87045
The region represented by \(|x-y| \leq 2\) and \(\mid x+\) \(\mathbf{y} \mid \leq \mathbf{2}\) is bounded by a
1 Rhombus of side length 2 units
2 Rhombus of area \(8 \sqrt{2}\) sq units
3 Square of side length \(2 \sqrt{2}\) units
4 Square of area 16 sq units
Explanation:
(C) : We have obtained a square \(A B C D\) with length of the side at \(2 \sqrt{2}\) Length of \(C D=\sqrt{(0-2)^{2}+(-2-0)^{2}}\) \(=\sqrt{4+4}=8\) \(=2 \sqrt{2}\)
JEE Main-2019-10.04.2019
Application of the Integrals
87046
The area (in sq units ) of the region \(A\{x, y) \in R\) \(\left.\times R \mid 0 \leq x \leq 3,0 \leq y \leq 4 y \leq x^{2}+3 x\right\}\) is
1 \(\frac{53}{6}\)
2 8
3 \(\frac{59}{6}\)
4 \(\frac{26}{3}\)
Explanation:
(C) : Given, \(y \leq x^{2}+3 x, \quad 0 \leq x \leq 3\) Required region \(=\int_{0}^{1}\left[x^{2}+3 x\right] d x+2 \times 4\) \(=\left|\frac{x^{3}}{3}+\frac{3 x^{2}}{2}\right|_{0}^{1}+8=\left[\frac{1}{3}+\frac{3}{2}-0\right]+8\) \(=\frac{11}{6}+8=\frac{59}{6}\)
JEE Main-2019-08.04.2019
Application of the Integrals
87015
The area enclosed by \(2|x|+3|y| \leq 6\) is
1 \(3 \mathrm{sq} \cdot\) units
2 12 sq . units
3 \(9 \mathrm{sq}\). units
4 \(24 \mathrm{sq}\). units
Explanation:
(B) : We have, \(\quad 2|x|+3|y| \leq 6\) \(2 \mathrm{x}+3 \mathrm{y} \leq 6\) \(-2 \mathrm{x}-3 \mathrm{y} \leq 6\) \(2 \mathrm{x}-3 \mathrm{y} \leq 6\) \(-2 x+3 y \leq 6\) \(\therefore\) Required area \(=\) Area of the shaded region \(=4(\text { area of } \triangle \mathrm{OAB})\) \(=4 \cdot \frac{1}{2} \cdot(3) \cdot(2)=12 \mathrm{sq} \cdot \text { units. }\)
AMU-2012
Application of the Integrals
87016
The area bounded by the curves \(y=\ln x, y=\ln\) \(|x|, y=|\ln x|\) and \(y=|\ln | x||\), is
1 5 sq. units
2 2 sq. units
3 4 sq. untis
4 none of these
Explanation:
(C) : \(\therefore\) required area \(=4 \int_{-\infty}^{1} \mathrm{e}^{\mathrm{y}} \mathrm{dy}\) \(=4\left[\mathrm{e}^{\mathrm{y}}\right]_{-\infty}^{0}=4\left[\mathrm{e}^{0}-\mathrm{e}^{-\infty}\right]=4[1-0]=4\) sq. unit. Hence, option (c) is correct.
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Application of the Integrals
87045
The region represented by \(|x-y| \leq 2\) and \(\mid x+\) \(\mathbf{y} \mid \leq \mathbf{2}\) is bounded by a
1 Rhombus of side length 2 units
2 Rhombus of area \(8 \sqrt{2}\) sq units
3 Square of side length \(2 \sqrt{2}\) units
4 Square of area 16 sq units
Explanation:
(C) : We have obtained a square \(A B C D\) with length of the side at \(2 \sqrt{2}\) Length of \(C D=\sqrt{(0-2)^{2}+(-2-0)^{2}}\) \(=\sqrt{4+4}=8\) \(=2 \sqrt{2}\)
JEE Main-2019-10.04.2019
Application of the Integrals
87046
The area (in sq units ) of the region \(A\{x, y) \in R\) \(\left.\times R \mid 0 \leq x \leq 3,0 \leq y \leq 4 y \leq x^{2}+3 x\right\}\) is
1 \(\frac{53}{6}\)
2 8
3 \(\frac{59}{6}\)
4 \(\frac{26}{3}\)
Explanation:
(C) : Given, \(y \leq x^{2}+3 x, \quad 0 \leq x \leq 3\) Required region \(=\int_{0}^{1}\left[x^{2}+3 x\right] d x+2 \times 4\) \(=\left|\frac{x^{3}}{3}+\frac{3 x^{2}}{2}\right|_{0}^{1}+8=\left[\frac{1}{3}+\frac{3}{2}-0\right]+8\) \(=\frac{11}{6}+8=\frac{59}{6}\)
JEE Main-2019-08.04.2019
Application of the Integrals
87015
The area enclosed by \(2|x|+3|y| \leq 6\) is
1 \(3 \mathrm{sq} \cdot\) units
2 12 sq . units
3 \(9 \mathrm{sq}\). units
4 \(24 \mathrm{sq}\). units
Explanation:
(B) : We have, \(\quad 2|x|+3|y| \leq 6\) \(2 \mathrm{x}+3 \mathrm{y} \leq 6\) \(-2 \mathrm{x}-3 \mathrm{y} \leq 6\) \(2 \mathrm{x}-3 \mathrm{y} \leq 6\) \(-2 x+3 y \leq 6\) \(\therefore\) Required area \(=\) Area of the shaded region \(=4(\text { area of } \triangle \mathrm{OAB})\) \(=4 \cdot \frac{1}{2} \cdot(3) \cdot(2)=12 \mathrm{sq} \cdot \text { units. }\)
AMU-2012
Application of the Integrals
87016
The area bounded by the curves \(y=\ln x, y=\ln\) \(|x|, y=|\ln x|\) and \(y=|\ln | x||\), is
1 5 sq. units
2 2 sq. units
3 4 sq. untis
4 none of these
Explanation:
(C) : \(\therefore\) required area \(=4 \int_{-\infty}^{1} \mathrm{e}^{\mathrm{y}} \mathrm{dy}\) \(=4\left[\mathrm{e}^{\mathrm{y}}\right]_{-\infty}^{0}=4\left[\mathrm{e}^{0}-\mathrm{e}^{-\infty}\right]=4[1-0]=4\) sq. unit. Hence, option (c) is correct.
87045
The region represented by \(|x-y| \leq 2\) and \(\mid x+\) \(\mathbf{y} \mid \leq \mathbf{2}\) is bounded by a
1 Rhombus of side length 2 units
2 Rhombus of area \(8 \sqrt{2}\) sq units
3 Square of side length \(2 \sqrt{2}\) units
4 Square of area 16 sq units
Explanation:
(C) : We have obtained a square \(A B C D\) with length of the side at \(2 \sqrt{2}\) Length of \(C D=\sqrt{(0-2)^{2}+(-2-0)^{2}}\) \(=\sqrt{4+4}=8\) \(=2 \sqrt{2}\)
JEE Main-2019-10.04.2019
Application of the Integrals
87046
The area (in sq units ) of the region \(A\{x, y) \in R\) \(\left.\times R \mid 0 \leq x \leq 3,0 \leq y \leq 4 y \leq x^{2}+3 x\right\}\) is
1 \(\frac{53}{6}\)
2 8
3 \(\frac{59}{6}\)
4 \(\frac{26}{3}\)
Explanation:
(C) : Given, \(y \leq x^{2}+3 x, \quad 0 \leq x \leq 3\) Required region \(=\int_{0}^{1}\left[x^{2}+3 x\right] d x+2 \times 4\) \(=\left|\frac{x^{3}}{3}+\frac{3 x^{2}}{2}\right|_{0}^{1}+8=\left[\frac{1}{3}+\frac{3}{2}-0\right]+8\) \(=\frac{11}{6}+8=\frac{59}{6}\)
JEE Main-2019-08.04.2019
Application of the Integrals
87015
The area enclosed by \(2|x|+3|y| \leq 6\) is
1 \(3 \mathrm{sq} \cdot\) units
2 12 sq . units
3 \(9 \mathrm{sq}\). units
4 \(24 \mathrm{sq}\). units
Explanation:
(B) : We have, \(\quad 2|x|+3|y| \leq 6\) \(2 \mathrm{x}+3 \mathrm{y} \leq 6\) \(-2 \mathrm{x}-3 \mathrm{y} \leq 6\) \(2 \mathrm{x}-3 \mathrm{y} \leq 6\) \(-2 x+3 y \leq 6\) \(\therefore\) Required area \(=\) Area of the shaded region \(=4(\text { area of } \triangle \mathrm{OAB})\) \(=4 \cdot \frac{1}{2} \cdot(3) \cdot(2)=12 \mathrm{sq} \cdot \text { units. }\)
AMU-2012
Application of the Integrals
87016
The area bounded by the curves \(y=\ln x, y=\ln\) \(|x|, y=|\ln x|\) and \(y=|\ln | x||\), is
1 5 sq. units
2 2 sq. units
3 4 sq. untis
4 none of these
Explanation:
(C) : \(\therefore\) required area \(=4 \int_{-\infty}^{1} \mathrm{e}^{\mathrm{y}} \mathrm{dy}\) \(=4\left[\mathrm{e}^{\mathrm{y}}\right]_{-\infty}^{0}=4\left[\mathrm{e}^{0}-\mathrm{e}^{-\infty}\right]=4[1-0]=4\) sq. unit. Hence, option (c) is correct.
87045
The region represented by \(|x-y| \leq 2\) and \(\mid x+\) \(\mathbf{y} \mid \leq \mathbf{2}\) is bounded by a
1 Rhombus of side length 2 units
2 Rhombus of area \(8 \sqrt{2}\) sq units
3 Square of side length \(2 \sqrt{2}\) units
4 Square of area 16 sq units
Explanation:
(C) : We have obtained a square \(A B C D\) with length of the side at \(2 \sqrt{2}\) Length of \(C D=\sqrt{(0-2)^{2}+(-2-0)^{2}}\) \(=\sqrt{4+4}=8\) \(=2 \sqrt{2}\)
JEE Main-2019-10.04.2019
Application of the Integrals
87046
The area (in sq units ) of the region \(A\{x, y) \in R\) \(\left.\times R \mid 0 \leq x \leq 3,0 \leq y \leq 4 y \leq x^{2}+3 x\right\}\) is
1 \(\frac{53}{6}\)
2 8
3 \(\frac{59}{6}\)
4 \(\frac{26}{3}\)
Explanation:
(C) : Given, \(y \leq x^{2}+3 x, \quad 0 \leq x \leq 3\) Required region \(=\int_{0}^{1}\left[x^{2}+3 x\right] d x+2 \times 4\) \(=\left|\frac{x^{3}}{3}+\frac{3 x^{2}}{2}\right|_{0}^{1}+8=\left[\frac{1}{3}+\frac{3}{2}-0\right]+8\) \(=\frac{11}{6}+8=\frac{59}{6}\)
JEE Main-2019-08.04.2019
Application of the Integrals
87015
The area enclosed by \(2|x|+3|y| \leq 6\) is
1 \(3 \mathrm{sq} \cdot\) units
2 12 sq . units
3 \(9 \mathrm{sq}\). units
4 \(24 \mathrm{sq}\). units
Explanation:
(B) : We have, \(\quad 2|x|+3|y| \leq 6\) \(2 \mathrm{x}+3 \mathrm{y} \leq 6\) \(-2 \mathrm{x}-3 \mathrm{y} \leq 6\) \(2 \mathrm{x}-3 \mathrm{y} \leq 6\) \(-2 x+3 y \leq 6\) \(\therefore\) Required area \(=\) Area of the shaded region \(=4(\text { area of } \triangle \mathrm{OAB})\) \(=4 \cdot \frac{1}{2} \cdot(3) \cdot(2)=12 \mathrm{sq} \cdot \text { units. }\)
AMU-2012
Application of the Integrals
87016
The area bounded by the curves \(y=\ln x, y=\ln\) \(|x|, y=|\ln x|\) and \(y=|\ln | x||\), is
1 5 sq. units
2 2 sq. units
3 4 sq. untis
4 none of these
Explanation:
(C) : \(\therefore\) required area \(=4 \int_{-\infty}^{1} \mathrm{e}^{\mathrm{y}} \mathrm{dy}\) \(=4\left[\mathrm{e}^{\mathrm{y}}\right]_{-\infty}^{0}=4\left[\mathrm{e}^{0}-\mathrm{e}^{-\infty}\right]=4[1-0]=4\) sq. unit. Hence, option (c) is correct.
