Function f(x)= ll ( _2 2 x )^ _x 8 ; & x 1 (k-1)^3 ; & x=1 . is continuous at x=1, then k = __________

Function f(x)= ll ( _2 2 x )^ _x 8 ; & x 1 (k-1)^3 ; & x=1 . is c…

MCQ

Function $f(x)=\left\{\begin{array}{ll}\left(\log _2 2 x\right)^{\log _x 8} ; & x \neq 1 \\ (k-1)^3 ; & x=1\end{array}\right.$ is continuous at $x=1$, then $k =$ __________

✓
$e+1$
B
$e^{1 / 3}$
C
$e ^3$
D
$e-1$

✓

Answer

Correct option: A.

$e+1$

(A)
$\lim _{x \rightarrow 1}\left(\log _2 2 x\right)^{\log _x 8}$
$=\lim _{x \rightarrow 1}\left[\log _2 2+\log _2 x\right]^{\log _x 2^3}$
$=\lim _{x \rightarrow 1}\left[1+\log _2 x\right]^{3 \log _x 2}=\lim _{x \rightarrow 1}\left[1+\log _2 x\right]^{\frac{3}{\log _2 x}}$
$= e ^{\lim _{x \rightarrow 1} \log _2 x \times \frac{3}{\log _2 x}}= e ^3$
Since the function is continuous at $x=1$,
$\therefore \quad \lim _{x \rightarrow 1} f(x)=f(1)$
$\Rightarrow e ^3=( k -1)^3$
$\Rightarrow e = k -1$
$\Rightarrow k = e + l$

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