MCQ
If $f(x) = 2x + {\cot ^{ - 1}}x + \log (\sqrt {1 + {x^2}} - x)$, then $f(x)$
- ✓Increases in $ [0 ,\infty )$
- BDecreases in $[0 , \infty $)
- CNeither increases nor decreases in $ (0 , \infty $)
- DNone of these
$\therefore f'(x) = 2 - \frac{1}{{1 - {x^2}}} + \frac{1}{{\sqrt {1 + {x^2}} - x}}\left( {\frac{x}{{\sqrt {1 - {x^2}} }} - 1} \right)$
$ = \frac{{1 + 2{x^2}}}{{1 + {x^2}}} - \frac{1}{{\sqrt {1 + {x^2}} }} = \frac{{1 + 2{x^2}}}{{1 + {x^2}}} - \frac{{\sqrt {(1 + {x^2})} }}{{1 + {x^2}}}$
$ = \frac{{{x^2} + \sqrt {1 + {x^2}} (\sqrt {1 + {x^2}} - 1)}}{{1 + {x^2}}} \ge 0$ for all $x$
Hence $ f(x) $ is an increasing function on $( - \infty ,\,\infty )$ and
in particular on $[0,\;\infty )$.
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Consider $f(x)=k e^x-x$ for all real $x$ where $k$ is a real constant.
$1.$ The line $\mathrm{y}=\mathrm{x}$ meets $\mathrm{y}=k e^{\mathrm{x}}$ for $\mathrm{k} \leq 0$ at
$(A)$ no point $(B)$ one point
$(C)$ two points $(D)$ more than two points
$2.$ The positive value of $\mathrm{k}$ for which $\mathrm{ke}^{\mathrm{x}}-\mathrm{x}=0$ has only one root is
$(A)$ $1 / \mathrm{e}$ $(B)$ $1$ $(C)$ e $(D)$ $\log _e 2$
$3.$ For $k>0$, the set of all values of $k$ for which $k e^x-x=0$ has two distinct roots is
$(A)$ $\left(0, \frac{1}{\mathrm{e}}\right)$ $(B)$ $\left(\frac{1}{\mathrm{e}}, 1\right)$ $(C)$ $\left(\frac{1}{e}, \infty\right)$ $(D)$ $(0,1)$
Give the answer question $1,2$ and $3.$