MCQ
વિધેય ${{x - 2} \over {x + 1}},(x \ne - 1)$ એ . . .. અંતરાલમાં વધતું છે.
- A$( - \infty ,\,\,\,0]$
- B$[0, \infty )$
- ✓$R$
- Dએકપણ નહીં.
therefore $f(x) = \frac{{x - 2}}{{x + 1}}$ is increasing in interval
$( - \infty ,\,\infty )$ or $R.$
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$\sin \left(2 x^{2}\right) \log _{e}\left(\tan x^{2}\right) d y+\left(4 x y-4 \sqrt{2} x \sin \left(x^{2}-\frac{\pi}{4}\right)\right) d x=0$,$0 < x < \sqrt{\frac{\pi}{2}}$ નો ઉકેલ વક્ર $y=y(x)$ છે. જે બિંદુ $\left(\sqrt{\frac{\pi}{6}}, 1\right)$ માંથી પસાર થાય છે. તો $\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|=$ ..............
$A\left[ {\begin{array}{*{20}{c}}
1&2&3 \\
0&2&3 \\
0&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0&1 \\
1&0&0 \\
0&1&0
\end{array}} \right]$
તો $A^{-1}$ મેળવો.