Let $f(x)=\left\{\begin{array}{rl}x^{2}\left|\cos \frac{\pi}{x}\right|, & x \neq 0 \\ 0, & x=0\end{array} \quad, x \in \mathbb{R}\right.$, then $f$ is

(B) differentiable at $x=0$ but not differentiable at $x=2$

(A) differentiable both at $x=0$ and at $x=2$

(C) not differentiable at $x=0$ but differentiable at $x=2$

(D) differentiable neither at $x=0$ nor at $x=2$

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Continuity and Differentiability

Let

f (x) = {x^{2} cos \frac{π}{x}, 0, x \neq = 0 x = 0, x \in R

, then

f

Answer: (B)

x = 0

f^{'} (0) = lim_{h \to 0} \frac{h ^{2} ∣ c o s ( π / h ) ∣}{h} = lim_{h \to 0} h ∣ cos (π / h) ∣ = 0

, so

f

is differentiable at

x = 0

.

At

x = 2

f^{'} (2^{+}) = lim_{h \to 0} \frac{( 2 + h ) ^{2} ∣ c o s ( π / ( 2 + h )) ∣}{h} = lim_{h \to 0} \frac{( 2 + h ) ^{2} s i n ( π h / ( 2 ( 2 + h )))}{h} = π

. Similarly,

f^{'} (2^{-}) = - π

. Since the left and right derivatives differ,

f

is not differentiable at

x = 2

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option