MHT CET · Maths · Continuity and Differentiability
Let \(\mathrm{S}=\left\{\mathrm{t} \in \mathrm{R} / \mathrm{f}(x)=|x-\pi|\left(\mathrm{e}^{|x|}-1\right) \sin |x|\right.\) is not differentiable at \(\mathrm{t}\}\), then \(\mathrm{S}\) is
- A \(\phi\) (an empty set)
- B \(\{0\}\)
- C \(\{\pi\}\)
- D \(\{0, \pi\}\)
Answer & Solution
Correct Answer
(A) \(\phi\) (an empty set)
Step-by-step Solution
Detailed explanation
Differentiability at \(x=\pi\) :
L.h.lim
\(=\lim _{h \rightarrow 0} \frac{|\pi-h-\pi|\left(e^{|\pi-h|}-1\right) \sin |\pi-h|-0}{-h}=0\)
R.h.lim
\(=\lim _{h \rightarrow 0} \frac{|\pi+h-\pi|\left(e^{|\pi+h|}-1\right) \sin |\pi+h|-0}{h}=0\)
Differentiability at \(\boldsymbol{x}=\mathbf{0}\) :
L.h.lim \(=\lim _{\mathrm{h} \rightarrow 0} \frac{|-\mathrm{h}-\pi|\left(\mathrm{e}^{|-\mathrm{h}|}-1\right) \sin |-\mathrm{h}|-0}{-\mathrm{h}}=0\)
R.h.lim \(=\lim _{h \rightarrow 0} \frac{|h-\pi|\left(e^{|h|}-1\right) \sin |h|-0}{h}=0\)
The function \(\mathrm{f}(x)\) is differentiable at \(x=0, \pi\).
\(\Rightarrow\) Set \(\mathrm{S}\) is an empty set.
L.h.lim
\(=\lim _{h \rightarrow 0} \frac{|\pi-h-\pi|\left(e^{|\pi-h|}-1\right) \sin |\pi-h|-0}{-h}=0\)
R.h.lim
\(=\lim _{h \rightarrow 0} \frac{|\pi+h-\pi|\left(e^{|\pi+h|}-1\right) \sin |\pi+h|-0}{h}=0\)
Differentiability at \(\boldsymbol{x}=\mathbf{0}\) :
L.h.lim \(=\lim _{\mathrm{h} \rightarrow 0} \frac{|-\mathrm{h}-\pi|\left(\mathrm{e}^{|-\mathrm{h}|}-1\right) \sin |-\mathrm{h}|-0}{-\mathrm{h}}=0\)
R.h.lim \(=\lim _{h \rightarrow 0} \frac{|h-\pi|\left(e^{|h|}-1\right) \sin |h|-0}{h}=0\)
The function \(\mathrm{f}(x)\) is differentiable at \(x=0, \pi\).
\(\Rightarrow\) Set \(\mathrm{S}\) is an empty set.
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