MCQ
The solution of $\frac{{dy}}{{dx}} = {2^{y - x}}$ is
- A${2^x} + {2^y} = c$
- B${2^x} - {2^y} = c$
- ✓$\frac{1}{{{2^x}}} - \frac{1}{{{2^y}}} = c$
- D$x + y = c$
Integrating both sides, $\int {\frac{{dy}}{{{2^y}}} = \int {\frac{{dx}}{{{2^x}}}} } $
$ - {2^{ - y}}\log 2 = - {2^{ - x}}\log 2 + {c_1}$
$\frac{{\log 2}}{{{2^x}}} - \frac{{\log 2}}{{{2^y}}} = {c_1}$; $\frac{1}{{{2^x}}} - \frac{1}{{{2^y}}} = \frac{{{c_1}}}{{\log 2}} = c$.
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$y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{1 / 9}+\ldots . .$ upto $\infty$ terms and $\frac{2+4+6+\ldots+2 \mathrm{y}}{3+6+9+\ldots+3 \mathrm{y}}=\frac{4}{\log _{10} \mathrm{x}}$, then the ordered pair $(x, y)$ is equal to :
Statement $-2:$ The number of solutions of the equation, $2\,cos^2\,\theta - 3\,sin\,\theta = 0$ in the interval $[0, \pi ]$ is two.