Question 219 of 401MediumMCQ
Let S={t∈R:f(x)=∣x−π∣(e∣x∣−1)sin∣x∣ is not differentiable at t}. Then the set S is equal to:
Solutions · 1
Official solution
Check differtiability at x = \[\pi \] and x = 0
at x = 0 :
We have L. H. D = \[\mathop {\lim }\limits_{h \to 0} {{f\left( {0 - h} \right) - f\left( 0 \right)} \over { - h}}\]
= \[\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| h \right| - 0} \over { - h}} = 0\]
R. H. D = \[\mathop {\lim }\limits_{h \to 0} {{f\left( {0 + h} \right) - f\left( 0 \right)} \over h}\]
\[ = \mathop {\lim }\limits_{h \to 0} {{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - o} \over h}\]
= 0
\[\therefore,\,\,\] LHD = RHD
Therefore, function is differentiable at x = \[\pi \].
at x = \[\pi \] :
L. H. D = \[\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi - h} \right) - f\left( \pi \right)} \over { - h}}\]
= \[\mathop {\lim }\limits_{h \to 0} {{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {\pi - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0.} \over { - h}}\]
= 0
RH. D = \[\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi + h} \right) - f\left( \pi \right)} \over h}\]
= \[\mathop {\lim }\limits_{h \to 0} {{\left| {\pi + h - \pi } \right|\left( {{e^{\left| {\pi + h} \right|}} - 1} \right)\sin \left| {\pi + h} \right| - 0} \over h}\]
= 0
\[\therefore\,\,\,\] L. H. D = R H D
Therefore, function is differentiable at x = \[\pi \],
Since, the function f(x) is differentiable at all the points including 0 and \[\pi \].
So, f(x) is differentiable everywhere.
Therefore, set S is empty set.
S = \[\phi \]
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