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JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 8)

Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$
Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$
Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$
If $$f\left( { - 10\sqrt 2 } \right) = 2\sqrt 2 ,$$ then $$f''\left( { - 10\sqrt 2 } \right) = $$
$${{4\sqrt 2 } \over {{7^3}{3^2}}}$$
$$-{{4\sqrt 2 } \over {{7^3}{3^2}}}$$
$${{4\sqrt 2 } \over {{7^3}3}}$$
$$-{{4\sqrt 2 } \over {{7^3}3}}$$

Explanation

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

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

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