A Level Mathematics Questionbanks



1. a) Area of R = [pic] M1 A1 A1

[3]

b) Volume generated = [pic] M1 A1 A1

= [pic] M1 A1 A1

[6]

2. a) u = sinx

du = cos x dx M1 A1

[pic]

= eu + c

= esinx + c A1 A1

[4]

b) [pic]

u = x + 1 M1

du = dx

x = u – 1

[pic] =[pic] M1 A1

= ln|u| +[pic]+ C A1

= ln(x + 1( + [pic]+ C A1

[5]

c) [pic] u = lnx [pic]= x

[pic] v = [pic]

[pic] = [pic]lnx – [pic] M1

= [pic]lnx – [pic]+ C A1 A1

[5]

3. a) [pic]dx = - [pic] = -3[cos[pic] – cos 0] M1 M1

= -3[-1 – 1] = 6 A1

[3]

b) [pic][pic] = [pic] M1 A1

= [pic] – [pic] M1

= [pic] A1

= [pic]

= [pic] = [pic] = [pic] A1

[5]

4. a) [pic] [pic] = [pic][pic] M1 [pic]

Let u = cos[pic] then [pic] = -sin[pic] M1 (substitution)

So[pic] = [pic]

Hence [pic][pic] = [pic] = -[pic]du = -ln u + C A1 (correct integrand)

= - ln[pic] + C A1

= ln[pic] + C

= ln [pic] + C A1

[5]

b) [pic] = ln[pic] – ln[pic] M1

= ln[pic] – ln[pic] A1

= 0 – ln[pic] M1

= -[pic]ln2 A1

[4]

5. a) [pic][pic] Let u = sin[pic], then [pic] = cos[pic] M1

So [pic] = [pic]

Hence [pic][pic] = [pic] = [pic]du A1

= 2[pic] + C = [pic] + C A1

Thus [pic][pic] = [pic][pic] = [pic] M1

= [pic] = [pic] A1

[5]

b) sin3( sin2( = [pic][cos( – cos5(] M1 A1

[pic] M1 A1

= [pic]([pic] = [pic] M1 A1

[6]

6. a) [pic]dx = [pic] M1 A1

= -[pic] = -[pic] M1

= [pic] A1

[4]

b) P: (0, 0) Q: [pic]

Area required = Area under curve – Area under line M1

= [answer to a)] – [pic] M1 A1

= [pic] – [pic] [pic] A1 f.t.

[4]

7. The two curves intersect when sin x = cos x M1

i.e. [pic] = 1

i.e. tan x = 1 A1

( x = arctan 1 = [pic] A1 (not 45o)

The area enclosed by the two curves and the y-axis is shown sketched below.

Shaded area = [pic] dx = [pic] M1 (subtraction) A1

= [pic] A1

= [pic] – 1 = [pic] – 1 A1

[7]

8. a) [pic]dt

cos 2A[pic] 2 cos2A – 1 M1

( cos2A[pic] [pic](1 + cos 2A)

and cos24t [pic] [pic](1 + cos 8t) A1

Hence [pic]dt = [pic] dt M1

= 4t + [pic]sin 8t + C A1 A1

[5]

b) [pic] dt = [pic] dt M1 A1

= [pic] A1

= [pic] – 0 M1

= [pic] A1

[5]

9. a) [pic][pic][pic][pic][pic] + [pic][pic][pic] M1

Hence 11 – 3x [pic] A(x – 1) + B(x + 3) A1

Let x = 1: 8 = 4B ( B = 2

M1 A1

Let x = -3: 20 = -4A ( A = -5

Thus [pic] = [pic] – [pic]

[4]

b) [pic] = [pic] = 2 ln[pic] – 5ln[pic] + C M1 A1

2 ln[pic] = ln[pic] 5 ln[pic] = ln[pic] M1 A1

So 2 ln[pic] – 5ln[pic] = ln[pic]

So integral = ln[pic] + C A1

Let C = ln A; integral = ln[pic] + ln A = ln[pic] M1 A1

[7]

10. a) Using integration by parts:

[pic] dx

Let u = ln[pic] then [pic] = [pic] M1 A1

[pic], so v = x. A1

Hence [pic]dx = [pic](x) – [pic] M1

= x ln[pic] – [pic]dx

[pic] dx = x ln[pic] – x + C A1

[5]

b) [pic] dx = [pic]

= 2e2ln(e2) – 2e2 – 2ln1 + 2 M1 A1

= 4e2 – 2e2 + 2

= 2e2 + 2 A1

[3]

c) Area required = area in b) + area under line. M1

Area under line is a triangle

[pic] Area under line = [pic](20 – 2e2)(2) M1

= 20 – 2e2 A1

So total area = 2e2 + 2 + 20 – 2e2

= 22 A1

[4]

11. a) y = 2t2ln t – t2

[pic] = 4tln t + [pic] – 2t M1 A1 A1

= 4tln t + 2t – 2t

= 4tln t A1

[4]

b) [pic]dt = [pic] M1 A1

= [pic]

= 2ln 2 – 1– 0 +[pic] A1

= ln 22 – [pic] M1

= ln 4 – [pic] A1

[5]

12. a) 4[pic] + 2x = 5

4[pic] = 5 – 2x

[pic]dy = [pic]dx

4y = 5x – x2 + C A1 A1

x = 2, y = [pic] [pic] 10 = 10 – 4 + C [pic] C = 4 M1 A1

4y = 5x – x2 + 4

y = [pic]x – [pic] + 1 A1

[7]

b) Since ([pic]– 1) [pic] = [pic] then [pic]d[pic] = [pic] dt M1 A1

( [pic] d[pic] = [pic] dt M1

( [pic] – ln [pic] = 2t + C A1 A1

[pic] = 1 when t = [pic] ( [pic] – ln 1 = [pic] + C ( C = -1 M1 A1

( particular solution is: [pic] – ln [pic] = 2t – 1 B1

[8]

13. a) 2xy [pic] = y2 – 1

[pic] [pic] =[pic] M1

[pic] ln[pic] = ln[pic] + C M1 A1 A1

Let C = ln A: ln[pic] = ln[pic] + ln A M1

y > 1; x > 0 [pic] modulus signs not needed:

ln (y2 – 1) = ln x + ln A

[pic] ln (y2 – 1) = ln Ax A1

y2 – 1 = Ax

y2 = Ax + 1 A1

[7]

b) y = 2, x = 1 [pic] 4 = A + 1 ( A = 3 M1 A1

[2]

c)

[pic]dy =[pic] M1 A1 A1

= [pic] A1

= [pic] – [pic]= [pic] M1 A1

OR: 2 – [pic] M1 A1 A1

= 2 - [pic] A1

= 2 - [pic] = [pic] M1 A1

[6]

14.a) Rate of decrease of C = -[pic] B1 B1

This is proportional to C [pic] - [pic] = kC M1

[pic] [pic] = -kC

[3]

b) [pic] = [pic]dt M1

ln C = -kt + A A1 A1

t = 0, C = 2 [pic] A = ln 2 M1 A1

ln C = -kt + ln 2

[pic] C = [pic] = 2[pic] M1 A1

[7]

c) [pic] = 2[pic] M1

[pic] = [pic] ( ln [pic] = -k A1

-ln 4 = -k ( ln 4 = k A1

[3]

d) C0 = 2[pic] M1

ln [pic] = -toln 4 ( to = -[pic] A1

[pic] = 2e[pic] ( t1 = -[pic] A1

[3]

e) t1 – t2 M1

= [pic]

= [pic] M1 A1

=[pic] A1

=[pic] B1

[5]

15. a) [pic]= -k([pic]) B1 B1 (-1 if no “-”)

[2]

b) [pic]=[pic]dt M1

ln[pic] = -kt + C A1

[2]

c) [pic][pic] [pic] = [pic] B1 (expln. for removing

mod. signs)

a)[pic][pic] = [pic] M1 A1

t = 0, [pic] [pic] [pic] = [pic] M1

[pic] = [pic] A1

[pic] = [pic] [pic] θ = [pic] A1

[6]

d)

G1 (shape)

G1 (2θo)

G1 (Asymptote [pic])

[3]

e) [pic]= [pic] M1

[pic] = 1 + [pic]

[pic] = [pic] A1

3 = ekT M1

kT = ln3 M1

T = [pic]ln3 A1

[5]

16. a) At P, x is maximum, so x = A [pic] (A, 0) B1

At Q, y is maximum, so y = B [pic] (0, B) B1

[2]

b) At Q, 0 = A cos t; B = B sin t [pic] sin t = 1 M1 (may be gained for P

or Q)

[pic] t = [pic] A1

At P, 0 = B sin t; A = Acos t [pic] cos t = 1

[pic] t = 0 A1

[3]

c) Area = 4[pic] dx = [pic] M1 (use of “4”)

[pic] = -A sin t [pic] dx = -A sin t dt M1 A1

So area = 4[pic] dt M1

= -4AB[pic] dt

= 4AB[pic]dt B1

[5]

d) Area = 4AB[pic]dt M1 A1

= 2AB[pic]dt

= 2AB[pic] M1 A1

= 2AB[pic] M1

= [pic] A1

[6]

17. a) u = x [pic]

[pic] v = [pic]

[pic]dx = [pic] – [pic] M1 (use of parts)

A1 A1

=[pic] – [pic] + C A1

[4]

b) u = x2 [pic]

[pic] v = [pic]

[pic]dx = [pic] – [pic] M1 (correct use of parts)

= [pic] – [pic]dx

= [pic] – [pic] M1 (use of a)) A1 f.t.

= [pic] – [pic] + [pic] + K A1

[4]

c) y = 0 [pic] (x2 – 3x + 2)[pic] = 0

(x – 2)(x – 1)[pic]= 0 M1

x = 1, 2 A1 (both)

x = 0 [pic] y = 2 B1

[3]

d) [pic] dx M1 (limits)

=[pic]dx – 3[pic]dx + 2[pic]dx M1 (splitting)

=[pic] A1f.t. A1f.t. A1

=[pic]

= e[pic][pic] – [pic] =[pic] M1 A1

But this is negative (area below axis). So area = [pic] B1

[8]

18. a) [pic]dx = [pic] – [pic]dx M1 A1 A1

[pic]dx = [pic] – [pic] dx M1 A1

So [pic]dx = [pic] – [pic] – [pic]dx M1 A1

[pic]dx = [pic] M1

[pic]dx = [pic] + C A1

[9]

b) [pic]dx M1

= [pic]

= [pic] – [pic] A1 A1

= [pic] A1

[4]

19.a) u = 1 – x2

[pic] = – 2x M1

dx =[pic]

I =[pic] M1

=[pic] A1

= – [pic] + C A1

= – [pic]ln[pic] + C A1

OR: [pic] M1 A1

[pic]

=[pic]ln(1 – x([pic]ln(1 + x( + C M1

= [pic]ln((1 + x)(1 – x)( + C A1

= [pic]ln((1 – x2)( + C A1

[5]

b) x = cos [pic] [pic] [pic] = -sin [pic] ( dx = -sin [pic] [pic] M1

[pic] [pic] M1

= [pic] [pic] = [pic][pic] M1 A1

= -ln[pic] + C A1

= -ln[pic] + C A1

[6]

c) The two expressions must be the same. M1

So -ln[pic]= -[pic]

= [pic] M1

= (ln[pic]

So sin(cos-1x) = [pic] A1

[3]

20. a) u = ln x [pic] [pic] M1 (substitution)

dx = x dy M1 (finding dx)

[pic] =[pic] M1

=[pic] A1

= ln [pic] A1

= ln[pic] + C A1 A1

[pic] = [pic] M1

= ln[pic] – ln[pic] A1

= ln 4 – ln 2 A1

= ln[pic]

= ln 2 A1

[11]

b) u = 2x: [pic] [pic] = 2 [pic] dx = [pic] M1 M1

[pic] [pic]dx = [pic][pic]du

= [pic]ln[pic] + C

= [pic] + C A1

[3]

c) u = sin x M1

[pic] = cos x M1

dx = [pic]

[pic] = [pic] M1

= [pic] A1

= ln[pic] + C A1

= ln[pic] + C A1

[5]

21.a) [pic]dx = [pic] M1

= [pic]dx – [pic]dx

= -cos x + [pic] + C A1 M1 A1

[4]

b) [pic]dx = [pic]dx M1

= tan x – x + C A1 A1

[3]

c) [pic]dx = [pic] dx M1

= [pic]dx – [pic]dx A1

= [pic] – ln(sec x( M1 A1 A1

[5]

d) [pic]dx = [pic]dx M1 A1

= -[pic] + C M1 A1

[4]

22. a) y = (x + 1)[pic]

[pic] = (x + 1)4x[pic] + [pic] M1 A1 A1

= [pic](4x2 + 4x + 1) M1

= [pic](2x + 1)2 A1

[5]

b) [pic](2x + 1)2 = [pic](4x2 + 4x + 1)

[pic]dx = [pic]– 4x[pic]dx M1 A1

= (x+1) [pic] – [pic] + C B1 M1 A1 A1

= [pic](x + 1 –1) + C M1

= x[pic] + C A1

[8]

c) i) P: y = 1 B1

Q: y = 5e2 B1

[2]

ii) Area under curve = [pic] M1

= e2 A1

Area under line = [pic] M1 A1

Required area = area under line – area under curve M1

= [pic]

= [pic] A1

[6]

23. a) u = ( [pic] M1

[pic] v = -cos(

Hence [pic] = -( cos( + [pic] A1

= sin( – ( cos( + C A1

[3]

b) u = 4θ [pic] ( dθ = [pic] M1

[pic] = [pic] A1

= [pic] M1 A1

[4]

24. a)

G2 (shapes)

G2(all intersection

points)

[4]

b) x2 = 18 – x2 M1

x2 = 9 [pic] x = [pic]3 A2

y = 9 for both A1

So, (3,9) and (-3,9) [4]

c) Volume required = [pic] B1

Consider the two curves separately. M1

y = 18 – x2 between y = 9 and y = 18

y = x2 between y = 0 and y = 9 A1 (correct splits)

“top” part: y = 18 – x2 [pic] x2 = 18 – y

So [pic] M1 A1

= [pic]

= [pic] M1

= [pic] A1

“bottom” part: [pic]

= [pic] B1

= [pic] B1

So total is 81[pic] A1 f.t.

[10]

25. v = [pic] M1

= [pic] A1 A1

= [pic] A1

= [pic] M1 (limits)

= 54[pic] A1

[6]

26. a) x – 6 = [pic] M1

x2 – 12x + 36 = x

x2 – 13x + 36 = 0 M1

(x – 9)(x – 4) = 0

x = 9, 4 A1 (both or just x = 9)

y = 3, -2 A1 ( both or just y = 3)

But y must be positive, since y = [pic]

So (9, 3) B1

[5]

b) V = [pic] B1

Curve and line separately:

Curve: [pic] M1 A1 A1

=[pic]

= [pic] A1

Line: lower limit is where it crosses the x-axis, at x = 6 B1

So [pic] M1 A1

= [pic]

= 9[pic] A1

Required volume is difference between the two. M1

So V = [pic] = [pic] A1

[12]

-----------------------

2θ0

θ = θ0

y = cosx

y = sinx

M1 (sep.variables) A1

(0, 1)

(1, 2)

y = 2

y2 = 3x + 1

(-(18, 0)

M1 A1

y =x2

y = 18– x2

(0, 18)

((18, 0)

A1 (or equivalent using volume of cone)

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