Question

Function $f\left(x\right)=\{\begin{array}{cc}({\log }_{2}2x{)}^{{\log }_{x}8};& x\ne 1\\ (k-1{)}^3;& x=1\end{array}$ is continuous at $x=1$, then $k=$ _______.

• A. $e+1$
• B. $e^{1/3}$
• C. $e^3$
• D. $e-1$

Answer 1

The correct option is A $e+1$

Given function is $f\left(x\right)=\{\begin{array}{cc}({\log }_{2}2x{)}^{{\log }_{x}8};& x\ne 1\\ (k-1{)}^3;& x=1\end{array}$

$f$ is continuous at $x=1$

$\therefore f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)$

Now,

$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}{e}^{\log f\left(x\right)}$
$=\underset{x\to 1}{lim}{e}^{\left({\log }_{2}2x\right){\log }_{x}8}$
$=\underset{x\to 1}{lim}{e}^{\left({\log }_{2}2x\right)\frac{{\log }_{2}8}{{\log }_{2}x}}$

$=e^3$
$\Rightarrow e^3=(k-1{)}^3$
$k=e+1$.