Calculus Standard Review Ms

IB Questionbank Maths SL 1

1. (a) p = 3 (A1) (C1)

(b) Area = ∫π

20

dcos3 xx (M1)

= 20]sin3[π

x (A1) = 3 square units (A1) (C3)

[4]

2. (a) f ʹ′(x) = 3(2x + 5)2 × 2 (M1)(A1) Note: Award (M1) for an attempt to use the chain rule.

= 6(2x + 5)2 (C2)

(b) ∫ ×

+=

24)52(d)(4xxxf + c (A2) (C2)

Note: Award (A1) for (2x + 5)4 and (A1) for /8. [4]

3. (a) x-intercepts at –3, 0, 2 A2 N2

(b) –3 < x < 0, 2 < x < 3 A1A1 N2

(c) correct reasoning R2 e.g. the graph of f is concave-down (accept convex), the first derivative is decreasing therefore the second derivative is negative AG

[6]

4. (a) 10 A1 N1


(b) ( ) ( ) xxfxxxxfx dd3d33

1

3

1

23

1

2 ∫∫∫ +=+

[ ] 127d331

33

1

2 −==∫ xxx (A1)

= 26 (may be seen later) A1

Splitting the integral (seen anywhere) M1

( )∫ ∫+ xxfxxge dd3.. 2

Using ( ) 5d3

1=∫ xxf (M1)

eg ( ) 526d33

1

2 +=+∫ xxfx

( ) 31d33

1

2 =+∫ xxfx A1 N3

[6]

5. (a) fʹ′ (x) = 5e5x A1A1 N2

(b) gʹ′ (x) = 2 cos 2x A1A1 N2

(c) hʹ′ = fgʹ′ + gf ′ (M1)

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

6. (a) When t = 0, (M1) h = 2 + 20 × 0 – 5 × 02 = 2 h = 2 (A1) 2

(b) When t = 1, (M1) h = 2 + 20 × 1 – 5 × 12 (A1) = 17 (AG) 2


(c) (i) h = 17 ⇒ 17 = 2 + 20t – 5t2 (M1)

(ii) 5t2 – 20t + 15 = 0 (M1)

⇔ 5(t2 – 4t + 3) = 0 ⇔ (t – 3)(t – 1) = 0 (M1)

Note: Award (M1) for factorizing or using the formula

⇔ t = 3 or 1 (A1) 4 Note: Award (A1) for t = 3

(d) (i) h = 2 + 20t – 5t2

⇒ thdd

= 0 + 20 – 10t

= 20 – 10t (A1)(A1)

(ii) t = 0 (M0)

⇒ thdd

= 20 – 10 × 0 = 20 (A1)

(iii) thdd

= 0 (M1)

⇔ 20 – 10t = 0 ⇔ t = 2 (A1)

(iv) t = 2 (M1) ⇒ h = 2 + 20 × 2 – 5 × 22 = 22 ⇒ h = 22 (A1) 7

[15]

7. evidence of integrating the acceleration function (M1)

e.g. ttt

d2sin31∫ ⎟

⎠

⎞⎜⎝

⎛+

correct expression ln t – 23

cos 2t + c A1A1

evidence of substituting (1, 0) (M1)

e.g. 0 = ln 1 – 23

cos 2 + c

c = –0.624 ⎟⎠

⎞⎜⎝

⎛−= 2cos

23or 1ln2cos

23

(A1)

v = ln t – ⎟⎠

⎞⎜⎝

⎛−+−+−=− 1ln2cos

232cos

23lnor 2cos

232cos

23ln624.02cos

23 ttttt (A1)

v(5) = 2.24 (accept the exact answer ln 5 – 1.5 cos 10 + 1.5 cos 2) A1 N3 [7]


8. (a) f ʺ″(x) = 2x – 2 ⇒f ʹ′(x) = x2 – 2x + c (M1)(M1) = 0 when x = 3 ⇒ 0 = 9 – 6 + c c = –3 (A1) f ʹ′(x) = x2 – 2x – 3 (AG)

f (x) = 3

3x – x2 – 3x + d (M1)

When x = 3, f (x) = –7 ⇒ –7 = 9 – 9 – 9 + d (M1) ⇒ d = 2 (A1) 6

⇒ f (x) = 3

3x – x2 – 3x + 2

(b) f (0) = 2 (A1)

f (–1) = –31

– 1 + 3 + 2

= 332

(A1)

f ʹ′(–1) = 1 + 2 – 3 = 0 (A1) 3

(c) f ʹ′(–1) = 0 ⇒ ⎟⎠

⎞⎜⎝

⎛−

323,1 is a stationary point

y

x

(3, –7)

–1, 3 23

2

(A4) 4 Note: Award (A1) for maximum, (A1) for (0, 2) (A1) for (3, –7), (A1) for cubic.

[13]

9. (a) y = ex/2 at x = 0 y = e0 = 1 P(0, 1) (A1)(A1) 2


(b) V = π ∫2ln

0

22/ d)( xe x (A4) 4

Notes: Award (A1) for π (A1) for each limit (A1) for (ex/2)2.

(c) V = ∫2ln

0dxe x (A1)

= π 2ln0][ xe (A1)

= π[eln2 – e0] (A1) = π[2 – 1] = π (A1)(A1) = π (AG) 5

[11]

10. METHOD 1

evidence of antidifferentiation (M1)

e.g. ∫(10e2x – 5)dx

y = 5e2x – 5x + C A2A1

Note: Award A2 for 5e2x, A1 for –5x. If “C” is omitted, award no further marks.

substituting (0, 8) (M1)

e.g. 8 = 5 + C

C = 3 (y = 5e2x – 5x + 3) (A1)

substituting x = 1 (M1)

y = 34.9 (5e2 – 2) A1 N4 8

METHOD 2

evidence of definite integral function expression (M2)

e.g. ( ) ( ) ( ) ( )∫∫ −−=x xx

afxfttf0

2

a

' 5e10,d

initial condition in definite integral function expression (A2)

e.g. ( ) ( ) 8d5e10,8d5e100 0

22 +−−=−∫ ∫ xytx x xt

correct definite integral expression for y when x =1 (A2)

e.g. ( ) 8d5e101

0

2 +−∫ xx

y = 34.9 (5e2 – 2) A2 N4 8


11. (a) y = x(x – 4)2

(i) y = 0 ⇔ x = 0 or x = 4 (A1)

(ii) xydd

= 1(x – 4)2 + x × 2(x – 4) = (x – 4)(x – 4 + 2x)

= (x – 4)(3x – 4) (A1)

xydd

= 0 ⇒ x = 4 or x = 34

(A1)

34

04)2)(2(dd2

03)1)(3(dd1

⇒

⎪⎪⎭

⎪⎪⎬

⎫

<−=−=⇒=

>=−−=⇒=

xyx

xyx

is a maximum (R1)

Note: A second derivative test may be used.

x = 34

⇒ y = 27256

964

34

38

344

34

34 22

=×=⎟⎠

⎞⎜⎝

⎛ −×=⎟

⎠

⎞⎜⎝

⎛−×

⎟⎠

⎞⎜⎝

⎛27256,

34

(A1)

Note: Proving that ⎟⎠

⎞⎜⎝

⎛27256,

34 is a maximum is not necessary to

receive full credit of [4 marks] for this part.

(iii) ( )x

xxxx

ydd)43)(4(

dd

dd

2

2

=−−= (3x2 – 16x + 16) = 6x – 16 (A1)

2

2

ddxy

= 0 ⇔ 6x – 16 = 0 (M1)

⇔ x = 38

(A1)

x = 38

⇒ y = 27128

916

38

34

384

38

38 22

=×=⎟⎠

⎞⎜⎝

⎛ −=⎟

⎠

⎞⎜⎝

⎛−

⎟⎠

⎞⎜⎝

⎛27128,

38

(A1) 9

Note: GDC use is likely to give the answer (1.33, 9.48). If this answer is given with no explanation, award (A2), If the answer is given with the explanation “used GDC” or equivalent, award full credit.


(b)

x–intercepts

10

5

00 1 2 3 4 x

ymax pt.

pt. of inflexion

(A3) 3 Note: Award (A1) for intercepts, (A1) for maximum and (A1) for point of inflexion.

(c) (i) See diagram above (A1)

(ii) 0 < y < 10 for 0 ≤ x ≤ 4 (R1)

So ∫ ∫ ∫∫ <<⇒<<4

0

4

0

4

0

4

040d0d10dd0 xyxxyx (R1) 3

[15]

12. (a) tvadd= (M1)

= –10 A1 3

(b) s = ∫vdt (M1) = 50t – 5t2 + c A1 40 = 50(0) – 5(0) + c ⇒ c = 40 A1

s = 50t – 5t2 + 40 A1 3 Note: Award (M1) and the first (A1) in part (b) if c is missing, but do not award the final 2 marks.

[6]


13. (a) y = ln x ⇒ xx

y 1dd

= (A1)

when x = e, e1

dd

=xy

tangent line: y = ⎟⎠

⎞⎜⎝

⎛e1

(x – e) + 1 (M1)

y = e1

(x) – 1 + 1 = ex

(A1)

x = 0 ⇒ y = e0

= 0 (M1)

(0, 0) is on line (AG) 4

(b) xdd

(x ln x – x) = (1) × ln x + x × ⎟⎠

⎞⎜⎝

⎛x1

– 1 = ln x (M1)(A1)(AG) 2

Note: Award (M1) for applying the product rule, and (A1) for

(1) × ln x + x × ⎟⎠

⎞⎜⎝

⎛x1 .

(c) Area = area of triangle – area under curve (M1)

= ∫−⎟⎠⎞

⎜⎝

⎛××

e

1dln1e

21 xx (A1)

= e1]ln[

2e xxx −− (A1)

= 2e

– {(e ln e – 1 ln1) – (e – 1)} (A1)

= 2e

– {e – 0 – e + 1}

= 21

e – 1. (AG) 4

[10]

14. (a) (i) x = –25

(A1)

(ii) y = 23

(A1) 2


(b) By quotient rule (M1)

2)52()2)(23()3)(52(

dd

+

−−+=

xxx

xy

(A1)

= 2)52(19+x

(A1) 3

(c) There are no points of inflexion. (A1) 1 [6]

15. (a) s = 25t − ct +334 (M1)(A1)(A1)

Note: Award no further marks if “c” is missing.

Substituting s = 10 and t = 3 (M1)

10 = 25 × 3 − c+3(3)34

10 = 75 − 36 + c

c = − 29 (A1)

s = 25t − 2934 3 −t (A1) (N3)

(b) METHOD 1

s is a maximum when v = 0dd

=ts

(may be implied) (M1)

25 − 4t2 = 0 (A1)

t2 = 425

t = 25 (A1) (N2)

METHOD 2

Using maximum of s (3212 , may be implied) (M1)

25t − 321229

34 3 =−t (A1)

t = 2.5 (A1) (N2)

(c) 25t − 02934 3 >−t (accept equation) (M1)

m = 1.27, n = 3.55 (A1)(A1) (N3)


[12]

16. (a) gradient is 0.6 A2 N2

(b) at R, y = 0 (seen anywhere) A1 at x = 2, y = ln 5 (= 1.609...) (A1) gradient of normal = –1.6666... (A1) evidence of finding correct equation of normal A1

e.g. y – ln 5 = 35

− (x – 2), y = – 1.67x + c

x = 2.97 (accept 2.96) A1

coordinates of R are (2.97, 0) N3 [7]

17. y = sin (2x – 1)

xydd

= 2 cos (2x – 1) (A1)(A1)

At ⎟⎠

⎞⎜⎝

⎛ 0,21

, the gradient of the tangent = 2 cos 0 (A1)

= 2 (A1) (C4) [4]

18. (a) xxxx

xx

2dd,1ln

dd 2 == (seen anywhere) A1A1

attempt to substitute into the quotient rule (do not accept product rule) M1

e.g.4

2 ln21

x

xxx

x −⎟⎠

⎞⎜⎝

⎛

correct manipulation that clearly leads to result A1

e.g. ( )4444

ln2,ln21,ln2xxx

xx

xxx

xxxx

−−−

( )3

ln21x

xxg −=ʹ′ AG N0 4


(b) evidence of setting the derivative equal to zero (M1)

e.g. g′(x) = 0, 1– 2ln x = 0

21ln =x A1

21

e=x A1 N2 3 [7]

19. METHOD 1 (quotient)

derivative of numerator is 6 (A1)

derivative of denominator is –sin x (A1)

attempt to substitute into quotient rule (M1)

correct substitution A1

e.g. ( )( ) ( )( )

( )2cossin66cos

xxxx −−

substituting x = 0 (A1)

e.g( )( ) ( )( )

( )20cos0sin0660cos −×−

h′(0) = 6 A1 N2 [6]

METHOD 2 (product)

h(x) = 6x × (cos x)–1

derivative of 6x is 6 (A1)

derivative of (cos x)–1 is (–(cos x)–2 (–sin x)) (A1)

attempt to substitute into product rule (M1)

correct substitution A1

e.g. (6x) (–(cos x)–2 (–sin x)) + (6) (cos x)–1

substituting x = 0 (A1)

e.g. (6× 0) (–(cos 0)–2 (–sin 0)) + (6) (cos 0)–1

h′(0) = 6 A1 N2 [6]


20. evidence of finding intersection points (M1)

e.g. f (x) = g (x), cos x2 = ex, sketch showing intersection

x = –1.11, x = 0 (may be seen as limits in the integral) A1A1

evidence of approach involving integration and subtraction (in any order)(M1)

e.g. ∫∫ ∫ −−−−

fgxxx xx ,d)e(cos,ecos0

11.1

22

area = 0.282 A2 N3 [6]

21. attempt to set up integral expression M1

e.g. xxxxx d416,)416(π2,d4162

22

0

22

2 ∫∫∫ −−−π

34d4,16d16

32 xxxxx == ∫∫ (seen anywhere) A1A1

evidence of substituting limits into the integrand (M1)

e.g. 36464,

33232

33232 −⎟

⎠

⎞⎜⎝

⎛+−−⎟

⎠

⎞⎜⎝

⎛−

volume = 3π128

A2 N3

[6]

22. (a) B, D A1A1 N2 2

(b) (i) f′(x) = 2

e2 xx −− A1A1 N2

Note: Award A1 for 2xe− and A1 for –2x.

(ii) finding the derivative of –2x, i.e. –2 (A1)

evidence of choosing the product rule (M1)

e.g. 22

e22e2 xx xx −− −×−− 22

e4e2 2 xx x −− +− A1

f ′′(x) = (4x2 – 2)2

e x− AG N0 5


(c) valid reasoning R1

e.g. f ′′(x) = 0

attempting to solve the equation (M1)

e.g. (4x2 – 2) = 0, sketch of f ′′(x)

p = 0.707 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

21707.0,

21 q A1A1 N3 4

(d) evidence of using second derivative to test values on either side of POI M1

e.g. finding values, reference to graph of f′′, sign table

correct working A1A1

e.g. finding any two correct values either side of POI,

checking sign of f ′′ on either side of POI

reference to sign change of f ′′(x) R1 N0 4 [15]

Calculus Standard Review Ms

Documents

Transcript of Calculus Standard Review Ms