Question

The value of $k$ for which the function $f\left(x\right)=\frac{{(e^x-1)}^4}{\sin \left(\frac{x^2}{k^2}\right)\log \{1+\left(\frac{x^2}{2}\right)\}},x\ne 0;f\left(0\right)=8$ may be continuous function is

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (C), (D)

The correct options are
C 2
D -2

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

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

$\Rightarrow 2k^2=8\Rightarrow k=\pm 2$