(i) (b) log x
(ii) Given; y = (x - 2)2 ⇒ \(\frac{dy}{dx}\) = 2(x - 2)
Slope of the chord = \(\frac{4-0}{4-2}\) = 2
⇒ 2 = 2(x - 2) ⇒ x = 3 ⇒ y = (3 - 2)2 = 1
Therefore the required point is (3,1)
(iii) Given; p(x) = 41 – 24x – 6x2
p’(x) = – 24 – 12x
p”(x) = – 12
For turning points p’(x) = – 24 – 12x = 0
⇒ x = -2
Since p”(x) = – 12 always maximum Therefore maximum value
p(- 2) = 41 – 24(- 2) 6(- 2)2 = 65