87033
If a point \(A(x, y)\) lies in the region bounded by the \(y\)-axis, straight lines \(2 y+x=6\) and \(5 x-6 y\) \(=30\), then the probability that \(\mathrm{y}\lt 1\) is :
1 \(\frac{1}{6}\)
2 \(\frac{5}{6}\)
3 \(\frac{2}{3}\)
4 \(\frac{6}{7}\)
Explanation:
(B) : Given, Now, The probability that \(\mathrm{y}\lt 1\) \(=1-\frac{\operatorname{area}(\mathrm{BDE})}{\operatorname{area}(\mathrm{ABC})}=1-\frac{\frac{1}{2} \times 2 \times 4}{\frac{1}{2} \times 8 \times 6}=1-\frac{1}{6}=\frac{5}{6}\)
Shift-II
Application of the Integrals
87054
The area of the region \(R=\left\{(x, y): 5 x^{2} \leq y \leq 2 x^{2}\right.\) \(+9\}\) is
1 \(11 \sqrt{3}\) sq units
2 \(12, \sqrt{3}\) sq units
3 \(9 \sqrt{3}\) sq units
4 \(6 \sqrt{3}\) sq units
Explanation:
(B) : Required area \(=2 \int_{0}^{\sqrt{3}}\left(2 x^{2}+9-5 x^{2}\right) d x\) \(=2 \int_{0}^{\sqrt{3}}\left(9-3 x^{2}\right) d x=2\left[9 x-x^{3}\right]_{0}^{\sqrt{3}}\) \(=2\left[9 \sqrt{3}-(\sqrt{3})^{3}-0\right]=2[9 \sqrt{3}-3 \sqrt{3}]\) \(=2[6 \sqrt{3}]=12 \sqrt{3} \text { Square units }\)
JEE Main-2021-25.02.2021
Application of the Integrals
87063
If the area of the circle \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda\) \(=0\) is \(9 \pi\) sq units, then the value of \(\lambda\) is
1 4
2 -4
3 16
4 -16
5 -8
Explanation:
(D) : Given, \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda=0\) \(x^{2}+y^{2}-2 x+4 y+\frac{\lambda}{4}=0\) Radius of circle \((\mathrm{r})=\sqrt{(-1)^{2}+(2)^{2}-\frac{\lambda}{4}}=\sqrt{\frac{20-\lambda}{4}}\) Area \((\mathrm{A})=\pi \mathrm{r}^{2}\) \(9 \pi=\pi \mathrm{r}^{2}\) \(9 \pi=\pi\left(\sqrt{\frac{20-\lambda}{4}}\right)^{2} \Rightarrow 9 \pi=\pi \frac{20-\lambda}{4}\) \(20-\lambda=36 \Rightarrow \lambda=-16\)
Kerala CEE-2016
Application of the Integrals
87034
The area enclosed by the curves \(y=x|x| \cdot x=-1\) and \(x=1\) is sq. units.
1 \(3 / 2\)
2 \(2 / 3\)
3 \(5 / 3\)
4 \(7 / 3\)
Explanation:
(B): Given, The curve are \(y=x|x|=\left\{\begin{array}{c}x^{2}: x \geq 0 \\ -x^{2}: x\lt 0\end{array}\right.\) \(x \pm 1\) The area enclosed by the curves and of , first part equal to area of second part \(\mathrm{A}_{1}=\int_{0}^{1} \mathrm{x}^{2} \mathrm{dx}=\frac{\mathrm{x}^{3}}{3}\) Similary \(\mathrm{A}_{2}=\frac{1}{3} \Rightarrow \mathrm{A}=\frac{1}{3}+\frac{1}{3} \Rightarrow \mathrm{A}=2 / 3\)
Shift-2]
Application of the Integrals
87044
The area (in sq units) of the region \(\mathbf{A}=\) \(\left\{(x, y): \frac{y^{2}}{2} \leq x \leq y+4\right\}\) is
1 30
2 \(\frac{53}{3}\)
3 16
4 18
Explanation:
(D) : Given, We know that, The point of intersection \(x-y-4=0 \tag{i}\) \(y^{2}-y-4=0\) \(y^{2}-2 y-8=0\) \(y^{2}-4 y+2 y-8=0\) \(y(y-4)+2(y-4)=0\) \((y-4)(y+2)=0\) \(y=4,-2\) \(\therefore \quad \mathrm{x}=8,2\) Now area \(A=\int_{-2}^{4}\left\{(y+4)-\frac{y^{2}}{2}\right\} d y\) \(=\left\{\frac{y^{2}}{2}+4 y-\frac{y^{3}}{6}\right\}_{-2}^{4}=\left\{8+16-\frac{64}{6}-\left(2-8+\frac{8}{6}\right)\right\}\) \(=24-\frac{64}{6}+6-\frac{8}{6}=30-\frac{72}{6}=18\)
87033
If a point \(A(x, y)\) lies in the region bounded by the \(y\)-axis, straight lines \(2 y+x=6\) and \(5 x-6 y\) \(=30\), then the probability that \(\mathrm{y}\lt 1\) is :
1 \(\frac{1}{6}\)
2 \(\frac{5}{6}\)
3 \(\frac{2}{3}\)
4 \(\frac{6}{7}\)
Explanation:
(B) : Given, Now, The probability that \(\mathrm{y}\lt 1\) \(=1-\frac{\operatorname{area}(\mathrm{BDE})}{\operatorname{area}(\mathrm{ABC})}=1-\frac{\frac{1}{2} \times 2 \times 4}{\frac{1}{2} \times 8 \times 6}=1-\frac{1}{6}=\frac{5}{6}\)
Shift-II
Application of the Integrals
87054
The area of the region \(R=\left\{(x, y): 5 x^{2} \leq y \leq 2 x^{2}\right.\) \(+9\}\) is
1 \(11 \sqrt{3}\) sq units
2 \(12, \sqrt{3}\) sq units
3 \(9 \sqrt{3}\) sq units
4 \(6 \sqrt{3}\) sq units
Explanation:
(B) : Required area \(=2 \int_{0}^{\sqrt{3}}\left(2 x^{2}+9-5 x^{2}\right) d x\) \(=2 \int_{0}^{\sqrt{3}}\left(9-3 x^{2}\right) d x=2\left[9 x-x^{3}\right]_{0}^{\sqrt{3}}\) \(=2\left[9 \sqrt{3}-(\sqrt{3})^{3}-0\right]=2[9 \sqrt{3}-3 \sqrt{3}]\) \(=2[6 \sqrt{3}]=12 \sqrt{3} \text { Square units }\)
JEE Main-2021-25.02.2021
Application of the Integrals
87063
If the area of the circle \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda\) \(=0\) is \(9 \pi\) sq units, then the value of \(\lambda\) is
1 4
2 -4
3 16
4 -16
5 -8
Explanation:
(D) : Given, \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda=0\) \(x^{2}+y^{2}-2 x+4 y+\frac{\lambda}{4}=0\) Radius of circle \((\mathrm{r})=\sqrt{(-1)^{2}+(2)^{2}-\frac{\lambda}{4}}=\sqrt{\frac{20-\lambda}{4}}\) Area \((\mathrm{A})=\pi \mathrm{r}^{2}\) \(9 \pi=\pi \mathrm{r}^{2}\) \(9 \pi=\pi\left(\sqrt{\frac{20-\lambda}{4}}\right)^{2} \Rightarrow 9 \pi=\pi \frac{20-\lambda}{4}\) \(20-\lambda=36 \Rightarrow \lambda=-16\)
Kerala CEE-2016
Application of the Integrals
87034
The area enclosed by the curves \(y=x|x| \cdot x=-1\) and \(x=1\) is sq. units.
1 \(3 / 2\)
2 \(2 / 3\)
3 \(5 / 3\)
4 \(7 / 3\)
Explanation:
(B): Given, The curve are \(y=x|x|=\left\{\begin{array}{c}x^{2}: x \geq 0 \\ -x^{2}: x\lt 0\end{array}\right.\) \(x \pm 1\) The area enclosed by the curves and of , first part equal to area of second part \(\mathrm{A}_{1}=\int_{0}^{1} \mathrm{x}^{2} \mathrm{dx}=\frac{\mathrm{x}^{3}}{3}\) Similary \(\mathrm{A}_{2}=\frac{1}{3} \Rightarrow \mathrm{A}=\frac{1}{3}+\frac{1}{3} \Rightarrow \mathrm{A}=2 / 3\)
Shift-2]
Application of the Integrals
87044
The area (in sq units) of the region \(\mathbf{A}=\) \(\left\{(x, y): \frac{y^{2}}{2} \leq x \leq y+4\right\}\) is
1 30
2 \(\frac{53}{3}\)
3 16
4 18
Explanation:
(D) : Given, We know that, The point of intersection \(x-y-4=0 \tag{i}\) \(y^{2}-y-4=0\) \(y^{2}-2 y-8=0\) \(y^{2}-4 y+2 y-8=0\) \(y(y-4)+2(y-4)=0\) \((y-4)(y+2)=0\) \(y=4,-2\) \(\therefore \quad \mathrm{x}=8,2\) Now area \(A=\int_{-2}^{4}\left\{(y+4)-\frac{y^{2}}{2}\right\} d y\) \(=\left\{\frac{y^{2}}{2}+4 y-\frac{y^{3}}{6}\right\}_{-2}^{4}=\left\{8+16-\frac{64}{6}-\left(2-8+\frac{8}{6}\right)\right\}\) \(=24-\frac{64}{6}+6-\frac{8}{6}=30-\frac{72}{6}=18\)
87033
If a point \(A(x, y)\) lies in the region bounded by the \(y\)-axis, straight lines \(2 y+x=6\) and \(5 x-6 y\) \(=30\), then the probability that \(\mathrm{y}\lt 1\) is :
1 \(\frac{1}{6}\)
2 \(\frac{5}{6}\)
3 \(\frac{2}{3}\)
4 \(\frac{6}{7}\)
Explanation:
(B) : Given, Now, The probability that \(\mathrm{y}\lt 1\) \(=1-\frac{\operatorname{area}(\mathrm{BDE})}{\operatorname{area}(\mathrm{ABC})}=1-\frac{\frac{1}{2} \times 2 \times 4}{\frac{1}{2} \times 8 \times 6}=1-\frac{1}{6}=\frac{5}{6}\)
Shift-II
Application of the Integrals
87054
The area of the region \(R=\left\{(x, y): 5 x^{2} \leq y \leq 2 x^{2}\right.\) \(+9\}\) is
1 \(11 \sqrt{3}\) sq units
2 \(12, \sqrt{3}\) sq units
3 \(9 \sqrt{3}\) sq units
4 \(6 \sqrt{3}\) sq units
Explanation:
(B) : Required area \(=2 \int_{0}^{\sqrt{3}}\left(2 x^{2}+9-5 x^{2}\right) d x\) \(=2 \int_{0}^{\sqrt{3}}\left(9-3 x^{2}\right) d x=2\left[9 x-x^{3}\right]_{0}^{\sqrt{3}}\) \(=2\left[9 \sqrt{3}-(\sqrt{3})^{3}-0\right]=2[9 \sqrt{3}-3 \sqrt{3}]\) \(=2[6 \sqrt{3}]=12 \sqrt{3} \text { Square units }\)
JEE Main-2021-25.02.2021
Application of the Integrals
87063
If the area of the circle \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda\) \(=0\) is \(9 \pi\) sq units, then the value of \(\lambda\) is
1 4
2 -4
3 16
4 -16
5 -8
Explanation:
(D) : Given, \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda=0\) \(x^{2}+y^{2}-2 x+4 y+\frac{\lambda}{4}=0\) Radius of circle \((\mathrm{r})=\sqrt{(-1)^{2}+(2)^{2}-\frac{\lambda}{4}}=\sqrt{\frac{20-\lambda}{4}}\) Area \((\mathrm{A})=\pi \mathrm{r}^{2}\) \(9 \pi=\pi \mathrm{r}^{2}\) \(9 \pi=\pi\left(\sqrt{\frac{20-\lambda}{4}}\right)^{2} \Rightarrow 9 \pi=\pi \frac{20-\lambda}{4}\) \(20-\lambda=36 \Rightarrow \lambda=-16\)
Kerala CEE-2016
Application of the Integrals
87034
The area enclosed by the curves \(y=x|x| \cdot x=-1\) and \(x=1\) is sq. units.
1 \(3 / 2\)
2 \(2 / 3\)
3 \(5 / 3\)
4 \(7 / 3\)
Explanation:
(B): Given, The curve are \(y=x|x|=\left\{\begin{array}{c}x^{2}: x \geq 0 \\ -x^{2}: x\lt 0\end{array}\right.\) \(x \pm 1\) The area enclosed by the curves and of , first part equal to area of second part \(\mathrm{A}_{1}=\int_{0}^{1} \mathrm{x}^{2} \mathrm{dx}=\frac{\mathrm{x}^{3}}{3}\) Similary \(\mathrm{A}_{2}=\frac{1}{3} \Rightarrow \mathrm{A}=\frac{1}{3}+\frac{1}{3} \Rightarrow \mathrm{A}=2 / 3\)
Shift-2]
Application of the Integrals
87044
The area (in sq units) of the region \(\mathbf{A}=\) \(\left\{(x, y): \frac{y^{2}}{2} \leq x \leq y+4\right\}\) is
1 30
2 \(\frac{53}{3}\)
3 16
4 18
Explanation:
(D) : Given, We know that, The point of intersection \(x-y-4=0 \tag{i}\) \(y^{2}-y-4=0\) \(y^{2}-2 y-8=0\) \(y^{2}-4 y+2 y-8=0\) \(y(y-4)+2(y-4)=0\) \((y-4)(y+2)=0\) \(y=4,-2\) \(\therefore \quad \mathrm{x}=8,2\) Now area \(A=\int_{-2}^{4}\left\{(y+4)-\frac{y^{2}}{2}\right\} d y\) \(=\left\{\frac{y^{2}}{2}+4 y-\frac{y^{3}}{6}\right\}_{-2}^{4}=\left\{8+16-\frac{64}{6}-\left(2-8+\frac{8}{6}\right)\right\}\) \(=24-\frac{64}{6}+6-\frac{8}{6}=30-\frac{72}{6}=18\)
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Application of the Integrals
87033
If a point \(A(x, y)\) lies in the region bounded by the \(y\)-axis, straight lines \(2 y+x=6\) and \(5 x-6 y\) \(=30\), then the probability that \(\mathrm{y}\lt 1\) is :
1 \(\frac{1}{6}\)
2 \(\frac{5}{6}\)
3 \(\frac{2}{3}\)
4 \(\frac{6}{7}\)
Explanation:
(B) : Given, Now, The probability that \(\mathrm{y}\lt 1\) \(=1-\frac{\operatorname{area}(\mathrm{BDE})}{\operatorname{area}(\mathrm{ABC})}=1-\frac{\frac{1}{2} \times 2 \times 4}{\frac{1}{2} \times 8 \times 6}=1-\frac{1}{6}=\frac{5}{6}\)
Shift-II
Application of the Integrals
87054
The area of the region \(R=\left\{(x, y): 5 x^{2} \leq y \leq 2 x^{2}\right.\) \(+9\}\) is
1 \(11 \sqrt{3}\) sq units
2 \(12, \sqrt{3}\) sq units
3 \(9 \sqrt{3}\) sq units
4 \(6 \sqrt{3}\) sq units
Explanation:
(B) : Required area \(=2 \int_{0}^{\sqrt{3}}\left(2 x^{2}+9-5 x^{2}\right) d x\) \(=2 \int_{0}^{\sqrt{3}}\left(9-3 x^{2}\right) d x=2\left[9 x-x^{3}\right]_{0}^{\sqrt{3}}\) \(=2\left[9 \sqrt{3}-(\sqrt{3})^{3}-0\right]=2[9 \sqrt{3}-3 \sqrt{3}]\) \(=2[6 \sqrt{3}]=12 \sqrt{3} \text { Square units }\)
JEE Main-2021-25.02.2021
Application of the Integrals
87063
If the area of the circle \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda\) \(=0\) is \(9 \pi\) sq units, then the value of \(\lambda\) is
1 4
2 -4
3 16
4 -16
5 -8
Explanation:
(D) : Given, \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda=0\) \(x^{2}+y^{2}-2 x+4 y+\frac{\lambda}{4}=0\) Radius of circle \((\mathrm{r})=\sqrt{(-1)^{2}+(2)^{2}-\frac{\lambda}{4}}=\sqrt{\frac{20-\lambda}{4}}\) Area \((\mathrm{A})=\pi \mathrm{r}^{2}\) \(9 \pi=\pi \mathrm{r}^{2}\) \(9 \pi=\pi\left(\sqrt{\frac{20-\lambda}{4}}\right)^{2} \Rightarrow 9 \pi=\pi \frac{20-\lambda}{4}\) \(20-\lambda=36 \Rightarrow \lambda=-16\)
Kerala CEE-2016
Application of the Integrals
87034
The area enclosed by the curves \(y=x|x| \cdot x=-1\) and \(x=1\) is sq. units.
1 \(3 / 2\)
2 \(2 / 3\)
3 \(5 / 3\)
4 \(7 / 3\)
Explanation:
(B): Given, The curve are \(y=x|x|=\left\{\begin{array}{c}x^{2}: x \geq 0 \\ -x^{2}: x\lt 0\end{array}\right.\) \(x \pm 1\) The area enclosed by the curves and of , first part equal to area of second part \(\mathrm{A}_{1}=\int_{0}^{1} \mathrm{x}^{2} \mathrm{dx}=\frac{\mathrm{x}^{3}}{3}\) Similary \(\mathrm{A}_{2}=\frac{1}{3} \Rightarrow \mathrm{A}=\frac{1}{3}+\frac{1}{3} \Rightarrow \mathrm{A}=2 / 3\)
Shift-2]
Application of the Integrals
87044
The area (in sq units) of the region \(\mathbf{A}=\) \(\left\{(x, y): \frac{y^{2}}{2} \leq x \leq y+4\right\}\) is
1 30
2 \(\frac{53}{3}\)
3 16
4 18
Explanation:
(D) : Given, We know that, The point of intersection \(x-y-4=0 \tag{i}\) \(y^{2}-y-4=0\) \(y^{2}-2 y-8=0\) \(y^{2}-4 y+2 y-8=0\) \(y(y-4)+2(y-4)=0\) \((y-4)(y+2)=0\) \(y=4,-2\) \(\therefore \quad \mathrm{x}=8,2\) Now area \(A=\int_{-2}^{4}\left\{(y+4)-\frac{y^{2}}{2}\right\} d y\) \(=\left\{\frac{y^{2}}{2}+4 y-\frac{y^{3}}{6}\right\}_{-2}^{4}=\left\{8+16-\frac{64}{6}-\left(2-8+\frac{8}{6}\right)\right\}\) \(=24-\frac{64}{6}+6-\frac{8}{6}=30-\frac{72}{6}=18\)
87033
If a point \(A(x, y)\) lies in the region bounded by the \(y\)-axis, straight lines \(2 y+x=6\) and \(5 x-6 y\) \(=30\), then the probability that \(\mathrm{y}\lt 1\) is :
1 \(\frac{1}{6}\)
2 \(\frac{5}{6}\)
3 \(\frac{2}{3}\)
4 \(\frac{6}{7}\)
Explanation:
(B) : Given, Now, The probability that \(\mathrm{y}\lt 1\) \(=1-\frac{\operatorname{area}(\mathrm{BDE})}{\operatorname{area}(\mathrm{ABC})}=1-\frac{\frac{1}{2} \times 2 \times 4}{\frac{1}{2} \times 8 \times 6}=1-\frac{1}{6}=\frac{5}{6}\)
Shift-II
Application of the Integrals
87054
The area of the region \(R=\left\{(x, y): 5 x^{2} \leq y \leq 2 x^{2}\right.\) \(+9\}\) is
1 \(11 \sqrt{3}\) sq units
2 \(12, \sqrt{3}\) sq units
3 \(9 \sqrt{3}\) sq units
4 \(6 \sqrt{3}\) sq units
Explanation:
(B) : Required area \(=2 \int_{0}^{\sqrt{3}}\left(2 x^{2}+9-5 x^{2}\right) d x\) \(=2 \int_{0}^{\sqrt{3}}\left(9-3 x^{2}\right) d x=2\left[9 x-x^{3}\right]_{0}^{\sqrt{3}}\) \(=2\left[9 \sqrt{3}-(\sqrt{3})^{3}-0\right]=2[9 \sqrt{3}-3 \sqrt{3}]\) \(=2[6 \sqrt{3}]=12 \sqrt{3} \text { Square units }\)
JEE Main-2021-25.02.2021
Application of the Integrals
87063
If the area of the circle \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda\) \(=0\) is \(9 \pi\) sq units, then the value of \(\lambda\) is
1 4
2 -4
3 16
4 -16
5 -8
Explanation:
(D) : Given, \(4 x^{2}+4 y^{2}-8 x+16 y+\lambda=0\) \(x^{2}+y^{2}-2 x+4 y+\frac{\lambda}{4}=0\) Radius of circle \((\mathrm{r})=\sqrt{(-1)^{2}+(2)^{2}-\frac{\lambda}{4}}=\sqrt{\frac{20-\lambda}{4}}\) Area \((\mathrm{A})=\pi \mathrm{r}^{2}\) \(9 \pi=\pi \mathrm{r}^{2}\) \(9 \pi=\pi\left(\sqrt{\frac{20-\lambda}{4}}\right)^{2} \Rightarrow 9 \pi=\pi \frac{20-\lambda}{4}\) \(20-\lambda=36 \Rightarrow \lambda=-16\)
Kerala CEE-2016
Application of the Integrals
87034
The area enclosed by the curves \(y=x|x| \cdot x=-1\) and \(x=1\) is sq. units.
1 \(3 / 2\)
2 \(2 / 3\)
3 \(5 / 3\)
4 \(7 / 3\)
Explanation:
(B): Given, The curve are \(y=x|x|=\left\{\begin{array}{c}x^{2}: x \geq 0 \\ -x^{2}: x\lt 0\end{array}\right.\) \(x \pm 1\) The area enclosed by the curves and of , first part equal to area of second part \(\mathrm{A}_{1}=\int_{0}^{1} \mathrm{x}^{2} \mathrm{dx}=\frac{\mathrm{x}^{3}}{3}\) Similary \(\mathrm{A}_{2}=\frac{1}{3} \Rightarrow \mathrm{A}=\frac{1}{3}+\frac{1}{3} \Rightarrow \mathrm{A}=2 / 3\)
Shift-2]
Application of the Integrals
87044
The area (in sq units) of the region \(\mathbf{A}=\) \(\left\{(x, y): \frac{y^{2}}{2} \leq x \leq y+4\right\}\) is
1 30
2 \(\frac{53}{3}\)
3 16
4 18
Explanation:
(D) : Given, We know that, The point of intersection \(x-y-4=0 \tag{i}\) \(y^{2}-y-4=0\) \(y^{2}-2 y-8=0\) \(y^{2}-4 y+2 y-8=0\) \(y(y-4)+2(y-4)=0\) \((y-4)(y+2)=0\) \(y=4,-2\) \(\therefore \quad \mathrm{x}=8,2\) Now area \(A=\int_{-2}^{4}\left\{(y+4)-\frac{y^{2}}{2}\right\} d y\) \(=\left\{\frac{y^{2}}{2}+4 y-\frac{y^{3}}{6}\right\}_{-2}^{4}=\left\{8+16-\frac{64}{6}-\left(2-8+\frac{8}{6}\right)\right\}\) \(=24-\frac{64}{6}+6-\frac{8}{6}=30-\frac{72}{6}=18\)
