Question:
If
\[
f(x) =
\begin{cases}
\dfrac{x \log(\cos x)}{\log(1 + x^2)}, & x \ne 0 \\[2mm]
0, & x = 0
\end{cases}
\]
then \(f(x)\) is
If
\[ f(x) = \begin{cases} \dfrac{x \log(\cos x)}{\log(1 + x^2)}, & x \ne 0 \\[2mm] 0, & x = 0 \end{cases} \]
then \(f(x)\) is
\[ f(x) = \begin{cases} \dfrac{x \log(\cos x)}{\log(1 + x^2)}, & x \ne 0 \\[2mm] 0, & x = 0 \end{cases} \]
then \(f(x)\) is
Show Hint
Use series expansions to test continuity and differentiability at 0.
Updated On: Mar 23, 2026
- continuous as well as differentiable at x=0
- continuous but not differentiable at x=0
- differentiable but not continuous at x=0
- neither continuous nor differentiable at x=0
Show Solution
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The Correct Option is A
Solution and Explanation
Using series expansions:
\[ \log(\cos x) \sim -\frac{x^2}{2}, \quad \log(1 + x^2) \sim x^2 \]
Hence \[ f(x) \sim -\frac{x}{2} \to 0 \text{ as } x \to 0, \] so \(f\) is continuous.
Derivative at 0 also exists. Hence differentiable.
\[ \log(\cos x) \sim -\frac{x^2}{2}, \quad \log(1 + x^2) \sim x^2 \]
Hence \[ f(x) \sim -\frac{x}{2} \to 0 \text{ as } x \to 0, \] so \(f\) is continuous.
Derivative at 0 also exists. Hence differentiable.
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