JEE MAIN - Mathematics (2016 (Offline) - No. 2)
For $$x \in \,R,\,\,f\left( x \right) = \left| {\log 2 - \sin x} \right|\,\,$$
and $$\,\,g\left( x \right) = f\left( {f\left( x \right)} \right),\,\,$$ then :
and $$\,\,g\left( x \right) = f\left( {f\left( x \right)} \right),\,\,$$ then :
$$g$$ is not differentiable at $$x=0$$
$$g'\left( 0 \right) = \cos \left( {\log 2} \right)$$
$$g'\left( 0 \right) = - \cos \left( {\log 2} \right)$$
$$g$$ is differentiable at $$x=0$$ and $$g'\left( 0 \right) = - \sin \left( {\log 2} \right)$$
Explanation
$$g\left( x \right) = f\left( {f\left( x \right)} \right)$$
In the neighbourhood of $$x=0,$$
$$f\left( x \right) = \left| {\log 2 - \sin \,x} \right| = \left( {\log 2 - \sin x} \right)$$
$$\therefore$$ $$g\left( x \right) = \left| {\log 2 - \sin \left. {\left| {\log 2 - \sin x} \right|} \right|} \right.$$
$$ = \left( {\log 2 - \sin \left( {\log 2 - \sin x} \right)} \right)$$
$$\therefore$$ $$g(x)$$ is differentiable
and $$g'\left( x \right) = - \cos \left( {\log 2 - \sin x} \right)\left( { - \cos x} \right)$$
$$ \Rightarrow g'\left( 0 \right) = \cos \left( {\log 2} \right)$$
In the neighbourhood of $$x=0,$$
$$f\left( x \right) = \left| {\log 2 - \sin \,x} \right| = \left( {\log 2 - \sin x} \right)$$
$$\therefore$$ $$g\left( x \right) = \left| {\log 2 - \sin \left. {\left| {\log 2 - \sin x} \right|} \right|} \right.$$
$$ = \left( {\log 2 - \sin \left( {\log 2 - \sin x} \right)} \right)$$
$$\therefore$$ $$g(x)$$ is differentiable
and $$g'\left( x \right) = - \cos \left( {\log 2 - \sin x} \right)\left( { - \cos x} \right)$$
$$ \Rightarrow g'\left( 0 \right) = \cos \left( {\log 2} \right)$$
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