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  • . If      <img alt="\mathrm{f(x)=\left\{\begin{array}{cc} \frac{x \log \cos x}{\log \left(1+x^2\right)}, & x \neq 0 \\ 0, & x=0 \end{array}\right. \text { then }}" src="https

. If      \mathrm{f(x)=\left\{\begin{array}{cc} \frac{x \log \cos x}{\log \left(1+x^2\right)}, & x \neq 0 \\ 0, & x=0 \end{array}\right. \text { then }}

Option: 1

f(x) is not continuous at x=0


Option: 2

f(x) is continuous at x=1


Option: 3

f(x) is continuous at x=0 but not differentiable at x=0


Option: 4

f(x) is differentiable at x=0


Answers (1)

best_answer

We have,   \mathrm{\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{\log \cos x}{\log \left(1+x^2\right)}}

\mathrm{ =\lim _{x \rightarrow 0} \frac{\log (1-1+\cos x)}{\log \left(1+x^2\right)} \cdot \frac{1-\cos x}{1-\cos x} }
\mathrm{ =\lim _{x \rightarrow 0} \frac{\log |1-(1-\cos x)|}{1-\cos x} \cdot \frac{1-\cos x}{\log \left(1+x^2\right)}}

\mathrm{ =-\lim _{x \rightarrow 0} \log \frac{[1-(1-\cos x)]}{-(1-\cos x)} \cdot \frac{2 \sin ^2 \frac{x}{2}}{4\left(\frac{x}{2}\right)^2} \cdot \frac{x^2}{\log \left(1+x^2\right)}=-\frac{1}{2} . }
Hence, f(x) is differentiable at x=0 and hence continuous at x=0. at x=0.

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