AP EAMCET · Maths · Continuity and Differentiability
If the function \(f(x)\), defined below, is continuous everywhere, then \(k\) equals \(f(x)=\left\{\begin{array}{cc}\frac{2^x-1}{\sqrt{1+x}-1}, & -1 \leq x < \infty \\ k, & x=0\end{array}\right.\)
- A \(\frac{1}{2} \log _e 2\)
- B \(\log _e 4\)
- C \(\log _e 8\)
- D \(\log _e 2\)
Answer & Solution
Correct Answer
(B) \(\log _e 4\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\left\{\begin{array}{ccc}\frac{2^x-1}{\sqrt{1+x}-1}, & -1 \leq x < \infty \\ k, & x=0\end{array}\right.\) is continuous everywhere. Since, \(f(x)\) is continuous everywhere \(\Rightarrow \quad f(x)\) will be continuous at \(x=0\)…
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