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 \[x=\pi \].You need to login to perform this action.
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