QJEEM18Ma390a: JEE Main 2018 Mathematics - Class 12 JEEM18Ma390a Solution: Let g(x) = cosx 2 , f(x) = and be the roots of the quadratic…

Question

Verified Solution • 2025–26 Session moderate

Let g(x) = cosx², f(x) =

\sqrt x

and

\alpha ,\beta \left( {\alpha &lt; \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 :

Accepted Answer

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 &lt; 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

JEE Main 2018 (Offline) Mathematics - Area Under The Curves Question 127 English Explanation

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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