Question

If $f\left(x\right)=\{\begin{array}{cc}\log ({\sec }^{2}x{)}^{{\cot }^{2}x},& \text{for}x\ne 0\\ K,& x=0\end{array}$
is continuous at $x=0$ then $K$ is

• A. $e^{-1}$
• B. $1$
• C. $e$
• D. $0$

Answer 1

The correct option is B $1$

Since, $f$ is continuous at $x=0$

$\therefore f\left(0\right)=\underset{x\to 0}{lim}\log ({\sec }^{2}x{)}^{{\cot }^{2}x}$

Therefore we have,

$K=\underset{x\to 0}{lim}{\cot }^{2}x.\log \left({\sec }^{2}x\right)$

$=\underset{x\to 0}{lim}{\cot }^{2}x.\log (1+{\tan }^{2}x)$

$=\underset{x\to 0}{lim}\frac{\log (1+{\tan }^{2}x)}{{\tan }^{2}x}$

$\therefore K=1$.......$\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1$