QJEEM22Ma1cd1: JEE Main 2022 Mathematics - Class 12 JEEM22Ma1cd1 Solution: For the curve , the value of , at the point , , on C, is equal…

$\because$ $C:({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$ for point ( $\alpha$ , $\alpha$ )

${\alpha ^2} + {\alpha ^2} - 3 + {({\alpha ^2} - {\alpha ^2} - 1)^5} = 0$

$\therefore$ $\alpha = \sqrt 2$

On differentiating $({x^2} + {y^2} - 3) + {({x^2} - {y^2} - 1)^5} = 0$ we get

$x + yy' + 5{({x^2} - {y^2} - 1)^4}(x - yy') = 0$ ...... (i)

When $x = y = \sqrt 2$ then $y' = {3 \over 2}$

Again on differentiating eq. (i) we get :

$1 + {(y')^2} + yy'' + 20({x^2} - {y^2} - 1)(2x - 2yy')(x - y'y) + 5{({x^2} - {y^2} - 1)^4}(1 - y{'^2} - yy'') = 0$

For $x = y = \sqrt 2$ and $y' = {3 \over 2}$ we get $y'' = - {{23} \over {4\sqrt 2 }}$

$\therefore$ $3y' - {y^3}y'' = 3\,.\,{3 \over 2} - {\left( {\sqrt 2 } \right)^3}\,.\,\left( { - {{23} \over {4\sqrt 2 }}} \right) = 16$