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JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 6)

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 $$g(0)=0$$
$$\int\limits_{ - 1}^1 {g'\left( x \right)dx = } $$
$$2g(-1)$$
$$0$$
$$-2g(1)$$
$$2g(1)$$

Explanation

$$y' = {1 \over {3[1 - f{{(x)}^2}]}}$$

Clearly $$f(x)$$ is an odd function then $$g'(x)$$ is an even function

So,

$$\int\limits_{ - 1}^1 {g'(x) = 2\int\limits_0^1 {g'(x)dx} } $$

$$ = 2[g(x)]_0^1 = 2[g(1) - g(0)]$$

$$ = 2g(1)$$

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