d (5 cos 2 x) = 10 cos x sin x x x d y = (cos x)(e d (x 2 + 1) 2 d (ln(3x 1)) = (3) (M1)(M1) (C2) Differentiation Practice Answers 1.

Transcription

1 . (a) y x ( x) Differentiation Practice Answers dy ( x) ( ) (A)(A) (C) Note: Award (A) for each element, to a maximum of [ marks]. y e sin x d y (cos x)(e sin x ) (A)(A) (C) Note: Award (A) for each element. []. (a) d (x + ) (x + ) (x) (M)(M) (C) x(x + ) d (ln(x )) () (M)(M) (C) x x [] x x d. e e d (5 cos x) 0 cos x sin x (A)(A) (A)(A)(A) fʹ (x) x e 0 cos x sin x (A) (C6)

2 . METHOD f( x) 6x (A) fʹ ( x) x x x (A)(A) (C6) METHOD f( x) 6( x ) (A) fʹ ( x) 6 ( x ) x (A)(A) fʹ () x x (A) (C6) 5. (a) d y cos x A N dy x + tan x accept x sec x + tan x AA N cos x (c) METHOD Evidence of using the quotient rule (M) x ln x dy x x AA dy ln x x N METHOD y x In x Evidence of using the product rule (M) dy x + ln x x x ( )( ) AA dy ln x N x x

3 6. (a) x or 5x 0 (A) (N) 5 fʹ ( x) (5x )(6 x) ( x )(5) (5x ) (M)(A) 0x 6x 5x (5x ) (may be implied) (A) 5x 6x (5x ) (accept a 5, b 6) (A) (N) [5] 7. (a) fʹ (x) 5e 5x AA N gʹ (x) cos x AA N (c) hʹ fgʹ + gf (M) e 5x ( cos x) + sin x (5e 5x ) A N 8. (a) f ʹ (x) x x 0 (A)(A)(A) x x (C) Gradient f ʹ () (M) (A) (A) (C) 9. (a) 0 p 00e (M) 00 (A) (C) Rate of increase is dp d e dp dt 0.05t t 0.05t 5e When t 0 (M) (A)(A) dp 5e dt 0.05(0) 0.5 5e ( 8. 5 e ) (A) (C)

4 0. f () b + c + 0 (M) f ʹ (x) x b, f ʹ () 6 b 0 b 6, b (M) (A) () + c + 0, c (A) Note: In the event of no working shown, award (C) for correct answer. []. (a) f (5 + h) h f (5) (5.) (or 76.5 to sf) (A) (C) lim h 0 f (5 + h) h f (5) f ʹ (5) (M) (5) (A) 75 (A) (C) []. (a) (i) Vertical asymptote x l (A) (ii) Horizontal asymptote y 0 (A) (iii) 6 y 0 x Note: Award (A) for each branch. (A)(A)

5 (i) f ' (x) 6x ( + x ) ( ) ( ) ( + x x + 6x ()( + x ) x ) f '' (x) ( + x ) ( + x )( x ) x ( + x ) ( + x ) x( x ) ( + x ) + 6x + 6x (M) (A) (A) (AG) (ii) Point of inflexion > f " (x) 0 (M) > x 0 or x x 0 or x 0.79 ( sf) (A)(A) OR x 0, x 0.79 (G)(G) 6 (c) (i) Approximate value of f ( x), h b a n 5 (A) 5 [ ] (A) 5 (.805) (A) (ii) f ( x) y 0 x (A) Between and, the graph is 'concave up', so that the straight lines forming the trapezia are all above the graph. (R) 5 [5] 5

6 . (a) x (A) (i) f ( 000).0 (A) (ii) y (A) (c) f ʹ (x) ( x ) (x ) ( x )(x (x 9x 7 ( x ) ( x ) ( x ) 7x + ) (x 6x + 0) x + 0) (A)(A) (A) Notes: Award (M) for the correct use of the quotient rule, the first (A) for the placement of the correct expressions into the quotient rule. Award the second (A) for doing sufficient simplification to make the given answer reasonably obvious. (AG) (d) f () 0 stationary (or turning) point (R) 8 f ʺ () > 0 minimum (R) 6 (e) Point of inflexion f ʺ (x) 0 x (A) x y 0 Point of inflexion (, 0) (A) OR Point of inflexion (, 0) (G) [0]. (a) y 0 (A) x fʹ ( x) ( + x ) (A)(A)(A) (c) 6x ( + x ) 0 (or sketch of fʹ ( x) showing the maximum) (M) 6x 0 (A) x ± (A) x ( 0.577) (A) (N) (d) d d d x x x x x x (A)(A) [0] 6

7 5. (a) f ʹ (x) xe x x e x ( (x x )e x x ( x)e x ) AA N Maximum occurs at x (A) Exact maximum value e A N ± 6 8 x x + 0, x, etc. M (c) For inflexion, f ʺ (x) 0 ( ) ( + ) + 8 x A N 6. (a) x (A) Using quotient rule (M) ( x ) () ( x )[( x )] Substituting correctly gʹ (x) ( x ) A ( x ) (x ) ( x ) (A) x (Accept a, n ) ( x ) A (c) Recognizing at point of inflexion gʺ (x) 0 M x A Finding corresponding y-value 0. ie P, A 9 9 [8] 7. (a) y e x cos x d y e x ( sin x) + cos x (e x ) (A)(M) e x ( cos x sin x) (AG) d y e x ( cos x sin x) + e x ( sin x cos x) (A)(A) e x ( cos x sin x sin x cos x) (A) e x ( cos x sin x) (A) 7

8 (c) (i) d y At P, 0 cos x sin x tan x At P, x a, ie tan a (R) (M) (A) (ii) The gradient at any point e x ( cos x sin x) (M) Therefore, the gradient at P e a ( cos a sin a) When tan a, cos a, sin a 5 5 (A)(A) (by drawing a right triangle, or by calculator) Therefore, the gradient at P e a (A) e a (A) 8 [] 8. (a) x (A) EITHER The gradient of g( x ) goes from positive to negative OR g( x ) goes from increasing to decreasing OR (R) (R) when x, gʹ ʹ ( x) is negative (R) < x< and < x< (A) gʹ ( x) is negative (R) (c) x (A) EITHER gʹ ʹ ( x) changes from positive to negative (R) OR concavity changes (R) 8

9 (d) (A) [9] 9. (a) A B E f ʹ (x) negative 0 negative AAA N A B E f ʹ ʹ (x) positive positive negative AAA N 0. (a) Interval gʹ gʹ ʹ a < x < b positive positive e < x < f negative negative AA AA N Conditions gʹ (x) 0, gʹ ʹ (x) < 0 gʹ (x) < 0, gʹ ʹ (x) 0 Point C D A A N N 9

10 . y (, ) 0 x (0, ) AAAAAA N6 Notes: On interval [,0], award A for decreasing, A for concave up. On interval [0,], award A for increasing, A for concave up. On interval [,], award A for change of concavity, A for concave down. 0

11 . y x + d y x Slope of tangent at any point Therefore at point where x, slope Slope of normal (M) (M)(A) Equation of normal: y (x ) y 6 x + x + y 7 0 Note: Accept equivalent forms eg y x + (A) (C) []. (a) y x(x ) (i) y 0 x 0 or x (A) (ii) d y (x ) + x (x ) (x )(x + x) (x )(x ) (A) d y 0 x or x (A) dy x ( )( ) > 0 is a maximum dy x ( )() < 0 (R) Note: A second derivative test may be used x y , 7 (A) 56 Note: Proving that, is a maximum is not necessary to 7 receive full credit of [ marks] for this part.

12 d y d d (x 6x + 6) 6x 6 (A) d y 0 6x 6 0 (M) 8 x (A) (iii) (( x )(x ) ) 8 8, x y 9 7 Note: GDC use is likely to give the answer (., 9.8). If this answer is given with no explanation, award (A), If the answer is given with the explanation used GDC or equivalent, award full credit. (A) 9 0 y max pt. pt. of inflexion x x intercepts Note: Award (A) for intercepts, (A) for maximum and (A) for point of inflexion. (A) (c) (i) See diagram above (A) (ii) 0 < y < 0 for 0 x (R) So 0 < y < 0 0 < y < 0 (R) [5]

13 . y x x d y x gradient at any point. (M) Line parallel to y 5x x 5 (M) x (A) y 6 (A) Point (, 6) (C)(C) [] 5. (a) π (.) (accept (π, 0), (., 0)) A N (i) For using the product rule (M) f ʹ (x) e x cos x + e x sin x e x (cos x + sin x) AA N (ii) At B, f ʹ (x) 0 A N (c) f ʺ (x) e x cos x e x sin x + e x sin x + e x cos x AA e x cos x AG N0 (d) (i) At A, f ʺ (x) 0 A N (ii) Evidence of setting up their equation (may be seen in part (d)(i)) eg e x cos x 0, cos x 0 A π π x (.57), y e (.8) AA π Coordinates are, e (.57,.8 ) π N π π x 0 0 (e) (i) e sin x or f ( x) A N (ii) Area. A N [5]

14 6. y sin (x ) d y cos (x ) (A)(A) At, 0, the gradient of the tangent cos 0 (A) (A) (C) [] 7. (a) f ʹ (x) 6x 5 A N f ʹ (p) 7 (or 6p 5 7) M p A N (c) Setting y () f () (M) Substituting y () 7 9 ( 5), and f () 5 + k ( k + ) A k + 5 k A N 8. (a) METHOD f ʹ (x) 6 sin x + sin x cos x AAA 6 sin x + sin x A 5 sin x AG N0 METHOD cos sin x x (A) f (x) cos x + f (x) 5 cos x A cos x + A 5 f ʹ (x) ( sin x) f ʹ (x) 5 sin x AG N0 A π k ( ).57 A N

15 9. (a) EITHER Recognizing that tangents parallel to the x-axis mean maximum and minimum (may be seen on sketch) Sketch of graph of f R M OR Evidence of using fʹ (x) 0 Finding fʹ (x) x 6x x 6x 0 Solutions x or x THEN Coordinates are P(, 9) and Q(, 79) M A AA NN P N N Q (i) (, 9) A N (ii) (, 79) A N 5

16 0. METHOD l + w 60 l 60 w (M) (A) A w(60 w) ( 60w w ) (A) da dw 60 w (A) Using w 5 da dw 0 (60 w 0) (M) (A) (C6) METHOD w + l 60 w 60 l (A) (A) A l(60 l) ( 60l l ) (A) da dl 60 l (A) Using l 5 da dl 0 (60 l 0) (M) w 0 (A) (C6) 6