Let y = y(x) be the solution of the differential equation ${\rm{x}}\,\,{\rm{tan}}\left( {\frac{{\rm{y}}}{{\rm{x}}}} \right){\rm{dy = }}\left( {{\rm{y tan}}\left( {\frac{{\rm{y}}}{{\rm{x}}}} \right){\rm{ - x}}} \right){\rm{dx}}$, ${\rm{ - 1}} \le {\rm{x}} \le 1,{\rm{ y}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}$ Then, the area of the region bounded by the curves x = 0, ${\rm{x = }}\frac{1}{{\sqrt 2 }}$ and y = y(x) in the upper half plane is

$\frac{1}{{12}}\left( {\pi - 3} \right)$

$\frac{1}{4}\left( {\pi - 2} \right)$

$\frac{1}{6}\left( {\pi - 1} \right)$

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy

1 of 1

Application +4 | -1

Let y = y(x) be the solution of the differential equation

\mathrm{x}\,\,\mathrm{tan}\left( {\frac{\mathrm{y}}{\mathrm{x}}} \right)\mathrm{dy = }\left( {\mathrm{y tan}\left( {\frac{\mathrm{y}}{\mathrm{x}}} \right)\mathrm{ - x}} \right)\mathrm{dx}

\mathrm{ - 1} \le \mathrm{x} \le 1,\mathrm{ y}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}

Then, the area of the region bounded by the curves x = 0,

\mathrm{x = }\frac{1}{{\sqrt 2 }}

and y = y(x) in the upper half plane is

View answer and explanation

Correct option: A

\frac{1}{8}\left( {\pi - 1} \right)

Option A (

\frac{1}{8}\left( {\pi - 1} \right)

) is correct because solving the differential equation and integrating the resulting function over the given interval yields this area.