MCQ
${d \over {dx}}\left( {{{\tan }^{ - 1}}{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$ is equal to
- A${1 \over {1 + {x^2}}}$
- ✓${1 \over {2(1 + {x^2})}}$
- C${{{x^2}} \over {2\sqrt {1 + {x^2}} (\sqrt {1 + {x^2}} - 1)}}$
- D${2 \over {1 + {x^2}}}$
Put $x = \tan \theta ,$ then
$y = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}} \right)$
$y = {\tan ^{ - 1}}\left( {\frac{{\sec \theta - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)$
$y = {\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right) = {\tan ^{ - 1}}\tan \frac{\theta }{2}$
$y = \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$, $(\because \theta ={{\tan }^{-1}}x)$
Hence $\frac{{dy}}{{dx}} = \frac{1}{{2(1 + {x^2})}}$.
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$f(x)=\left\{\begin{array}{rc}x^5+5 x^4+10 x^3+10 x^2+3 x+1, & x<0 \\ x^2-x+1, & 0 \leq x<1 \\ \frac{2}{3} x^3-4 x^2+7 x-\frac{8}{3}, & 1 \leq x<3 \\ (x-2) \log _e(x-2)-x+\frac{10}{3}, & x \geq 3\end{array}\right.$
Then which of the following options is/are correct?
$(1)$ $f^{\prime}$ has a local maximum at $x =1$ $(2)$ $f$ is onto
$(3)$ $f$ is increasing on $(-\infty, 0)$ $(4)$ $f^{\prime}$ is $NOT$ differentiable at $x =1$
$(i)$ $f (x)$ is bounded on $a \le x \le b.$
$(ii)$ The equation $f (x) = 0$ has at least one solution in $a < x < b.$
$(iii)$ The maximum and minimum values of $f (x)$ on $a \le x \le b$ occur at points where $f ' (c) = 0$.
$(iv)$ There is at least one point $c$ with $a < c < b$ where $f ' (c) > 0$.
$(v)$ There is at least one point $d$ with $a < d < b$ where $f ' (c) < 0.$