JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 17)
Let $$y = p(x)$$ be the parabola passing through the points $$( - 1,0),(0,1)$$ and $$(1,0)$$. If the area of the region $$\{ (x,y):{(x + 1)^2} + {(y - 1)^2} \le 1,y \le p(x)\} $$ is A, then $$12(\pi - 4A)$$ is equal to ___________.
Answer
16
Explanation
Let, $y=p(x)$ be the parabola passing through the points $(-1,0)(0,1)(1,0)$.
Now, to find the area of the region
$$ \left\{(x, y) ;(x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\right\} $$
_10th_April_Morning_Shift_en_17_1.png)
Thus, area of shaded region is
$$ \begin{aligned} & A =\int\limits_{-1}^0\left\{\left(1-x^2\right)-\left(1-\sqrt{1-(x+1)^2}\right\} d x\right. \\\\ & =\int\limits_{-1}^0\left\{-x^2+\sqrt{1-(x+1)^2}\right\} d x \\\\ & =\int\limits_{-1}^0-x^2 d x+\int_{-1}^0 \sqrt{1-(x+1)^2} d x \\\\ & =-\left(\frac{x^3}{3}\right)_{-1}^0+\left[\frac{\sqrt{1-(x+1)^2}}{2}+\frac{1}{2} \sin ^{-1}(x+1)\right]_{-1}^0 \\\\ & =-\frac{1}{3}+\left[\frac{1}{2}+\frac{\pi}{4}-\frac{1}{2}\right] \\\\ & A =\frac{\pi}{4}-\frac{1}{3} \end{aligned} $$
Now, $12(\pi-4 A)$
$$ \begin{aligned} & =12\left[\pi-4\left(\frac{\pi}{4}-\frac{1}{3}\right)\right] \\\\ & =12\left(\pi-\pi+\frac{4}{3}\right)=16 \end{aligned} $$
Now, to find the area of the region
$$ \left\{(x, y) ;(x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\right\} $$
Thus, area of shaded region is
$$ \begin{aligned} & A =\int\limits_{-1}^0\left\{\left(1-x^2\right)-\left(1-\sqrt{1-(x+1)^2}\right\} d x\right. \\\\ & =\int\limits_{-1}^0\left\{-x^2+\sqrt{1-(x+1)^2}\right\} d x \\\\ & =\int\limits_{-1}^0-x^2 d x+\int_{-1}^0 \sqrt{1-(x+1)^2} d x \\\\ & =-\left(\frac{x^3}{3}\right)_{-1}^0+\left[\frac{\sqrt{1-(x+1)^2}}{2}+\frac{1}{2} \sin ^{-1}(x+1)\right]_{-1}^0 \\\\ & =-\frac{1}{3}+\left[\frac{1}{2}+\frac{\pi}{4}-\frac{1}{2}\right] \\\\ & A =\frac{\pi}{4}-\frac{1}{3} \end{aligned} $$
Now, $12(\pi-4 A)$
$$ \begin{aligned} & =12\left[\pi-4\left(\frac{\pi}{4}-\frac{1}{3}\right)\right] \\\\ & =12\left(\pi-\pi+\frac{4}{3}\right)=16 \end{aligned} $$
Comments (0)
