JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(\mathrm{g}(\mathrm{x})\) be a linear function and \(f(x)=\left\{\begin{array}{cl}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{array}\right.\), is continuous at \(x=0\). If \(f^{\prime}(1)=f(-1)\), then the value of \(g(3)\) is
- A \(\frac{1}{3} \log _e\left(\frac{4}{9 e^{1 / 3}}\right)\)
- B \(\frac{1}{3} \log _e\left(\frac{4}{9}\right)+1\)
- C \(\log _e\left(\frac{4}{9}\right)-1\)
- D \(\log _e\left(\frac{4}{9 e^{1 / 3}}\right)\)
Answer & Solution
Correct Answer
(D) \(\log _e\left(\frac{4}{9 e^{1 / 3}}\right)\)
Step-by-step Solution
Detailed explanation
Let \(g(x)=a x+b\) Now function \(f(x)\) in continuous at \(x=0\) \(\therefore \lim _{x \rightarrow 0^{+}} f(x)=f(0)\) \(\Rightarrow \lim _{x \rightarrow 0}\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}=b\) \(\Rightarrow 0=b\) \(\therefore g(x)=a x\) Now, for \(\mathrm{x}>0\)…
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