A) Both statement I and II are false.
B) Both statement I and II are true.
C) Statement I is true, statement II is false.
D) Statement I is false, statement II is true.
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{matrix} x\sin \left( \frac{1}{x} \right),x\ne 0 & x\ne 0 \\ 0, & x=0 \\ \end{matrix} \right.\]and \[g(x)=xf(x)\] LHL\[=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left\{ -h\sin \left( -\frac{1}{h} \right) \right\}\] = 0 × a finite quantity between ?1 and 1 = 0 \[RHL=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,h\sin \frac{1}{h}=0\] Also, x = 0 Thus LHL = RHL = f(0) \[\therefore \]f (x) is continuous at x = 0 \[g(x)=\left\{ \begin{matrix} {{x}^{2}}\sin \frac{1}{x}, & x\ne 0 \\ 0, & x=0 \\ \end{matrix} \right.\] For g(x) LHL \[=\underset{h\to {{o}^{-}}}{\mathop{\lim }}\,\left\{ -{{h}^{2}}\sin \left( \frac{1}{h} \right) \right\}\] \[={{0}^{2}}\times a\]a finite quantity between ?1 and 1 = 0 RHL\[=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,{{h}^{2}}\sin \left( \frac{1}{h} \right)=0\] Also g(0) = 0 \[\therefore \] (x) is continuous at x = 0
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