Let gx-cosx2-fx-x- and - - be the roots of the quadratic equation 18x2-9-x-2-0- Then the area in sq- units bounded by the curve y-gofx and the lines x- x- and y-0- is-

Here, nbsp;18x^2 9πx+π^2=0 ⇒(3x π)(6x π)=0 ⇒α=π6,β=π3 Also, gof(x)=cosx ∴ Required area =∫_π/6^π/3cosxdx=√(3) 12 ...

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Byju's Answer

Standard XII

Physics

Definite Integrals

Let gx=cosx2,...

Question

Let $g (x) = cos x^{2}, f (x) = \sqrt{x},$ 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} (\sqrt{3} + 1)

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

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

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

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Solution

The correct option is D $\frac{1}{2} (\sqrt{3} - 1)$
Here, $18 x^{2} - 9 π x + π^{2} = 0$
$\Rightarrow (3 x - π) (6 x - π) = 0$
$\Rightarrow α = \frac{π}{6}, β = \frac{π}{3}$
Also, $g o f (x) = cos x$
$∴$ Required area $= \frac{π}{3} \int \frac{π}{6} cos x d x = \frac{\sqrt{3} - 1}{2}$

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Similar questions

Q. Let

g (x) = cos x^{2}, f (x) = \sqrt{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

s q . u n i t s

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and the lines

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and

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Then which of the following statement is false?

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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:

The correct option is D 12(√3−1)Here, 18x2−9πx+π2=0 ⇒(3x−π)(6x−π)=0 ⇒α=π6,β=π3 Also, gof(x)=cosx ∴ Required area =π3∫π6cosxdx=√3−12

Let $g (x) = cos x^{2}, f (x) = \sqrt{x},$ 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:

The correct option is D $\frac{1}{2} (\sqrt{3} - 1)$
Here, $18 x^{2} - 9 π x + π^{2} = 0$
$\Rightarrow (3 x - π) (6 x - π) = 0$
$\Rightarrow α = \frac{π}{6}, β = \frac{π}{3}$
Also, $g o f (x) = cos x$
$∴$ Required area $= \frac{π}{3} \int \frac{π}{6} cos x d x = \frac{\sqrt{3} - 1}{2}$