| Assertion [A]: \[f\left( x \right)=x\text{ }sin\left( \frac{1}{x} \right)\]is differentiable at x = 0. |
| Reason [R]: f(x) is continuous at x = 0. |
A) Both A and R are individually true and R is the correct explanation of A.
B) Both A and R are individually true and R is not the correct explanation of A.
C) 'A' is true but 'R' is false
D) 'A' is false but 'R' is true
E) Both A and R are false.
Correct Answer: D
Solution :
Given \[f\left( x \right)=x\text{ }sin\left( \frac{1}{x} \right)\] At \[x\,=\,0,\,\,L.H.D.\,=\underset{h\to 0}{\mathop{Lt}}\,\frac{f(0-h)-f(0)}{-h}\] \[=\underset{h\to 0}{\mathop{Lt}}\,\frac{(0-h)sin\left( \frac{1}{0-h} \right)-0}{-h}\] \[=\underset{h\to 0}{\mathop{Lt}}\,\frac{(-h)sin\left( -\frac{1}{h} \right)}{-h}=\underset{h\to 0}{\mathop{Lt}}\,-\sin \left( \frac{1}{h} \right)\] = -finite value say\[a=-a\]. R.H.D. =\[=\underset{h\to 0}{\mathop{Lt}}\,\frac{f(0+h)-f(0)}{h}\] \[=\underset{h\to 0}{\mathop{Lt}}\,\frac{(0+h)sin\left( \frac{1}{0+h} \right)-0}{h}\] \[=\underset{h\to 0}{\mathop{Lt}}\,\sin \left( \frac{1}{h} \right)\]= A finite value lies between -1 and 1 which cannot qualified exactly say a = a. Clearly\[LHD\ne RHD\ne a\]definite value \[\therefore \,\,f\left( x \right)\]is not differentiable at \[x\text{ }=\text{ }0\] Hence Given Assertion [A] is not true. Also LH Lt =\[\underset{x\to {{0}^{-}}}{\mathop{Lt}}\,f(x)\,=\,\underset{h\to 0}{\mathop{Lt}}\,(0-h)sin\left( \frac{1}{0-h} \right)\] \[=\underset{h\to 0}{\mathop{Lt}}\,(-h)sin\left( -\frac{1}{h} \right)=\underset{h\to 0}{\mathop{Lt}}\,h\sin \left( \frac{1}{h} \right)=0\] RH \[Lt=\underset{x\to 0}{\mathop{Lt}}\,f(x)=\underset{h\to 0}{\mathop{Lt}}\,(0+h)sin\left( \frac{1}{0+h} \right)\] \[=\underset{h\to 0}{\mathop{Lt}}\,h\sin \left( \frac{1}{h} \right)=0\] \[\therefore \] L.H. Lt = R.H. Lt \[\Rightarrow \,\,f\left( x \right)\]is continuous at \[x=0\] \[\Rightarrow \]Reason (R) is true \[\therefore \] Assertion [A] is false but Reason [R] is true Hence option [D] is the correct answer.
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