MCQ
$4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}} = . . ..$
- A$\pi $
- B$\frac{\pi }{2}$
- C$\frac{\pi }{3}$
- ✓$\frac{\pi }{4}$
$\therefore$ $4{\tan ^{ - 1}}\frac{1}{5} = 2\,\left[ {2{{\tan }^{ - 1}}\frac{1}{5}} \right] = 2{\tan ^{ - 1}}\frac{{\frac{2}{5}}}{{1 - \frac{1}{{25}}}}$
$ = 2{\tan ^{ - 1}}\frac{{10}}{{24}} = {\tan ^{ - 1}}\frac{{\frac{{20}}{{24}}}}{{1 - \frac{{100}}{{576}}}} = {\tan ^{ - 1}}\frac{{120}}{{119}}$
So, $4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}} = {\tan ^{ - 1}}\frac{{120}}{{119}} - {\tan ^{ - 1}}\frac{1}{{239}}$
$ = {\tan ^{ - 1}}\frac{{\frac{{120}}{{119}} - \frac{1}{{239}}}}{{1 + \frac{{120}}{{119}}.\frac{1}{{239}}}} = {\tan ^{ - 1}}\frac{{(120 \times 239) - 119}}{{(119 \times 239) + 120}}$
==> ${\tan ^{ - 1}}1 = \frac{\pi }{4}$.
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$f(\mathrm{x})= -\frac{4}{3} x^{3}+2 x^{2}+3 x ,\quad x>0$
$\quad\quad\quad\quad 3 x e^{x}, \quad\quad\quad\quad\quad\quad\mathrm{x} \leq 0$
તો $\mathrm{f}$ એ . . . . અંતરાલમાં વધે છે .
$(A)$ $ \equiv \frac{{x + 1}}{1} = \frac{{y - 2}}{{ - 2}} = \frac{{z - 0}}{1}$
$(B)$ $ \equiv \frac{x}{1} = \frac{y}{{ - 2}} = \frac{{z - 1}}{1}$
$(C)$ $ \frac{{x + 1/2}}{1} = \frac{{y - 1}}{{ - 2}} = \frac{{z - 1/2}}{1}$
વિધાન -$2$ : જો બે રેખાઓ માંથી સમતલ પસાર ન થાય તો તે બે રેખાઓ વિસમિત રેખાઓ છે