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. $e^{\frac{1}{3}}$
• C. $e^3$
• D. $e-1$

Answer 1

The correct option is A $e+1$

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

is continuous at $x=1$.
$\Rightarrow \underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$
$\Rightarrow \underset{x\to 1}{lim}({\log }_{2}2x{)}^{{\log }_{x}8}=(k-1{)}^3$

$\underset{x\to 1}{lim}({\log }_{2}2x{)}^{{\log }_{x}8}$
$=\underset{x\to 1}{lim}(1+{\log }_{2}x{)}^{1/{\log }_{8}x}[1^{\infty }\text{form}]$
$=\exp \left(\underset{x\to 1}{lim}\frac{{\log }_{2}x}{{\log }_{8}x}\right)$

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

$=\exp \left(\underset{x\to 1}{lim}{\log }_{2}8\right)$
$=e^3$

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