MCQ
If ${2^x} + {2^y} = {2^{x + y}}$, then ${{dy} \over {dx}} = $
- A${2^{x - y}}{{{2^y} - 1} \over {{2^x} - 1}}$
- ✓${2^{x - y}}{{{2^y} - 1} \over {1 - {2^x}}}$
- C${{{2^x} + {2^y}} \over {{2^x} - {2^y}}}$
- DNone of these
$ = {2^x}{.2^y}\frac{{dy}}{{dx}}.\log 2 + {2^y}{.2^x}\log 2$
==> ${2^x} + {2^y}\frac{{dy}}{{dx}} = {2^{x + y}}\frac{{dy}}{{dx}} + {2^{x + y}}$
==> $\frac{{dy}}{{dx}} = \frac{{{2^{x + y}} - {2^x}}}{{{2^y} - {2^{x + y}}}} = {2^{x - y}}\frac{{{2^y} - 1}}{{1 - {2^x}}}$.
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$x - cy - cz = 0 \,\,;\,\, cx - y + cz = 0 \,\,;\,\, cx + cy - z = 0 $ has a non -trivial solution, is
$\frac{\pi}{\sqrt{\text{a}^2-\text{b}^2}}$
$\frac{\pi}{\text{ab}}$
$\frac{\pi}{\text{a}^2-\text{b}^2}$
$({\text{a}+\text{b}})\pi$