Question

If the function
$f(x)= \begin{cases} \frac{\log _e\left(1-x+x^2\right)+\log_e\left(1+x+x^2\right)}{\sec x-\cos x}, x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\} \\k \,\,\,\,\, x=0\end{cases}$
is continuous at $x=0$, then $k$ is equal to :

• A. 1
• B. -1
• C. e
• D. 0

Answer 1

Answer: (A)

$ \displaystyle\lim _{x \rightarrow 0} \frac{\left(\ln \left(1+x^2+x^4\right)\right) \cos x}{1-\cos ^2 x} $
$ \displaystyle\lim _{x \rightarrow 0} \frac{\left(\frac{\ln \left(1+x^2+x^4\right)}{x^2+x^4}\right) x^2\left(1+x^2\right) \cos x}{\left(\frac{\sin ^2 x}{x^2}\right) x^2}=1 $
$ \therefore k=1$