find the locus of P if Re Let z = +iy z z iy ) iy z iy i i( y i) = PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., ) i y i( y ) i( y ) i( y ) ( y( y ) i y ( )( y ) ( y) z ( ) y( y ) Re z ( y ) It is given that Re ( ) y( y ) ( y) ie ( ) y( y ) ie y y z z Locus of y y Z6={[],[],[],[],[4],[5]} O[]= O[]=6 O[]= []+[]+[]= O[]= []+[]= O[4]= [4]+[4]+[4] O[5]=6 []+[]+[]+[]+[]+[]= [5]+[5]+[5]+[5]+[5]+[5]= LHSa b, b c, c a a b b c c a.. a b b c a c c a a b. b c b a c c c a a b. b c a b c a a. b ca bc a b. b ca bc a a. bc a. ab a. ca b. bc b. ab b. ca ab c bc a ab c cb a ab c ab c RHS Page 4

5 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., Page 5 MARCH 9 JUNE 9 OCTOBER 9 P presents the variable comple number z. Solve the following system of linear Show that the points A (,,), B (, -, ), C (-,, Find the locus of p, if z i z i. Let z iy z i z i iy i iy i i( y ) i( y ) ( y ) ( y ) Squaring both sides y y y y y y The locus of P is ais. equations by determinant method -y=7; 4-6y= y Since and y. But atleast one aij is not equal to zero. The system is consistent and has infinitely many solutions. Put y = k () k 7 7 k 7 k 7 k Thesolution is, y k where k R ) and D (-6, -4, ) are laying on the same plane. OA i jk OB i jk OC i jk OD 6i 4 j k AB OB OA i j k AC OC OA i jk AD OD OA 5i jk AB, AC, AD 5 6 = (--) (-) (-) (8-5) = (-) + () () = = = AB, AC, AD are Co planar Hence the points A,B,C,D are coplanar.

6 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., Page 6 MARCH JUNE OCTOBER T T A A for the matri With usual symbol, prove that ( Z5{}, 5) 4 Find the rank is a group Let G = { [], []. [],[4]} A Verify that A adj A A A Adj( A) 5 A T 6...() 7 5 T A 5 6 T ( ) T Adj A A T A T A 5 7 T 6 5 adj A 6 5 T A...() 7 T T From () and () A A 5 [] [] [] [4] [] [] [] [] [4] [] [] [4] [] [] [] [] [] [4] [] [4] [4] [] [] [] From the table i. All the elements of the composition ii. table are the elements of G. The closure aiom is true. Multiplication modulo 5 is always associative. iii. The identity elements [] G and it satisfies the identity aiom. iv. Inverse of [] is [] Inverse of [] is [] ; Inverse of [] is [] Inverse of [4] is [4] and it satisfies the inverse aiom. The given set forms a group under multiplication modulo 5. 7 ~ R R 4 7 ~ R R R 4 7 ~ R R R 4 7 ~ R R 4 5 ~ R R 5 ~ R R R ~ Rankof thematriis

7 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., Evaluate MARCH JUNE OCTOBER P represents the variable comple number z. Solve ( D 5D 4) ysin 5 lim cos Find the locus of P it z5 z Page 7 Given lim cos lim cos cos = = Let z = +iy z 5 ( iy) 5 5) yi z 5 5) ( y) y...() z iy ) i y ) y y...() z 5 z 9 5 9y y Squaring both sides we get by() and( 9 9y 5 9( y ) 9 9 y y 8 9 Locus of P is a straight line Characteristic equation is P P P P 5p4 4 4 P (P+4) + + (P+4)= (P+4) (P+) = P +4 = or P+ = P = -4 or P = - 4 Complementary function is Ae Be P. I Sin5 D 5D D 4 Sin 5 5D Sin 5D Sin5 (5D ) (5D ) 5D Sin5 5 ( 5 ) 44 (5D ) sin [ 5.5 cos sin 5 ] 66 (5 cos5 sin 5 ) 66 General Solution is 4 (5 cos5 sin 5 ) y Ae Be 66

8 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., MARCH JUNE OCTOBER Find the order of each element of the 4 dy ( y) group, ( Z7 {[]}, 7) Find the rank of matri 4 d 4 G = {[], [], [], [4], [5], [6]} Page 8 i). The identify element is [] ii). iii). iv). v). vi). So its order is ([]) = [] = [] 7 [] 7 [] = [8] = [] ([]) = 6 [] = [] 7 [] 7 [] 7 [] 7 [] 7 [] 7 = [79] = [] ([] = 6 [4] = [4] 7 [4] 7 [4] 7 = [64] = [] ([4] = 6 [5] = [5] 7 [5] 7 [5] 7 [5] 7 [5] 7 [5] 7 = [565] = [] ([5]) = 6 [6] = [6] 7 [6] = [6] = [] ([6]) = 4 LetA 4 4 ~ R R R ~ R R R ( A) dy Let Z = + y () d ( y) () Differentiate w.r, t dz dy d d dy dz d d () dz Put (), () in () () z d dz dz dz z d z d z d z z dz z d ( z ) dz z d dz d z dz dz d C z ztan z c ytan ( y) c ytan ( y) c

9 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., Page 9 MARCH JUNE Solve the following system of linear If ai j, b j k, c k j, then find equations by determinant method +y+z=4; +y+6z =7; +y+z=. 6 = (-6) (-) + (-4) = (-4) (-) + (-) = = - + = = = 4 (-6) (7-6) + (7-) = 4 (-4) - (-5) + (-) = = = - = - and and so The given equations are in consistent and hence no solution. a b, b c, c, a a b ( i j) j k) a b i j j k a b i j k b ci j k c a i j k a b, b c, c a [ i j k, i j k, i j k] = (+) (-) (-4) + (+) = () + (-) + () = -6+ = 6-6 = a b, b c, c a

10 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., TEN MARKS Page MARCH 6 JUNE 6 OCTOBER 6 Find the volume of the solid Find the vector and Cartesian equation of the A poster is to have an area of 8 cm with cm generated by the revolution of the plane passing through the point margins at the bottom and sides and a cm loop of the curve t and t yt (-, -, ) and perpendicular to two planes margin on the top. What dimension will give the +y+4z+7= and -y+z+=. largest printed area?(diagram also needed) t ie y t t yt t t Lety or or Volume y d d d 9 d () The vector normal to the planes +y+4z+7 = and - y+z+= are i j 4k and i j k The required plane is r to the planes +y+4z+7 = and -y+z+= The required plane parallel to the above two vectors i j 4k and i j k and passes through the point (-, -, -). Vector equation is r a sutv Iet r i j k s i j 4k t i j k Its Cartesian equation is ( y y z z, y, z) (,,) m n (, m, n ) ) m n (, m, n ) (, Thus the Cartesian equation is y z 4 (+) [6+4] (y+) [-8] + (z-) [--4] = ++5y+-5z+5 = +5y-5z+5 = +y-z+5 =. Let the length of printed portion ABCD be can and its breadth be y cm (+) cm and its breadth is (y+) cm. Poster area is given as 8 (+) (y+) = 8 y 8 cm 8 y () Let the area of the printed portion be A A = y 8 A 8 A da 6 d ( ) 6 ( ) d A 7 d 6 ( ) ( ) ( ) da d 6 by ( ) 6, ( ) ( ) d A 7 7 Negative Printed d ( ) ( ) area is maimum when 8 () 9 9 y = The he printed area are cm and cm.

11 MARCH 7 JUNE 7 OCTOBER 7 Find the common area between y Find the ais focus, latusreetum, equation of and PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., Page Find the eccentricity, centre, foci and vertices of the y. the latus rectum, verte and directri for the parabola y y 8 7 hyperbola 4y 4 y7 and draw the diagram. y () y y y 8 7 y y 8 7 4y 4 y7 () Solve () and () to get the point of intersection Put () in () ( ) 4 ( ) = = = in (), y = = in (), y = The point of intersection are (,) and (,) b Required area [ f ( ) g( )] d f ( ) y, d a g( ): y.. ). sq. units y y 8 7 ( y) 8 7 ( y ) 8 7 Y ( y) 8( ) 8Xwhere Y=y-X=- 4a 8 a About X,Y Referred to.y Ais Y= Y= y-=y= Focus (a,) ie (,) X= -==+ =4 Latus rectum X=a ie X= Y= y-= y= F(4,) X= -==+ =4 Verte (,) X= -= = Directri X = -a X = - Graph referred to,y Y= y-= y=v (,) X ( ) 4 ( y 8y 4 4 ) 7 [( ) ] 4[( y 4) 6] 7 ( ) 4( y 4) 7 64 X Y Where X Y y a b 75 4 e 4 5 e e e e 4 4 Referred to X, Y Centre C (, ) C (,4) Foci F (ae, ) = (5,) Vertics 5 A, F (-a e, ) = (-5,) Graph Referred to, y 5 A, Referred to,y F (6,4)(-4, 4) 7, 4 A,4

12 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., MARCH 8 JUNE 8 OCTOBER 8 Page Find the vector and Cartesian equation of Find the common area between y and Find the local minimum and maimum values of the plane passing through the point (-, -, ) and perpendicular to two planes +y+4z+7= and -y+z+=. y. f ( ) 6 The vector normal to the planes +y+4z+7 = and -y+z+= are i j 4k and i j k. The required plane is r to the planes +y+4z+7 = and -y+z+= The required plane parallel to the above two vectors i j 4k and i j k and passes through the point (-, -, -). Vector equation is r a sutv r i j k s i j 4k t i j k Iet Its Cartesian equation is y y z z m n m n Thus the Cartesian equation is y z 4 (+) [6+4] (y+) [-8] + (z-) [--4] = ++5y+-5z+5 = +5y-5z+5 = +y-z+5 = (, y, z) (,,) (, m, n) ) (, m, n ) (, y () y () Solve () and () to get the point of intersection Put () in () ( ) 4 ( ) = = = in (), y = = in (), y = The point of intersection are (,) and (,) b Required area [ f ( ) g( )] d f ( ) y, d a g( ): y.. ). sq. units f ( ) 6 f '( ) f "( ) 6 Let f '( ) i. e, ( ) ( ) ( ) ( ),, [ f "( )] ( ) negative f() attains maimum when = maimum value is ( ) ( ) 6 ( ) = ( 7) (9) [ f "( )] () positive f( ) attains minimum when = Minimum value is () () 6() ie (8) (4) 7 ie6 7 ie 4 Minimum value is -4.

13 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., MARCH 9 JUNE 9 OCTOBER 9 P represents the variable Z Find the eccentricity, centre, foci and Solve : ( D 5D 6) y Sin e z Find the locus of P if Re z vertices of the ellipse and draw the diagram. 6 9y 6y 9 Page Let z = +iy z iy ) iy z iy i i( y i) = ) i y i( y ) i( y ) i( y ) ( y( y ) i y ( )( y ) ( y) z ( ) y( y ) Re z ( y ) It is given that Re ( ) y( y ) ( y) ie ( ) y( y ) ie y y z z Locus of y y 6 9y 6y 9 6( ) 9 ( y 4y 4 4) 9 6 ( ) 9 ( y ) 44 ing by 44 y a 9 6 w.r. to X, Y Centre (, ) Foci (, ae) (, 7) Vertices (, ±a)(, ±4) ( ) ( y) 9 6 a b 6, b 9 e a w.r to,y, y y (, -) 7 4 (, 7) and(, 7) (, ) and( 6) Characteristic equation is (p-) (p-) = P = or p = P 5p Complementary function is Ae Be PI Sin D 5D 6 5D 6 Sin 5D 6 Sin 5D 5 Sin D D Sin Sin 5 ( D ) ( D ) 5 D D Sin D 5 Sin D ) Sin 5 DSin Sin ] Cos Sin ] PI e e e D 5D PI e e ( D ) ( D ) ( D ) ( ) e e e D. D General Solution is y = C.F + P.I + P.I y Ae Be [ Cos Sin ] e

14 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., MARCH JUNE OCTOBER Find the volume of the solid obtained by ( ) dy Find the vector and Cartesian equations of the 6 y Cosec revolving the area of the triangle whose sides d plane, through the point (,-) and parallel to are having the equation y =, =4 and -4y=, about 4y 4 the line y z and perpendicular to 4 the plane +y-z=8. Page 4 about 4y y = = = = to 4 4y = 4y = 9 y y 4 6 Volume b y d a d d cu. units 6 4 ( ) dy 6 sec d y Co dy 6 Cosec, d 6 sec P Q Co 6 Pd log ( ) Pd log ( ) e e Solution is y( IF) Q( IF) d C Cosec y( ) ( ) dc y( ) Cosec d C y( ) Cot C y( ) Cot C The normal vector to the plane +y+z =8 is i j k. This vector is parallel to the required plan. The required plane passes through (,-) and parallel to u i j 4k and v i j k. The vector equation of the required plane is ra su tv i.e. ri j k s( i j.4 k) t ( i j k) y y z z m n m n Cartesian from (, y, z ) (,, ) (, m, n) ) (, m, n ) (, y z 4 i.e. (-) (-6+) (y-) (9+8) + (z+) (9+4) = 6 (-) 7 (y-) + (z+) = y+4+z+6 = 6-7y+ z+ 54 =

15 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., MARCH JUNE OCTOBER Find the verte, ais, focus, equation of latus rectum, equation of directri and length of latus Find the verte, ais, focus, equation of latus rectum, equation of directri and length of latus rectum of the Solve the system to equations. X+y+z=, +4y+z=4, -y-z= rectum of the parabola y 4y 4 8 and parabola 4 4y hence. Draw the diagram. Page 5 y y y y y ) 4 4 ) 4 Y 4X Where Y = y + ; X = + ; 4a = 4 ; a = The type is open left ward. Referred to X, Y Referred to,y X = + Y = y+ Verte (, ) (-,-) ais Y = Y = - Focus (-a,)i.e (-,) Focus = = (-, -) Equation of latus rectum X = -a i.e X = - X = - + = - = - Equation of directri X =a i.e X = X = + = = Length of L.R 4a = 4 4a =4 4 4y 4 4y ( ) 4( y ) X Y Where Y y 4 4a 4 X a The type is open downward. Referred tox, Y Referred to,y X = + Y = y+ ais X = - = Verte (,) (, ) Focus (, -) Focus = (, ) Equ.ofL.R Y = y = Equ.of dir. Y = - y = Len. L.R 4a=4 4a = 4 4 ( 4 4) ( ) ( 4 4) ( 4) ( 8) ( 4 4) ( 4 ) ( 8 ) = 8 4 = -6 y 4 ( 4 ) ( ) ( 4) = = -8 z 4 4 ( 8) ( 4) ( 4 4) = = 6 y 8 y 6 6 z z 6, y, z y

16 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., MARCH JUNE OCTOBER Find the area between the curves Find the eccentricity, centre, foci and Find the vector and Cartesian equations of the y ais and the lines = - and =5. vertices of the hyperbola y z plane which contains the line and (9 6 ) (7 y 4 y) 9and draw the perpendicular to the plane -y+z-=. diagram. Page 6 Required area A A A yd ( y) d yd ( ) ( ( d d sq.unit = (9 6 ) (7 y 4 y) 9 9 ( 4 ) 7 ( y y) 9 9[( ) 4] 7[( y) ] 9 7 ( y) 9( ) 6 7( y) 9( ) 6 6 ( y) ( ) ie. 9 7 Y X Where X ; Y y 9 7 a 9a b 7 e Referred to X, Y b a Centre C (, ) (-, ) Referred to,y X ; Y y Foci F (, 6 ) (-, 7) & (-, - 5) Vertices V(, ) (-, 4) and (-, -) y z The required plane contains the line i.e, the line y z ( ) The plane passes through the point (,, -) and parallel to the vector u i j k This plane is perpendicular to -y+z- =. That is the plane is parallel to i j k. The required plane passes through (,, -) whose postion vector ai j and parallel to two vector ui j k and Lj k. Vector equation r a su tv i.e, r i j s(i j k) t( i j k) Its Cartesian form is Ie y z y y z z m n m n (-) (-9+) y (6-) + (z+) (-4+) = (-7) (-7) y(5) + (z+) (-) = -7 5y z +6 = 7+5y+z-6 =

17 PREPARED BY :S.MANIKANDAN., VICE PRINCIPAL., JOTHI VIDHYALAYA MHSS., ELAMPILLAI., SALEM., MARCH JUNE u u u u y y Verify for the function n y Cos if y y y y y u sin y usin Pr ove. y y u sin y y usin y y i. esin ( u) y u cos y f Sin ( u) y y y Page 7 u cos. Sin y y y y y y cos Sin...() y y y y u cos y y y u cos sin y y y y y y cos Sin y y y y u u y y t ty Put = t, y = ty f t ty t( y) f ( y) ft t ( y) y f is a homogeneous function in and y of degree f f &By Euler s theorem y. f y y (sin ) y ( Sin u ) Sin u u u u u y.. y Sin u u u.. y y u Sin u y y Sin. Sin Sin y y y y Cos. y y y y Cos y y

PRACTICE PAPER 6 SOLUTIONS

PRACTICE PAPER 6 SOLUTIONS SECTION A I.. Find the value of k if the points (, ) and (k, 3) are conjugate points with respect to the circle + y 5 + 8y + 6. Sol. Equation of the circle is + y 5 + 8y + 6