JEE MAIN - Mathematics (2018 (Offline) - No. 16)
Let S = { t $$ \in R:f(x) = \left| {x - \pi } \right|.\left( {{e^{\left| x \right|}} - 1} \right)$$$$\sin \left| x \right|$$ is not differentiable at t}, then the set S is equal to
{0, $$\pi $$}
$$\phi $$ (an empty set)
{0}
{$$\pi $$}
Explanation
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 = $$\phi $$
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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