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Application of the Integrals

87020 The area enclosed by \(y^{2}=8 x\) and \(y=\sqrt{2} x\) that lies outside the triangle formed by \(y=\sqrt{2} x, x=\) \(1, y=2 \sqrt{2}\), is equal to :

1 \(\frac{16 \sqrt{2}}{6}\)

2 \(\frac{11 \sqrt{2}}{6}\)

3 \(\frac{13 \sqrt{2}}{6}\)

4 \(\frac{5 \sqrt{2}}{6}\)

JEE Main-2022-29.06.2022

Application of the Integrals

87021 A point \(P\) moves so that the sum of squares of its distances from the points \((1,2)\) and \((-2,1)\) is 14. Let \(f(x, y)=0\) be the locus of \(P\), which intersects the \(x\)-axis at the point \(A, B\) and the \(y\)-axis at the point \(C, D\). Then the area of the quadrilateral ACBD is equal to :

1 \(\frac{9}{2}\)

2 \(\frac{3 \sqrt{17}}{2}\)

3 \(\frac{3 \sqrt{17}}{4}\)

4 9

Explanation:

(B) : Let points \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is a point s.t.
\((\mathrm{h}-1)^{2}+(\mathrm{k}-2)^{2}+(\mathrm{h}+2)^{2}+(\mathrm{k}-1)^{2}=14\)
So,
\(\mathrm{h}^{2}-2 \mathrm{~h}+1+\mathrm{k}^{2}+4-4 \mathrm{k}+\mathrm{h}^{2}+4 \mathrm{~h}+4+\mathrm{k}^{2}+1-2 \mathrm{k}=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k+1+4+4+1=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k=4\)
Or \(\quad h^{2}+h+k^{2}-3 k=2\)
Or \(\quad \mathrm{h}^{2}+\mathrm{h}+\frac{1}{(2)^{2}}+\mathrm{k}^{2}-3 \mathrm{k}+\left(\frac{3}{2}\right)^{2}=2+\frac{1}{4}+\frac{9}{4}\)
Or \(\quad\left(h+\frac{1}{2}\right)^{2}+\left(\mathrm{k}-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
Put \(\mathrm{h}=\mathrm{x}\) and \(\mathrm{k}=\mathrm{y}\), we get
\(\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
At \(\mathrm{x}\)-axis, \(\mathrm{y}=0\)
\(\left(x+\frac{1}{2}\right)^{2}+\frac{9}{4}=\frac{9}{2}\)
Or \(\quad\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}\) or \(\left(x+\frac{1}{2}\right)= \pm \frac{3}{2}\)
\(\therefore \quad \mathrm{x}=\frac{3}{2}-\frac{1}{2}=1\) and \(\mathrm{x}+\frac{1}{2}=\frac{-3}{2} \Rightarrow \mathrm{x}=-2\)
So, point \(A(-2,0)\) and point \(B\) is \((1,0)\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0\)
\(\therefore \quad\left(0+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
\(\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}-\frac{1}{4}=\frac{18}{4}-\frac{1}{4}=\frac{17}{4}\)
\(\therefore \quad\left(\mathrm{y}-\frac{3}{2}\right)= \pm \frac{\sqrt{17}}{2}\)
\(\therefore \quad y=\frac{3}{2} \pm \frac{\sqrt{17}}{2}=\frac{3+\sqrt{17}}{2}, \frac{3-\sqrt{17}}{2}\)
So, point \(\mathrm{C}\) is \(\left(0, \frac{3+\sqrt{17}}{2}\right)\) and point \(\mathrm{D}\) is \(\left(0, \frac{3-\sqrt{17}}{2}\right)\)
Now, area of quadrilateral is given by
\(A=\frac{1}{2}\left[\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{4}+x_{4} y_{1}\right]\right.\)
\(=-\left[\left(x_{2} y_{1}+x_{3} y_{2}+x_{4} y_{3}+x_{1} y_{4}\right)\right]\)
\(=\frac{1}{2}\left[(-2) \cdot\left(\frac{3-\sqrt{17}}{2}\right)+1 \cdot \frac{3+\sqrt{17}}{2}+0+0\right]\)
\(\quad-\left(0+1 \times \frac{3-\sqrt{17}}{2}+0+(-2) \cdot \frac{3+\sqrt{17}}{2}\right)\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3+\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3}{2}+\frac{\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]=\frac{3 \sqrt{17}}{2}\)

JEE Main-2022-26.07.2022

Application of the Integrals

87022 Area enclosed by the curve
\(\pi\left[4(x-\sqrt{2})^{2}+y^{2}\right]=8 \text { is }\)

1 \(\pi\)

2 2

3 \(3 \pi\)

4 4

VITEEE-2008

Application of the Integrals

87023 The odd natural number a, such that the area of the region bounded by \(y=1, y=3, x=0, x=y^{a}\)
is \(\frac{364}{3}\), is equal to :

1 3

2 5

3 7

4 9

JEE Main-2022-26.07.2022

Application of the Integrals

87024 If \(f(x)\) be continuous function such that the area bounded by the curve \(\mathrm{y}=f(\mathrm{x})\), the \(\mathrm{x}\)-axis and the lines \(x=a\) and \(x=0\) is \(\frac{a^{2}}{2}+\frac{a}{2} \sin a+\frac{\pi}{2}\) cos a. Value of \(f\left(\frac{\pi}{2}\right)\) is

1 \(1 / 2\)

2 \(\mathrm{a} / 2\)

3 \(\mathrm{a}^{2} / 2\)

4 \(\pi / 2\)

AMU-2009

Application of the Integrals

87020 The area enclosed by \(y^{2}=8 x\) and \(y=\sqrt{2} x\) that lies outside the triangle formed by \(y=\sqrt{2} x, x=\) \(1, y=2 \sqrt{2}\), is equal to :

1 \(\frac{16 \sqrt{2}}{6}\)

2 \(\frac{11 \sqrt{2}}{6}\)

3 \(\frac{13 \sqrt{2}}{6}\)

4 \(\frac{5 \sqrt{2}}{6}\)

JEE Main-2022-29.06.2022

Application of the Integrals

87021 A point \(P\) moves so that the sum of squares of its distances from the points \((1,2)\) and \((-2,1)\) is 14. Let \(f(x, y)=0\) be the locus of \(P\), which intersects the \(x\)-axis at the point \(A, B\) and the \(y\)-axis at the point \(C, D\). Then the area of the quadrilateral ACBD is equal to :

1 \(\frac{9}{2}\)

2 \(\frac{3 \sqrt{17}}{2}\)

3 \(\frac{3 \sqrt{17}}{4}\)

4 9

Explanation:

(B) : Let points \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is a point s.t.
\((\mathrm{h}-1)^{2}+(\mathrm{k}-2)^{2}+(\mathrm{h}+2)^{2}+(\mathrm{k}-1)^{2}=14\)
So,
\(\mathrm{h}^{2}-2 \mathrm{~h}+1+\mathrm{k}^{2}+4-4 \mathrm{k}+\mathrm{h}^{2}+4 \mathrm{~h}+4+\mathrm{k}^{2}+1-2 \mathrm{k}=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k+1+4+4+1=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k=4\)
Or \(\quad h^{2}+h+k^{2}-3 k=2\)
Or \(\quad \mathrm{h}^{2}+\mathrm{h}+\frac{1}{(2)^{2}}+\mathrm{k}^{2}-3 \mathrm{k}+\left(\frac{3}{2}\right)^{2}=2+\frac{1}{4}+\frac{9}{4}\)
Or \(\quad\left(h+\frac{1}{2}\right)^{2}+\left(\mathrm{k}-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
Put \(\mathrm{h}=\mathrm{x}\) and \(\mathrm{k}=\mathrm{y}\), we get
\(\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
At \(\mathrm{x}\)-axis, \(\mathrm{y}=0\)
\(\left(x+\frac{1}{2}\right)^{2}+\frac{9}{4}=\frac{9}{2}\)
Or \(\quad\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}\) or \(\left(x+\frac{1}{2}\right)= \pm \frac{3}{2}\)
\(\therefore \quad \mathrm{x}=\frac{3}{2}-\frac{1}{2}=1\) and \(\mathrm{x}+\frac{1}{2}=\frac{-3}{2} \Rightarrow \mathrm{x}=-2\)
So, point \(A(-2,0)\) and point \(B\) is \((1,0)\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0\)
\(\therefore \quad\left(0+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
\(\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}-\frac{1}{4}=\frac{18}{4}-\frac{1}{4}=\frac{17}{4}\)
\(\therefore \quad\left(\mathrm{y}-\frac{3}{2}\right)= \pm \frac{\sqrt{17}}{2}\)
\(\therefore \quad y=\frac{3}{2} \pm \frac{\sqrt{17}}{2}=\frac{3+\sqrt{17}}{2}, \frac{3-\sqrt{17}}{2}\)
So, point \(\mathrm{C}\) is \(\left(0, \frac{3+\sqrt{17}}{2}\right)\) and point \(\mathrm{D}\) is \(\left(0, \frac{3-\sqrt{17}}{2}\right)\)
Now, area of quadrilateral is given by
\(A=\frac{1}{2}\left[\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{4}+x_{4} y_{1}\right]\right.\)
\(=-\left[\left(x_{2} y_{1}+x_{3} y_{2}+x_{4} y_{3}+x_{1} y_{4}\right)\right]\)
\(=\frac{1}{2}\left[(-2) \cdot\left(\frac{3-\sqrt{17}}{2}\right)+1 \cdot \frac{3+\sqrt{17}}{2}+0+0\right]\)
\(\quad-\left(0+1 \times \frac{3-\sqrt{17}}{2}+0+(-2) \cdot \frac{3+\sqrt{17}}{2}\right)\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3+\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3}{2}+\frac{\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]=\frac{3 \sqrt{17}}{2}\)

JEE Main-2022-26.07.2022

Application of the Integrals

87022 Area enclosed by the curve
\(\pi\left[4(x-\sqrt{2})^{2}+y^{2}\right]=8 \text { is }\)

1 \(\pi\)

2 2

3 \(3 \pi\)

4 4

VITEEE-2008

Application of the Integrals

87023 The odd natural number a, such that the area of the region bounded by \(y=1, y=3, x=0, x=y^{a}\)
is \(\frac{364}{3}\), is equal to :

1 3

2 5

3 7

4 9

JEE Main-2022-26.07.2022

Application of the Integrals

87024 If \(f(x)\) be continuous function such that the area bounded by the curve \(\mathrm{y}=f(\mathrm{x})\), the \(\mathrm{x}\)-axis and the lines \(x=a\) and \(x=0\) is \(\frac{a^{2}}{2}+\frac{a}{2} \sin a+\frac{\pi}{2}\) cos a. Value of \(f\left(\frac{\pi}{2}\right)\) is

1 \(1 / 2\)

2 \(\mathrm{a} / 2\)

3 \(\mathrm{a}^{2} / 2\)

4 \(\pi / 2\)

AMU-2009

Application of the Integrals

87020 The area enclosed by \(y^{2}=8 x\) and \(y=\sqrt{2} x\) that lies outside the triangle formed by \(y=\sqrt{2} x, x=\) \(1, y=2 \sqrt{2}\), is equal to :

1 \(\frac{16 \sqrt{2}}{6}\)

2 \(\frac{11 \sqrt{2}}{6}\)

3 \(\frac{13 \sqrt{2}}{6}\)

4 \(\frac{5 \sqrt{2}}{6}\)

JEE Main-2022-29.06.2022

Application of the Integrals

87021 A point \(P\) moves so that the sum of squares of its distances from the points \((1,2)\) and \((-2,1)\) is 14. Let \(f(x, y)=0\) be the locus of \(P\), which intersects the \(x\)-axis at the point \(A, B\) and the \(y\)-axis at the point \(C, D\). Then the area of the quadrilateral ACBD is equal to :

1 \(\frac{9}{2}\)

2 \(\frac{3 \sqrt{17}}{2}\)

3 \(\frac{3 \sqrt{17}}{4}\)

4 9

Explanation:

(B) : Let points \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is a point s.t.
\((\mathrm{h}-1)^{2}+(\mathrm{k}-2)^{2}+(\mathrm{h}+2)^{2}+(\mathrm{k}-1)^{2}=14\)
So,
\(\mathrm{h}^{2}-2 \mathrm{~h}+1+\mathrm{k}^{2}+4-4 \mathrm{k}+\mathrm{h}^{2}+4 \mathrm{~h}+4+\mathrm{k}^{2}+1-2 \mathrm{k}=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k+1+4+4+1=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k=4\)
Or \(\quad h^{2}+h+k^{2}-3 k=2\)
Or \(\quad \mathrm{h}^{2}+\mathrm{h}+\frac{1}{(2)^{2}}+\mathrm{k}^{2}-3 \mathrm{k}+\left(\frac{3}{2}\right)^{2}=2+\frac{1}{4}+\frac{9}{4}\)
Or \(\quad\left(h+\frac{1}{2}\right)^{2}+\left(\mathrm{k}-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
Put \(\mathrm{h}=\mathrm{x}\) and \(\mathrm{k}=\mathrm{y}\), we get
\(\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
At \(\mathrm{x}\)-axis, \(\mathrm{y}=0\)
\(\left(x+\frac{1}{2}\right)^{2}+\frac{9}{4}=\frac{9}{2}\)
Or \(\quad\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}\) or \(\left(x+\frac{1}{2}\right)= \pm \frac{3}{2}\)
\(\therefore \quad \mathrm{x}=\frac{3}{2}-\frac{1}{2}=1\) and \(\mathrm{x}+\frac{1}{2}=\frac{-3}{2} \Rightarrow \mathrm{x}=-2\)
So, point \(A(-2,0)\) and point \(B\) is \((1,0)\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0\)
\(\therefore \quad\left(0+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
\(\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}-\frac{1}{4}=\frac{18}{4}-\frac{1}{4}=\frac{17}{4}\)
\(\therefore \quad\left(\mathrm{y}-\frac{3}{2}\right)= \pm \frac{\sqrt{17}}{2}\)
\(\therefore \quad y=\frac{3}{2} \pm \frac{\sqrt{17}}{2}=\frac{3+\sqrt{17}}{2}, \frac{3-\sqrt{17}}{2}\)
So, point \(\mathrm{C}\) is \(\left(0, \frac{3+\sqrt{17}}{2}\right)\) and point \(\mathrm{D}\) is \(\left(0, \frac{3-\sqrt{17}}{2}\right)\)
Now, area of quadrilateral is given by
\(A=\frac{1}{2}\left[\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{4}+x_{4} y_{1}\right]\right.\)
\(=-\left[\left(x_{2} y_{1}+x_{3} y_{2}+x_{4} y_{3}+x_{1} y_{4}\right)\right]\)
\(=\frac{1}{2}\left[(-2) \cdot\left(\frac{3-\sqrt{17}}{2}\right)+1 \cdot \frac{3+\sqrt{17}}{2}+0+0\right]\)
\(\quad-\left(0+1 \times \frac{3-\sqrt{17}}{2}+0+(-2) \cdot \frac{3+\sqrt{17}}{2}\right)\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3+\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3}{2}+\frac{\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]=\frac{3 \sqrt{17}}{2}\)

JEE Main-2022-26.07.2022

Application of the Integrals

87022 Area enclosed by the curve
\(\pi\left[4(x-\sqrt{2})^{2}+y^{2}\right]=8 \text { is }\)

1 \(\pi\)

2 2

3 \(3 \pi\)

4 4

VITEEE-2008

Application of the Integrals

87023 The odd natural number a, such that the area of the region bounded by \(y=1, y=3, x=0, x=y^{a}\)
is \(\frac{364}{3}\), is equal to :

1 3

2 5

3 7

4 9

JEE Main-2022-26.07.2022

Application of the Integrals

87024 If \(f(x)\) be continuous function such that the area bounded by the curve \(\mathrm{y}=f(\mathrm{x})\), the \(\mathrm{x}\)-axis and the lines \(x=a\) and \(x=0\) is \(\frac{a^{2}}{2}+\frac{a}{2} \sin a+\frac{\pi}{2}\) cos a. Value of \(f\left(\frac{\pi}{2}\right)\) is

1 \(1 / 2\)

2 \(\mathrm{a} / 2\)

3 \(\mathrm{a}^{2} / 2\)

4 \(\pi / 2\)

AMU-2009

Application of the Integrals

87020 The area enclosed by \(y^{2}=8 x\) and \(y=\sqrt{2} x\) that lies outside the triangle formed by \(y=\sqrt{2} x, x=\) \(1, y=2 \sqrt{2}\), is equal to :

1 \(\frac{16 \sqrt{2}}{6}\)

2 \(\frac{11 \sqrt{2}}{6}\)

3 \(\frac{13 \sqrt{2}}{6}\)

4 \(\frac{5 \sqrt{2}}{6}\)

JEE Main-2022-29.06.2022

Application of the Integrals

87021 A point \(P\) moves so that the sum of squares of its distances from the points \((1,2)\) and \((-2,1)\) is 14. Let \(f(x, y)=0\) be the locus of \(P\), which intersects the \(x\)-axis at the point \(A, B\) and the \(y\)-axis at the point \(C, D\). Then the area of the quadrilateral ACBD is equal to :

1 \(\frac{9}{2}\)

2 \(\frac{3 \sqrt{17}}{2}\)

3 \(\frac{3 \sqrt{17}}{4}\)

4 9

Explanation:

(B) : Let points \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is a point s.t.
\((\mathrm{h}-1)^{2}+(\mathrm{k}-2)^{2}+(\mathrm{h}+2)^{2}+(\mathrm{k}-1)^{2}=14\)
So,
\(\mathrm{h}^{2}-2 \mathrm{~h}+1+\mathrm{k}^{2}+4-4 \mathrm{k}+\mathrm{h}^{2}+4 \mathrm{~h}+4+\mathrm{k}^{2}+1-2 \mathrm{k}=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k+1+4+4+1=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k=4\)
Or \(\quad h^{2}+h+k^{2}-3 k=2\)
Or \(\quad \mathrm{h}^{2}+\mathrm{h}+\frac{1}{(2)^{2}}+\mathrm{k}^{2}-3 \mathrm{k}+\left(\frac{3}{2}\right)^{2}=2+\frac{1}{4}+\frac{9}{4}\)
Or \(\quad\left(h+\frac{1}{2}\right)^{2}+\left(\mathrm{k}-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
Put \(\mathrm{h}=\mathrm{x}\) and \(\mathrm{k}=\mathrm{y}\), we get
\(\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
At \(\mathrm{x}\)-axis, \(\mathrm{y}=0\)
\(\left(x+\frac{1}{2}\right)^{2}+\frac{9}{4}=\frac{9}{2}\)
Or \(\quad\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}\) or \(\left(x+\frac{1}{2}\right)= \pm \frac{3}{2}\)
\(\therefore \quad \mathrm{x}=\frac{3}{2}-\frac{1}{2}=1\) and \(\mathrm{x}+\frac{1}{2}=\frac{-3}{2} \Rightarrow \mathrm{x}=-2\)
So, point \(A(-2,0)\) and point \(B\) is \((1,0)\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0\)
\(\therefore \quad\left(0+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
\(\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}-\frac{1}{4}=\frac{18}{4}-\frac{1}{4}=\frac{17}{4}\)
\(\therefore \quad\left(\mathrm{y}-\frac{3}{2}\right)= \pm \frac{\sqrt{17}}{2}\)
\(\therefore \quad y=\frac{3}{2} \pm \frac{\sqrt{17}}{2}=\frac{3+\sqrt{17}}{2}, \frac{3-\sqrt{17}}{2}\)
So, point \(\mathrm{C}\) is \(\left(0, \frac{3+\sqrt{17}}{2}\right)\) and point \(\mathrm{D}\) is \(\left(0, \frac{3-\sqrt{17}}{2}\right)\)
Now, area of quadrilateral is given by
\(A=\frac{1}{2}\left[\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{4}+x_{4} y_{1}\right]\right.\)
\(=-\left[\left(x_{2} y_{1}+x_{3} y_{2}+x_{4} y_{3}+x_{1} y_{4}\right)\right]\)
\(=\frac{1}{2}\left[(-2) \cdot\left(\frac{3-\sqrt{17}}{2}\right)+1 \cdot \frac{3+\sqrt{17}}{2}+0+0\right]\)
\(\quad-\left(0+1 \times \frac{3-\sqrt{17}}{2}+0+(-2) \cdot \frac{3+\sqrt{17}}{2}\right)\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3+\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3}{2}+\frac{\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]=\frac{3 \sqrt{17}}{2}\)

JEE Main-2022-26.07.2022

Application of the Integrals

87022 Area enclosed by the curve
\(\pi\left[4(x-\sqrt{2})^{2}+y^{2}\right]=8 \text { is }\)

1 \(\pi\)

2 2

3 \(3 \pi\)

4 4

VITEEE-2008

Application of the Integrals

87023 The odd natural number a, such that the area of the region bounded by \(y=1, y=3, x=0, x=y^{a}\)
is \(\frac{364}{3}\), is equal to :

1 3

2 5

3 7

4 9

JEE Main-2022-26.07.2022

Application of the Integrals

87024 If \(f(x)\) be continuous function such that the area bounded by the curve \(\mathrm{y}=f(\mathrm{x})\), the \(\mathrm{x}\)-axis and the lines \(x=a\) and \(x=0\) is \(\frac{a^{2}}{2}+\frac{a}{2} \sin a+\frac{\pi}{2}\) cos a. Value of \(f\left(\frac{\pi}{2}\right)\) is

1 \(1 / 2\)

2 \(\mathrm{a} / 2\)

3 \(\mathrm{a}^{2} / 2\)

4 \(\pi / 2\)

AMU-2009

Application of the Integrals

87020 The area enclosed by \(y^{2}=8 x\) and \(y=\sqrt{2} x\) that lies outside the triangle formed by \(y=\sqrt{2} x, x=\) \(1, y=2 \sqrt{2}\), is equal to :

1 \(\frac{16 \sqrt{2}}{6}\)

2 \(\frac{11 \sqrt{2}}{6}\)

3 \(\frac{13 \sqrt{2}}{6}\)

4 \(\frac{5 \sqrt{2}}{6}\)

JEE Main-2022-29.06.2022

Application of the Integrals

87021 A point \(P\) moves so that the sum of squares of its distances from the points \((1,2)\) and \((-2,1)\) is 14. Let \(f(x, y)=0\) be the locus of \(P\), which intersects the \(x\)-axis at the point \(A, B\) and the \(y\)-axis at the point \(C, D\). Then the area of the quadrilateral ACBD is equal to :

1 \(\frac{9}{2}\)

2 \(\frac{3 \sqrt{17}}{2}\)

3 \(\frac{3 \sqrt{17}}{4}\)

4 9

Explanation:

(B) : Let points \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is a point s.t.
\((\mathrm{h}-1)^{2}+(\mathrm{k}-2)^{2}+(\mathrm{h}+2)^{2}+(\mathrm{k}-1)^{2}=14\)
So,
\(\mathrm{h}^{2}-2 \mathrm{~h}+1+\mathrm{k}^{2}+4-4 \mathrm{k}+\mathrm{h}^{2}+4 \mathrm{~h}+4+\mathrm{k}^{2}+1-2 \mathrm{k}=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k+1+4+4+1=14\)
Or \(\quad 2 h^{2}+2 h+2 k^{2}-6 k=4\)
Or \(\quad h^{2}+h+k^{2}-3 k=2\)
Or \(\quad \mathrm{h}^{2}+\mathrm{h}+\frac{1}{(2)^{2}}+\mathrm{k}^{2}-3 \mathrm{k}+\left(\frac{3}{2}\right)^{2}=2+\frac{1}{4}+\frac{9}{4}\)
Or \(\quad\left(h+\frac{1}{2}\right)^{2}+\left(\mathrm{k}-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
Put \(\mathrm{h}=\mathrm{x}\) and \(\mathrm{k}=\mathrm{y}\), we get
\(\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
At \(\mathrm{x}\)-axis, \(\mathrm{y}=0\)
\(\left(x+\frac{1}{2}\right)^{2}+\frac{9}{4}=\frac{9}{2}\)
Or \(\quad\left(x+\frac{1}{2}\right)^{2}=\frac{9}{4}\) or \(\left(x+\frac{1}{2}\right)= \pm \frac{3}{2}\)
\(\therefore \quad \mathrm{x}=\frac{3}{2}-\frac{1}{2}=1\) and \(\mathrm{x}+\frac{1}{2}=\frac{-3}{2} \Rightarrow \mathrm{x}=-2\)
So, point \(A(-2,0)\) and point \(B\) is \((1,0)\)
At \(\mathrm{y}\)-axis, \(\mathrm{x}=0\)
\(\therefore \quad\left(0+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}\)
\(\left(y-\frac{3}{2}\right)^{2}=\frac{9}{2}-\frac{1}{4}=\frac{18}{4}-\frac{1}{4}=\frac{17}{4}\)
\(\therefore \quad\left(\mathrm{y}-\frac{3}{2}\right)= \pm \frac{\sqrt{17}}{2}\)
\(\therefore \quad y=\frac{3}{2} \pm \frac{\sqrt{17}}{2}=\frac{3+\sqrt{17}}{2}, \frac{3-\sqrt{17}}{2}\)
So, point \(\mathrm{C}\) is \(\left(0, \frac{3+\sqrt{17}}{2}\right)\) and point \(\mathrm{D}\) is \(\left(0, \frac{3-\sqrt{17}}{2}\right)\)
Now, area of quadrilateral is given by
\(A=\frac{1}{2}\left[\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{4}+x_{4} y_{1}\right]\right.\)
\(=-\left[\left(x_{2} y_{1}+x_{3} y_{2}+x_{4} y_{3}+x_{1} y_{4}\right)\right]\)
\(=\frac{1}{2}\left[(-2) \cdot\left(\frac{3-\sqrt{17}}{2}\right)+1 \cdot \frac{3+\sqrt{17}}{2}+0+0\right]\)
\(\quad-\left(0+1 \times \frac{3-\sqrt{17}}{2}+0+(-2) \cdot \frac{3+\sqrt{17}}{2}\right)\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3+\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]\)
\(=\frac{1}{2}\left[-3+\sqrt{17}+\frac{3}{2}+\frac{\sqrt{17}}{2}-\frac{3}{2}+\frac{\sqrt{17}}{2}+3+\sqrt{17}\right]=\frac{3 \sqrt{17}}{2}\)

JEE Main-2022-26.07.2022

Application of the Integrals

87022 Area enclosed by the curve
\(\pi\left[4(x-\sqrt{2})^{2}+y^{2}\right]=8 \text { is }\)

1 \(\pi\)

2 2

3 \(3 \pi\)

4 4

VITEEE-2008

Application of the Integrals

87023 The odd natural number a, such that the area of the region bounded by \(y=1, y=3, x=0, x=y^{a}\)
is \(\frac{364}{3}\), is equal to :

1 3

2 5

3 7

4 9

JEE Main-2022-26.07.2022

Application of the Integrals

87024 If \(f(x)\) be continuous function such that the area bounded by the curve \(\mathrm{y}=f(\mathrm{x})\), the \(\mathrm{x}\)-axis and the lines \(x=a\) and \(x=0\) is \(\frac{a^{2}}{2}+\frac{a}{2} \sin a+\frac{\pi}{2}\) cos a. Value of \(f\left(\frac{\pi}{2}\right)\) is

1 \(1 / 2\)

2 \(\mathrm{a} / 2\)

3 \(\mathrm{a}^{2} / 2\)

4 \(\pi / 2\)

AMU-2009

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