Cambridge International AS and A Level Mathematics Pure Mathematics 1

Answers

296

P1^

  7 (i) ddyx = 3(x + 3)(x − 1)

(ii) x = −3 or 1

(v)

  8 (i) ddyx = −3(x + 1)(x − 3)

(ii) Minimum when x = −1,

maximum when x = 3

(iii) When x = −1, y = −5;

when x = 3, y = 27

(iv)

  9 (i) Maximum at ()–,^2341327 ,

minimum at (2, −5)

(ii)

10 (i) Maximum at (0, 300),

minimum at (3, 165),

minimum at (−6, −564)

(ii)

11 (i) d

d

y

x

= 3(x^2 + 1)

(ii) There are no stationary

points.

(iii)

x − 3        − 2     − 1 0 1        2       3

y − 36 − 14  − 4 0 4 14 36

(iv)

12 (i) ddyx

= 6 x^2 + 6 x − 72

(ii) y = 18

(iii) ddyx

= 48; y = 48 x − 174

(iv) (−4, 338) and (3, −5)

13 (i) (^) (^12 , 4) and (−^12 , − (^4) )
(ii) −^12  x ^12
14 (i) ddyx = (2x − 3)^2 − 4
(ii) 2 y + 9 = 10 x
(iii) x  212 or x  (^12)
15 (ii) x  1.5
(iii) (−1, 8) and (2, 2)
(iv) (^334)
16 (i) x = 112 and x = 2
(ii) (2, 1) is the stationary point
Activity 5.7 (Page 155)
At P (max.) the gradient of d
d
y
x
is
negative.
At Q (min.) the gradient of ddyx is
positive.
Exercise 5F (Page 158)
  1 (i) ddyx
= 3 x^2 ; dd
2
2
y
x
= 6 x
(ii) ddyx
= 5 x^4 ; dd
2
2
y
x
= 20 x^3
(iii) ddyx
= 8 x; dd
2
2
y
x^ =^8
(iv) ddyx
= − 2 x−^3 ; d
d
2
2
y
x
= 6 x−^4
(v) ddyx== 23 x ddxy^34 x−
(^12212)
; 2
(vi) ddyx
= 4 x^3 + (^) x^64 ;
dd
2
2
y
x^ =^12 x
(^2) − (^245)
x
  2 (i) (−1, 3), minimum
(ii) (3, 9), maximum
(iii) (−1, 2), maximum and
(1, −2), minimum
(iv) (0, 0), maximum and
(1, −1), minimum
(v) (−1, 2), minimum;
(^) ( –^34 , 2.02), maximum;
(1, −2), minimum
x
y
 2 1

1
x
y
–1 O 3
2
–
x
y
O 2
–5

-^2 – 3

41327 –

x

y

±^23

±54

300

15

x

y

O

O x

y

O x

dy

dx

O x

gradient

of dy

dx

Q

P