question_answer

Direction: For the following questions. Choose the correct answer from the codes [a], [b], [c] and [d] defined as follows.

For each of the following questions, one out of given options is correct.

Let consider \[f(x)=\sin \,x\]and \[g(x)=\,|\,f(x)|\].

Statement I Then, the function \[f(x)\,g(x)\] is not differentiable in \[[0,2\pi ]\].

Statement II \[f(x)\] is differentiable and \[g(x)\]is not differentiable in \[[0,\,2\pi ]\].

A)  Statement I is true, Statement II is also true and Statement II is the correct explanation of Statement I.

B)  Statement I is true. Statement II is also true and Statement II is not the correct explanation of Statement I.

C)  Statement I is true, Statement II is false.

D)  Statement I is false. Statement II is true.

Correct Answer: D

Solution :

 \[\because \] \[f(x)=\sin \,x\] is differentiable in \[[0,\,\,2\pi ]\] and \[g(x)=\,|sin\,x|\] is not differentiable at \[x=\pi \]. Let \[h(x)=f(x)\,g(x)=|\sin \,x|\,\sin \,x\] \[\therefore \]    \[h'(x)=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{|\sin \,x|\,\sin \,x}{x-\pi }\]            \[(\text{let}\,x-\pi =h)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{|\sin \,(\pi +h)\,\sin \,(\pi +h)}{h}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,|\sin \,h|\,\frac{(-\sin \,h)}{h}=0\] Hence, \[f(x)\] is differentiable but \[g(x)\] is not differentiable at \[x=\pi \].