(i) Consider f(x) = {(3x-8, x≤5)(2k, x>5) Find the Value of K if f(x) is continuous at x =5.

Question

(i) Consider f(x) = \( \begin{cases} 3x -8, & x≤5 \\\ 2k, & x> 5 \end{cases} \)Find the Value of K if f(x) is continuous at x =5.

(ii) Find \(\frac{dy}{dx}\), if y = (sin x)^logx, sin x >0

(iii) if y = (sin - 1x) 2 , then show that

(1 - x²) \(\frac{d^2y}{dx^2}\) - x \(\frac{dy}{dx}\) = 2.

Answer 1

(i) since f is continuous at x = 5, we have;

(ii) Given ; y = (sin x)^logx

Take log on both sides;

logy = log x x log (sin x)

Differentiating w.r.t x;

(iii) Given : y = (sin^-1x)² ; Differentiating w.r.t x;