If \[ f(x)= \begin{cases}-x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} 1,\end{cases} \] then

(A) $f(x)$ is continuous at $x=-\frac{\pi}{2}$; (B) $f(x)$ is not differentiable at $x=0$; (C) $f(x)$ is differentiable at $x=1$; (D) $f(x)$ is differentiable at $x=-\frac{3}{2}$

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Continuity and Differentiability

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f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1,

then

(A) $f (x)$ is continuous at $x = - \frac{π}{2}$ ✓ Correct
(B) $f (x)$ is not differentiable at $x = 0$ ✓ Correct
(C) $f (x)$ is differentiable at $x = 1$ ✓ Correct
(D) $f (x)$ is differentiable at $x = - \frac{3}{2}$ ✓ Correct

Answer: (A), (B), (C), (D)

Solution

Checking continuity at

x = - π /2

x \to (- π /2)^{-} lim f (x) = - (- π /2) - π /2 = 0

x \to (- π /2)^{+} lim f (x) = - cos (- π /2) = 0

f (- π /2) = 0

f

is continuous at

x = - π /2

. (A) is correct.

Checking differentiability at

x = 0

f^{'} (0^{-}) = sin (0) = 0

f^{'} (0^{+}) = 1

Since left and right derivatives differ,

f

is not differentiable at

x = 0

. (B) is correct.

Checking differentiability at

x = 1

f^{'} (1^{-}) = 1

f^{'} (1^{+}) = \frac{1}{1} = 1

f

is differentiable at

x = 1

. (C) is correct.

At

x = - 3/2

, the function is

f (x) = - x - π /2

, which is differentiable everywhere in its domain. (D) is correct.

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JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Continuity and Differentiability

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