87013
The area (in sq. units) of the circle \(x^{2}+y^{2}=36\), which is outside the parabola \(y^{2}=9 x\). is
1 \(24 \pi+3 \sqrt{3}\)
2 \(24 \pi-3 \sqrt{3}\)
3 \(12 \pi+3 \sqrt{3}\)
4 \(12 \pi-3 \sqrt{3}\)
Explanation:
(B) : Given that, the circle, \(x^{2}+y^{2}=36 \tag{i}\) \(y^{2}=9 x \tag{ii}\) Putting the value of \(y^{2}=9 x\) in equation (i), we get - \(x^{2}+9 x-36=0\) \((x+12)(x-3)=0\) So, \(x=3,-12\) \(\therefore\) Area \(=\pi(\mathrm{r})^{2}-2 \int_{0}^{3} \sqrt{9 \mathrm{x}} \mathrm{dx}-\int_{3}^{6} \sqrt{36-\mathrm{x}^{2}} \mathrm{dx}\) \(=\pi(6)^{2}-12 \sqrt{3}-2\left[\frac{x}{2} \sqrt{36-x^{2}}+18 \sin ^{-1} \frac{x}{6}\right]_{3}^{6}\) \(=36 \pi-12 \sqrt{3}-2\left(9 \pi-3 \pi-\frac{9 \sqrt{3}}{2}\right)=24 \pi-3 \sqrt{3}\)
JEE Main-2021-24.02.2021
Application of the Integrals
87014
The area of the region bounded by the parabola \((y-2)^{2}=(x-1)\), the tangent to it at the point whose ordinate is 3 and then -axis is
1 9
2 10
3 4
4 6
Explanation:
(A) : Given that, the parabola, \((\mathrm{y}-2)^{2}=(\mathrm{x}-1)\) Differentiate w.r.t. \(x\), we get \(-2(y-2) y^{\prime}=1\) \(y^{\prime}=\frac{1}{2(y-2)}\) \(\therefore \quad \mathrm{y}=3\) then \(\mathrm{x}=2\) So, point \((2,3)\) Thus, \(\mathrm{y}_{(2,3)}^{\prime}=\frac{1}{2}\) \(\therefore \quad \frac{y-3}{x-2}=\frac{1}{2}\) \(x-2 y+4=0\) So, area \(=\int_{0}^{3}\left[(y-2)^{2}+1-(2 y-4)\right] d y=9\) square unit.
JEE Main-2021-27.08.2021
Application of the Integrals
86983
Area lying between the parabola \(y^{2}=4 a x\) and its latus rectum is
1 \(\frac{1}{3}\) a sq.units
2 \(\frac{1}{3} \mathrm{a}^{2}\) sq.units
3 \(\frac{8}{3}\) a sq.units
4 \(\frac{8}{3} \mathrm{a}^{2}\) sq.units
Explanation:
(D) : We have, \(\mathrm{y}^{2}=4 \mathrm{ax}\), parabola with vertex, \((0,0)\) and focus \((a, 0)\) and latus rectum \(4 \mathrm{a}\). Required area \(=\) area of shaded region \(\mathrm{A}=2 \int_{0}^{\mathrm{a}} 2 \sqrt{\mathrm{a}} \sqrt{\mathrm{x}} \mathrm{dx}=2\left[2 \sqrt{\mathrm{a}} \mathrm{x}^{3 / 2} \times \frac{2}{3}\right]_{0}^{\mathrm{a}}=\frac{8}{3} \mathrm{a}^{2}\) sq.units \(\frac{8}{3} \mathrm{a}^{2}\) sq.units
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of the Integrals
87013
The area (in sq. units) of the circle \(x^{2}+y^{2}=36\), which is outside the parabola \(y^{2}=9 x\). is
1 \(24 \pi+3 \sqrt{3}\)
2 \(24 \pi-3 \sqrt{3}\)
3 \(12 \pi+3 \sqrt{3}\)
4 \(12 \pi-3 \sqrt{3}\)
Explanation:
(B) : Given that, the circle, \(x^{2}+y^{2}=36 \tag{i}\) \(y^{2}=9 x \tag{ii}\) Putting the value of \(y^{2}=9 x\) in equation (i), we get - \(x^{2}+9 x-36=0\) \((x+12)(x-3)=0\) So, \(x=3,-12\) \(\therefore\) Area \(=\pi(\mathrm{r})^{2}-2 \int_{0}^{3} \sqrt{9 \mathrm{x}} \mathrm{dx}-\int_{3}^{6} \sqrt{36-\mathrm{x}^{2}} \mathrm{dx}\) \(=\pi(6)^{2}-12 \sqrt{3}-2\left[\frac{x}{2} \sqrt{36-x^{2}}+18 \sin ^{-1} \frac{x}{6}\right]_{3}^{6}\) \(=36 \pi-12 \sqrt{3}-2\left(9 \pi-3 \pi-\frac{9 \sqrt{3}}{2}\right)=24 \pi-3 \sqrt{3}\)
JEE Main-2021-24.02.2021
Application of the Integrals
87014
The area of the region bounded by the parabola \((y-2)^{2}=(x-1)\), the tangent to it at the point whose ordinate is 3 and then -axis is
1 9
2 10
3 4
4 6
Explanation:
(A) : Given that, the parabola, \((\mathrm{y}-2)^{2}=(\mathrm{x}-1)\) Differentiate w.r.t. \(x\), we get \(-2(y-2) y^{\prime}=1\) \(y^{\prime}=\frac{1}{2(y-2)}\) \(\therefore \quad \mathrm{y}=3\) then \(\mathrm{x}=2\) So, point \((2,3)\) Thus, \(\mathrm{y}_{(2,3)}^{\prime}=\frac{1}{2}\) \(\therefore \quad \frac{y-3}{x-2}=\frac{1}{2}\) \(x-2 y+4=0\) So, area \(=\int_{0}^{3}\left[(y-2)^{2}+1-(2 y-4)\right] d y=9\) square unit.
JEE Main-2021-27.08.2021
Application of the Integrals
86983
Area lying between the parabola \(y^{2}=4 a x\) and its latus rectum is
1 \(\frac{1}{3}\) a sq.units
2 \(\frac{1}{3} \mathrm{a}^{2}\) sq.units
3 \(\frac{8}{3}\) a sq.units
4 \(\frac{8}{3} \mathrm{a}^{2}\) sq.units
Explanation:
(D) : We have, \(\mathrm{y}^{2}=4 \mathrm{ax}\), parabola with vertex, \((0,0)\) and focus \((a, 0)\) and latus rectum \(4 \mathrm{a}\). Required area \(=\) area of shaded region \(\mathrm{A}=2 \int_{0}^{\mathrm{a}} 2 \sqrt{\mathrm{a}} \sqrt{\mathrm{x}} \mathrm{dx}=2\left[2 \sqrt{\mathrm{a}} \mathrm{x}^{3 / 2} \times \frac{2}{3}\right]_{0}^{\mathrm{a}}=\frac{8}{3} \mathrm{a}^{2}\) sq.units \(\frac{8}{3} \mathrm{a}^{2}\) sq.units
87013
The area (in sq. units) of the circle \(x^{2}+y^{2}=36\), which is outside the parabola \(y^{2}=9 x\). is
1 \(24 \pi+3 \sqrt{3}\)
2 \(24 \pi-3 \sqrt{3}\)
3 \(12 \pi+3 \sqrt{3}\)
4 \(12 \pi-3 \sqrt{3}\)
Explanation:
(B) : Given that, the circle, \(x^{2}+y^{2}=36 \tag{i}\) \(y^{2}=9 x \tag{ii}\) Putting the value of \(y^{2}=9 x\) in equation (i), we get - \(x^{2}+9 x-36=0\) \((x+12)(x-3)=0\) So, \(x=3,-12\) \(\therefore\) Area \(=\pi(\mathrm{r})^{2}-2 \int_{0}^{3} \sqrt{9 \mathrm{x}} \mathrm{dx}-\int_{3}^{6} \sqrt{36-\mathrm{x}^{2}} \mathrm{dx}\) \(=\pi(6)^{2}-12 \sqrt{3}-2\left[\frac{x}{2} \sqrt{36-x^{2}}+18 \sin ^{-1} \frac{x}{6}\right]_{3}^{6}\) \(=36 \pi-12 \sqrt{3}-2\left(9 \pi-3 \pi-\frac{9 \sqrt{3}}{2}\right)=24 \pi-3 \sqrt{3}\)
JEE Main-2021-24.02.2021
Application of the Integrals
87014
The area of the region bounded by the parabola \((y-2)^{2}=(x-1)\), the tangent to it at the point whose ordinate is 3 and then -axis is
1 9
2 10
3 4
4 6
Explanation:
(A) : Given that, the parabola, \((\mathrm{y}-2)^{2}=(\mathrm{x}-1)\) Differentiate w.r.t. \(x\), we get \(-2(y-2) y^{\prime}=1\) \(y^{\prime}=\frac{1}{2(y-2)}\) \(\therefore \quad \mathrm{y}=3\) then \(\mathrm{x}=2\) So, point \((2,3)\) Thus, \(\mathrm{y}_{(2,3)}^{\prime}=\frac{1}{2}\) \(\therefore \quad \frac{y-3}{x-2}=\frac{1}{2}\) \(x-2 y+4=0\) So, area \(=\int_{0}^{3}\left[(y-2)^{2}+1-(2 y-4)\right] d y=9\) square unit.
JEE Main-2021-27.08.2021
Application of the Integrals
86983
Area lying between the parabola \(y^{2}=4 a x\) and its latus rectum is
1 \(\frac{1}{3}\) a sq.units
2 \(\frac{1}{3} \mathrm{a}^{2}\) sq.units
3 \(\frac{8}{3}\) a sq.units
4 \(\frac{8}{3} \mathrm{a}^{2}\) sq.units
Explanation:
(D) : We have, \(\mathrm{y}^{2}=4 \mathrm{ax}\), parabola with vertex, \((0,0)\) and focus \((a, 0)\) and latus rectum \(4 \mathrm{a}\). Required area \(=\) area of shaded region \(\mathrm{A}=2 \int_{0}^{\mathrm{a}} 2 \sqrt{\mathrm{a}} \sqrt{\mathrm{x}} \mathrm{dx}=2\left[2 \sqrt{\mathrm{a}} \mathrm{x}^{3 / 2} \times \frac{2}{3}\right]_{0}^{\mathrm{a}}=\frac{8}{3} \mathrm{a}^{2}\) sq.units \(\frac{8}{3} \mathrm{a}^{2}\) sq.units
