87050
The area (in sq. units) of the region \(A=\{x, y)\) : \((x-1)[x] \leq y \leq 2 \sqrt{x}, \quad 0 \leq x \leq 2\}\), where \([t]\) denotes the greatest integer function, is
87051
Consider region \(R=\left\{(x, y) \in R^{2} ; x^{2} \leq y \leq 2 x\right\}\) if a line \(y=\alpha\) divides the area of region \(R\) into two equal parts, then which of the following is true?
1 \(\alpha^{3}-6 \alpha^{2}+16=0\)
2 \(3 \alpha^{2}-8 \alpha^{3 / 2}+8=0\)
3 \(3 \alpha^{2}-8 \alpha+8=0\)
4 \(\alpha^{3}-6 \alpha^{3 / 2}-16=0\)
Explanation:
(B) : Given, \(y \geq x^{2} \Rightarrow\) upper region of \(y=x^{2}\) \(y \leq 2 x \Rightarrow\) lower region of \(y=2 x\) So, Area of \(\mathrm{OABC}=2\) area of \(\mathrm{OAC}\) \(\int_{0}^{4}\left(\sqrt{y}-\frac{y}{2}\right) d x=2 \int_{0}^{\alpha}\left(\sqrt{y}-\frac{y}{2}\right) d y\) \({\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{4}=2\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{\alpha}}\) \(\frac{2}{3}(4)^{3 / 2}-\frac{(4)^{2}}{4}=2\left[\frac{2}{3}(\alpha)^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(\frac{2}{3} \times 8-4=2\left[\frac{2}{3} \alpha^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(\frac{4}{3}=2\left[2 \alpha^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(3 \alpha^{2}-8 \alpha^{3 / 2}+8=0\)
JEE Main-2020-02.09.2020
Application of the Integrals
87052
The area (in sq. units) of the region \(\{(x, y): 0 \leq\) \(\left.y \leq x+1,0 \leq y \leq x+1 \frac{1}{2} \leq x \leq 2\right\}\) is
1 \(\frac{23}{16}\)
2 \(\frac{79}{24}\)
3 \(\frac{79}{16}\)
4 \(\frac{23}{6}\)
Explanation:
(B) : Given, \(0 \leq y \leq x^{2}+1\) \(0 \leq y \leq x+1\) \(\frac{1}{2} \leq x \leq 2\) So, required area \(=\int_{\frac{1}{2}}^{1}\left(\mathrm{x}^{2}+1\right) \mathrm{dx}+\frac{1}{2}(2+3) \times 1\) \(=\left[\frac{\mathrm{x}^{3}}{3}+\mathrm{x}\right]_{1 / 2}^{1}+\frac{5}{2}=\frac{1}{3}+1-\frac{1}{24}-\frac{1}{2}+\frac{5}{2}\) \(=\frac{8+24-1-12+60}{24}=\frac{79}{24}\)
JEE Main-2020-03.09.2020
Application of the Integrals
87053
The area (in sq. units) of the part of the region \(A=\left\{(x, y):|x|+|y| \leq 1,2 \mathbf{y}^{2} \geq|\mathbf{x}|\right\}\) is
1 \(\frac{1}{3}\)
2 \(\frac{7}{6}\)
3 \(\frac{1}{6}\)
4 \(\frac{5}{6}\)
Explanation:
(D) : Given, unsolved \(|x|+|y| \leq 1\) \(2 y^{2} \geq|x|\) by the figure due to symmetry \(A=4 \int_{0}^{1 / 2}\left(1-x-\sqrt{\frac{x}{2}}\right) d x\) \(\because\left[2 y^{2}+y-1=0 \Rightarrow(2 y-1)(y+1)=0 \Rightarrow y=1 / 2\right]\) \(=4\left[x-\frac{x^{2}}{2}-\frac{2 x^{3 / 2}}{3 / \sqrt{2}}\right]_{0}^{1 / 2}\) \(=4\left[\frac{1}{2}-\frac{1}{8}-\frac{2}{12}\right]=\frac{5}{6}\) square units.
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Application of the Integrals
87050
The area (in sq. units) of the region \(A=\{x, y)\) : \((x-1)[x] \leq y \leq 2 \sqrt{x}, \quad 0 \leq x \leq 2\}\), where \([t]\) denotes the greatest integer function, is
87051
Consider region \(R=\left\{(x, y) \in R^{2} ; x^{2} \leq y \leq 2 x\right\}\) if a line \(y=\alpha\) divides the area of region \(R\) into two equal parts, then which of the following is true?
1 \(\alpha^{3}-6 \alpha^{2}+16=0\)
2 \(3 \alpha^{2}-8 \alpha^{3 / 2}+8=0\)
3 \(3 \alpha^{2}-8 \alpha+8=0\)
4 \(\alpha^{3}-6 \alpha^{3 / 2}-16=0\)
Explanation:
(B) : Given, \(y \geq x^{2} \Rightarrow\) upper region of \(y=x^{2}\) \(y \leq 2 x \Rightarrow\) lower region of \(y=2 x\) So, Area of \(\mathrm{OABC}=2\) area of \(\mathrm{OAC}\) \(\int_{0}^{4}\left(\sqrt{y}-\frac{y}{2}\right) d x=2 \int_{0}^{\alpha}\left(\sqrt{y}-\frac{y}{2}\right) d y\) \({\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{4}=2\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{\alpha}}\) \(\frac{2}{3}(4)^{3 / 2}-\frac{(4)^{2}}{4}=2\left[\frac{2}{3}(\alpha)^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(\frac{2}{3} \times 8-4=2\left[\frac{2}{3} \alpha^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(\frac{4}{3}=2\left[2 \alpha^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(3 \alpha^{2}-8 \alpha^{3 / 2}+8=0\)
JEE Main-2020-02.09.2020
Application of the Integrals
87052
The area (in sq. units) of the region \(\{(x, y): 0 \leq\) \(\left.y \leq x+1,0 \leq y \leq x+1 \frac{1}{2} \leq x \leq 2\right\}\) is
1 \(\frac{23}{16}\)
2 \(\frac{79}{24}\)
3 \(\frac{79}{16}\)
4 \(\frac{23}{6}\)
Explanation:
(B) : Given, \(0 \leq y \leq x^{2}+1\) \(0 \leq y \leq x+1\) \(\frac{1}{2} \leq x \leq 2\) So, required area \(=\int_{\frac{1}{2}}^{1}\left(\mathrm{x}^{2}+1\right) \mathrm{dx}+\frac{1}{2}(2+3) \times 1\) \(=\left[\frac{\mathrm{x}^{3}}{3}+\mathrm{x}\right]_{1 / 2}^{1}+\frac{5}{2}=\frac{1}{3}+1-\frac{1}{24}-\frac{1}{2}+\frac{5}{2}\) \(=\frac{8+24-1-12+60}{24}=\frac{79}{24}\)
JEE Main-2020-03.09.2020
Application of the Integrals
87053
The area (in sq. units) of the part of the region \(A=\left\{(x, y):|x|+|y| \leq 1,2 \mathbf{y}^{2} \geq|\mathbf{x}|\right\}\) is
1 \(\frac{1}{3}\)
2 \(\frac{7}{6}\)
3 \(\frac{1}{6}\)
4 \(\frac{5}{6}\)
Explanation:
(D) : Given, unsolved \(|x|+|y| \leq 1\) \(2 y^{2} \geq|x|\) by the figure due to symmetry \(A=4 \int_{0}^{1 / 2}\left(1-x-\sqrt{\frac{x}{2}}\right) d x\) \(\because\left[2 y^{2}+y-1=0 \Rightarrow(2 y-1)(y+1)=0 \Rightarrow y=1 / 2\right]\) \(=4\left[x-\frac{x^{2}}{2}-\frac{2 x^{3 / 2}}{3 / \sqrt{2}}\right]_{0}^{1 / 2}\) \(=4\left[\frac{1}{2}-\frac{1}{8}-\frac{2}{12}\right]=\frac{5}{6}\) square units.
87050
The area (in sq. units) of the region \(A=\{x, y)\) : \((x-1)[x] \leq y \leq 2 \sqrt{x}, \quad 0 \leq x \leq 2\}\), where \([t]\) denotes the greatest integer function, is
87051
Consider region \(R=\left\{(x, y) \in R^{2} ; x^{2} \leq y \leq 2 x\right\}\) if a line \(y=\alpha\) divides the area of region \(R\) into two equal parts, then which of the following is true?
1 \(\alpha^{3}-6 \alpha^{2}+16=0\)
2 \(3 \alpha^{2}-8 \alpha^{3 / 2}+8=0\)
3 \(3 \alpha^{2}-8 \alpha+8=0\)
4 \(\alpha^{3}-6 \alpha^{3 / 2}-16=0\)
Explanation:
(B) : Given, \(y \geq x^{2} \Rightarrow\) upper region of \(y=x^{2}\) \(y \leq 2 x \Rightarrow\) lower region of \(y=2 x\) So, Area of \(\mathrm{OABC}=2\) area of \(\mathrm{OAC}\) \(\int_{0}^{4}\left(\sqrt{y}-\frac{y}{2}\right) d x=2 \int_{0}^{\alpha}\left(\sqrt{y}-\frac{y}{2}\right) d y\) \({\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{4}=2\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{\alpha}}\) \(\frac{2}{3}(4)^{3 / 2}-\frac{(4)^{2}}{4}=2\left[\frac{2}{3}(\alpha)^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(\frac{2}{3} \times 8-4=2\left[\frac{2}{3} \alpha^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(\frac{4}{3}=2\left[2 \alpha^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(3 \alpha^{2}-8 \alpha^{3 / 2}+8=0\)
JEE Main-2020-02.09.2020
Application of the Integrals
87052
The area (in sq. units) of the region \(\{(x, y): 0 \leq\) \(\left.y \leq x+1,0 \leq y \leq x+1 \frac{1}{2} \leq x \leq 2\right\}\) is
1 \(\frac{23}{16}\)
2 \(\frac{79}{24}\)
3 \(\frac{79}{16}\)
4 \(\frac{23}{6}\)
Explanation:
(B) : Given, \(0 \leq y \leq x^{2}+1\) \(0 \leq y \leq x+1\) \(\frac{1}{2} \leq x \leq 2\) So, required area \(=\int_{\frac{1}{2}}^{1}\left(\mathrm{x}^{2}+1\right) \mathrm{dx}+\frac{1}{2}(2+3) \times 1\) \(=\left[\frac{\mathrm{x}^{3}}{3}+\mathrm{x}\right]_{1 / 2}^{1}+\frac{5}{2}=\frac{1}{3}+1-\frac{1}{24}-\frac{1}{2}+\frac{5}{2}\) \(=\frac{8+24-1-12+60}{24}=\frac{79}{24}\)
JEE Main-2020-03.09.2020
Application of the Integrals
87053
The area (in sq. units) of the part of the region \(A=\left\{(x, y):|x|+|y| \leq 1,2 \mathbf{y}^{2} \geq|\mathbf{x}|\right\}\) is
1 \(\frac{1}{3}\)
2 \(\frac{7}{6}\)
3 \(\frac{1}{6}\)
4 \(\frac{5}{6}\)
Explanation:
(D) : Given, unsolved \(|x|+|y| \leq 1\) \(2 y^{2} \geq|x|\) by the figure due to symmetry \(A=4 \int_{0}^{1 / 2}\left(1-x-\sqrt{\frac{x}{2}}\right) d x\) \(\because\left[2 y^{2}+y-1=0 \Rightarrow(2 y-1)(y+1)=0 \Rightarrow y=1 / 2\right]\) \(=4\left[x-\frac{x^{2}}{2}-\frac{2 x^{3 / 2}}{3 / \sqrt{2}}\right]_{0}^{1 / 2}\) \(=4\left[\frac{1}{2}-\frac{1}{8}-\frac{2}{12}\right]=\frac{5}{6}\) square units.
87050
The area (in sq. units) of the region \(A=\{x, y)\) : \((x-1)[x] \leq y \leq 2 \sqrt{x}, \quad 0 \leq x \leq 2\}\), where \([t]\) denotes the greatest integer function, is
87051
Consider region \(R=\left\{(x, y) \in R^{2} ; x^{2} \leq y \leq 2 x\right\}\) if a line \(y=\alpha\) divides the area of region \(R\) into two equal parts, then which of the following is true?
1 \(\alpha^{3}-6 \alpha^{2}+16=0\)
2 \(3 \alpha^{2}-8 \alpha^{3 / 2}+8=0\)
3 \(3 \alpha^{2}-8 \alpha+8=0\)
4 \(\alpha^{3}-6 \alpha^{3 / 2}-16=0\)
Explanation:
(B) : Given, \(y \geq x^{2} \Rightarrow\) upper region of \(y=x^{2}\) \(y \leq 2 x \Rightarrow\) lower region of \(y=2 x\) So, Area of \(\mathrm{OABC}=2\) area of \(\mathrm{OAC}\) \(\int_{0}^{4}\left(\sqrt{y}-\frac{y}{2}\right) d x=2 \int_{0}^{\alpha}\left(\sqrt{y}-\frac{y}{2}\right) d y\) \({\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{4}=2\left[\frac{2}{3} y^{3 / 2}-\frac{y^{2}}{4}\right]_{0}^{\alpha}}\) \(\frac{2}{3}(4)^{3 / 2}-\frac{(4)^{2}}{4}=2\left[\frac{2}{3}(\alpha)^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(\frac{2}{3} \times 8-4=2\left[\frac{2}{3} \alpha^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(\frac{4}{3}=2\left[2 \alpha^{3 / 2}-\frac{\alpha^{2}}{4}\right]\) \(3 \alpha^{2}-8 \alpha^{3 / 2}+8=0\)
JEE Main-2020-02.09.2020
Application of the Integrals
87052
The area (in sq. units) of the region \(\{(x, y): 0 \leq\) \(\left.y \leq x+1,0 \leq y \leq x+1 \frac{1}{2} \leq x \leq 2\right\}\) is
1 \(\frac{23}{16}\)
2 \(\frac{79}{24}\)
3 \(\frac{79}{16}\)
4 \(\frac{23}{6}\)
Explanation:
(B) : Given, \(0 \leq y \leq x^{2}+1\) \(0 \leq y \leq x+1\) \(\frac{1}{2} \leq x \leq 2\) So, required area \(=\int_{\frac{1}{2}}^{1}\left(\mathrm{x}^{2}+1\right) \mathrm{dx}+\frac{1}{2}(2+3) \times 1\) \(=\left[\frac{\mathrm{x}^{3}}{3}+\mathrm{x}\right]_{1 / 2}^{1}+\frac{5}{2}=\frac{1}{3}+1-\frac{1}{24}-\frac{1}{2}+\frac{5}{2}\) \(=\frac{8+24-1-12+60}{24}=\frac{79}{24}\)
JEE Main-2020-03.09.2020
Application of the Integrals
87053
The area (in sq. units) of the part of the region \(A=\left\{(x, y):|x|+|y| \leq 1,2 \mathbf{y}^{2} \geq|\mathbf{x}|\right\}\) is
1 \(\frac{1}{3}\)
2 \(\frac{7}{6}\)
3 \(\frac{1}{6}\)
4 \(\frac{5}{6}\)
Explanation:
(D) : Given, unsolved \(|x|+|y| \leq 1\) \(2 y^{2} \geq|x|\) by the figure due to symmetry \(A=4 \int_{0}^{1 / 2}\left(1-x-\sqrt{\frac{x}{2}}\right) d x\) \(\because\left[2 y^{2}+y-1=0 \Rightarrow(2 y-1)(y+1)=0 \Rightarrow y=1 / 2\right]\) \(=4\left[x-\frac{x^{2}}{2}-\frac{2 x^{3 / 2}}{3 / \sqrt{2}}\right]_{0}^{1 / 2}\) \(=4\left[\frac{1}{2}-\frac{1}{8}-\frac{2}{12}\right]=\frac{5}{6}\) square units.
