JEE MAIN - Mathematics (2018 (Offline) - No. 15)
Let g(x) = cosx2, f(x) = $$\sqrt x $$ and $$\alpha ,\beta \left( {\alpha < \beta } \right)$$ be the roots of the quadratic equation 18x2 - 9$$\pi $$x + $${\pi ^2}$$ = 0. Then the area (in sq. units) bounded by the curve
y = (gof)(x) and the lines $$x = \alpha $$, $$x = \beta $$ and y = 0 is :
y = (gof)(x) and the lines $$x = \alpha $$, $$x = \beta $$ and y = 0 is :
$${1 \over 2}\left( {\sqrt 2 - 1} \right)$$
$${1 \over 2}\left( {\sqrt 3 - 1} \right)$$
$${1 \over 2}\left( {\sqrt 3 + 1} \right)$$
$${1 \over 2}\left( {\sqrt 3 - \sqrt 2 } \right)$$
Explanation
Given quadratic equation,
$$18{x^2} - 9\pi x + {\pi ^2} = 0$$
$$ \Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0$$
$$ \Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0$$
$$ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0$$
$$\therefore$$ $$\,\,\,\,x = {\pi \over 3},{\pi \over 6}$$
as $$\,\,\,\, \propto < B$$
$$\therefore$$ $$\,\,\,\, \propto = {\pi \over 6}$$ $$\,\,\,\,$$ and $$\,\,\,\,$$ $$\beta = {\pi \over 3}$$
Given, $$g\left( x \right) = \cos {x^2}$$ and $$ + \left( x \right) = \sqrt x $$
$$y = \left( {gof} \right)x$$
$$ = \,\,\,\,\,g\left( {f\left( x \right)} \right)$$
$$ = \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)$$
$$ = \,\,\,\,\cos {\left( {\sqrt x} \right)^2}$$
$$ = \,\,\,\,\cos x$$
So, the required area in the curve is
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Area $$ = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx} $$
$$ = \left[ {\sin x} \right]_{{\pi \over 6}}^{{\pi \over 3}}$$
$$ = \sin {\pi \over 3} - \sin {\pi \over 6}$$
$$ = {{\sqrt 3} \over 2} - {1 \over 2}$$
$$ = {{\sqrt 3 - 1} \over 2}$$
$$18{x^2} - 9\pi x + {\pi ^2} = 0$$
$$ \Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0$$
$$ \Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0$$
$$ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0$$
$$\therefore$$ $$\,\,\,\,x = {\pi \over 3},{\pi \over 6}$$
as $$\,\,\,\, \propto < B$$
$$\therefore$$ $$\,\,\,\, \propto = {\pi \over 6}$$ $$\,\,\,\,$$ and $$\,\,\,\,$$ $$\beta = {\pi \over 3}$$
Given, $$g\left( x \right) = \cos {x^2}$$ and $$ + \left( x \right) = \sqrt x $$
$$y = \left( {gof} \right)x$$
$$ = \,\,\,\,\,g\left( {f\left( x \right)} \right)$$
$$ = \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)$$
$$ = \,\,\,\,\cos {\left( {\sqrt x} \right)^2}$$
$$ = \,\,\,\,\cos x$$
So, the required area in the curve is
Area $$ = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx} $$
$$ = \left[ {\sin x} \right]_{{\pi \over 6}}^{{\pi \over 3}}$$
$$ = \sin {\pi \over 3} - \sin {\pi \over 6}$$
$$ = {{\sqrt 3} \over 2} - {1 \over 2}$$
$$ = {{\sqrt 3 - 1} \over 2}$$
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