Question:

Let \( f(x) = x^2 \log(\cos x) \log(1 + x) \) for \( x \neq 0 \), and \( f(0) = 0 \). Determine the behavior of \( f(x) \) at \( x = 0 \).

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To check if a function is differentiable at a point, verify both its continuity and the existence of the derivative at that point.
Updated On: June 02, 2025
  • not continuous
  • continuous but not differentiable
  • differentiable
  • not continuous, but differentiable
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The Correct Option is C

Approach Solution - 1

Step 1: {Check for continuity at \( x = 0 \)}
We need to find the limit of \( f(x) \) as \( x \to 0 \): \[ \lim_{x \to 0} \frac{x^2 \log(\cos x)}{\log(1 + x)} = \lim_{x \to 0} x^2 \cdot \log(\cos x) = 0 \cdot \log(1) = 0 \] Thus, \( f(x) \) is continuous at \( x = 0 \). 
Step 2: {Check for differentiability at \( x = 0 \)}
We now check if \( f(x) \) is differentiable at \( x = 0 \) by finding the derivative at this point: \[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \log(\cos h)}{h \log(1 + h)} = 0 \] Thus, \( f(x) \) is differentiable at \( x = 0 \).

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Approach Solution -2

Step 1: Understanding the function \( f(x) \)
The function is defined as:
\[ f(x) = \begin{cases} x^2 \log(\cos x) \log(1 + x), & \text{for } x \neq 0, \\ 0, & \text{for } x = 0. \end{cases} \]
We are asked to determine the behavior of \( f(x) \) at \( x = 0 \). Specifically, we need to check whether the function is continuous and differentiable at \( x = 0 \).

Step 2: Check continuity at \( x = 0 \)
For continuity at \( x = 0 \), we need to check if the following limit exists:
\[ \lim_{x \to 0} f(x) = f(0). \] First, we compute the limit of \( f(x) = x^2 \log(\cos x) \log(1 + x) \) as \( x \to 0 \).
- As \( x \to 0 \), \( \cos x \to 1 \), so \( \log(\cos x) \to \log(1) = 0 \). - Also, \( \log(1 + x) \to \log(1) = 0 \) as \( x \to 0 \). Thus, both \( \log(\cos x) \) and \( \log(1 + x) \) tend to 0 as \( x \to 0 \). Therefore, \( f(x) \) tends to: \[ f(x) \approx x^2 \cdot 0 \cdot 0 = 0 \quad \text{as} \quad x \to 0. \] Since \( f(0) = 0 \), we have: \[ \lim_{x \to 0} f(x) = f(0) = 0. \] Therefore, \( f(x) \) is continuous at \( x = 0 \).

Step 3: Check differentiability at \( x = 0 \)
For differentiability at \( x = 0 \), we need to compute the derivative of \( f(x) \) at \( x = 0 \). The derivative is given by:
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}. \] Since \( f(0) = 0 \), this simplifies to:
\[ f'(0) = \lim_{h \to 0} \frac{f(h)}{h}. \] For \( x \neq 0 \), we have \( f(h) = h^2 \log(\cos h) \log(1 + h) \). So: \[ f'(0) = \lim_{h \to 0} \frac{h^2 \log(\cos h) \log(1 + h)}{h}. \] Simplifying this expression: \[ f'(0) = \lim_{h \to 0} h \log(\cos h) \log(1 + h). \] As \( h \to 0 \), \( \log(\cos h) \to 0 \) and \( \log(1 + h) \to 0 \). Therefore, the product \( h \log(\cos h) \log(1 + h) \to 0 \). Hence: \[ f'(0) = 0. \] Therefore, \( f(x) \) is differentiable at \( x = 0 \), and \( f'(0) = 0 \).

Step 4: Conclusion
Since \( f(x) \) is both continuous and differentiable at \( x = 0 \), the correct answer is:
differentiable
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