Given: f(x) = \ array ll x, 0 ≤ x (1)/(2), x = (1)/(2) array 1 - x, (1)/(2)) and g(x) = (x - (1)/(2)) ^ 2 , x ∈ R . Then, the area (in sq. units) of the region bounded by the curves, y = f(x) and y = g (x) between the

Correct answer: B. 3 4 - (1)/(3) Required area = Area of trapezium ABCD - ∫ _ 1/2 ^ 3 / 2 (x - (1)/(2)) ^ 2 dx = (1)/(2)( 3 - 1 2 )((1)/(2) + 1 - 3 2 ) - (1)/(3) ( (x - (1)/(2)) ^ 3 ) _ (1)/(2) ^ 3 2 = 3 4 - (1)/(3)

JEE Main 20209 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2020 · 9 January · Evening Shift · Mathematics · Area Under Curves

Given: f(x) = \ array ll x, 0 ≤ x (1)/(2), x = (1)/(2) array 1 - x,...

Given:

f\left(x\right) = \left\{

\begin{array}{ll}x, 0 \le x \\ \dfrac{1}{2}, x = \dfrac{1}{2}\end{array}

1 - x, \dfrac{1}{2}\right)<br />and

g\left(x\right) = {\left(x - \dfrac{1}{2}\right)}^{2}, x \in R .

Then, the area (in sq. units) of the region bounded by the curves,

y = f\left(x\right)

and

y = g

\left(x\right)

between the lines

2x = 1

and

2x = \sqrt{3},

is: $Given: f(x) = \ array ll x, 0 ≤ x (1)/(2), x = (1)/(2) array 1 - x, (1)/(2)) and g(x) = (x - (1)/(2)) ^ 2 , x ∈ R .$

Options

A $\dfrac{1}{3} + \dfrac{\sqrt{3}}{4}$
B $\dfrac{\sqrt{3}}{4} - \dfrac{1}{3}$
C $\dfrac{1}{2} - \dfrac{\sqrt{3}}{4}$
D $\dfrac{1}{2} + \dfrac{\sqrt{3}}{4}$

Correct answer

B. $\dfrac{\sqrt{3}}{4} - \dfrac{1}{3}$

Step-by-step solution

$Given: f(x) = \ array ll x, 0 ≤ x (1)/(2), x = (1)/(2) array 1 - x, (1)/(2)) and g(x) = (x - (1)/(2)) ^ 2 , x ∈ R .$
Required area = Area of trapezium ABCD

- \mathop{∫}_{1/2}^{\sqrt{3} / 2}{\left(x - \dfrac{1}{2}\right)}^{2}dx

= \dfrac{1}{2}\left(\dfrac{\sqrt{3} - 1}{2}\right)\left(\dfrac{1}{2} + 1 - \dfrac{\sqrt{3}}{2}\right) - \dfrac{1}{3}{\left({\left(x - \dfrac{1}{2}\right)}^{3}\right)}_{\dfrac{1}{2}}^{\dfrac{\sqrt{3}}{2}}

= \dfrac{\sqrt{3}}{4} - \dfrac{1}{3}

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