Question

If $f\left(x\right)=\{\begin{array}{c}\frac{(1-\cos px)(2^x-1)}{(\sqrt{1+x^2}-1)\sin x};x\ne 0\\ \frac{1}{2};x=0\end{array}$
is continuous, then the value of $e^{\frac{p^2+1}{p^2}}$ is

• A. $2e$
• B. $4e$
• C. $8e$
• D. $e$

Answer 1

The correct option is B $4e$

$f\left(x\right)$ is continuous.
So at $x=0,$
$\underset{x\to 0}{lim}\frac{(1-\cos px)(2^x-1)}{(\sqrt{1+x^2}-1)\sin x}=\frac{1}{2}$

$\Rightarrow \underset{x\to 0}{lim}\frac{\left(2{\sin }^2\frac{px}{2}\right)(2^x-1)}{(\sqrt{1+x^2}-1)\sin x}=\frac{1}{2}$

$\Rightarrow \underset{x\to 0}{lim}\frac{\frac{\left(2{\sin }^2\frac{px}{2}\right)}{{\left(\frac{px}{2}\right)}^2}\cdot {\left(\frac{px}{2}\right)}^2\cdot \frac{(2^x-1)}{x}.x.(\sqrt{1+x^2}+1)}{(\sqrt{1+x^2}-1)(\sqrt{1+x^2}+1)\cdot \frac{\sin x}{x}.x}=\frac{1}{2}$

$\Rightarrow \underset{x\to 0}{lim}\frac{2\times \frac{p^2{x}^2}{4}\times \ln 2\times x\times 2}{(1+x^2-1)\times 1\times x}=\frac{1}{2}\Rightarrow p^2\ln 2=\frac{1}{2}\Rightarrow \frac{1}{p^2}=\ln 4\Rightarrow 1+\frac{1}{p^2}=\ln 4+1=\ln 4e$

Hence, the value of $e^{\frac{p^2+1}{p^2}}=4e$