JEE Math Practice Question
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 :
- A.(1, 2)
- B.$$\left( {{1 \over 2},1} \right]$$
- C.(2, 3)
- D.$$\left( {0,{1 \over 2}} \right]$$
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