MCQ
If $\text{f(x)}=\log_{\text{x}^2}(\log\text{x}),$ the $f(x)$ at $x = e$ is:
  • A
    $0$
  • B
    $1$
  • C
    $\frac{1}{\text{e}}$
  • $\frac{1}{2\text{e}}$

Answer

Correct option: D.
$\frac{1}{2\text{e}}$
We have, $\text{f(x)}=\log_{\text{x}^2}(\log\text{x})$
$\Rightarrow\text{f(x)}=\frac{\log(\log\text{x})}{\log\text{x}^2}$
$\Rightarrow\text{f(x)}=\frac{\log(\log\text{x})}{2\log\text{x}}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\frac{\text{d}}{\text{dx}}\bigg\{\frac{\log(\log\text{x})}{\log\text{x}}\bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\log\text{x}}\times\frac{1}{\text{x}}\times\log\text{x}-\frac{\log(\log)\text{x}}{\text{x}}}{(\log\text{x})^2}\Bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\text{x}}-\frac{\log(\log)\text{x}}{\text{x}}}{(\log\text{x})^2}\Bigg\}$
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\Bigg\{\frac{\frac{1}{\text{e}}-\frac{\log(\log)\text{e}}{\text{e}}}{(\log\text{e})^2}\Bigg\}$
$[$Putting $x = e]$
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\bigg\{\frac{\frac{1}{\text{e}}}{1}\bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2\text{e}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free