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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