TS EAMCET · Maths · Continuity and Differentiability
If the function defined by \(f(x)=\left\{\begin{array}{cl}\frac{2^x-2^{-x}}{x}, & x \neq 0 \text { is continuous at } \ k, & x=0\end{array}\right.\) \(x=0\), then \(e^k\) is equal to
- A \(\log \left(\frac{2}{e}\right)\)
- B \(\log 4\)
- C 4
- D 1
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\left\{\begin{array}{cc}\frac{2^x-2^{-x}}{x}, & x \neq 0 \\ k, & x=0\end{array}\right.\) Given, \(f(x)\) is continuous at \(x=0\). Hence, \(\lim _{x \rightarrow 0} f(x)=k\)…
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