TS EAMCET · Maths · Continuity and Differentiability
If the function \(f(x)=\left\{\begin{array}{cc}\frac{\left(e^{k x}-1\right) \sin k x}{4 \tan x}, & x \neq 0 \ P, & x=0\end{array}\right.\) is differentiable at \(x=0\), then
- A \(P=0, f^{\prime}(0)=\frac{k^2}{4}\)
- B \(P=0, f^{\prime}(0)=-\frac{1}{2}\)
- C \(P=k, f^{\prime}(0)=-\frac{k^2}{4}\)
- D \(P=k, f^{\prime}(0)=-\frac{1}{4}\)
Answer & Solution
Correct Answer
(A) \(P=0, f^{\prime}(0)=\frac{k^2}{4}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \quad f(x)=\left\{\begin{array}{c}\frac{\left(e^{k x}-1\right) \sin k x}{4 \tan x} \\ \mathrm{P}\end{array}, x \neq 0\right. \\ & \text { is differentiable ait } x=0, x=0 \\ & \text { So, } f(x) \text { is continuous at } x=0 \\ & \Rightarrow \lim _{x…
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