The value of a for which the function f(x)= (4^x-1)^3 ( x a ) (1+… - Vidyadip

MCQ

The value of a for which the function $\text{f(x)}=\begin{cases}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}},&\text{x}\neq0\\12(\log4)^3,&\text{x}=0\end{cases}$ may be continuous at $x = 0$ is :

A
$1$
B
$2$
C
$3$
✓
None of these.

✓

Answer

Correct option: D.

None of these.

$\lim\limits_{\text{x}\rightarrow0}\frac{(4^\text{x}-1)^3}{\sin\Big(\frac{\text{x}}{\text{a}}\Big)\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^2}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\frac{(4^\text{x}-1)^3}{\text{x}^3}}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\text{x}}}\times\frac{1}{\frac{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}{\text{x}^2}}=12(\log4)^3$
$\lim\limits_{\text{x}\rightarrow0}\frac{\Big(\frac{(4^\text{x}-1}{\text{x}^3}\Big)}{\frac{\sin\Big(\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}}\text{a}\text{x}\ {\times}\frac{\frac{1}{\log\Big\{\Big(1+\frac{\text{x}^2}{3}\Big)\Big\}}}{\frac{\text{x}^3}{3}}\text{x}^3=12(\log4)^3$
$3(\log4)^3=12(\log4)^3$
$3\text{a}=12$
$\text{a}=12$
Note : The question is incorrect, so it has been modified.

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