NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
87017
The area of the plane figure bounded by lines \(y\) \(=\sqrt{x}, x \in[0,1], y=x^{2}, x \in[1,2]\) and \(y=-x^{2}+\) \(2 x+4, x \in[0,2]\) is
1 \(10 / 7\)
2 \(19 / 3\)
3 \(3 / 5\)
4 \(4 / 3\)
Explanation:
(B) : Required area \(=\int_{0}^{2}\left(-x^{2}+2 x+4\right) d x-\int_{0}^{1} \sqrt{x} d x-\int_{1}^{2} x^{2} d x\) \(=\left[\frac{-x^{3}}{3}+x^{2}+4 x\right]_{0}^{2}-\left|\frac{x^{3 / 2}}{3 / 2}\right|_{0}^{1}-\left|\frac{x^{3}}{3}\right|_{1}^{2}\) \(=\left(-\frac{8}{3}+4+8-0\right)-\frac{2}{3}(1-0)-\frac{1}{3}(8-1)\) \(=-\frac{8}{3}+12-\frac{2}{3}-\frac{7}{3}=\frac{-8+36-2-7}{3}=\frac{19}{3}\) hence, option (d) is correct .
SRM JEE-2014
Application of the Integrals
87018
The line \(y=m x\) bisects the area enclosed by lines \(x=0, y=0\) and \(x=3 / 2\) and the curve \(y=1\) \(+4 x-x^{2}\). Then the value of \(m\) is
1 \(\frac{13}{6}\)
2 \(\frac{13}{2}\)
3 \(\frac{13}{5}\)
4 \(\frac{13}{7}\)
Explanation:
Exp:(a) \(\mathrm{y}=1+4 \mathrm{x}-\mathrm{x}^2=5-(\mathrm{x}-2)^2\) We have \(\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) d x=2 \int_{0}^{3 / 2} m x d x\) \(\frac{3}{2}+2\left(\frac{9}{4}\right)-\frac{1}{3}\left(\frac{27}{8}\right)=\mathrm{m} \cdot \frac{9}{4}\) \(\frac{3}{2}+\frac{9}{2}-\frac{9}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \frac{12}{2}-\frac{9}{8}=\frac{9}{4} \mathrm{~m}\) \(6-\frac{9}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \frac{39}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \mathrm{m}=\frac{13}{6}\)
BITSAT-2020
Application of the Integrals
87019
Let the straight line \(x=b\) divide the area enclosed by \(y=(1-x)^{2}, y=0\) and \(x=0\) into two parts \(R_{1}(0 \leq x \leq b)\) and \(R_{2}(b \leq x \leq 1)\) such that \(R_{1}-R_{2}=\frac{1}{4}\). Then \(b\) equals
87017
The area of the plane figure bounded by lines \(y\) \(=\sqrt{x}, x \in[0,1], y=x^{2}, x \in[1,2]\) and \(y=-x^{2}+\) \(2 x+4, x \in[0,2]\) is
1 \(10 / 7\)
2 \(19 / 3\)
3 \(3 / 5\)
4 \(4 / 3\)
Explanation:
(B) : Required area \(=\int_{0}^{2}\left(-x^{2}+2 x+4\right) d x-\int_{0}^{1} \sqrt{x} d x-\int_{1}^{2} x^{2} d x\) \(=\left[\frac{-x^{3}}{3}+x^{2}+4 x\right]_{0}^{2}-\left|\frac{x^{3 / 2}}{3 / 2}\right|_{0}^{1}-\left|\frac{x^{3}}{3}\right|_{1}^{2}\) \(=\left(-\frac{8}{3}+4+8-0\right)-\frac{2}{3}(1-0)-\frac{1}{3}(8-1)\) \(=-\frac{8}{3}+12-\frac{2}{3}-\frac{7}{3}=\frac{-8+36-2-7}{3}=\frac{19}{3}\) hence, option (d) is correct .
SRM JEE-2014
Application of the Integrals
87018
The line \(y=m x\) bisects the area enclosed by lines \(x=0, y=0\) and \(x=3 / 2\) and the curve \(y=1\) \(+4 x-x^{2}\). Then the value of \(m\) is
1 \(\frac{13}{6}\)
2 \(\frac{13}{2}\)
3 \(\frac{13}{5}\)
4 \(\frac{13}{7}\)
Explanation:
Exp:(a) \(\mathrm{y}=1+4 \mathrm{x}-\mathrm{x}^2=5-(\mathrm{x}-2)^2\) We have \(\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) d x=2 \int_{0}^{3 / 2} m x d x\) \(\frac{3}{2}+2\left(\frac{9}{4}\right)-\frac{1}{3}\left(\frac{27}{8}\right)=\mathrm{m} \cdot \frac{9}{4}\) \(\frac{3}{2}+\frac{9}{2}-\frac{9}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \frac{12}{2}-\frac{9}{8}=\frac{9}{4} \mathrm{~m}\) \(6-\frac{9}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \frac{39}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \mathrm{m}=\frac{13}{6}\)
BITSAT-2020
Application of the Integrals
87019
Let the straight line \(x=b\) divide the area enclosed by \(y=(1-x)^{2}, y=0\) and \(x=0\) into two parts \(R_{1}(0 \leq x \leq b)\) and \(R_{2}(b \leq x \leq 1)\) such that \(R_{1}-R_{2}=\frac{1}{4}\). Then \(b\) equals
87017
The area of the plane figure bounded by lines \(y\) \(=\sqrt{x}, x \in[0,1], y=x^{2}, x \in[1,2]\) and \(y=-x^{2}+\) \(2 x+4, x \in[0,2]\) is
1 \(10 / 7\)
2 \(19 / 3\)
3 \(3 / 5\)
4 \(4 / 3\)
Explanation:
(B) : Required area \(=\int_{0}^{2}\left(-x^{2}+2 x+4\right) d x-\int_{0}^{1} \sqrt{x} d x-\int_{1}^{2} x^{2} d x\) \(=\left[\frac{-x^{3}}{3}+x^{2}+4 x\right]_{0}^{2}-\left|\frac{x^{3 / 2}}{3 / 2}\right|_{0}^{1}-\left|\frac{x^{3}}{3}\right|_{1}^{2}\) \(=\left(-\frac{8}{3}+4+8-0\right)-\frac{2}{3}(1-0)-\frac{1}{3}(8-1)\) \(=-\frac{8}{3}+12-\frac{2}{3}-\frac{7}{3}=\frac{-8+36-2-7}{3}=\frac{19}{3}\) hence, option (d) is correct .
SRM JEE-2014
Application of the Integrals
87018
The line \(y=m x\) bisects the area enclosed by lines \(x=0, y=0\) and \(x=3 / 2\) and the curve \(y=1\) \(+4 x-x^{2}\). Then the value of \(m\) is
1 \(\frac{13}{6}\)
2 \(\frac{13}{2}\)
3 \(\frac{13}{5}\)
4 \(\frac{13}{7}\)
Explanation:
Exp:(a) \(\mathrm{y}=1+4 \mathrm{x}-\mathrm{x}^2=5-(\mathrm{x}-2)^2\) We have \(\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) d x=2 \int_{0}^{3 / 2} m x d x\) \(\frac{3}{2}+2\left(\frac{9}{4}\right)-\frac{1}{3}\left(\frac{27}{8}\right)=\mathrm{m} \cdot \frac{9}{4}\) \(\frac{3}{2}+\frac{9}{2}-\frac{9}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \frac{12}{2}-\frac{9}{8}=\frac{9}{4} \mathrm{~m}\) \(6-\frac{9}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \frac{39}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \mathrm{m}=\frac{13}{6}\)
BITSAT-2020
Application of the Integrals
87019
Let the straight line \(x=b\) divide the area enclosed by \(y=(1-x)^{2}, y=0\) and \(x=0\) into two parts \(R_{1}(0 \leq x \leq b)\) and \(R_{2}(b \leq x \leq 1)\) such that \(R_{1}-R_{2}=\frac{1}{4}\). Then \(b\) equals
87017
The area of the plane figure bounded by lines \(y\) \(=\sqrt{x}, x \in[0,1], y=x^{2}, x \in[1,2]\) and \(y=-x^{2}+\) \(2 x+4, x \in[0,2]\) is
1 \(10 / 7\)
2 \(19 / 3\)
3 \(3 / 5\)
4 \(4 / 3\)
Explanation:
(B) : Required area \(=\int_{0}^{2}\left(-x^{2}+2 x+4\right) d x-\int_{0}^{1} \sqrt{x} d x-\int_{1}^{2} x^{2} d x\) \(=\left[\frac{-x^{3}}{3}+x^{2}+4 x\right]_{0}^{2}-\left|\frac{x^{3 / 2}}{3 / 2}\right|_{0}^{1}-\left|\frac{x^{3}}{3}\right|_{1}^{2}\) \(=\left(-\frac{8}{3}+4+8-0\right)-\frac{2}{3}(1-0)-\frac{1}{3}(8-1)\) \(=-\frac{8}{3}+12-\frac{2}{3}-\frac{7}{3}=\frac{-8+36-2-7}{3}=\frac{19}{3}\) hence, option (d) is correct .
SRM JEE-2014
Application of the Integrals
87018
The line \(y=m x\) bisects the area enclosed by lines \(x=0, y=0\) and \(x=3 / 2\) and the curve \(y=1\) \(+4 x-x^{2}\). Then the value of \(m\) is
1 \(\frac{13}{6}\)
2 \(\frac{13}{2}\)
3 \(\frac{13}{5}\)
4 \(\frac{13}{7}\)
Explanation:
Exp:(a) \(\mathrm{y}=1+4 \mathrm{x}-\mathrm{x}^2=5-(\mathrm{x}-2)^2\) We have \(\int_{0}^{3 / 2}\left(1+4 x-x^{2}\right) d x=2 \int_{0}^{3 / 2} m x d x\) \(\frac{3}{2}+2\left(\frac{9}{4}\right)-\frac{1}{3}\left(\frac{27}{8}\right)=\mathrm{m} \cdot \frac{9}{4}\) \(\frac{3}{2}+\frac{9}{2}-\frac{9}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \frac{12}{2}-\frac{9}{8}=\frac{9}{4} \mathrm{~m}\) \(6-\frac{9}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \frac{39}{8}=\frac{9}{4} \mathrm{~m} \Rightarrow \mathrm{m}=\frac{13}{6}\)
BITSAT-2020
Application of the Integrals
87019
Let the straight line \(x=b\) divide the area enclosed by \(y=(1-x)^{2}, y=0\) and \(x=0\) into two parts \(R_{1}(0 \leq x \leq b)\) and \(R_{2}(b \leq x \leq 1)\) such that \(R_{1}-R_{2}=\frac{1}{4}\). Then \(b\) equals
