Limits, Continuity & Differentiability
Differential Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
If $f(x) = \begin{cases} \frac{x \cdot \ln(\cos x)}{\ln(1+x^2)} & x \neq 0 \\ 0 & x = 0 \end{cases}$ then:
f is continuous at x = 0
f is continuous at x = 0 but not differentiable at x = 0
f is differentiable at x = 0
f is not continuous at x = 0
Step-by-Step Solution
Key Concept: A function is differentiable at a point only if the right and left derivatives exist and are equal.
For the right derivative: $f'(0^+) = \lim_{h \to 0^+} \frac{h \ln(\cosh h)}{h \cdot \frac{\ln(1+h^2)}{h^2}} = \lim_{h \to 0^+} \frac{h^2 \ln(\cosh h)}{\ln(1+h^2)} = \lim_{h \to 0^+} \frac{-\sinh}{(\cosh) \times 2h} = \frac{1}{2}$. For the left derivative: $f'(0^-) = \lim_{h \to 0^-} \frac{-h \ln(\cosh h)}{-h \cdot \frac{\ln(1+h^2)}{h^2}} = -\frac{1}{2}$. Since the right and left derivatives are not equal, the function is not differentiable at $x=0$.
Correct Answer: 1,3