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Let $f: R \rightarrow R$ be defined by
$f(x)=\left\{\begin{array}{ccc}\alpha+\frac{\sin [x]}{x}, & \text { if } & x>0 \\ 2, & \text { if } & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text { if } & x < 0\end{array}\right.$
equal to
Options:
$f(x)=\left\{\begin{array}{ccc}\alpha+\frac{\sin [x]}{x}, & \text { if } & x>0 \\ 2, & \text { if } & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text { if } & x < 0\end{array}\right.$
equal to
Solution:
1401 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{4}$
Given,
$f(x)=\log \left[e^x\left(\frac{x-2}{x+2}\right)^{3 / 4}\right]$
$\Rightarrow f(x)=x+\frac{3}{4}[\log (x-2)-\log (x+2)]$
On differentiating w.r.t. $x$, we get
$\begin{aligned} f^{\prime}(x) & =1+\frac{3}{4}\left(\frac{1}{x-2}-\frac{1}{x+2}\right) \\ & =1+\frac{3}{4}\left(\frac{4}{x^2-4}\right)=1+\frac{3}{x^2-4} \\ \Rightarrow f^{\prime}(0) & =1+\frac{3}{0-4}=\frac{1}{4}\end{aligned}$
$f(x)=\log \left[e^x\left(\frac{x-2}{x+2}\right)^{3 / 4}\right]$
$\Rightarrow f(x)=x+\frac{3}{4}[\log (x-2)-\log (x+2)]$
On differentiating w.r.t. $x$, we get
$\begin{aligned} f^{\prime}(x) & =1+\frac{3}{4}\left(\frac{1}{x-2}-\frac{1}{x+2}\right) \\ & =1+\frac{3}{4}\left(\frac{4}{x^2-4}\right)=1+\frac{3}{x^2-4} \\ \Rightarrow f^{\prime}(0) & =1+\frac{3}{0-4}=\frac{1}{4}\end{aligned}$
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