The area (in sq. units) of the region described by \(x, y): y^2 \leq 2 x , and y \geq 4 x-1\ is

Area Under the Curves PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q38 with Solution

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Correct answer: A

The region is given by $y^2 \le 2x \quad \text{and} \quad y \ge 4x-1.$

From $y^2 \le 2x$ , we get $x \ge \frac{y^2}{2},$ which is the region to the right of the parabola $y^2=2x$ .
From $y \ge 4x-1$ , we get $x \le \frac{y+1}{4},$ which is the region to the left of the line $y=4x-1$ .

So for a given $y$ , the region lies between $x=\frac{y^2}{2} \quad \text{and} \quad x=\frac{y+1}{4}.$

This is possible only when $\frac{y^2}{2} \le \frac{y+1}{4}.$

Solve $\frac{y^2}{2} = \frac{y+1}{4}.$ Multiply by $4$ : $2y^2 = y+1$ $2y^2-y-1=0$ $(2y+1)(y-1)=0.$ Thus, $y=-\frac12, \quad y=1.$

Hence the required area is $A=\int_{-1/2}^{1}\left(\frac{y+1}{4}-\frac{y^2}{2}\right)dy.$

$A=\int_{-1/2}^{1}\left(\frac y4+\frac14-\frac{y^2}{2}\right)dy.$

Antiderivative:

=\frac{y^2}{8}+\frac y4-\frac{y^3}{6}.$$ So, $$A=\left[\frac{y^2}{8}+\frac y4-\frac{y^3}{6}\right]_{-1/2}^{1}.$$ At $y=1$: $$\frac{1}{8}+\frac{1}{4}-\frac{1}{6} =\frac{3+6-4}{24} =\frac{5}{24}.$$ At $y=-\frac12$: $$\frac{(1/4)}{8}+\frac{-1/2}{4}-\frac{(-1/8)}{6} =\frac{1}{32}-\frac{1}{8}+\frac{1}{48}.$$ Taking LCM $96$: $$\frac{3-12+2}{96}=-\frac{7}{96}.$$ Therefore, $$A=\frac{5}{24}-\left(-\frac{7}{96}\right) =\frac{20}{96}+\frac{7}{96} =\frac{27}{96} =\frac{9}{32}.$$ 4. **Match with the options** $$\boxed{\frac{9}{32}}$$ So the correct option is **A**.