A Level Mathematics Questionbanks



1. a) 2sinx cosx = sinx B1

sinx (2cosx − 1) = 0 M1 A1

sinx = 0 or cosx = [pic]

x = 0, 180, 60, 300 A3 (-1 eeoo)

[6]

b) From (a), have 2x = 0, 180, 60, 300 M1

x = 0, 90, 30, 150 A2 ft

[3]

2. a) cos2x + sin2x ( 1 ( cosA =[pic], cosB = [pic] M1 A1

sin(A+B) ( sinA cosB + cosA sinB B1

= [pic] M1 A1

[5]

b) cos2A ( 2cos2A − 1 B1

= [pic] M1 A1 cao

[3]

3. a) cos2θ ( 2cos2θ − 1 B1

[pic] M1

a2 = 2(b+1) A1

[3]

b) cos2θ ( 1 − 2sin2θ B1

a = 1 − 2(b−2)2 M1 A1

[3]

c) sin2θ ( 2sinθ cosθ B1

sin22θ ( 4sin2θ cos2θ M1

cos2θ ( 1 − sin2θ M1

(a − 2)2 = 4b2(1 − b2) A1

[4]

4. a) cos 3A ( cos(2A+A) M1

cos3A ( cos2A cosA − sin2A sinA B1

But cos2A ( cos2A − sin2A and sin2A ( 2sinAcosA B1

( cos3A ( (cos2A − sin2A)cosA − 2sinAcosAsinA A1

But sin2A ( 1 − cos2A B1

( cos3A ( (2cos2A − 1)cosA − 2cosA(1 − cos2A) M1

( 4cos3A − 3cosA A1

[7]

b) 4cos3A − 3cosA + 2cosA = 0 M1

4cos3A − cosA = 0

cosA (4cos2A − 1) = 0 M1

cosA = 0, [pic], -[pic]

A = [pic],[pic],[pic] A3 ( -1 if degrees)

[5]

5. a) 2cosx cos60 − 2sinx sin60 = 1

cosx cos60 − sinx sin60 = [pic] M1 A1

cos(x + 60) = [pic] M1

0(x(360 ( 60(x+60(420

x + 60 = 60, 300, 420 A2 (-1 eeoo)

x = 0, 240, 360 A1

[6]

b) i) From above, [pic]x = 0, 240, 360 M1

x = 0 only A1 ft

[2]

ii) x − 30 = 0, 240, 360 M1

x = 30, 270, 390. But 390 outside range

x = 30, 270 only A1 ft

[2]

6. a) sin (50 + 10) =sin60=[pic] M1 A1

[2]

b) cos(75 + 15) =cos90 = 0 M1 A1

[2]

c) tan(80 + 55) = tan135= -1 M1 A1

[2]

d) sin(2(75) = sin150 = [pic] M1 A1

[2]

e) cos(2(15) = cos30 =[pic] M1 A1

[2]

7. a) 6sin [pic]x cos[pic]x ( 3sinx B1

3sinx + cosx = 0 M1

tanx = -[pic] A1

x = 161.6, 341.6 A2

[5]

b) cos2x ( 1 − 2sin2x B1

3 (1 − 2sin2x) + sinx = 1 M1

6sin2x − sinx − 2 = 0

(3sinx − 2)(2sinx + 1) = 0 M1 A1

sinx = [pic]or -[pic]

x = 41.8, 138.2, -30, -150 A3 (-1 eeoo)

[7]

8. a) [pic] M1

[pic] M1

But sin2x + cos2x (1 and sinxcosx ( [pic] sin2x B1

[pic] M1

So tanx + cotx ( 2cosec 2x A1

[5]

b) [pic] M1A1

But sec2x − tan2x ( 1 B1

Hence [pic] A1

[4]

9. a) i) [pic] M1

[pic] M1 A1

[pic] A1

[4]

ii) [pic] M1 A1

( tan ([pic]x) A1

[3]

b) sin2θ − cos2θ = sinθ + cosθ

sin2θ − sinθ = cos2θ + cosθ M1

[pic] M1

tan([pic]θ) = 1 M1

0(θ(π B1

[pic]θ ’ [pic] ( θ ’ [pic] A1

[5]

10. a) 2cos2x − 1=cosx B1

(2cosx + 1)(cosx − 1) = 0 M1 A1

cosx = -[pic] or 1

x = 120, 240, 0 A2 (-1 eeoo)

[5]

b) From a), have [pic]x = 120, 240, 0 M1

x = 240, 480, 0 A2 ft

[3]

11. [pic] M1 A1

[pic] M1

[pic] A1

= tanA −tanB A1

[5]

12. a) 2cos(x − 45) = 2(cosx cos45 + sinx sin45) M1

= 2 ( cosx + sinx) A1

(2

= (2(cosx + sinx) A1

[3]

b) i) [pic]cos(x−45)=1 M1

cos(x−45) =[pic]

x−45 = (45 A1

x= 0, 90 M1 A1

[4]

ii) 4 [ (2cos(x − 45)]4 = 1 M1

4( 4cos4(x − 45) = 1

cos4 (x − 45) = [pic] A1

cos(x − 45) = ([pic] A1 (both)

x − 45 = 60, 300, 120, 240 A2

x = 105, 345, 165, 285 A1 ft

[6]

13.a) i) sin3θ + sinθ ( 2sin2θ cosθ M1 A1 A1

[3]

ii) cos 3θ +cosθ ( 2cos2θcosθ B1 B1

[2]

b) [pic] M1

[pic] M1

( tan2θ A1

[3]

14. a) [pic] B1

2tanx + tanx (1 − tan2x) = 0 M1

tanx (tan2x − 3) = 0

tanx = 0, ((3 A1

x = 0, 60, -60 A2 (-1 eeoo)

[5]

b) [pic] M1 M1 A1

4sinxcosx = 3sinx

sinx(4cosx – 3) =0 M1

sinx = 0 or cosx = ¾ A1 (both)

x= 41.4o A1 (only answer)

[6]

15. a) sin15 = sin(45 − 30) M1

= sin45 cos30 − sin30 cos45 B1

=[pic] A1

=[pic] A1 cao

[4]

b) cos 165 = cos(120 + 45)

= cos120 cos45 − sin120 sin45 M1

= [pic] A1

= [pic] (or [pic]) A1 cao

OR: cos165 = -cos15

= -(cos45cos30 + sin45sin30) M1 A1

=[pic] A1

[3]

c) tan75 = tan(45 + 30)

=[pic] M1

=[pic] A1

= [pic] M1

= 2 + (3 A1

OR: tan75 = [pic] M1

=[pic] M1

= [pic] A1 ft

= 2 + (3 A1

[4]

16. a) 2sinx cosx ( cosx + sin2x = 1 M1

2cos2x sinx + (1 − cos2x) = 1 M1

cos2x (2sinx − 1) = 0 A1

x = [pic],[pic],[pic],[pic] A3 (-1 eeoo)

[6]

b) tanx ( [pic] = 1 M1

2tan2x = 1 − tan2x A1

tanx =[pic] A1

x = ([pic], ([pic] A2

[5]

17. a) Since A + B + C = 180, C = 180 − (A+B) M1

sinC = sin [180 − (A+B)]

sinC = sin(A+B) A1

[2]

b) sin(A − B) + sinC = sin(A − B) + sin (A+B)

= sinAcosB − cosAsinB + sinAcosB + cosAsinB M1 A1

= 2sinA cosB A1

[3]

c) sinC = sin(B+60+B) M1

= sin(2B + 60)

= sin2Bcos60 + cos2Bsin60 M1

= [pic] ( sin2B + [pic]( cos2B A1

= [pic] ( sin2B + (3cos2B) A1

[4]

18 a) (tanx + secx)2 ( tan2x + 2tanxsecx + sec2x M1

[pic]

([pic] A1

[pic] [pic][pic]

[pic] A1

[5]

b) [pic] M1 (using identity)

4sinx + 4 = 5 – 5sinx M1

9sinx = 1

sinx = [pic] A1

x= 6.38o, 1730 A1 A1

[5]

19. 3cosx + 4sinx (Rcos(x−α)( Rcosxcosα + Rsinxsinα M1

3 = Rcosα 4 = Rsinα

R2 = 32 + 42 ( R=5 M1 A1

tanα = [pic]( α = 53.1o A1

5cos(x −53.1) = 2

cos(x −53.1) = 0.4 M1

x −53.1 = 66.4, 294 A1

x= 120, 347 A1 A1

[8]

20. a) cosx + 7sinx (Rcos(x−α) ( Rcosxcosα + Rsinxsinα Μ1

1 ’ Rcosα 7 = Rsinα

R2 = 12 + 72 ( R=(50 M1 A1

tanα = 7( α = 81.9o A1

[4]

b)(50cos(x – 81.9) = 5

cos(x−81.9) =[pic] M1

x – 81.9 = (45 A1

x= 36.9, 127 A1 A1

[4]

21. a) 2sinx + 3cosx (Rsin(x+α) ( Rsinxcosα + Rcosxsinα Μ1

2 ’ Rcosα 3 = Rsinα

R2 = 22 + 32 ( R=(13 M1 A1

tanα = 1.5 ( α = 56.3o A1

[4]

b) -1(sin(x+56.3)(1 M1

So -(13((13sin(x+56.3)((13 A1

So 4 −(13( 2sinx + 3cosx + 4 ( 4 + (13 A1

[3]

22. a) cos(x+40) = 3sin(x+50)

cosxcos40 – sinxsin40 = 3sinxcos50 + 3cosxsin50 M1 A1

cosx(cos40 – 3sin50) = sinx(3cos50 + sin40)

[pic] M1

-0.5959 = tanx A1

[4]

b) tan-1(-0.5959) = -30.8 M1

149.2, 329.2 A1 A1

[3]

23. a) cos2x − (3sin2x (Rcos(2x+α) ( Rcos2xcosα + Rsin2xsinα Μ1

1 ’ Rcosα (3 = Rsinα

R2 = 12 + ((3)2 ( R=2 M1 A1

tanα = (3 ( α = 60o A1

[4]

b) G1 (shape)

G1 (2 & -2)

G2 (intersections)

G1 (max & min)

[5]

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

(0,1)

(285,0)

(195,0)

(105,0)

(15,0)

(330, 2)

(240, -2)

(150, 2)

(60, -2)

y=2cos(2x+60)

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