Cambridge International AS and A Level Mathematics Pure Mathematics 1

Answers

P1^

 6 (i) h + 4 r = 100,

2 πrh + 2 πr^2 = 1400 π

(ii) 6000 π or^9800027 π cm^3

 7 (i) (3x + 2 y)(2x + y) m^2

(iii) xy=^12 , =^14

Exercise 1H (Page 37)

 1  (i) a  6

(ii) b  2

(iii) c  − 2

(iv) d  −^43

(v) e  7

(vi) f  − 1

(vii) g  1.4

(viii) h  0

 2  (i) 1  p  4

(ii) p  1 or p  4

(iii) − 2  x  − 1

(iv) x  −2 or x  − 1

(v) y  −1 or y  3

(vi) − 4  z  5

(vii) q  2

(viii) y  −2 or y  4

(ix) –2  x ^13

(x) y  −^12 or y  6

(xi) 1  x  3

(xii) y  −^12 or y ^35

 3 (i) k ^98

(ii) k  − 4

(iii) k  10 or k  − 10

(iv) k  0 or k  3

 4 (i) k  9

(ii) k  −^18

(iii) − 8  k  8

(iv) 0  k  8

Chapter  2

Activity 2.1 (Page 40)

A:^12 ; B: −1; C: 0; D: ∞

●?^ (Page^ 40)

No, the numerator and denominator

of the gradient formula would have

the same magnitude but the opposite

sign, so m would be unchanged.

Activity 2.2 (Page 41)

An example of L 2 is the line joining

(4, 4) to (6, 0).

m 1 = 12 , m 2 = − 2 ⇒ m 1 m 2 = −1.

Activity 2.3 (Page 41)

ABE BCD

AB = BC

AEB = BDC

BAE = CBD

⇒ Triangles ABE and BCD are

congruent so BE = CD and AE = BD.

⇒

mm

mm

1 2

12 1

==

=× =

BE

AE

BD

CD

BE

AE

BD

CD

;–

––

Exercise 2A (Page 44)

1  (i) (a) − 2

(b) (1, −1)

(c) 20

(d) 12

(ii) (a) − 3

(b) (^) () 312 ,^12
(c) 10
(d) (^13)
(iii) (a) 0
(b) (0, 3)
(c)  12
(d) Infinite
(iv) (a)  103 &
(b) (^) () 3312 ,–
(c) 109 &
(d) – (^103) &
(v) (a) (^32)
(b) (^) () 31 ,^12
(c) 13
(d) –^23
(vi) (a) Infinite
(b) (1, 1)
(c)  6
(d)  0
  2 5
  3 1
  4 (i) AB:B 21 ,,C:^32 CD:D^12 , A: 23
(ii) Parallelogram
(iii)
5 (i) 6
(ii) AB= 20 ,BC= 5
(iii) 5 square units
y
(^0) 1 2 3 4 5 6 x
1
2
3
(^4) L 1
L 2
y
0 2 4 6 8 x
2
4
6
D
A
B
8 C