JEE MAIN - Mathematics (2016 (Offline) - No. 6)
If a curve $$y=f(x)$$ passes through the point $$(1,-1)$$ and satisfies the differential equation, $$y(1+xy) dx=x$$ $$dy$$, then $$f\left( { - {1 \over 2}} \right)$$ is equal to :
$${2 \over 5}$$
$${4 \over 5}$$
$$-{2 \over 5}$$
$$-{4 \over 5}$$
Explanation
$$y\left( {1 + xy} \right)dx = xdy$$
$${{xdy - ydx} \over {{y^2}}} = xdx \Rightarrow \int { - d\left( {{x \over y}} \right) = \int {xdx} } $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ - {x \over y} = {{{x^2}} \over 2} + C\,\,$$
as $$\,\,\,y\left( 1 \right) = - 1 \Rightarrow C = {1 \over 2}$$
Hence, $$y = {{ - 2x} \over {{x^2} + 1}} \Rightarrow f\left( {{{ - 1} \over 2}} \right) = {4 \over 5}$$
$${{xdy - ydx} \over {{y^2}}} = xdx \Rightarrow \int { - d\left( {{x \over y}} \right) = \int {xdx} } $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ - {x \over y} = {{{x^2}} \over 2} + C\,\,$$
as $$\,\,\,y\left( 1 \right) = - 1 \Rightarrow C = {1 \over 2}$$
Hence, $$y = {{ - 2x} \over {{x^2} + 1}} \Rightarrow f\left( {{{ - 1} \over 2}} \right) = {4 \over 5}$$
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