QUESTION PAPER CODE 65/2/2/F EXPECTED ANSWER/VALUE POINTS

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1 QUESTION PAPER CODE EXPECTED ANSWER/VALUE POINTS SECTION A. P 6 (A A ) P 6 9. (a b c) (a b c) 0 a b c (a b b c c a) 0 a b b c c a. a b sin θ a b cos θ 400 b 4 4. x z 5 or x z 5 mark for dc's of normal Expanding we get x 8 x SECTION B 7. f x 5 f b f f b 5 lim f (x) f () lim f (x) x x 4 a b a 8. Let u x tan x Put x tan θ θ tan x u sec θ tan tan θ cos θ tan sin θ ()

2 θ tan tan θ tan x du ( x ) x v sin x tan x dv x du dv du / dv / 4 x sin t cos t dt d sin pt p cos pt dt d p cos pt cos t d cos t ( p sin pt) p cos pt ( sin t) dt cos t p sin pt cos t p cos pt sin t cos t Now d d ( x ) x p 0 9. Eqn of given curves 4ax and x 4b Their point of intersections are (0, 0) and ( ) d d Substituting values of, & 4a b, 4a b d a a 4ax, slope...(i) b x 4b d x, slope a b /...(ii) b At (0, 0), angle between two curves is 90 or Acute angle θ between (i) and (ii) is a b θ tan a b () /

3 0. I I I I I ( x) 0 sin α sin ( x) 0 sin α sin x / 0 sin α sin x / 0 x tan sin α x tan dt 0 Put tan x t t t sin α dt 0 (t sin α ) cos α t sin α tan cos α cos α 0 I α cos α. I (x 5) 0 4x x ( 4 6x) 0 4x x 0 4x x 4 (0 4x x ) x 9 4 x x 7 x (0 4x x ) sin C x (sa) ( ) ( 4) ( ) ( 5) A B 5 using partial fraction we get A, B (x ) (x 4). (x ) (x 5) 7 4 x 4 x 5 x 7 x 5 x tan log C x 5 (4)

4 . I x sin x x put sin x t x dt t sin t dt t cos t sin t c x sin x x c. (x x ) d 0 d (x x ) put x v d dv v d dv v d (v v ) dv v d Integrating both sides tan v log c tan x log c x 4. d cot I.F e cot e d ( ) x e cot d cot cot t e Integrating, we get x e cot cot cot e d put cot t t t e dt ( t) e t c x ( cot ) ce cot (5)

5 5. a b c d...(i) a c b d...(ii) () () a (b c) d (b c) (a d) (b c) 0 (a d) (b c) 6. Equation of line AB r ( ˆj k) ˆ λ (4i ˆ 6j ˆ k) ˆ Equation of line CD r (i ˆ 9j ˆ 4k) ˆ µ ( 7i ˆ 5j) ˆ a a i ˆ 0 ˆ j 5k ˆ ˆi ˆj kˆ b b 4 6 0i ˆ 4ˆj kˆ (a a ) (b b ) Lines intersect 7. Let selection of defective pen be considered success p, q Reqd probabilit P(x 0) P(x ) P(x ) C0 C C P(x i ) i 0 Σ 8k k 8 (i) P(x ) 8 (6)

6 (ii) P(at most colleges) P(0) P() P() 5 8 (iii) P(atleast colleges) [P(x 0) P(x )] LHS cot x x x x cos sin cos sin x x x x cos sin cos sin cot cot x x RHS tan x x x x x x x x 4 x 4 tan 4 x 7 ± 9. Let each poor child pa ` x per month and each rich child pa ` per month. 0x x In matrix form, 0 5 x AX B X A B A x x 00, 000 Value: Compassion or an relevant value (7)

7 SECTION C 0. Their point of intersection (, ) Correct Figure Required Area 0 () (x ) x (x ) 4x x x x 4 x x sin sin 0 5 Sq. units. Equation of famil of planes passing through two given planes (x z 4) k (x z 5) 0 ( k) x ( k) ( k) z 4 5k...(i) x z 4 5 k 4 5 k 4 5 k k k k As per condition 4 5 k k (4 5 k) ( k) k 4 or 5 5 For k, Eqn. of plance is 7x 4z 5 5 For k 4 5, Eqn. of plane is x 4 z 0 Equation of plane passing through (,, ) and parallel to the plane is: 7(x ) ( ) 4(z ) 0 7x 4z Vector form: r (7i ˆ j ˆ 4k) ˆ. Let H be the event red balls are transferred H be the event red and balck ball, transferred H be the event black and balck ball transferred E be the event that ball drawn from B is red. P(H ) P(H ) C 8 C P(E/H 8 ) C 0 C C 5 P(E/H 8 ) 5 0 (8)

8 P(H ) 5 C 0 8 C P(E/H 8 ) 4 0 P(H /E) Let x tablets of tpe X and tablets of tpe Y are taken Minimise C x subjected to 6x 8 x x 4 6 x, 0 Correct Graph C A(0, 9) 9 C B (, 6) 8 Minimum value C C (6, ) C D (8, 0) 6 x < 8 does not pass through unbounded region Thus, minimum value of C 8 at x, f(x) x x, g(x) x x V x R (fog) (x) f(g(x)) x x x (gof) (x) g(f(x)) x x x x (fog) ( ) 6 (fog) (5) 0 (gof) ( ) 5. a b c abc a b c 0 a b c (9)

9 C C C C a b c b c abc a b c b c a b c b c 0 b c abc a b c b c b c 0 R R R, R R R abc a b c 0 a, b, c, 0 0 a b c A adj A cos α sin α 0 sin cos 0 α α 0 0 A(adj A) I A I 0 0 I S 6x 4r r S (i) V x 4 r 4 S 6x x 4 x (S 6x ) 6 (0)

10 dv x x S 6x dv 0 x x S 6x r x [using (i)] d V d V r x x ( x) 4x S 6x S 6x > 0 V is minimum at x r i.e. r x Equatioin of given curve cos (x ) x Minimum value of sum of volume 6x cubic units...(i) d d sin (x ) d sin (x ) sin (x ) given line x 0, its slope condition of lines sin (x ) sin (x ) sin (x ) cos (x ) 0 0 using (i) cos x 0 x (n ), n I x, [, ] Thus tangents are to the line x 0 onl at pts, 0 and, 0 Required equation of tangents are 0 x x x x 4 0 ()