The function $f(x)=\left\{\begin{array}{ll}\frac{\pi}{4}+\tan^{-1}x, & |x|\leq 1 \\ \frac{1}{2}(|x|-1), & |x|>1\end{array}\right.$ is:

(C) continuous on $\mathbb{R}-\{1\}$ and differentiable on $\mathbb{R}-\{-1,1\}$.

(A) both continuous and differentiable on $\mathbb{R}-\{1\}$.

(B) both continuous and differentiable on $\mathbb{R}-\{-1\}$.

(D) continuous on $\mathbb{R}-\{-1\}$ and differentiable on $\mathbb{R}-\{-1,1\}$.

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Continuity and Differentiability

The function

f (x) = {\frac{π}{4} + tan^{- 1} x, \frac{1}{2} (∣ x ∣ - 1), ∣ x ∣ \leq 1 ∣ x ∣ > 1

is:

Answer: (C)

x = 1

the left-hand limit is

\frac{π}{4} + tan^{- 1} 1 = \frac{π}{2}

while the right-hand limit is

\frac{1}{2} (1 - 1) = 0

, so

f

is discontinuous there. At

x = - 1

both one-sided limits equal

0

, so

f

is continuous. Its derivative, however, is

\frac{1}{2}

from the right of

- 1

but

- \frac{1}{2}

from the left, so

f

fails to be differentiable at

- 1

as well as at

1

. Hence

f

is continuous on

R - {1}

and differentiable on

R - {- 1, 1}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option