JEE Advanced 2022 Paper 2 · Q08 · Area Under Curves

Question

Consider the functions $f, g:\mathbb{R}	o\mathbb{R}$ defined by $f(x)=x^{2}+\dfrac{5}{12}$ and $g(x)=\begin{cases}2\left(1-\dfrac{4|x|}{3}ight), & |x|\le \dfrac{3}{4}\ 0, & |x|>\dfrac{3}{4}\end{cases}$. If $\alpha$ is the area of the region $\left\{(x,y)\in\mathbb{R}	imes\mathbb{R}: |x|\le\dfrac{3}{4},\ 0\le y\le \min\{f(x),g(x)\}ight\}$, then the value of $9\alpha$ is _____.

SolveKar AI · Accepted Answer

Solve $f(x)=g(x)$ on $0\le x\le 3/4$: $x^2+5/12 = 2-8x/3$, i.e. $x^2+8x/3-19/12=0$, multiply by 12: $12x^2+32x-19=0$, so $x=\dfrac{-32\pm\sqrt{1024+912}}{24}=\dfrac{-32\pm\sqrt{1936}}{24}=\dfrac{-32\pm 44}{24}$. Positive root $x=\dfrac{12}{24}=\dfrac{1}{2}$. By symmetry, intersection on the negative side is at $x=-1/2$. For $|x|\le 1/2$, $f

JEE Advanced 2022 Paper 2 · Q08 · Area Under Curves

Solution