Question

The function $f(x) = \begin{cases} \frac{\pi}{4} + tan^{-1} x, |x| \le 1 \\[2ex] \frac{1}{2}(|x| - 1), |x| > 1 \end{cases}$

• A. continuous on $R -\{1\}$ and differentiable on $R -\{-1,1\}$
• B. both continuous and differentiable on $R-\{-1\}$
• C. continuous on $R -\{-1\}$ and differentiable on $R -\{-1,1\}$.
• D. both continuous and differentiable on $R-\{1\}$

Answer 1

Answer: (A)

$f(x) = \begin{cases} \frac{\pi}{4}+tan^{-1}\,x, & x \in(-\infty,-1] \cup[1, \infty) \\ -\frac{(x+1)}{2} , & x \in(-1,0]\\ \frac{x-1}{2}, & x \in(0,1) \end{cases} $
for continuity at $x=-1$
$L.H.L. =\frac{\pi}{4}-\frac{\pi}{4}=0$
$R.H.L. =0$
so, continuous at $x=-1$
for continuity at $x=1$
$L.H.L. =0$
$R . H.L. =\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$
so, not continuous at $x=1$
For differentiability at $x=-1$
$L.H.D. =\frac{1}{1+1}=\frac{1}{2}$
$R . H.D. =-\frac{1}{2}$
so, non differentiable at $x=-1$