JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 7)
If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval :
(1, 2)
$$\left( {{1 \over 2},1} \right]$$
(2, 3)
$$\left( {0,{1 \over 2}} \right]$$
Explanation
$${{dy} \over {dx}} = {{{2^x}(y + {2^y})} \over {{2^x}(1 + {2^y}\ln 2)}}$$
$$ \Rightarrow \int {{{(1 + {2^y})\ln 2} \over {(y + {2^y})}}dy = \int {dx} } $$
$$ \Rightarrow \ln \left| {y + {2^y}} \right| = x + c$$
x = 0; y = 0 $$\Rightarrow$$ c = 0
$$ \Rightarrow x = \ln \left| {y + {2^y}} \right|$$
$$\Rightarrow$$ at y = 1, x = ln3
$$\because$$ $$3 \in (e,{e^2}) \Rightarrow x \in (1,2)$$
$$ \Rightarrow \int {{{(1 + {2^y})\ln 2} \over {(y + {2^y})}}dy = \int {dx} } $$
$$ \Rightarrow \ln \left| {y + {2^y}} \right| = x + c$$
x = 0; y = 0 $$\Rightarrow$$ c = 0
$$ \Rightarrow x = \ln \left| {y + {2^y}} \right|$$
$$\Rightarrow$$ at y = 1, x = ln3
$$\because$$ $$3 \in (e,{e^2}) \Rightarrow x \in (1,2)$$
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