JEE MAIN - Mathematics (2008 - No. 2)
Let $$f\left( x \right) = \left\{ {\matrix{
{\left( {x - 1} \right)\sin {1 \over {x - 1}}} & {if\,x \ne 1} \cr
0 & {if\,x = 1} \cr
} } \right.$$
Then which one of the following is true?
Then which one of the following is true?
$$f$$ is neither differentiable at x = 0 nor at x = 1
$$f$$ is differentiable at x = 0 and at x = 1
$$f$$ is differentiable at x = 0 but not at x = 1
$$f$$ is differentiable at x = 1 but not at x = 0
Explanation
We have $$f\left( x \right) = \left\{ {\matrix{
{\left( {x - 1} \right)\sin \left( {{1 \over {x - 1}}} \right),} & {if\,\,x \ne 1} \cr
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {if\,\,x = 1} \cr
} } \right.$$
$$Rf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{h\,\sin {1 \over h} - 0} \over h} = \mathop {\lim }\limits_{h \to 0} \,\,\sin {1 \over h} = a$$ finite number
Let this finite number be $$l$$
$$L$$ $$f'(1)$$ $$ = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{ - h\,\sin \left( {{1 \over { - h}}} \right)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} \,\,\sin \left( {{1 \over { - h}}} \right)$$
$$ = - \mathop {\lim }\limits_{h \to 0} \sin \left( {{1 \over h}} \right) = - (\,$$ a finite number $$\,)$$ $$=-l$$
Thus $$Rf'\left( 1 \right) \ne Lf'\left( 1 \right)$$
$$\therefore$$ $$f$$ is not differentiable at $$x=1$$
Also, $$f'\left( 0 \right) = \sin {1 \over {\left( {x - 1} \right)}} - {{x - 1} \over {{{\left( {x - 1} \right)}^2}}}\cos {\left. {\left( {{1 \over {x - 1}}} \right)} \right]_{x = 0}}$$
$$ = - \sin 1 + \cos \,1$$
$$\therefore$$ $$f$$ is differentiable at $$x=0$$
$$Rf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{h\,\sin {1 \over h} - 0} \over h} = \mathop {\lim }\limits_{h \to 0} \,\,\sin {1 \over h} = a$$ finite number
Let this finite number be $$l$$
$$L$$ $$f'(1)$$ $$ = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}$$
$$ = \mathop {\lim }\limits_{h \to 0} {{ - h\,\sin \left( {{1 \over { - h}}} \right)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} \,\,\sin \left( {{1 \over { - h}}} \right)$$
$$ = - \mathop {\lim }\limits_{h \to 0} \sin \left( {{1 \over h}} \right) = - (\,$$ a finite number $$\,)$$ $$=-l$$
Thus $$Rf'\left( 1 \right) \ne Lf'\left( 1 \right)$$
$$\therefore$$ $$f$$ is not differentiable at $$x=1$$
Also, $$f'\left( 0 \right) = \sin {1 \over {\left( {x - 1} \right)}} - {{x - 1} \over {{{\left( {x - 1} \right)}^2}}}\cos {\left. {\left( {{1 \over {x - 1}}} \right)} \right]_{x = 0}}$$
$$ = - \sin 1 + \cos \,1$$
$$\therefore$$ $$f$$ is differentiable at $$x=0$$
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