Verified Solution • 2025–26 Session moderate

QJEEM20Ma80bf: JEE Main 2020 Mathematics - Class 12 JEEM20Ma80bf Solution: Let y = y(x) be a solution of the differential equation, ,

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

1−x2dydx+1−y2=0\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0 , |x| < 1.

If y(12)=32y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2} , then y(−12)y\left( { - {1 \over {\sqrt 2 }}} \right) is equal to :

Multiple Choice Options

  • A −32- {{\sqrt 3 } \over 2}
  • B None of those
  • C 12{{1 \over {\sqrt 2 }}}
  • D −12-{{1 \over {\sqrt 2 }}}

Step-by-Step Verified Solution

Given 1−x2dydx+1−y2=0\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0

⇒\Rightarrow dydx=−1−y21−x2{{dy} \over {dx}} = - {{\sqrt {1 - {y^2}} } \over {\sqrt {1 - {x^2}} }}

⇒\Rightarrow dy1−y2+dx1−x2=0{{dy} \over {\sqrt {1 - {y^2}} }} + {{dx} \over {\sqrt {1 - {x^2}} }} = 0

⇒\Rightarrow sin-1 y + sin-1 x = c

Given that y(12)=32y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}

So when x = 12{1 \over 2} then y = 32{{\sqrt 3 } \over 2}

∴\therefore sin-1 32{{\sqrt 3 } \over 2} + sin-1 12{1 \over 2} = c

⇒\Rightarrow c = π3{\pi \over 3} + π6{\pi \over 6} = π2{\pi \over 2}

So sin-1 y + sin-1 x = π2{\pi \over 2}

⇒\Rightarrow sin-1 y = π2{\pi \over 2} - sin-1 x

⇒\Rightarrow sin-1 y = cos-1 x

Now y(−12)y\left( { - {1 \over {\sqrt 2 }}} \right) means x = −12- {1 \over {\sqrt 2 }} and find y.

Putting x = −12- {1 \over {\sqrt 2 }}

sin-1 y = cos-1 (−12)\left( { - {1 \over {\sqrt 2 }}} \right) = 3π4{{3\pi } \over 4}

y (at x = −12- {1 \over {\sqrt 2 }} ) = sin( 3π4{{3\pi } \over 4} ) = 12{{1 \over {\sqrt 2 }}}

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