Calculus#Practice#06907#(MarkScheme)#Calculus#Practice#06907#(MarkScheme)# 1. (a) f′ (x) = 5e 5x A1A1 N2 (b) g′ (x) = 2 cos 2x A1A1 N2 (c) h′ = fg′ + gf ′ (M1) = e 5x (2

IB Math – Standard Level – Calculus Practice 06-‐07 -‐ MarkScheme Alei -‐ Desert Academy

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Calculus Practice 06-‐07 (MarkScheme)

1. (a) f′ (x) = 5e5x A1A1 N2 (b) g′ (x) = 2 cos 2x A1A1 N2 (c) h′ = fg′ + gf ′ (M1)

= e5x (2 cos 2x) + sin 2x (5e5x) A1 N2 [6]

2. (a) A B E

f ′(x) negative 0 negative A1A1A1N3

(b) A B E

f ′′(x) positive positive negative A1A1A1N3

[6]

3. Finding anti-derivative of 4t3 − 2t (M1)

s = t4 − t2 + c A1A1

Substituting correctly 8 = 24 − 22 + c A1 Note: Exception to the FT rule. Allow full FT on incorrect integration.

c = −4 (A1)

s = t4 − t2 − 4 A1 N3 [6]

4. (a) Interval g′ g′′ a < x < b positive positive e < x < f negative negative A1A1 A1A1 N4

(b) Conditions Point

g′ (x) = 0, g′′ (x) < 0 C g′ (x) < 0, g′′ (x) = 0 D

A1 N1 A1 N1

[6]

5. (a) Attempting to use the formula V = (M1)

Volume = A2 N3

(b) Volume = (A1)

= (A1)

= or 3.35 (accept 1.07π) A1 N3

[6] 6. (a) f ′(x) = 6x − 5 A1 N1

(b) f ′(p) = 7 (or 6p −5 = 7) M1



p = 2 A1 N1 (c) Setting y (2) = f (2) (M1)

Substituting y (2) = 7 × 2 − 9 (= 5), and f (2) = 3 × 22 − 5 × 2 + k (= k + 2) A1 k + 2 = 5 k = 3 A1 N2

[6]

7. (a) f ′ (x) = 2xe−x − x2e−x (= (2x − x2)e−x = x (2 − x)e−x) A1A1 N2 (b) Maximum occurs at x = 2 (A1)

Exact maximum value = 4e−2 A1 N2

(c) For inflexion, f ″(x) = 0 M1

A1 N1

[6] 8. (a) Smin = 6.05 (accept (1, 6.05)) A1 N1

(b) A1

a = (M1)

a = = − 45 cos 3t + 2 (Exception to FT rule : allow FT

from ) A1 N2

(c) EITHER Maximum value of a when cos 3t is minimum ie cos 3t = −1 (A1) OR

At maximum (135 sin 3t = 0) (A1)

THEN

t = (accept 1.05; do not accept 60°) A1 N2

[6] 9.

Graph Diagram (a) f′ (x) I A3 N3 (b) f ″ (x) IV A3 N3

[6]

10. (M1)

A1A1

Substituting t = 0.5

c = 9.5 (A1) Substituting t = 1 M1

s = A1 N3



[6]

11. Using V = (M1)

Correctly integrating A1

V = π A1

= (A1)

Setting up their equation M1

a2 = 1.69 a = 1.3 A1 N2

[6] 12. Evidence of integration (M1)

s = −0.5 e−2t + 6t2 + c A1A1 Substituting t = 0, s = 2 (M1) eg 2 = −0.5 + c c = 2.5 (A1)

s = −0.5 e−2t + 6t2 + 2.5 A1 N4 [6]

13. (a) (i) 0 A1 N1

(ii) A1 N1

(b)

A2 N2

(c)

A2 N2

[6]



14. (a) EITHER Recognizing that tangents parallel to the x-axis mean maximum and minimum (may be seen on sketch) R1 Sketch of graph of f M1

OR Evidence of using f′ (x) = 0 M1

Finding f′ (x) = 3x2 − 6x − 24 A1

3x2 − 6x − 24 = 0 Solutions x = −2 or x = 4 THEN Coordinates are P(−2, 29) and Q(4, −79) A1A1N1N1

(b)

(i) (4, 29) A1 N1 (ii) (−2, −79) A1 N1

[6] 15. (a) 10 A1 N1

(b)

(A1)

= 26 (may be seen later) A1 Splitting the integral (seen anywhere) M1

Using (M1)

eg

A1 N3

[6]

16. f (x) = (M1)

f (x) = 4x3 −2x + c A1A1 Substituting x = −1, y = 1 (M1)

eg 1 = 4(−1)3 − 2(−1) + c



c = 3 (A1)

f (x) = 4x3 − 2x + 3 A1 N4 [6]

17. (a) Evidence of using (M1)

eg 3e3t − 2 a(1) = 3e (= 8.15) A1 N2

(b) Attempt to solve e3t − 2 = 22.3 (M1) eg (3t − 2) (ln e) = ln 22.3, sketch t = 1.70 A1 N2

(c) Evidence of using s = (limits not required) M1

A1 N1

[6] 18. (a) METHOD 1

f ′(x) = −6 sin 2x + 2 sin x cos x A1A1A1 = −6 sin 2x + sin 2x A1 = −5 sin 2x AG N0 METHOD 2

(A1)

f (x) = 3 cos 2x + A1

f (x) = A1

f ′(x) = A1

f ′(x) = − 5 sin 2x AG N0

(b) k = A2 N2

[6]

19. (a) A1 N1

(b) Area of A = 1 A1 N1 (c) Evidence of attempting to find the area of B (M1)

eg , − 0.134

Evidence of recognising that area B is under the curve/integral is negative (M1)

eg

Area of B = 0.134 (A1)

Total Area = 1 + 0.134



= 1.13 A1 N4

[6]

20. f ′(x) = 12x2 + 2 A1A1 When x = 1, f (1) = 6 (seen anywhere) (A1) When x = 1, f ′ (1) = 14 (A1) Evidence of taking the negative reciprocal (M1)

eg

Equation is y − 6 = A1 N4

[6]

21. (a) A1 N1

(b) accept x sec2 x + tan x A1A1 N2

(c) METHOD 1 Evidence of using the quotient rule (M1)

A1A1

N3

METHOD 2

y = x−1 In x Evidence of using the product rule (M1)

A1A1

N3

[6] 22.

A1A1A1A1A1A1 N6

Notes: On interval [− 2,0], award A1 for decreasing, A1 for concave up. On interval [0,1], award A1 for increasing, A1 for concave up. On interval [1,2], award A1 for change of concavity, A1 for concave down.

[6]



23. (a) (i) f ′(x) = A1A1 N2

(ii) For using the derivative to find the gradient of the tangent (M1) f ′(2) = − 2 (A1)

Using negative reciprocal to find the gradient of the normal M1

A1 N3

(iii) Equating (or sketch of graph) M1

3x2 − 2x − 8 = 0 (A1) (3x + 4)(x − 2) = 0

x = A1 N2

(b) (i) Any completely correct expression (accept absence of dx) A2

eg N2

(ii) Area = (accept 11.3) A1 N1

(iii) Attempting to use the formula for the volume (M1)

eg A2 N3

(c) A1A1A1

Note: Award A1 for , A1 for , A1 for 4x.

Substituting (M1)(A1)

= A1 N3

[21] 24. (a) METHOD 1

Attempting to interchange x and y (M1) Correct expression x = 3y − 5 (A1)

A1 N3

METHOD 2 Attempting to solve for x in terms of y (M1)

Correct expression (A1)

A1 N3

(b) For correct composition (g−1◦ f) (x) = (3x − 5) + 2 (A1)

(g−1◦ f) (x) = 3x − 3 A1 N2 (c) (A1)

A1 N2



(d) (i)

A1A1A1N3

Note: Award A1 for approximately correct x and y intervals, A1 for two branches of correct shape, A1 for both asymptotes.

(ii) (Vertical asymptote) x = 2, (Horizontal asymptote) y = 3 A1A1 N2 (Must be equations)

(e) (i) 3x + ln (x − 2) + C (3x + ln ⎜x − 2⎜ + C) A1A1 N2 (ii) (M1)

= (15 + ln 3) − (9 + ln1) A1 = 6 + ln 3 A1 N2

(f) Correct shading (see graph). A1 N1 [18]

25. (a)

A1A1A1N3

Note: Award A1 for the shape of the curve, A1 for correct domain, A1 for labelling both points P and Q in approximately correct positions.

(b) (i) Correctly finding derivative of 2x + 1 ie 2 (A1)

Correctly finding derivative of e−x ie −e−x (A1) Evidence of using the product rule (M1)

f ′ (x) = 2e−x + (2x + 1)(−e−x) A1

= (1 − 2x)e−x AG N0 (ii) At Q, f ′(x) = 0 (M1)

x = 0.5, y = 2e−0.5 A1A1

Q is (0.5, 2e−0.5) N3

(c) 1 ≤ k < 2e−0.5 A2 N2 (d) Using f ″ (x) = 0 at the point of inflexion M1



e−x (−3 + 2x) = 0 This equation has only one root. R1 So f has only one point of inflexion. AG N0

(e) At R, y = 7e−3 (= 0.34850 ...) (A1)

Gradient of (PR) is (A1)

Equation of (PR) is g (x) = A1

Evidence of appropriate method, involving subtraction of integrals or areas M2 Correct limits/endpoints A1

eg dx, area under curve − area under PR

Shaded area is

= 0.529 A1 N4 [21]

26. (a) (i) p = 1, q = 5 (or p = 5, q = 1) A1A1 N2 (ii) x = 3 (must be an equation) A1 N1

(b) y = (x − 1)(x − 5)

= x2 − 6x + 5 (A1)

= (x − 3)2 − 4 (accept h = 3, k = −4) A1A1 N3

(c) A1A1 N2

(d) When x = 0, (A1)

y − 5 = −6(x − 0) (y = −6x + 5 or equivalent) A1 N2 [10]

27. (a) π (3.14) (accept (π, 0), (3.14, 0)) A1 N1 (b) (i) For using the product rule (M1)

f ′(x) = ex cos x + ex sin x = ex(cos x + sin x) A1A1 N3 (ii) At B, f ′(x) = 0 A1 N1

(c) f ″(x) = ex cos x − ex sin x + ex sin x + ex cos x A1A1

= 2ex cos x AG N0 (d) (i) At A, f ″(x) = 0 A1 N1

(ii) Evidence of setting up their equation (may be seen in part (d)(i)) A1

eg 2ex cos x = 0, cos x = 0

A1A1

Coordinates are N2

(e) (i) A2 N2

(ii) Area = 12.1 A2 N2 [15]



28. (a) (i) p = 2 A1 N1 (ii) q = 1 A1 N1

(b) (i) f (x) = 0 (M1)

2 − = 0 (2x2 − 3x − 2 = 0) A1

x = x = 2

A1 N2

(ii) Using V = πy2dx (limits not required) (M1)

V = π A2

V = 2.52 A1 N2 (c) (i) Evidence of appropriate method M1

eg Product or quotient rule

Correct derivatives of 3x and x2 − 1 A1A1 Correct substitution A1

eg

f ′ (x) = A1

f ′ (x) = = AG N0

(ii) METHOD 1 Evidence of using f ′(x) = 0 at max/min (M1)

3 (x2 + 1) = 0 (3x2 + 3 = 0) A1 no (real) solution R1 Therefore, no maximum or minimum. AG N0 METHOD 2 Evidence of using f ′(x) = 0 at max/min (M1) Sketch of f ′(x) with good asymptotic behaviour A1 Never crosses the x-axis R1 Therefore, no maximum or minimum. AG N0 METHOD 3 Evidence of using f ′ (x) = 0 at max/min (M1) Evidence of considering the sign of f ′ (x) A1 f ′ (x) is an increasing function (f ′ (x) > 0, always) R1 Therefore, no maximum or minimum. AG N0

(d) For using integral (M1)

Area = A1

Recognizing that A2

Setting up equation (seen anywhere) (M1) Correct equation A1

eg dx = 2, − = 2, 2a2 + 3a − 2 = 0



a = a = − 2

a = A1 N2

[24] 29. (a)

A1A1A1N3

Notes: Award A1 for both asymptotes shown. The asymptotes need not be labelled. Award A1 for the left branch in approximately correct position, A1 for the right branch in approximately correct position.

(b) (i) y = 3, x = (must be equations) A1A1 N2

(ii) x = A1 N1

(iii) y = A1 N1

(c) (i)

A1A1A1

A1A1 N5

(ii) Evidence of using V = (M1)

Correct expression A1

eg

Substituting A1

Setting up an equation (M1)

Solving gives a = 4 A1 N2 [17]



30. (a)

A1A1A1N3

Note: Award A1 for the left branch asymptotic to the x-axis and crossing the y-axis, A1 for the right branch approximately the correct shape, A1 for a vertical asymptote at

approximately x = .

(b) (i) (must be an equation) A1 N1

(ii) A1 N1

(iii) Valid reason R1 N1 eg reference to area undefined or discontinuity

Note: GDC reason not acceptable.

(c) (i) V = π A2 N2

(ii) V = 105 (accept 33.3 π) A2 N2

(d) f ′(x) = 2e2x − 1 − 10(2x − 1)−2 A1A1A1A1 N4

(e) (i) x = 1.11 (accept (1.11, 7.49)) A1 N1 (ii) p = 0, q = 7.49 (accept 0 ≤ k < 7.49) A1A1 N2

[17]



31. Note: Accept exact answers given in terms of π. (a) Evidence of using l = rθ (M1)

arc AB = 7.85 (m) A1 N2

(b) Evidence of using (M1)

Area of sector AOB = 58.9 (m2) A1 N2 (c) METHOD 1

angle = (A1)

attempt to find 15 sin M1

height = 15 + 15 sin

= 22.5 (m) A1 N2 METHOD 2

angle = (A1)

attempt to find 15 cos M1

height = 15 + 15 cos

= 22.5 (m) A1 N2

(d) (i) (M1)

= 25.6 (m) A1 N2

(ii) h(0) = 15 − 15 cos (M1)

= 4.39(m) A1 N2 (iii) METHOD 1

Highest point when h = 30 R1

30 = 15 − 15 cos M1

cos = −1 (A1)

t = 1.18 A1 N2



METHOD 2

Sketch of graph of h M2 Correct maximum indicated (A1) t = 1.18 A1 N2 METHOD 3 Evidence of setting h′(t) = 0 M1

sin (A1)

Justification of maximum R1 eg reasoning from diagram, first derivative test, second

derivative test

t = 1.18 A1 N2

(e) h′(t) = 30 sin (may be seen in part (d)) A1A1 N2

(f) (i)

A1A1A1N3

Notes: Award A1 for range −30 to 30, A1 for two zeros. Award A1 for approximate correct sinusoidal shape.

(ii) METHOD 1 Maximum on graph of h′ (M1) t = 0.393 A1 N2 METHOD 2 Minimum on graph of h′ (M1) t = 1.96 A1 N2 METHOD 3 Solving h′′(t) = 0 (M1) One or both correct answers A1 t = 0.393, t = 1.96 N2

[22]

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Calculus#Practice#06907#(MarkScheme)#Calculus#Practice#06907#(MarkScheme)# 1. (a) f′ (x) = 5e 5x A1A1 N2 (b) g′ (x) = 2 cos 2x A1A1 N2 (c) h′ = fg′ + gf ′ (M1) = e 5x (2