Chapter(^5)
293
P1^
31 8 x − 1
32 4 x + 5
33 1
34 16 x^3 − 10 x
(^35 32)
(^12)
x
36 1
x
37 92 x −^1
x
38 34 x^2 −^12 x + 4
39 3
2
x
(^40 5432)
(^322)
xx
x
− −
Exercise 5C (Page 136)
1 (i) (a) − 2 x−^3
(b) − 128
(ii) (a) −x−^2 − 4 x−^5
(b) 3
(iii) (a) − 12 x−^4 − 10 x−^6
(b) − 22
(iv) (a) 12 x^3 + 24 x−^4
(b) 97.5
(v) (a) 1
2 x
- 3
(b) (^314)
(vi) (a) − 2 x−
(^32)
(b) − 272
2 (i)
(ii) (−2, 0), (2, 0)
(iii) (^) ddyx = 2 x
(iv) At (−2, 0), d
d
y
x
= −4;
at (2, 0), d
d
y
x
= 4
3 (i)
(ii) (^) ddyx = 2 x − 6
(iii) At (3, −9), ddyx = 0
(iv) Tangent is horizontal: curve
at a minimum.
4 (i)
(iii) d
d
y
x
= − 2 x : at (−1, 3), d
d
y
x
= 2
(iv) Yes: the line and the curve
both pass through (−1, 3)
and they have the same
gradient at that point.
(v) Yes, by symmetry.
5 (i)
(ii) (^) ddyx = 3 x^2 − 12 x + 11
(iii) x = 1: ddyx = 2; x = 2: ddyx = −1;
x = 3: d
d
y
x
= 2
The tangents at (1, 0) and
(3, 0) are therefore parallel.
6 (i)
(ii) (^) ddyx = 2 x + 3
(iii) (1, 3)
(iv) No, since the line does not
go through (1, 3).
7 (i)
(ii) (^) ddyx = 2 x
(iii) At (2, −5), ddyx = 4;
at (−2, −5), ddyx = − 4
(iv) At (2, 5), ddyx = −4;
at (−2, 5), ddyx = 4
(v) A rhombus
8 (i)
(ii) 4
(iv) y = x^2 + c, c ∈
9 (i) 4 a + b − 5 = 0
(ii) 12 a + b = 21
(iii) a = 2 and b = − 3
y
–4
–2 O 2 x
y
–
O x
y
O 2
5
x
4
–2
y
O x
–
2 3
y
O
–1
x
- –^32
y–3 O 3–99xyO
–13x