Question

If \[ f(x)= \begin{cases} \dfrac{x^2\log(\cos x)}{\log(1+x)}, & x\neq 0 \\ 0, & x=0 \end{cases} \] then at \(x=0\), \(f(x)\) is

• A. not continuous
• B. continuous but not differentiable
• C. differentiable
• D. not continuous, but differentiable

Hint:

For continuity and differentiability problems involving logarithmic and trigonometric functions, use standard expansions such as: \[ \log(1+x)\sim x,\qquad \log(\cos x)\sim -\frac{x^2}{2} \] near \(x=0\).

Answer 1

The Correct Option is C

Step 1: Check continuity at \(x=0\).
We evaluate the limit: \[ \lim_{x\to 0}\frac{x^2\log(\cos x)}{\log(1+x)} \] Using the standard approximations near \(x=0\), \[ \log(1+x)\sim x \] and \[ \cos x \sim 1-\frac{x^2}{2} \] Hence, \[ \log(\cos x) \sim \log\left(1-\frac{x^2}{2}\right) \sim -\frac{x^2}{2} \] Substituting these approximations, \[ \frac{x^2\log(\cos x)}{\log(1+x)} \sim \frac{x^2\left(-\dfrac{x^2}{2}\right)}{x} \] \[ = -\frac{x^3}{2} \] Therefore, \[ \lim_{x\to 0}\frac{x^2\log(\cos x)}{\log(1+x)}=0 \] Since \[ f(0)=0, \] the function is continuous at \(x=0\).

Step 2: Check differentiability at \(x=0\).
Using the definition of derivative, \[ f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x} \] Since \(f(0)=0\), \[ f'(0) = \lim_{x\to 0} \frac{1}{x} \cdot \frac{x^2\log(\cos x)}{\log(1+x)} \] \[ = \lim_{x\to 0} \frac{x\log(\cos x)}{\log(1+x)} \] Again using \[ \log(\cos x)\sim -\frac{x^2}{2} \] and \[ \log(1+x)\sim x, \] we get \[ \frac{x\log(\cos x)}{\log(1+x)} \sim \frac{x\left(-\dfrac{x^2}{2}\right)}{x} \] \[ = -\frac{x^2}{2} \] Thus, \[ f'(0)=0 \] Hence, \(f(x)\) is differentiable at \(x=0\).

Step 3: Final conclusion.
Therefore, the correct option is \[ \boxed{\text{Differentiable}} \]