Question

Let \( g(x) \) be a linear function and \[ f(x) = \begin{cases} g(x), & x \leq 0 \\ \left( \frac{1 + x}{2 + x} \right)^{\frac{1}{x}}, & x > 0 \end{cases} \] is continuous at \( x = 0 \). If \( f'(1) = f(-1) \), then the value of \( g(3) \) is

• A. \( \frac{1}{3} \log_e \left( \frac{4}{9e^{1/3}} \right) \)
• B. \( \frac{1}{3} \log_e \left( \frac{4}{9} \right) + 1 \)
• C. \( \log_e \left( \frac{4}{9} \right) - 1 \)
• D. \( \log_e \left( \frac{4}{9e^{1/3}} \right) \)

Answer 1

The Correct Option is D

Let \( g(x) = ax + b \).

Now function \( f(x) \) is continuous at \( x = 0 \).

\[ \therefore \lim_{x \to 0} f(x) = f(0) \] \[ \lim_{x \to 0} \left( 1 + x \over 2 + x \right)^{1 \over x} = b \] \[ \Rightarrow 0 = b \] \[ \therefore g(x) = ax \]

Now, for \( x > 0 \),

\[ f'(x) = \frac{1}{x} \left( 1 + x \over 2 + x \right)^{1 \over x} \cdot \frac{1}{(2 + x)^2} + \left( 1 + x \over 2 + x \right)^{1 \over x} \cdot \ln \left( 1 + x \over 2 + x \right) \cdot \frac{1}{x^2} \] \[ f'(1) = \frac{1}{9} - \frac{2}{3} \ln \left( \frac{2}{3} \right) \]

And \( f(-1) = g(-1) = -a \)

\[ a = 2 \ln \left( \frac{2}{3} \right) - \frac{1}{9} \] \[ g(3) = 2 \ln \left( \frac{2}{3} \right) - \frac{1}{3} \] \[ = \ln \left( \frac{4}{9 e^{-1/3}} \right) \]