Consider the functions f, g: ℝ → ℝ defined by f(x) = x ^ 2 + (5)/(12) and g(x) = \ cases 2(1 - (4|x|)/(3)), & |x| ≤ (3)/(4), 0, & |x| > (3)/(4) cases ) If α is the area of the region \ (x, y) ∈ ℝ × ℝ: |x| ≤ (3)/(4), 0 ≤

JEE Advanced 2022Paper 2

Exam: JEE Advanced
Year: 2022
Paper: Paper 2
Subject: Mathematics
Chapter: Area Under Curves

JEE Advanced 2022 · Paper 2 · Mathematics · Area Under Curves

Consider the functions f, g: ℝ → ℝ defined by f(x) = x ^ 2 +...

Consider the functions

f, g: \mathbb{R} \to \mathbb{R}

defined by

f\left(x\right) = {x}^{2} + \dfrac{5}{12} and

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

\begin{cases}2\left(1 - \dfrac{4\left|x\right|}{3}\right), & \left|x\right| \le \dfrac{3}{4}, \\ 0, & \left|x\right| > \dfrac{3}{4}\end{cases}

\right)
If

\alpha is the area of the region \left\{\left(x, y\right) \in \mathbb{R} \times \mathbb{R}: \left|x\right| \le \dfrac{3}{4}, 0 \le y \le min\left\{f\left(x\right), g\left(x\right)\right\}\right\}

, then the value of

9\alpha

is _____ . $Consider the functions f, g: ℝ → ℝ defined by f(x) = x ^ 2 + (5)/(12) and g(x) = \ cases 2(1 - (4|x|)/(3)), & |x| ≤$

Correct answer

Step-by-step solution

Plotting the diagram of the given function we have,
$Consider the functions f, g: ℝ → ℝ defined by f(x) = x ^ 2 + (5)/(12) and g(x) = \ cases 2(1 - (4|x|)/(3)), & |x| ≤$ Now by diagram we can see twice of ${A}_{1}$ and ${A}_{2}$ will be the required area,
So, required area $= 2 \cdot \left({A}_{1} + {A}_{2}\right)$
$⇒ \alpha = \left({∫}_{0}^{\dfrac{1}{2}}\left({x}^{2} + \dfrac{5}{12}\right)dx + \dfrac{1}{2}\left(\dfrac{3}{4} - \dfrac{1}{2}\right) \cdot \dfrac{2}{3}\right) \times 2$
$⇒ \alpha = \left({\left\left(\dfrac{{x}^{3}}{3} + \dfrac{5x}{12}\right)\right|}_{0}^{\dfrac{1}{2}} + \dfrac{1}{2} ⋅ \dfrac{1}{4} ⋅ \dfrac{2}{3}\right) \times 2$
$⇒ \alpha = \left(\dfrac{1}{4} + \dfrac{1}{12}\right) \times 2$
$⇒ \alpha = \dfrac{2}{3} ⇒ 9\alpha = 6$

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