The area (in sq. units) of the region A = (x, y) : |x| + |y| \le 1, 2y2 \ge |x|

Area Under the Curves PYQ — JEE Main 2020 · 6 Sep · Shift 1 · Q20 with Solution

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Correct answer: B

We need the area of $A=\{(x,y): |x|+|y|\le 1,\; 2y^2\ge |x|\}.$

So the point must satisfy both:

Because both conditions involve absolute values, the region is symmetric about both axes.

From $|x|+|y|\le 1,$ for a fixed $y$ with $|y|\le 1$ , we get $|x|\le 1-|y|.$

So for each $y$ , the allowed horizontal width inside the diamond is $-\,(1-|y|)\le x\le 1-|y|.$

From $2y^2\ge |x|,$ we get $|x|\le 2y^2.$

So for each fixed $y$ , this gives $-2y^2\le x\le 2y^2.$

For each fixed $y$ , both must hold, so $|x|\le \min(1-|y|,\,2y^2).$

Hence horizontal width of the region at height $y$ is $2\min(1-|y|,2y^2),$ provided $|y|\le 1$ .

Thus the area is $\text{Area}=\int_{-1}^{1} 2\min(1-|y|,2y^2)\,dy.$

Using symmetry about the $x$ -axis,

=4\int_0^1 \min(1-y,2y^2)\,dy.$$ --- 5. **Find where the two bounds switch** We solve $$2y^2=1-y.$$ That gives $$2y^2+y-1=0.$$ Factor: $$2y^2+y-1=(2y-1)(y+1)=0.$$ So $$y=\frac12 \quad \text{or} \quad y=-1.$$ In the interval $[0,1]$, the switching point is $$y=\frac12.$$ Now test: - for $0\le y<\frac12$, $$2y^2<1-y,$$ so minimum is $2y^2$, - for $\frac12<y\le 1$, $$1-y<2y^2,$$ so minimum is $1-y$. Therefore, $$\text{Area}=4\left(\int_0^{1/2}2y^2\,dy+\int_{1/2}^1(1-y)\,dy\right).$$ --- 6. **Evaluate the integrals** First integral: $$\int_0^{1/2}2y^2\,dy=2\cdot \left[\frac{y^3}{3}\right]_0^{1/2} =2\cdot \frac{1}{24}=\frac{1}{12}.$$ Second integral: $$\int_{1/2}^1(1-y)\,dy=\left[y-\frac{y^2}{2}\right]_{1/2}^1.$$ At $y=1$: $$1-\frac12=\frac12.$$ At $y=\frac12$: $$\frac12-\frac{1}{8}=\frac{3}{8}.$$ So, $$\int_{1/2}^1(1-y)\,dy=\frac12-\frac38=\frac18.$$ Therefore, $$\text{Area}=4\left(\frac{1}{12}+\frac{1}{8}\right) =4\left(\frac{2+3}{24}\right) =4\cdot \frac{5}{24} =\frac{5}{6}.$$ --- 7. **Match with options** $$\boxed{\frac{5}{6}}$$ So the correct option is **B**.