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QJEEM23Ma8db9: JEE Main 2023 Mathematics - Class 12 JEEM23Ma8db9 Solution: Let be the parabola passing through the points and . If the…

Let y=p(x)y = p(x) be the parabola passing through the points (−1,0),(0,1)( - 1,0),(0,1) and (1,0)(1,0) . If the area of the region {(x,y):(x+1)2+(y−1)2≤1,y≤p(x)}\{ (x,y):{(x + 1)^2} + {(y - 1)^2} \le 1,y \le p(x)\} is A, then 12(π−4A)12(\pi - 4A) is equal to ___________.

Step-by-Step Verified Solution

Let, y=p(x)y=p(x) be the parabola passing through the points (−1,0)(0,1)(1,0)(-1,0)(0,1)(1,0) .

Now, to find the area of the region

{(x,y);(x+1)2+(y−1)2≤1,y≤p(x)}\left\{(x, y) ;(x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\right\}

JEE Main 2023 (Online) 10th April Morning Shift Mathematics - Area Under The Curves Question 37 English Explanation

Thus, area of shaded region is

amp;A=∫−10{(1−x2)−(1−1−(x+1)2}dxamp;=∫−10{−x2+1−(x+1)2}dxamp;=∫−10−x2dx+∫−101−(x+1)2dxamp;=−(x33)−10+[1−(x+1)22+12sin⁡−1(x+1)]−10amp;=−13+[12+π4−12]amp;A=π4−13\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(π−4A)12(\pi-4 A)

amp;=12[π−4(π4−13)]amp;=12(π−π+43)=16\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}

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