Question

The function $f(x)=\{\begin{array}{l}\frac{\pi }{4}+ta{n}^{-1}x,|x|\le 1\\ \frac{1}{2}(|x|-1),|x|>1\end{array}$

• 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)

$f(x)=\{\begin{array}{ll}\frac{\pi }{4}+ta{n}^{-1}x,& x\in (-\infty ,-1]\cup [1,\infty )\\ -\frac{(x+1)}{2},& x\in (-1,0]\\ \frac{x-1}{2},& x\in (0,1)\end{array}$
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 $x=-1$