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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
The function $f(x)=\left\{\begin{array}{l}\frac{\pi}{4}+\tan ^{-1} x,|x| \leq 1 \\ \frac{1}{2}(|x|-1),|x|>1\end{array}\right.$
  • ✓
    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

Correct option: A.
continuous on $R -\{1\}$ and differentiable on $R-\{-1,1\}$
a
$f(x)=\left\{\begin{array}{ccc}\frac{\pi}{4}+\tan ^{-1} x & , & x \in(-\infty,-1] \cup[1, \infty) \\ -\frac{(x+1)}{2} & , & x \in(-1,0] \\ \frac{x-1}{2} & , & x \in(0,1)\end{array}\right.$

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 $x=-1$

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