Given, g(x)=cosx2,f(x)=x
Given quadratic is 18x2−9πx+π2=0
The roots are x=2×189π±81π2−72π2
⇒x=369π±9π2
⇒x=36(9±3)π
⇒x=3π,6π
y=gof(x)=cos(x)2=cosx
Required Area =∫6π3πcosxdx=[sinx]6π3π
⇒23−21
JEE Main 2018 — Mathematics Calculus
Let g(x)=cosx2,f(x)=x, and α,β(α<β) be the roots of the quadratic equation 18x2−9πx+π2=0. Then the area (in sq. units) bounded by the curve y=(gof)(x) and the lines x=α,x=β and y=0, is
Held on 8 Apr 2018 · Verified 6 Jul 2026.
21(2−1)
21(3−1)
21(3+1)
21(3−2)
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