Cambridge International AS and A Level Mathematics Pure Mathematics 1

(Michael S) #1
Chapter

(^5)
297
P1^
(vi) (1, 2), minimum and
(−1, −2), maximum
(vii) (^) ( 12 , 12) , minimum
(viii) ( 2 , 8 2 ) , minimum and
(− 2 , − 8 2 ), maximum
(ix) (16, 32), maximum
  3 (i) 4 x (x + 2)(x − 2)
(ii) 4(3x^2 − 4)
(iii) (−2, −16), minimum;
(0, 0), maximum;
(2, −16), minimum
(iv)
  4 (i) (^) ddyx
= (3x − 7)(x − 1)
(ii) Maximum at (1, 0);
minimum at () 2113 ,– 275
(iii)
5 (i) (^) ddyx
= 4 x(x − 1)(x − 2)
(ii) Minimum at (0, 0);
maximum at (1, 1);
minimum at (2, 0)
(iii)
6 (i) p + q = − 1
(ii) 3 p + 2 q = 0
(iii) p = 2 and q = − 3
  7 (i) f '(x) = 8 x − (^) x^12 ; f "(x) = 8 + (^) x^23
(ii) (^) (^12 , 3), minimum
  8 (i) 1 − 2
x
; x−
(^32)
(ii) (4, −4), minimum
  9 2
10  (i) 0, 10
(ii) −58.8
Exercise 5G (Page 162)
  1 (i) y = 60 − x
(ii) A = 60 x − x^2
(iii) ddAx
= 60 − 2 x;
dd
2
2
A
x
= − 2
Dimensions 30 m by 30 m,
area 900 m^2
  2 (i) V = 4 x^3 − 48 x^2 + 144 x
(ii) ddVx
= 12 x^2 − 96 x + 144;
dd
2
2
V
x
= 24 x − 96
  3 (i) y = 8 − x
(ii) S = 2 x^2 − 16 x + 64
(iii) 32
  4 (i)  2 x + y = 80
(ii) A = 80 x − 2 x^2
(iii) x = 20, y = 40
  5 (i) x(1 − 2 x)
(ii) V = x^2 − 2 x^3
(iii) ddVx
= 2 x − 6 x^2 ;
dd
2
2
V
x
= 2 − 12 x
(iv) All dimensions^13 m (a cube);
volume 271 m^3
  6 (i) (a) (4 − 2 x) cm
(b) (16 − 16 x + 4 x^2 ) cm^2
(iii) x = 1.143
(iv) A = 6.857
  7  (i) P = 2 πr, r = 15 – π^2 x
(iii) x = (^430) +π cm:
lengths  16.8 cm and
13.2 cm
  8 (i) h = (^125) r − r
(ii) V = 125 πr − πr^3
(iii) ddVr = 125 π − 3 πr^2 ;
dd
2
2
V
x^ =^ −^6 πr
(iv) r = 6.45 cm; h = 12.9 cm
(to 3 s.f.)
  9 (i) Area = xy = 18
(ii) T = 2 x + y
(iv) ddTx = 2 − (^18) x 2 ; dd
2
2
T
x^ =^
36
x^3
(v) x = 3 and y = 6
10 (i) V = x^2 y
(ii) A = x^2 + 4xy
(iii) A = x^2 +^2 x
(iv) ddAx = 2 x − (^) x^22 ; d
2
2
A
dx
= 2 + (^) x^43
(v) x = 1 and y = (^12)
11 (i) h = (^324) x 2
(iii) ddAx = 12 x − (^25922)
x
; stationary
point when x = 6 and h = 9
(iv) Minimum area = 648 cm^2
Dimensions:
6 cm × 18 cm × 9 cm
12 (i) y = (^24) x
(ii) A = 3 x + 30 + (^48) x
(iii) A = 54 m^2
13 (i) h = 12 − 2 r
(ii) 64 π or 201 cm^3
●?^ (Page^ 167)
d
d
V
h^ is the rate of change of the
volume with respect to the height of
the sand.
x
y
–2 O 2
–16
x
y
O
–3
1 3
x
y
O 1 2
1

Free download pdf