QJEEAdv08Ma8e4d: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Ma8e4d Solution: If then

We have

${y^3} - 3y + x = 0$

Differentiate both sides we get

$3{y^2}.y' - 3y' + 1 = 0$ ..... (i)

Put $y = 2\sqrt 2 ,x = - 10\sqrt 2$ then

$y'( - 10\sqrt 2 ) = {{ - 1} \over {21}}$

Differentiate eq. (i), we get

$3{y^2}y'' + 6y{(y')^2} - 3y'' = 0$

Put $y = 2\sqrt 2 ,x = - 10\sqrt 2 ,y' = {{ - 1} \over {21}}$ then

$y''( - 10\sqrt 2 ) = {{ - 4\sqrt 2 } \over {{7^3}\,.\,{3^2}}}$