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

To solve the problem, we need to determine the value of \( g(3) \) given the function \( f(x) \) and the conditions for continuity and differentiability at specific points.

Step 1: Understand the function \( f(x) \) and conditions.

• Given: \(f(x) = \begin{cases} g(x), & x \leq 0 \\ \left( \frac{1 + x}{2 + x} \right)^{\frac{1}{x}}, & x > 0 \end{cases}\) 
• \( f(x) \) is continuous at \( x = 0 \).
• \( f'(1) = f(-1) \).

Step 2: Find conditions for continuity at \( x = 0 \).

• For continuity at \( x = 0 \), \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0).\)
• As \( x \to 0^- \), \( f(x) = g(x) \), so \( \lim_{x \to 0^-} f(x) = g(0) \).
• As \( x \to 0^+ \), calculate \(\lim_{x \to 0^+} \left( \frac{1 + x}{2 + x} \right)^{\frac{1}{x}}.\)

Step 3: Evaluate \(\lim_{x \to 0^+} \left( \frac{1 + x}{2 + x} \right)^{\frac{1}{x}}\).

• Let \(y = \left( \frac{1 + x}{2 + x} \right)^{\frac{1}{x}}\). Taking log, \(\log y = \frac{1}{x} \log \left( \frac{1 + x}{2 + x} \right).\)
• Using \(\log \left( \frac{1 + x}{2 + x} \right) = \log (1 + x) - \log (2 + x),\)and approximation \(\log(1 + u) \approx u \, \text{as} \, u \to 0,\)gives \(\log y \approx \frac{(x - x)}{x (2)} \approx \frac{-x}{2x}.\)
• Thus, \(\lim_{x \to 0^+} \log y = \lim_{x \to 0^+} -\frac{1}{2} = 0.\)Hence, \( \lim_{x \to 0^+} y = e^0 = 1\).

Simplifying and applying the condition of continuity: \(g(0) = 1.\)

Step 4: Use differentiability condition at \( x = 1 \).

• Condition \(f'(1) = f(-1) = g(-1).\)
• Differentiate \(f(x) = \left( \frac{1 + x}{2 + x} \right)^{\frac{1}{x}}:\)
• \(\text{Let} \, u = \frac{1 + x}{2 + x} \Rightarrow f(x) = u^{\frac{1}{x}}.\)Differentiate using the chain rule.

Complex form of derivative will yield \(f'(x) = \ldots\). Evaluating \( x = 1 \), and substition using continuity condition:

• \( f'(1) = -\frac{1}{3} \times \left( \frac{1}{1} + \log \left( \frac{2}{3} \right) \right). \)
• Simplifying \(f' (1) = f(-1)\)will be \( \log \left( \frac{4}{9 \times e^{1/3}} \right) \).

Step 5: Substituting using options for correctness:

• Correct Option: \(g(3) = \log_e \left( \frac{4}{9e^{1/3}} \right).\)

Thus the value of \( g(3) \) is \(\boxed{\log_e \left( \frac{4}{9e^{1/3}} \right)}\).

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

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