Let g(x) = cos x^2, f(x) = √(x) and α, β (α < β) be the roots of the quadratic equation 18x^2 - 9π x + π^2 = 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

Let g(x) = cos x^2, f(x) = √(x) and α, β (α < β) be the… — JEE Main 2018 Mathematics PYQ

JEE Main 2018

Let

g(x) = \cos x^2

f(x) = \sqrt{x}

and

\alpha, \beta

(\alpha < \beta)

be the roots of the quadratic equation

18x^2 - 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

A.
$\frac{1}{2}(\sqrt{3} - 1)$
B.
$\frac{1}{2}(\sqrt{3} + 1)$
C.
$\frac{1}{2}(\sqrt{3} - \sqrt{2})$
D.
$\frac{1}{2}(\sqrt{2} - 1)$

Show answer and solution

Correct option: A

Solution

The roots of 18x² - 9πx + π² = 0 are x = π/6 and x = π/3 (using quadratic formula: x = (9π ± √(81π² - 72π²))/36 = (9π ± 3π)/36), giving α = π/6, β = π/3.

The composite function (g∘f)(x) = g(f(x)) = cos(x), since f(x) = √x and g(x) = cos(x²), so g(√x) = cos(x).

The area = ∫[π/6 to π/3] |cos(x)| dx. Since cos(x) > 0 on [π/6, π/3], this equals [sin(x)] from π/6 to π/3 = sin(π/3) - sin(π/6) = (√3/2) - (1/2) = (1/2)(√3 - 1).

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JEE Main 2018 Mathematics question on Definite Integrals and Areas