The value of k   ( k > 0 ) , for which the function f x = e x - 1 4 sin x 2 k 2 log 1 + x 2 2 ,…

Here, we can know for continuity, $\underset{x\to 0}{\lim }f\left(x\right)=8$

$\Rightarrow \underset{x\to 0}{\lim }\frac{{\left(e^x-1\right)}^4}{\sin \left(\frac{x^2}{k^2}\right)\log \left(1+\frac{x^2}{2}\right)}=8$

Now, $\underset{x\to 0}{\lim }\frac{{\left(e^x-1\right)}^4}{\sin \left(\frac{x^2}{k^2}\right)\log \left(1+\frac{x^2}{2}\right)}=\underset{x\to 0}{\lim }\frac{{\left({\displaystyle \frac{e^x-1}{x}}\right)}^4}{{\displaystyle \frac{\sin \left(\frac{x^2}{k^2}\right)}{x^2}}\times {\displaystyle \frac{\log \left(1+\frac{x^2}{2}\right)}{x^2}}}$ (dividing $x^4$, to the numerator and denominator)

$=\underset{x\to 0}{\lim }\frac{{\left({\displaystyle \frac{e^x-1}{x}}\right)}^4}{{\displaystyle \frac{\sin \left(\frac{x^2}{k^2}\right)}{k\times {\displaystyle \frac{x^2}{k^2}}}\times \frac{\log \left(1+\frac{x^2}{2}\right)}{{\displaystyle \frac{x^2}{2}}\times 2}}}=\underset{x\to 0}{\lim }\frac{{\left({\displaystyle \frac{e^x-1}{x}}\right)}^4}{{\displaystyle \frac{\sin \left(\frac{x^2}{k^2}\right)}{k^2\times {\displaystyle \frac{x^2}{k^2}}}\times \frac{\log \left(1+\frac{x^2}{2}\right)}{{\displaystyle \frac{x^2}{2}\times 2}}}}=\frac{1}{{\displaystyle \frac{1}{k^2}}\times {\displaystyle \frac{1}{2}}}=2k^2$

So, $2k^2=8 \Rightarrow k=2$ (ignoring the negative value)