MCQ
If $f(x)\, = {x^2} - x + 5,\,\,x > \frac{1}{2},$ and $g(x)$ is its inverse function, then $g'(7)$ equals
- A$-\frac {1}{3}$
- B$\frac {1}{13}$
- ✓$\frac {1}{3}$
- D$-\frac {1}{13}$
${x^2} - x + \frac{1}{4} - \frac{1}{4} + 5 = y$
${\left( {x - \frac{1}{2}} \right)^2} + \frac{{19}}{4} = y$
${\left( {x - \frac{1}{2}} \right)^2}\,\, = y - \frac{{19}}{4}$
$x - \frac{1}{2} = \pm \sqrt {y - \frac{{19}}{4}} $
$x = \frac{1}{2} \pm \sqrt {y - \frac{{19}}{4}} $
As $x > \frac{1}{2}$
$x = \frac{1}{2} + \sqrt {y - \frac{{19}}{4}} $
$g\left( x \right) = \frac{1}{2} + \sqrt {x - \frac{{19}}{4}} $
$g'\left( x \right) = \frac{1}{{2\sqrt {x - \frac{{19}}{4}} }}$
$g'\left( 7 \right) = \frac{1}{{2\sqrt {7 - \frac{{19}}{4}} }} = \frac{1}{{2\frac{{\sqrt {28 - 19} }}{2}}} = \frac{1}{3}$
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$(A)$ $\arg (-1- i )=\frac{\pi}{4}$, where $i =\sqrt{-1}$
$(B)$ The function $f: R \rightarrow(-\pi, \pi]$, defined by $f(t)=\arg (-1+i t)$ for all $t \in R$, is continuous at all points of $R$, where $i=\sqrt{-1}$
$(C)$ For any two non-zero complex numbers $z_1$ and $z_2$, $\arg \left(\left(\frac{z_1}{z_2}\right)-\arg \left(z_1\right)+\arg \left(z_2\right)\right.$ is an integer multiple of $2 \pi$.
$(D)$ For any three given distinct complex numbers, $z_1, z_2$ and $z_3$, the locus of the point $z$ satisfying the condition $\arg \left(\frac{\left( z - z _1\right)\left( z _2- z _3\right)}{\left( z - z _3\right)\left( z _2- z _1\right)}\right)=\pi$, lies on a straight line