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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$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$
$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and
$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.