Arrange $k, \alpha , \beta $ and $\mu $ in descending order :

A. If $y=\frac{1}{x}$ then $\left[\frac{d^2y}{dx^2}\right]_{x=\frac{1}{2}}=k$

A. If $y=\frac{2}{x}$ then $\left[\frac{d^2y}{dx^2}\right]_{x=2}=\alpha $

C. If $y=\frac{1}{x+1}$ then $\left[\frac{d^2y}{dx^2}\right]_{x=1}=\beta $

D. If $y=log_e(x+1)$ then $\left[\frac{d^2y}{dx^2}\right]_{x=1}=\mu $

Choose the correct answer from the option sgiven below :

$k > \alpha > \beta > \mu $

The correct answer is Option (4) → $k > \alpha > \beta > \mu $

(A) $y=\frac{1}{x}$

$⇒\frac{dy}{dx}=-\frac{1}{x^2}⇒\frac{d^2y}{dx^2}=\frac{2}{x^3}$

$\left[\frac{d^2y}{dx^2}\right]_{x=\frac{1}{2}}=16=k$

(B) $y=\frac{2}{x}$

$⇒\frac{d^2y}{dx^2}=\frac{4}{x^{-3}}$

$⇒\left[\frac{d^2y}{dx^2}\right]_{x=2}=\frac{4}{2^3}=\frac{1}{2}=\alpha $

(C) $y=\frac{1}{x+1}$

$⇒\frac{d^2y}{dx^2}=\frac{2}{(x+1)^3}$

$⇒\left[\frac{d^2y}{dx^2}\right]_{x=1}=\frac{2}{8}=\frac{1}{4}=\beta $

(D) $y=log_e(x+1)$

$⇒\frac{d^2y}{dx^2}=-\frac{1}{(x+1)^2}=-\frac{1}{4}=\mu$

$⇒k > \alpha > \beta > \mu $