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

Answer: (D)

$\log_e \left( \frac{4}{9e^{1/3}} \right)$

Answer 2

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 $f(-1) = g(-1) = -a$

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