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 \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2023 · 10 Apr · Shift 1 · Q40 with Solution

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Correct answer: 16

Find the parabola $y=p(x)$ .

Let $p(x)=ax^2+bx+c.$ Since it passes through $(-1,0),(0,1),(1,0)$ :

From $(0,1)$ : $c=1$
From $(1,0)$ : $a+b+1=0$
From $(-1,0)$ : $a-b+1=0$

Adding the last two equations: $2a+2=0\implies a=-1.$ Then $-1+b+1=0\implies b=0.$ So, $p(x)=1-x^2.$

Interpret the circle/region.

The inequality $(x+1)^2+(y-1)^2\le 1$ represents a circle of radius $1$ centered at $(-1,1)$ .

The required region is the part of this disk satisfying $y\le p(x)=1-x^2.$ So we need the area inside the circle and below the parabola.

Find intersection points of the circle and parabola.

Substitute $y=1-x^2$ into the circle: $(x+1)^2 + (1-x^2-1)^2 = 1$ $(x+1)^2 + x^4 = 1$ $x^2+2x+1+x^4=1$ $x^4+x^2+2x=0$ $x(x^3+x+2)=0.$ Now $x^3+x+2=(x+1)(x^2-x+2).$ Hence real roots are

\quad x=-1.$$ Corresponding points are $$(0,1),\quad (-1,0).$$ These are exactly the endpoints of the relevant arc. --- 4. **Write the circle as upper and lower semicircles.** From $$ (x+1)^2+(y-1)^2=1,$$ we get $$ y=1\pm \sqrt{1-(x+1)^2}. $$ Inside the disk, for each $x\in[-2,0]$, $$1-\sqrt{1-(x+1)^2}\le y\le 1+\sqrt{1-(x+1)^2}.$$ We need the portion with $$y\le 1-x^2.$$ So for each $x$, the vertical slice contributes from the lower circle up to the parabola, provided the parabola lies inside the disk. This happens on the interval of intersection: $$x\in[-1,0].$$ Thus $$A=\int_{-1}^{0}\Big[(1-x^2)-\big(1-\sqrt{1-(x+1)^2}\big)\Big]dx.$$ So $$A=\int_{-1}^{0}\left(\sqrt{1-(x+1)^2}-x^2\right)dx.$$ --- 5. **Evaluate the integral.** Split it: $$A=\int_{-1}^{0}\sqrt{1-(x+1)^2}\,dx-\int_{-1}^{0}x^2\,dx.$$ For the first integral, let $$u=x+1.$$ Then when $x=-1$, $u=0$, and when $x=0$, $u=1$. Thus $$\int_{-1}^{0}\sqrt{1-(x+1)^2}\,dx=\int_0^1\sqrt{1-u^2}\,du.$$ This is the area of a quarter circle of radius $1$: $$\int_0^1\sqrt{1-u^2}\,du=\frac{\pi}{4}.$$ Also, $$\int_{-1}^{0}x^2\,dx=\left[\frac{x^3}{3}\right]_{-1}^{0}=\frac{1}{3}.$$ Hence $$A=\frac{\pi}{4}-\frac{1}{3}.$$ --- 6. **Compute** $12(\pi-4A)$. First, $$4A=4\left(\frac{\pi}{4}-\frac{1}{3}\right)=\pi-\frac{4}{3}.$$ Therefore, $$\pi-4A=\pi-\left(\pi-\frac{4}{3}\right)=\frac{4}{3}.$$ So, $$12(\pi-4A)=12\cdot \frac{4}{3}=16.$$ --- 7. **Final answer** $$\boxed{16}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: 16

Find the parabola $y=p(x)$ .

Let $p(x)=ax^2+bx+c.$ Since it passes through $(-1,0),(0,1),(1,0)$ :

From $(0,1)$ : $c=1$
From $(1,0)$ : $a+b+1=0$
From $(-1,0)$ : $a-b+1=0$

Adding the last two equations: $2a+2=0\implies a=-1.$ Then $-1+b+1=0\implies b=0.$ So, $p(x)=1-x^2.$

Interpret the circle/region.

The inequality $(x+1)^2+(y-1)^2\le 1$ represents a circle of radius $1$ centered at $(-1,1)$ .

The required region is the part of this disk satisfying $y\le p(x)=1-x^2.$ So we need the area inside the circle and below the parabola.

Find intersection points of the circle and parabola.

Substitute $y=1-x^2$ into the circle: $(x+1)^2 + (1-x^2-1)^2 = 1$ $(x+1)^2 + x^4 = 1$ $x^2+2x+1+x^4=1$ $x^4+x^2+2x=0$ $x(x^3+x+2)=0.$ Now $x^3+x+2=(x+1)(x^2-x+2).$ Hence real roots are

\quad x=-1.$$ Corresponding points are $$(0,1),\quad (-1,0).$$ These are exactly the endpoints of the relevant arc. --- 4. **Write the circle as upper and lower semicircles.** From $$ (x+1)^2+(y-1)^2=1,$$ we get $$ y=1\pm \sqrt{1-(x+1)^2}. $$ Inside the disk, for each $x\in[-2,0]$, $$1-\sqrt{1-(x+1)^2}\le y\le 1+\sqrt{1-(x+1)^2}.$$ We need the portion with $$y\le 1-x^2.$$ So for each $x$, the vertical slice contributes from the lower circle up to the parabola, provided the parabola lies inside the disk. This happens on the interval of intersection: $$x\in[-1,0].$$ Thus $$A=\int_{-1}^{0}\Big[(1-x^2)-\big(1-\sqrt{1-(x+1)^2}\big)\Big]dx.$$ So $$A=\int_{-1}^{0}\left(\sqrt{1-(x+1)^2}-x^2\right)dx.$$ --- 5. **Evaluate the integral.** Split it: $$A=\int_{-1}^{0}\sqrt{1-(x+1)^2}\,dx-\int_{-1}^{0}x^2\,dx.$$ For the first integral, let $$u=x+1.$$ Then when $x=-1$, $u=0$, and when $x=0$, $u=1$. Thus $$\int_{-1}^{0}\sqrt{1-(x+1)^2}\,dx=\int_0^1\sqrt{1-u^2}\,du.$$ This is the area of a quarter circle of radius $1$: $$\int_0^1\sqrt{1-u^2}\,du=\frac{\pi}{4}.$$ Also, $$\int_{-1}^{0}x^2\,dx=\left[\frac{x^3}{3}\right]_{-1}^{0}=\frac{1}{3}.$$ Hence $$A=\frac{\pi}{4}-\frac{1}{3}.$$ --- 6. **Compute** $12(\pi-4A)$. First, $$4A=4\left(\frac{\pi}{4}-\frac{1}{3}\right)=\pi-\frac{4}{3}.$$ Therefore, $$\pi-4A=\pi-\left(\pi-\frac{4}{3}\right)=\frac{4}{3}.$$ So, $$12(\pi-4A)=12\cdot \frac{4}{3}=16.$$ --- 7. **Final answer** $$\boxed{16}$$ This matches the stored correct answer.

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