- $0$
- $1$
- $\frac{1}{\text{e}}$
- $\frac{1}{2\text{e}}$
Solution:
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]
$\Rightarrow\text{f}'\text{(e)}=\frac{1}{2}\times\bigg\{\frac{\frac{1}{\text{e}}}{1}\bigg\}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{2\text{e}}$
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$-x+2 y+5 z=b_1$
$2 x-4 y+3 z=b_2$
$x-2 y+2 z=b_3$
has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each$\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ $\in$ $S$ ?
$(A)$ $x+2 y+3 z=b_1, 4 y+5 z=b_2$ and $x+2 y+6 z=b_3$
$(B)$ $x+y+3 z=b_1, 5 x+2 y+6 z=b_2$ and $-2 x-y-3 z=b_3$
$(C)$ $-x+2 y-5 z=b_1, 2 x-4 y+10 z=b_2$ and $x-2 y+5 z=b_3$
$(D)$ $x+2 y+5 z=b_1, 2 x+3 z=b_2$ and $x+4 y-5 z=b_3$
If
$3\tan^{-1}\text{x}+\cot^{-1}\text{x}=\pi,$ then x equals:$0$
$1$
$-1$
$\frac{1}{2}$