A Level Mathematics Questionbanks



1. a) i) y = 3x2 (2(x ( y = 3x2− 2x[pic]½ M1

[pic] = 3(2x) − 2([pic]) = 6x −[pic] M1 A1

[3]

ii) When x = 4, [pic]= 6(4) − [pic] = 24 − [pic] = 23.5 M1 A1

[2]

b) i) y =[pic]x − 5x3 − [pic]= [pic]x − 5x3 – x-1 M1

[pic]= [pic] − 15x2 + x-2 ( = [pic]− 15x2 + [pic]) M1 A1

[3]

ii) When x = -[pic], [pic]= [pic]− [pic] + 4 = [pic] M1 A1

[2]

2. f(t) = 2t3− 2 ln t − 3et ( f ((t) = 6t2− [pic] − 3et M1 A1 A1

f ((2) = 6(2)2 − [pic] − 3e2 = 24 − 1 − 22.16717 = 0.833 M1 A1

[5]

3. a) v = [pic] B1

s = 3t2 – 6t + 3 M1

[pic] = 6t – 6 A1

t = 1.5, [pic] = 3 ms-1 A1

[4]

b) a = [pic] B1

= 6 ms-2 B1

[2]

4. a) y = 3x2− 2x + 1 Gradient = [pic] = 6x − 2 M1 A1

When x = 1, gradient = 6 − 2 = 4 B1

[3]

b) 6x – 2 = -14 M1

x = -2 A1

y = 17 B1 f.t.

[3]

5. a) y = 3 ln x −[pic]+ 5 = 3 ln x – 2x-2 + 5 M1

Gradient = [pic]= [pic] + 4x-3 = [pic] +[pic] M1 A1

x = 1, hence gradient = [pic] + [pic] = 7 B1

[4]

b) [pic] + [pic] = [pic] M1 (attempt to solve)

x > 0 so [pic]>0 B1

[pic]is positive for all x B1

So [pic] > 0 A1

[4]

6. a) y = 3[pic] + [pic] = 3[pic] + [pic]e-θ M1

[pic]= 6[pic]− [pic]e-θ (’ 6[pic] − [pic]) “1 “1

when θ ’ 1.5, [pic]= 6e2(1.5) − [pic] = 120.51 + 0.08925 M1

= 120.4 (4 SF) A1

[5]

b) 6e2θ − [pic]e-θ ’ 0

( 30e3θ − 2 = 0 M1 A1

e3θ − [pic]= 0 ( 3θ = ln [pic] (or 3( = -ln 15 ) M1

θ = [pic]ln [pic] (or -[pic] ln15) A1

[4]

7. y = 4 ln x + 2 ( [pic] = [pic] M1 A1

[pic] = 4e-2 ( x = e2 M1 A1

y = 4 lne2 + 2 = 10 M1 A1

[6]

8. a) y = 3x2 − 4x + 2

[pic]= 6x − 4 = 0 for a turning point M1 A1 [pic]

B1 [pic]

Hence 6x = 4 ( x = [pic] A1

When x =[pic], y = 3[pic] − 4[pic]+ 2 = [pic] B1

i.e. a turning point occurs at [pic]

[pic] = 6. Since this is positive, [pic]is a minimum point M1 A1

[7]

b) i) f(x) ( [pic] B1 f.t.

[1]

ii) No solutions B1 f.t.

[1]

9. a) y = x(6 − x) = 6x – x2 M1

[pic] = 6 − 2x = 0 for a maximum or minimum value A1 B1 [pic]

( x = 3 B1

When x = 3, y = 3(6 − 3) = 9 B1

Hence (3, 9) are the coordinates of the turning point.

[pic]= -2, hence (3, 9) is a maximum point. M1 A1

[7]

b)

G1 (general shape)

G1 (intercepts)

G1 (max pt)

[3]

c) x = 7, f(x) = -7 B1

Range is –7 ( f(x) ( 9 B1 B1 (-1 if ( not used)

[3]

10.a)

B1 intersection

G1 general shape

[2]

b) y = 3 ln 2x, ( gradient = [pic]= 3[pic] = [pic] M1 A1

When x = 1, gradient = [pic] = 3 B1 f.t

[3]

11.a) y =[pic] M1 A1 A1 A1

[4]

b) [pic]= 1 − 0 – x-2 = 1 − [pic] M1 A1

When x = 2, [pic]= 1 − [pic] = [pic] M1 A1

[4]

12. a) F = [pic]+ 4v = 324v-1 + 4v

Hence [pic] = -324v-2 + 4 = [pic]+ 4 = 0 for a maximum or minimum value M1 A1

( 4 = [pic]( v2 = [pic]= 81

Hence v = [pic]= 9 A1

When v = 9, F = [pic] + 4v = 72 B1

[4]

b) [pic] = 628v-3, and when v = 9, [pic]is positive M1

Hence F = 72 is a minimum value A1

[2]

13.a) (-2,0) (3,0) (0,-12) B1 B1 B1

[3]

b) y = (x2 + 4x + 4) (x − 3) M1 (expanding)

= x3 + x2 – 8x – 12 A1 A1

[pic] = 3x2 + 2x – 8 M1 (attempt to find [pic]) A1

0 = 3x2 + 2x – 8 M1

0 = (x + 2) (3x – 4)

x = -2, [pic] A1 (both)

y = 0, -18[pic] A1 (both)

[pic] = 6x + 2 M1 (or other method)

x = -2, [pic] = -10 ( max (-2, 0) A1

x = [pic],[pic] = 10 ( min ([pic], -18[pic]) A1

[11]

c)

G1 general shape

G2 intersections & max/min

[3]

d) From graph, 1 solution B1

[1]

14. a) P = 2r + rθ B1 B1

A = [pic]r2θ B1

[3]

b) 30 = 2r + rθ Μ1

( r( = 30 – 2r M1

A = [pic]r2θ ’ [pic]r ( r( M1

= [pic]r(30 – 2r)

= 15r – r2 A1

[4]

c) [pic]= 15 – 2r M1 A1

0 = 15 – 2r M1

r = 7[pic] A1

[pic]= -2

( maximum B1 (any method)

A = 15(7[pic]) – (7[pic])2 M1

= 56.25 cm2 A1

[7]

15.a)

Consider (BCD

M1 (attempt to use similar

triangles or other valid

method)

This is similar to (ACE

Hence [pic] M1

8 – H = [pic]

H = 8 − [pic] A1

[3]

b) V = (R2H

= (R2 [pic] M1

= (R2 ( [pic](6 – R)

= [pic](6 – R) A1

[2]

QUESTION 15 CONTINUED

c) V = [pic](6R2 – R3)

[pic](12R – 3R2) M1 (attempting to find [pic])

0 = [pic](12R – 3R2) A1

12R – 3R2 = 0 M1 (attempting to solve)

3R(4 –R) = 0

R = 0, 4 A1 (both or just R = 4)

R = 0 will obviously give zero value

[pic]= [pic](12 – 6R) M1

R= 4, [pic] = -16( A1

( maximum

V = [pic](6 ( 16 – 64) M1

= [pic] A1

[8]

16. a) Let length = L. Let shorter side of rectangle be x M1

Other 2 sides are [pic]L – x B1

Area = x([pic]L – x) M1

= [pic]Lx – x2

[pic] = [pic]L –2x A1

0 =[pic]L – 2x M1

[pic]L= x A1

[pic] = -2 ( maximum B1

So each side is [pic]L, so it is a square. B1

[8]

b)

other side is L – 2x B1

area = x(L – 2x) M1

= Lx – 2x2

[pic] = L – 4x M1

0 = L – 4x

x = [pic]L A1

[pic] = -4 ( max B1

so other side is [pic]L B1

[6]

17. a) Surface Area = 2 ( 3x2 + 2 ( hx + 2 ( 3xh M1

1240 = 6x2 + 8hx A1

[pic] = h M1 A1

V = 3x2h M1

= 3x2[pic] M1

= [pic](1240 – 6x2)

= [pic](620 – 3x2) A1

[7]

b) V = [pic](620x – 3x3) M1

[pic] = [pic](620 – 9x2) M1

0 = [pic](620 – 9x2) M1

x = [pic]

= 8.300 (3 DP) A1

[4]

c) V = [pic](620 – 3 ( 8.3002) M1

= 2572.98 cm3 (2 DP) A1

[2]

d) x = 8.300

3x = 24.90

h =[pic]

= 12.45 B1

diagonal = [pic] M1

= [pic]

= 29.1 cm (1 DP) A1

[3]

18.a) [pic]= 2 ( 0.02e0.02t M1

= 0.04e0.02t A1

t = 50ln 2, [pic]= 0.04e0.02 ( 50ln2

= 0.04eln2 = 0.08 B1 c.a.o

[3]

b) A = (r2 M1

= ((2e0.02t)2

= 4((e0.02t)2 M1 (attempt to multiply out)

= 4(e0.04t A1

[3]

c) [pic] = 4( ( 0.04e0.04t B1

= 4( ( 0.04e0.04 ( 3

= 0.567 (3 SF) B1

[2]

19. a) (r2h = 250( ( h = [pic] B1

A = 2(rh + (r2 B1

= (r2 + 2(r ([pic] M1

= (r2 + [pic] A1

[4]

b) [pic]= 2(r − [pic] M1 A1

0 = 2(r − [pic] M1

0 = 2((r3 – 250) M1 (attempt to solve)

r3 = 250 ( r = [pic] = 6.2996…

[pic] = 2( + [pic] M1

= 2( + [pic] = 6( > 0 A1

( minimum

A = (r2 + [pic] = ([pic] M1

= [pic] = 374 cm3 to 3 S.F. A1

[8]

20.a) x = [pic]t-2 –3t-1 + 2 − t[pic] M1 A1 A1 (-1 eeoo)

[pic]= -t-3 + 3t-2 − [pic]t[pic] [pic] M1 A1 A1 (-1 eeoo)

[6]

b) t = 3 : [pic]= [pic] = 0.0076 > 0 M1 A1

t = 4 : [pic]= [pic]

sign change, so [pic]= 0 in interval A1

[3]

c) As [pic]< 0, it moves back towards 0 B1

[1]

21. Gradient of line is 3 B1

For C, [pic]= 6x B1

at x = [pic], [pic]= 3 so tangent has gradient 3 also and is parallel to line M1 A1

[4]

22. a) Use of nxn-1 M1

( [pic] A1 A1

[3]

b) [pic] M1 A1

[pic] M1 A1 A1

[5]

23. [pic]= 6x2 – 6x – 12 = 6(x2 – x – 2) M1 A1

[pic] = 0 at x = 2 (and at x = -1) M1 A1

x = 2 ( y = 16 – 12 – 24 + 21 = 1 > 0 M1 A1

and this is a minimum point because y (((2) = 18 > 0 A1

Graph of y looks like

and y > 0 for x > 0. M1 A1

[9]

24. a) (1 ( x2)(1 ( 4x) = 0 ( x = (1, 0.25 M1 A1 (all)

Crosses y-axis at (0,1) B1

G1 (shape)

[4]

b) i)[pic]y = 4x3 – x2 – 4x + 1 M1

[pic]= 0 B1

x = [pic] A1

Points are [pic],[pic] A1

[4]

b) ii) [pic] M1 [pic]A1

Point is [pic] A1

[3]

25. a) [pic]= x – 2 M1

( x2 – 2x – 15 = 0

(x – 5)(x + 3) = 0 M1

A is (5, 3) B is (-3, -5) A1 A1

[4]

b) [pic] At A, [pic] M1

Tangent is y – 3 = [pic](x – 5) M1 A1

At A, normal has gradient [pic] M1

and equation y – 3 = [pic](x – 5) M1 A1

[6]

26. (r2h = 1000 B1

S.A. = 2(r2 + 2(rh M1 A1

= 2(r2 + 2(r[pic] M1

= 2(r2 + [pic] A1

[pic] = 4(r − [pic] M1 A1

4(r − [pic]= 0 M1

4(r3 = 2000

r = [pic]= 5.42 cm A1

A = 554 cm2 M1 A1

[11]

.

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

(-2, 0)

(0, 1)

(0.25, 0)

(1, 0)

(-1, 0)

(0, -12)

([pic], -18[pic])

y = (1 ( x2)(1( 4x)

y=3ln2x

(3, 0)

(0.5, 0)

(3, 9)

(0,4)

(2,1)

x

x

x

x

C

8 − H

8

D

A

E

12

C

12

2R

B

B

2R

C

y = x(6(x)

(6, 0)

E

H

A

D

8

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