Topic 6: Calculus Differentiation Markscheme 6.1 Product Quotient Chain Rules Paper 2

Transcription

1 Topic 6: Calculus Differentiation Marksceme 6. Product Quotient Cain Rules Paper. (a) attempt to expand (x + ) x + x + x + N evidence of substituting x + correct substitution ( x + ) ( x + ) + ( x x + ) e.g. f (x) lim 0 simplifying ( x + x + x + x + x + x ) e.g. factoring out (x + x + ) e.g. f (x) x AG N0 (c) f () () setting up an appropriate equation M e.g. x at Q, x, y (Q is (, )) N (d) recognizing tat f is decreasing wen f (x) < 0 R correct values for p and q (but do not accept p.5, q.5) e.g. p.5, q.5; ± ; an interval suc as.5 x.5 NN (e) f (x), y, [, [ A N. substituting x, y into f(x) p + q finding derivative f (x) px + q correct substitution, p + q 8 p 5, q. (a) (i) p, q 5 (or p 5, q ) N (ii) x (must be an equation) N y (x )(x 5) dx x 6x + 5 () (x ) (accept, k ) N (c) ( x ) ( x 6) (d) Wen x 0, 6 dx () y 5 6(x 0) (y 6x + 5 or equivalent) N NN N Standard Level [7] [0]

2 . (a) (i) p q (or p, q ) ()() (N)(N) (ii) y a( x+ )( x ) 8 a(6 + )(6 ) 8 6a a () (N) (iii) y ( x+ )( x ) (i) ( 8) y x x () (N) 5 y x x x () (N) dx (ii) x 7 x 8, y 0 P is (8, 0) ()() (N) ( ) (c) (i) wen x, gradient of tangent is (may be implied) () (ii) gradient of normal is y 0 ( x ) y x+ () x x x+ (or sketc/grap) 6 x x 0 x x 0 (may be implied) () (x+ 8)( x ) 0 () (N) 8 x or x 8 x (.67) () (N) 6 5. (a) f (x) x x 0 ()()() x x (C) Gradient f () () () (C) [6]

3 6. (a) (i) t 0 s 800 t 5 s distance in first 5 seconds m () (ii) v ds 00 8t () At t 5, velocity m s () (iii) Velocity 6 m s 00 8t 6 t 8 seconds after toucdown. () (iv) Wen t 8, s (8) (8) () m () If it touces down at P, it as m to stop. To come to rest, 00 8t 0 t.5 s Distance covered in.5 s 00(.5) (.5) () Since 65 < 656, it can stop safely. (R) 5 7. (a) t 50 5( ) () (c) (d) (e) (i) (ii) 5 t d d (50 5t ) 0 0t 0t () (A) d d (90 0t + 5t ) t 0 + 0t () Wen t (i) d d d 0() or () 0 0t 0(0 t ) or 0 + 0t 0( t 5) t 0 or t ()() (f) Wen t 90 0() + 5( ) ()

4 8. (a) evidence of coosing te product rule e.g. x ( sin x) + cos x f (x) cos x x sin x N 9. (a) (i).5,.5 N (ii) recognizing tat it occurs at P and Q e.g. x.5, x.5 k., k. N evidence of coosing te product rule e.g. uv + vu (c) derivative of x is x derivative of ln ( x x ) is x correct substitution x e.g. x + ln( x ) x x x g (x) + x ln( x ) x () () AG N N0 [7] (d) w.69, w < 0 A N N []

5 0. (a) f (x) xe x x e x ( (x x )e x x ( x)e x ) N Maximum occurs at x () Exact maximum value e N x ± 6 8 M (c) For inflexion, f (x) 0 ( x + ) 0, x, etc. ( + ) + 8 x N. (a) x () Using quotient rule ( x ) () ( x )[( x )] Substituting correctly g (x) ( x ) [6] ( x ) (x ) x ( x ) ( x ) () (Accept a, n ) (c) Recognizing at point of inflexion g (x) 0 M x Finding corresponding y-value 0. ie P, (a) (i) x () [8] (ii) y () By quotient rule (x + 5)() (x )() dx (x + 5) () 9 (x + 5) () (c) Tere are no points of inflexion. (). (a) (i) f ( x) 6sin x ()() [6] (ii) f ( x) sin xcos x 0 sin x 0 or cos x 0 THEN π x 0,, π ()()() (N) 6 (i) translation () in te y-direction of () (ii). (.0 from TRACE is subject to AP) (A) [0] 5

6 . (a) (i) At release(p), t 0 s cos 0 58 cm below ceiling () (ii) cos πt cos πt () t sec () (i) OR t sec (G) 5 ds 0π sin πt ()() Note: Award () for 0π, and () for sin πt. ds (ii) v 0π sin πt 0 sin πt 0 t 0,... (at least values) () s cos 0 or s 8 +0 cos π 58 cm (at P) 8 cm (0 cm above P) ()() 7 (c) 8 +0 cos πt cos πt t 0.6 secs () OR t 0.6 secs (G) (d) times (G) 5. (a) f () f (5) ()() EITHER distance between successive maxima period 5 () (AG) OR π Period of sin kx ; k π so period π () (AG) (c) EITHER A sin + B and A sin + B A + B, A + B ()() A, B (AG)() OR Amplitude A ( ) A A (AG) Midpoint value B + ( ) B B () 5 [6] 6

7 (d) f (x) sin x + f (x) cos x + 0 (A) π cos x (e) (i) y k πx is a tangent π π cos x cos x π x π or π or... () () x or 6... Since 0 x 5, we take x, so te point is (, ) () () (ii) Tangent line is: y π(x ) + y (π + ) πx k π + () 6 (f) f (x) sin x + () sin x () π π 5π π x or or x or or ()()() 5 [] 7