JEE Advanced 2022 · Paper 2 · Mathematics Q8

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MathematicsInteger+3 / -1Question 8 of 54

Consider the functions $f, g : \mathbb{R} \to \mathbb{R}$ defined by
$f(x) = x^2 + \frac{5}{12} \qquad\text{and}\qquad g(x) = \begin{cases} 2\left(1 - \dfrac{4|x|}{3}\right), & |x| \le \dfrac34,\\[2mm] 0, & |x| > \dfrac34.\end{cases}$
If $\alpha$ is the area of the region
$\left\{(x, y) \in \mathbb{R}\times\mathbb{R} : |x| \le \frac34,\ 0 \le y \le \min\{f(x), g(x)\}\right\},$
then the value of $9\alpha$ is _____.

Your answer

Answer and worked solution

Official answer: 6

Where they cross (for $x\ge0$ ):
$x^2+\frac5{12}=2-\frac{8x}3\implies12x^2+32x-19=0\implies x=\frac12.$
For $0\le x\le\frac12$ the smaller one is $f$ ; for $\frac12\le x\le\frac34$ it is $g$ .

Area for $x\ge0$ .

$\int_0^{1/2}\left(x^2+\frac5{12}\right)dx=\frac1{24}+\frac5{24}=\frac14$ .
$\int_{1/2}^{3/4}g\,dx$ is a triangle of base $\frac14$ and height $g(\frac12)=\frac23$ : area $\frac1{12}$ .

Total. Both functions are even, so
$\alpha=2\left(\frac14+\frac1{12}\right)=\frac23,\qquad9\alpha=6.$

Topics:Area between curves Area under a curve

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JEE Advanced 2022 · Paper 2, Mathematics question 8