Let g(x) = cos x2, f(x) = √x , and α , β (α

Question

Let g(x) = cos x2, f(x) = √x , and α , β (α < β) be the roots of the quadratic equation 18x2 - 9 π x + 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: (A) (1/2) ( √3 - 1) (B) (1/2) ( √3 + 1) (C) (1/2) ( √3 - √2) (D) (1/2) ( √2 - 1)

Tardigrade Admin · Accepted Answer

18 x2-9 π x+π2=0 (6 x-π)(3 x-π)=0 ∴ x=(π/6), (π/3) α=(π/6), β=(π/3) y=( gof )(x)= cos x Area =∫ limits(π/6)(π/3) cos x d x=( sin x)(π/6)(π/3) =(√3/2)-(1/2) =(1/2)(√3-1) sq. units

Q. Let $g (x) = cos x^{2}, f (x) = x$ , and $α, β (α < β)$ be the roots of the quadratic equation $18 x^{2} - 9 π x + π^{2} = 0$ . Then the area (in sq. units) bounded by the curve $y = (g o f) (x)$ and the lines $x = α, x = β$ and $y = 0$ , is:

$\frac{1}{2} (3 - 1)$

$\frac{1}{2} (3 + 1)$

$\frac{1}{2} (3 - 2)$

$\frac{1}{2} (2 - 1)$

Q. 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:

21​(3​−1)

21​(3​+1)

21​(3​−2​)

21​(2​−1)

18x2−9πx+π2=0 (6x−π)(3x−π)=0 ∴x=6π​,3π​ α=6π​,β=3π​ y=( gof )(x)=cosx Area =6π​∫3π​​cosxdx=(sinx)6π​3π​​ =23​​−21​ =21​(3​−1) sq. units

Q. Let $g (x) = cos x^{2}, f (x) = x$ , and $α, β (α < β)$ be the roots of the quadratic equation $18 x^{2} - 9 π x + π^{2} = 0$ . Then the area (in sq. units) bounded by the curve $y = (g o f) (x)$ and the lines $x = α, x = β$ and $y = 0$ , is:

$\frac{1}{2} (3 - 1)$

$\frac{1}{2} (3 + 1)$

$\frac{1}{2} (3 - 2)$

$\frac{1}{2} (2 - 1)$