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Let $f x = x 2 cos π x, x ≠ 0 0 x = 0$ , $x ∈ l$ . Then $f$ is

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differentiable neither at

x = 0

nor at

x = 2

differentiable both at

x = 0

and at

x = 2

differentiable at

x = 0

but not differentiable at

x = 2

not differentiable at

x = 0

but differentiable at

x = 2

answer is B.

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Detailed Solution

lim x → 0 x 2 cos π x = 0 x (any value between − 1 and 1) = 0

∴ f (x) i s c o n t i n u o u s a t x = 2 ⇒ f ′ 0 + = lim h → 0 h 2 cos π h − 0 h = lim h → 0 h cos π h = 0

Similarly f’ (0)=0 So, is differentiable at x=0 Clearly f(x)is continuous at x = 2.

f ′ 2 + = lim h → 0 (2 + h) 2 cos π h − 0 h = lim h → 0 (2 + h) 2 h sin π 2 − π 2 + h = π

Similarly, we can P.T

f ′ 2 − = − π

Hence, f(x) is not differentiable at x = 2.

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