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JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 7)

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 $$g(0)=0$$
The area of the region bounded by the curve $$y=f(x),$$ the
$$x$$-axis, and the lines $$x=a$$ and $$x=b$$, where $$ - \infty < a < b < - 2,$$ is :
$$\int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx + bf\left( b \right) - af\left( a \right)$$
$$ - \int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx + bf\left( b \right) - af\left( a \right)$$
$$\int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx - bf\left( b \right) + af\left( a \right)$$
$$ - \int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx - bf\left( b \right) + af\left( a \right)$$

Explanation

Required area $$\int\limits_a^b {ydx = \int\limits_a^b {f(x)dx} } $$

$$ = [f(x).x]_a^b - \int\limits_a^b {f'(x)xdx} $$

$$ = bf(b) - af(a) - \int\limits_a^b {f'(x)xdx} $$

$$ = bf(b) - af(a) + \int\limits_a^b {{{xdx} \over {3[{{\{ f(x)\} }^2} - 1]}}} $$

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

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