12 th Maths Way to Success

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1 th Maths Quarterly Eam-7-Answer Key Part - A Q.No Option Q.No Option Q.No Option Q.No Option Part B. A adj A A adja..() adja A () A I () From (), (),() we get A adja adja A A I. A 7 6 ~ 7 6 ~ ρ A. Let BC a, CA b, AB R R + R R R + R R R R By the area property of triangles a b b a a b b a ab sin(π C) b sin π A a sin(π B) Divide by ab ab sin C b sin A a sin B sin C Take the reiproals sin A sin B a b a b sin A sin B sin C. Let P be a point on the surfae of the sphere and AB be a diameter. Consider the great irle on the sphere passing through the points P, A and B. Take the entre O as the point of referene. PB OB OP. (i). (ii) AP OP OA OP + OB AP. PB OP + OB. OB OP OP OB Sine OP OB AB subtends a right angle at P on the surfae. a a i + a j + a k i a i i. i a i. a i a a i j a j j. j a j. a j a a j k a k k. k a k. a k a a k L.H.S. a (a i + a j + a k) a R.H.S sin θ b i j k b.n b n 6 9 θ sin 9 6. Let z + iy, z iy z + iy + + iy z iy iy z + + iy +iy z iy +iy + ( )+i + y iy ( )+y + y+ z + Re z + +y + y+ + + y + y + ways@gmail.om - -

2 The modules of sum of two omple 7. numbers is always less than or equal to the sum of their moduli. z + z z + z Let z and z be two omple 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 8. Sine i is a root, + i is also a root. Sum of the roots Produts of the roots The orresponding fator is ( + p ) Equating term, we get p + 8 p The other fator is 6, Thus the roots are ± i,, 9. y + y + k Centre (,) k Other asymptotes + y + Combined equation y + y Equation of R.H is y + y + Passes through, 7 Required equation y + y 7. (i) y k a( h) V(, ), y a( ) Fous (,) Required equation is y 8( ). + yy (ii) Point (, ) + y. F + df + d d F d +ve minimum For maimum or minimum df d + ± is not admissible The resistane is minimum When, the resistane F + Minimum value of resistane. f is a ontinuous funtion on [, π] f is a differentiable on (, π) f a f, f b f π f f sin f sin sin π (,π) When π, y ais at (π, ). Consider the differene f ( + ) n ( + n) f n + n n n[ + n ] Sine > and n > we have + n >, so f > Therefore f is stritly inreasing on [, ) > f > f() ( + ) n + n > + ( + ) ( + ) n + n > ( + ) n > + n. If T k l kl Then dt dl k l k l dl... m Error in T Approimate hange in T T dt dt dl Perentage error T T dl k l % (.) ways@gmail.om - -

3 . a. b k. l k l %. l %.. % % (.) 6..6% Hene the perentage error in the time of swing is a derease of.6% + y y y 8 6 Sine, and y and atleast one of the oeffiients a ij of the system is onsistent and has infinitely many solutions. All minor are zero and atleast minor is non zero. The system is redued to a single equation. We assign arbitrary value to (or y) and solve for y(or ) Suppose we assign t, from equation () We get y (8 t) The solution set is, y t, (8 t) t R u y sin y u t, ty (t)(ty) sin t ty t y sin y u is a homogeneous funtion of degree By Euler s theorem u u + y u y Part C 6. y z 7 A X B A A is a non-singular matri and hene A eists.the ofators are A 6, A 8, A A, A, A A, A, A The matri formed by the ofators is A ij A A 6 8 adj A adj A 6 8, y, z 7. The system of equations an be written as A X B μ y z The augmented matri is [A, B] μ ~ ~ μ μ 8 μ R R R R R R R R R Case(): If μ 8 then 8 μ and hene there are three non-zero rows. ρ A ρ A, B number of unknowns. The system has the trivial solution, y, z Case(): If μ 8 then ρ A, B and ρ A ρ A ρ A, B < number of unknowns The given system is equivalent to + y + z ; y + z ; y z z Taking z k, we get k, y k, z k, k R {} Whih are non-trivial solutions. Thus the system is onsistent and has infinitely many non-trivial solutions. 8. Let ABC be a triangle and let AD, BE be its two altitudes interseting at O. In order to prove that the altitudes are onurrent it is suffiient to prove that CO is perpendiular to AB. Taking O as the origin, let the position vetors of A, B, C be a, b, respetively. Then OA a, OB b, OC Now AD BC OA BC OA. BC a. b a. a. b BE CA OB CA OB. CA b. a b. a b.. Adding () and (), we get, ways@gmail.om - -

4 a. a. b + b. a b. a. b. (a b). BA. OC OC AB Hene, the three altitudes are onurrent. 9. The ondition for interseting is y y z z l m n l m n Compare with y y z z l m n y y z z l m n The determinant beomes + + The lines are interseting lines Point of intersetion: Take y+ z λ Any point on the line is of the form λ +, λ, λ Take y z+ μ Any point on the line is of the form μ +, μ +, μ and Sine they are interseting, for some λ, μ λ +, λ, λ μ +, μ +, μ μ, λ The point of intersetion (,) 6. p + p + q p ± iq, Let α p + iq, β p iq α β qi Given tan θ q q y+p y + p tan θ y + p q ot θ y q ot θ p y + α q y + α n Similarly y + β n q n sin n θ q n sin n θ os θ+i sin θ sin θ [os nθ + i sin nθ]..() [os nθ i sin nθ] () () () gives y + α n y + β n n sin nθ q sin n θ ; + 6. (i) os + i sin os k π + i sin kπ os kπ kπ + i sin k,,,, (ii) os π + i sin π os(k + ) π + i sin(k + )π os (k+)π + i sin (k+)π k,,, Thus we have 9 roots. As per the given information, we an take the parabola as open downwards ay Let P be the point on the flow path,.m below the line of the pipe and m beyond the vertial line through the end of the pipe. P(,.) Thus, 9 a(.) a 9 The equation of the parabola is 9 y Let be the distane between the bottom of the vertial line on the ground from the pipe end and the point on whih the water touhes the ground. But the height of the pipe from the ground is 7. m The point (, 7.) lies on the parabola The water strikes the ground m beyond the vertial line. 6. y y + 6 y + 6y ( ) (y y + ) {( + ) 9} y ( + ) y (+) y ways@gmail.om - -

5 X Y where X +, Y y The transverse ais is along X ais a, b, a, b e + b a ae Referred to X, Y Centre (,) Foi Verties (±ae, ) (±, ) (±a, ) (±,) Referred to, y X +, Y y X, Y +, y C(,), X, Y +, y +, y F ( +, ), X, Y +, y, y F (, ) (, ) X, Y +, y, y A(,) (,) X, Y +, y, y A (,) Thus the line y is a tangent to the hyperbola. It touhes the hyperbola The point of ontat is a m b () 9 The point of ontat is a m 9, b, 6 Let r, h respetively denote the base radius and height of the one of volume V, at time t min. Then we are given that r h To find dh when h ft and dv ft /min Volume of the one V πr h dv π dh dv πh dh π h π (h ) h h dh π π 6 π ft/ min The height of the one is inreasing at the rate of 6 π ft/min. 66 f sin, is ontinuous in [,π] f os f os os π or π The values of f at these ritial points are f π π sin π π 6 The ondition for y m + to be a tangent to a hyperbola y is a b a m b y y + y m, y 6 y a, b ; a m b (9) a m b f π π sin π π The values of f at the end points are f and f π π 6.8 Comparing these four numbers, the absolute minimum is f π π and the absolute maimum is f π π + 67 y dy d 6 d y d ( + ) d y d, ways@gmail.om - -

6 f < for (, ) and (, ) and f > for (, ) f is onave downward on (, ) (, ) and f is onave upward on (, ) And the points of infletions are, f, (, f()) That is (, 8) and (, 9) 68. w u e v, u, v y log y 69. u uev u y y v v u e v y. u +. v u v y y ( + y). u +. v y u y v y y y y [y log ] u y v y log 7. a 7. b Cartesian form: Let a, b and be the, y and z interepts of the plane respetively. The plane passes through the points a,,,, b,, (,, ) Here The equation is y y z z y y z z y y z z a y z a b a a b y a + z ( + ab) b + ya + zab ab b ab + ya + zab ab ab ab ab ab + y + z a b Vetor form: The equation of the plane passing through three given points is r ( s t)a + sb + t r ( s t)ai + sbj + tk i + yj + zk ( s t)ai + sbj + tk s t a; y sb; z t a s t, y b s, z t + y + z s t + s + t a b + y + z a b Take the midpoint of the base as the entre C(,). Sine the base is ft, the verties A and A are (,) and (,) Clearly a a, b 6 The orresponding equation + y 6 (9, y ) is a point on the equation 9 + y 6 y 6 9 y 9 The height of the arh 9ft from the right of the entre is 9 feet. Prepared by K.Dinesh M.S., M.Phil., PGDCA.,(Ph.D) ways@gmail.om

1 - a 1 - b 1 - c a) 1 b) 2 c) -1 d) The projection of OP on a unit vector OQ equals thrice the area of parallelogram OPRQ.

1 - a 1 - b 1 - c a) 1 b) 2 c) -1 d) The projection of OP on a unit vector OQ equals thrice the area of parallelogram OPRQ. Regter Number MODEL EXAMINATION PART III - MATHEMATICS [ENGLISH VERSION] Time : Hrs. Ma. Marks : 00 SECTION - A 0 = 0 Note :- (i) All questions are ompulsory. (ii) Eah question arries one mark. (iii) Choose