NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
86996
Let the locus of the centre \((\alpha, \beta), \beta>0\), of the circle which touches the circle \(x^{2}+(y-1)^{2}=1\) externally and also touches the \(x\)-axis be \(L\). Then the area bounded by \(L\) and the line \(y=4\) is :
1 \(\frac{32 \sqrt{2}}{3}\)
2 \(\frac{40 \sqrt{2}}{3}\)
3 \(\frac{64}{3}\)
4 \(\frac{32}{3}\)
Explanation:
(C) : Since, the locus of the centre \((\alpha, \beta), \beta>0\) of the circle which touches the circle \(x^{2}+(y-1)^{2}=1\) externally and also touches the \(\mathrm{x}\)-axis be \(\mathrm{L}\) Thus, \((\alpha-0)^{2}+(\beta-1)^{2}=(\beta+1)^{2}\) \(\alpha^{2}+\beta^{2}+1-2 \beta=\beta^{2}+1+2 \beta\) \(\alpha^{2}=4 \beta, \beta>0\) \(\mathrm{x}^{2}=4 \mathrm{y}, \mathrm{y}>0\) Hence the area bounded by L and the line \(\mathrm{y}=4\) is \(\mathrm{A}=2 \int_{0}^{4} 2 \sqrt{\mathrm{y}} \mathrm{dy}=4\left[\frac{2}{3} \mathrm{y}^{3 / 2}\right]_{0}^{4}=\frac{64}{3}\) square units For the next 02 (two) items that follow: Consider the ellipses \(4 x^{2}+y^{2}=1\) and \(x^{2}+4 y^{2}=1\)
JEE Main-2022-25.07.2022
Application of the Integrals
86997
What is the bounded area not common to both the ellipses?
1 \(\left(\pi-\tan ^{-1} 2\right)\) square units
2 \(\left(2 \pi-\tan ^{-1} 2\right)\) square units
3 \(\left(\pi-2 \tan ^{-1} 2\right)\) square units
4 None of the above
Explanation:
(D): We have to find the area not common to both ellipse:i.e. \(2[\text {Area of ellipse any - Area of common portion of ellipse}]\) \(=2\left[\pi(1)\left(\frac{1}{2}\right)-\left[\frac{\pi}{2}+\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)-\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)\right]\right.\) \(=2\left[\frac{\pi}{2}-\frac{\pi}{2}-\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)+\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)\right]\) \(=2\left[\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)-\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]\)
SCRA-2014
Application of the Integrals
86998
Let the straight line \(x=b\) divide the area enclosed by \(y=\left(1-x^{2}\right), y=0\) 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
1 \(\frac{3}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{4}\)
Explanation:
(B) : Given that straight line \(\mathrm{x}=\mathrm{b}\) Divide the area in closed by \(y=\left(1-x^{2}\right)\) \(y=0\) and \(x=0\) into two parts \(\mathrm{R}_{1}(0 \leq \mathrm{x} \leq \mathrm{b})\) and \(\mathrm{R}_{2}(\mathrm{~b} \leq \mathrm{x} \leq 1)\) Now \(R_{1}=\int_{0}^{1}(x-1)^{2} d x \Rightarrow\left[\frac{(1-x)^{3}}{3}\right]_{0}^{1}=\frac{(b-1)^{3}+1}{3}\) Now \(\mathrm{R}_{2} \Rightarrow \int_{0}^{1}(\mathrm{x}-1)^{2} \mathrm{dx}=\left[\left(\frac{\mathrm{x}-1}{3}\right)^{3}\right]_{0}^{1}=\frac{-(\mathrm{b}-1)^{3}}{3}\) \(\mathrm{R}_{1}-\mathrm{R}_{2}=\frac{1}{4}\) \(\frac{(\mathrm{b}-1)^{3}+1}{3}+\frac{(\mathrm{b}-1)^{3}}{3}=\frac{1}{4}\) \(\frac{2(\mathrm{~b}-1)^{3}}{3}=\frac{1}{4}-\frac{1}{3}=\frac{-1}{12}\) \((b-1)^{3}=\frac{-1}{8} \quad(1-b)^{3}=\frac{1}{8}\) \(1-\mathrm{b}=\frac{1}{2}\) or \(\mathrm{b}=\frac{1}{2}\)
AMU-2014
Application of the Integrals
87005
Area of the region bounded by the curve \(y^{2}=x\) and the lines \(x=1, x=4\) and \(X\)-axis in the first quadrant is
1 \(\frac{14}{3}\)
2 \(\frac{7}{3}\)
3 \(\frac{28}{3}\)
4 14
Explanation:
(A): According to question, Area of region \(\mathrm{OABC}=\int_{\mathrm{a}}^{b} \mathrm{ydx}\) \(\mathrm{A}=\int_{1}^{4} \sqrt{\mathrm{x}} \mathrm{dx} \Rightarrow \mathrm{A}=\frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{1}^{4}=\frac{2}{3}\left[(4)^{3 / 2}-(1)^{3 / 2}\right]\) \(\mathrm{A}=\frac{2}{3}[8-1]=\frac{2 \times 7}{3}\) So, \(\quad \mathrm{A}=\frac{14}{3}\) square unit.
86996
Let the locus of the centre \((\alpha, \beta), \beta>0\), of the circle which touches the circle \(x^{2}+(y-1)^{2}=1\) externally and also touches the \(x\)-axis be \(L\). Then the area bounded by \(L\) and the line \(y=4\) is :
1 \(\frac{32 \sqrt{2}}{3}\)
2 \(\frac{40 \sqrt{2}}{3}\)
3 \(\frac{64}{3}\)
4 \(\frac{32}{3}\)
Explanation:
(C) : Since, the locus of the centre \((\alpha, \beta), \beta>0\) of the circle which touches the circle \(x^{2}+(y-1)^{2}=1\) externally and also touches the \(\mathrm{x}\)-axis be \(\mathrm{L}\) Thus, \((\alpha-0)^{2}+(\beta-1)^{2}=(\beta+1)^{2}\) \(\alpha^{2}+\beta^{2}+1-2 \beta=\beta^{2}+1+2 \beta\) \(\alpha^{2}=4 \beta, \beta>0\) \(\mathrm{x}^{2}=4 \mathrm{y}, \mathrm{y}>0\) Hence the area bounded by L and the line \(\mathrm{y}=4\) is \(\mathrm{A}=2 \int_{0}^{4} 2 \sqrt{\mathrm{y}} \mathrm{dy}=4\left[\frac{2}{3} \mathrm{y}^{3 / 2}\right]_{0}^{4}=\frac{64}{3}\) square units For the next 02 (two) items that follow: Consider the ellipses \(4 x^{2}+y^{2}=1\) and \(x^{2}+4 y^{2}=1\)
JEE Main-2022-25.07.2022
Application of the Integrals
86997
What is the bounded area not common to both the ellipses?
1 \(\left(\pi-\tan ^{-1} 2\right)\) square units
2 \(\left(2 \pi-\tan ^{-1} 2\right)\) square units
3 \(\left(\pi-2 \tan ^{-1} 2\right)\) square units
4 None of the above
Explanation:
(D): We have to find the area not common to both ellipse:i.e. \(2[\text {Area of ellipse any - Area of common portion of ellipse}]\) \(=2\left[\pi(1)\left(\frac{1}{2}\right)-\left[\frac{\pi}{2}+\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)-\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)\right]\right.\) \(=2\left[\frac{\pi}{2}-\frac{\pi}{2}-\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)+\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)\right]\) \(=2\left[\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)-\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]\)
SCRA-2014
Application of the Integrals
86998
Let the straight line \(x=b\) divide the area enclosed by \(y=\left(1-x^{2}\right), y=0\) 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
1 \(\frac{3}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{4}\)
Explanation:
(B) : Given that straight line \(\mathrm{x}=\mathrm{b}\) Divide the area in closed by \(y=\left(1-x^{2}\right)\) \(y=0\) and \(x=0\) into two parts \(\mathrm{R}_{1}(0 \leq \mathrm{x} \leq \mathrm{b})\) and \(\mathrm{R}_{2}(\mathrm{~b} \leq \mathrm{x} \leq 1)\) Now \(R_{1}=\int_{0}^{1}(x-1)^{2} d x \Rightarrow\left[\frac{(1-x)^{3}}{3}\right]_{0}^{1}=\frac{(b-1)^{3}+1}{3}\) Now \(\mathrm{R}_{2} \Rightarrow \int_{0}^{1}(\mathrm{x}-1)^{2} \mathrm{dx}=\left[\left(\frac{\mathrm{x}-1}{3}\right)^{3}\right]_{0}^{1}=\frac{-(\mathrm{b}-1)^{3}}{3}\) \(\mathrm{R}_{1}-\mathrm{R}_{2}=\frac{1}{4}\) \(\frac{(\mathrm{b}-1)^{3}+1}{3}+\frac{(\mathrm{b}-1)^{3}}{3}=\frac{1}{4}\) \(\frac{2(\mathrm{~b}-1)^{3}}{3}=\frac{1}{4}-\frac{1}{3}=\frac{-1}{12}\) \((b-1)^{3}=\frac{-1}{8} \quad(1-b)^{3}=\frac{1}{8}\) \(1-\mathrm{b}=\frac{1}{2}\) or \(\mathrm{b}=\frac{1}{2}\)
AMU-2014
Application of the Integrals
87005
Area of the region bounded by the curve \(y^{2}=x\) and the lines \(x=1, x=4\) and \(X\)-axis in the first quadrant is
1 \(\frac{14}{3}\)
2 \(\frac{7}{3}\)
3 \(\frac{28}{3}\)
4 14
Explanation:
(A): According to question, Area of region \(\mathrm{OABC}=\int_{\mathrm{a}}^{b} \mathrm{ydx}\) \(\mathrm{A}=\int_{1}^{4} \sqrt{\mathrm{x}} \mathrm{dx} \Rightarrow \mathrm{A}=\frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{1}^{4}=\frac{2}{3}\left[(4)^{3 / 2}-(1)^{3 / 2}\right]\) \(\mathrm{A}=\frac{2}{3}[8-1]=\frac{2 \times 7}{3}\) So, \(\quad \mathrm{A}=\frac{14}{3}\) square unit.
86996
Let the locus of the centre \((\alpha, \beta), \beta>0\), of the circle which touches the circle \(x^{2}+(y-1)^{2}=1\) externally and also touches the \(x\)-axis be \(L\). Then the area bounded by \(L\) and the line \(y=4\) is :
1 \(\frac{32 \sqrt{2}}{3}\)
2 \(\frac{40 \sqrt{2}}{3}\)
3 \(\frac{64}{3}\)
4 \(\frac{32}{3}\)
Explanation:
(C) : Since, the locus of the centre \((\alpha, \beta), \beta>0\) of the circle which touches the circle \(x^{2}+(y-1)^{2}=1\) externally and also touches the \(\mathrm{x}\)-axis be \(\mathrm{L}\) Thus, \((\alpha-0)^{2}+(\beta-1)^{2}=(\beta+1)^{2}\) \(\alpha^{2}+\beta^{2}+1-2 \beta=\beta^{2}+1+2 \beta\) \(\alpha^{2}=4 \beta, \beta>0\) \(\mathrm{x}^{2}=4 \mathrm{y}, \mathrm{y}>0\) Hence the area bounded by L and the line \(\mathrm{y}=4\) is \(\mathrm{A}=2 \int_{0}^{4} 2 \sqrt{\mathrm{y}} \mathrm{dy}=4\left[\frac{2}{3} \mathrm{y}^{3 / 2}\right]_{0}^{4}=\frac{64}{3}\) square units For the next 02 (two) items that follow: Consider the ellipses \(4 x^{2}+y^{2}=1\) and \(x^{2}+4 y^{2}=1\)
JEE Main-2022-25.07.2022
Application of the Integrals
86997
What is the bounded area not common to both the ellipses?
1 \(\left(\pi-\tan ^{-1} 2\right)\) square units
2 \(\left(2 \pi-\tan ^{-1} 2\right)\) square units
3 \(\left(\pi-2 \tan ^{-1} 2\right)\) square units
4 None of the above
Explanation:
(D): We have to find the area not common to both ellipse:i.e. \(2[\text {Area of ellipse any - Area of common portion of ellipse}]\) \(=2\left[\pi(1)\left(\frac{1}{2}\right)-\left[\frac{\pi}{2}+\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)-\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)\right]\right.\) \(=2\left[\frac{\pi}{2}-\frac{\pi}{2}-\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)+\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)\right]\) \(=2\left[\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)-\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]\)
SCRA-2014
Application of the Integrals
86998
Let the straight line \(x=b\) divide the area enclosed by \(y=\left(1-x^{2}\right), y=0\) 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
1 \(\frac{3}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{4}\)
Explanation:
(B) : Given that straight line \(\mathrm{x}=\mathrm{b}\) Divide the area in closed by \(y=\left(1-x^{2}\right)\) \(y=0\) and \(x=0\) into two parts \(\mathrm{R}_{1}(0 \leq \mathrm{x} \leq \mathrm{b})\) and \(\mathrm{R}_{2}(\mathrm{~b} \leq \mathrm{x} \leq 1)\) Now \(R_{1}=\int_{0}^{1}(x-1)^{2} d x \Rightarrow\left[\frac{(1-x)^{3}}{3}\right]_{0}^{1}=\frac{(b-1)^{3}+1}{3}\) Now \(\mathrm{R}_{2} \Rightarrow \int_{0}^{1}(\mathrm{x}-1)^{2} \mathrm{dx}=\left[\left(\frac{\mathrm{x}-1}{3}\right)^{3}\right]_{0}^{1}=\frac{-(\mathrm{b}-1)^{3}}{3}\) \(\mathrm{R}_{1}-\mathrm{R}_{2}=\frac{1}{4}\) \(\frac{(\mathrm{b}-1)^{3}+1}{3}+\frac{(\mathrm{b}-1)^{3}}{3}=\frac{1}{4}\) \(\frac{2(\mathrm{~b}-1)^{3}}{3}=\frac{1}{4}-\frac{1}{3}=\frac{-1}{12}\) \((b-1)^{3}=\frac{-1}{8} \quad(1-b)^{3}=\frac{1}{8}\) \(1-\mathrm{b}=\frac{1}{2}\) or \(\mathrm{b}=\frac{1}{2}\)
AMU-2014
Application of the Integrals
87005
Area of the region bounded by the curve \(y^{2}=x\) and the lines \(x=1, x=4\) and \(X\)-axis in the first quadrant is
1 \(\frac{14}{3}\)
2 \(\frac{7}{3}\)
3 \(\frac{28}{3}\)
4 14
Explanation:
(A): According to question, Area of region \(\mathrm{OABC}=\int_{\mathrm{a}}^{b} \mathrm{ydx}\) \(\mathrm{A}=\int_{1}^{4} \sqrt{\mathrm{x}} \mathrm{dx} \Rightarrow \mathrm{A}=\frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{1}^{4}=\frac{2}{3}\left[(4)^{3 / 2}-(1)^{3 / 2}\right]\) \(\mathrm{A}=\frac{2}{3}[8-1]=\frac{2 \times 7}{3}\) So, \(\quad \mathrm{A}=\frac{14}{3}\) square unit.
86996
Let the locus of the centre \((\alpha, \beta), \beta>0\), of the circle which touches the circle \(x^{2}+(y-1)^{2}=1\) externally and also touches the \(x\)-axis be \(L\). Then the area bounded by \(L\) and the line \(y=4\) is :
1 \(\frac{32 \sqrt{2}}{3}\)
2 \(\frac{40 \sqrt{2}}{3}\)
3 \(\frac{64}{3}\)
4 \(\frac{32}{3}\)
Explanation:
(C) : Since, the locus of the centre \((\alpha, \beta), \beta>0\) of the circle which touches the circle \(x^{2}+(y-1)^{2}=1\) externally and also touches the \(\mathrm{x}\)-axis be \(\mathrm{L}\) Thus, \((\alpha-0)^{2}+(\beta-1)^{2}=(\beta+1)^{2}\) \(\alpha^{2}+\beta^{2}+1-2 \beta=\beta^{2}+1+2 \beta\) \(\alpha^{2}=4 \beta, \beta>0\) \(\mathrm{x}^{2}=4 \mathrm{y}, \mathrm{y}>0\) Hence the area bounded by L and the line \(\mathrm{y}=4\) is \(\mathrm{A}=2 \int_{0}^{4} 2 \sqrt{\mathrm{y}} \mathrm{dy}=4\left[\frac{2}{3} \mathrm{y}^{3 / 2}\right]_{0}^{4}=\frac{64}{3}\) square units For the next 02 (two) items that follow: Consider the ellipses \(4 x^{2}+y^{2}=1\) and \(x^{2}+4 y^{2}=1\)
JEE Main-2022-25.07.2022
Application of the Integrals
86997
What is the bounded area not common to both the ellipses?
1 \(\left(\pi-\tan ^{-1} 2\right)\) square units
2 \(\left(2 \pi-\tan ^{-1} 2\right)\) square units
3 \(\left(\pi-2 \tan ^{-1} 2\right)\) square units
4 None of the above
Explanation:
(D): We have to find the area not common to both ellipse:i.e. \(2[\text {Area of ellipse any - Area of common portion of ellipse}]\) \(=2\left[\pi(1)\left(\frac{1}{2}\right)-\left[\frac{\pi}{2}+\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)-\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)\right]\right.\) \(=2\left[\frac{\pi}{2}-\frac{\pi}{2}-\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)+\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)\right]\) \(=2\left[\operatorname{Sin}^{-1}\left(\frac{2}{\sqrt{5}}\right)-\operatorname{Sin}^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]\)
SCRA-2014
Application of the Integrals
86998
Let the straight line \(x=b\) divide the area enclosed by \(y=\left(1-x^{2}\right), y=0\) 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
1 \(\frac{3}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{1}{3}\)
4 \(\frac{1}{4}\)
Explanation:
(B) : Given that straight line \(\mathrm{x}=\mathrm{b}\) Divide the area in closed by \(y=\left(1-x^{2}\right)\) \(y=0\) and \(x=0\) into two parts \(\mathrm{R}_{1}(0 \leq \mathrm{x} \leq \mathrm{b})\) and \(\mathrm{R}_{2}(\mathrm{~b} \leq \mathrm{x} \leq 1)\) Now \(R_{1}=\int_{0}^{1}(x-1)^{2} d x \Rightarrow\left[\frac{(1-x)^{3}}{3}\right]_{0}^{1}=\frac{(b-1)^{3}+1}{3}\) Now \(\mathrm{R}_{2} \Rightarrow \int_{0}^{1}(\mathrm{x}-1)^{2} \mathrm{dx}=\left[\left(\frac{\mathrm{x}-1}{3}\right)^{3}\right]_{0}^{1}=\frac{-(\mathrm{b}-1)^{3}}{3}\) \(\mathrm{R}_{1}-\mathrm{R}_{2}=\frac{1}{4}\) \(\frac{(\mathrm{b}-1)^{3}+1}{3}+\frac{(\mathrm{b}-1)^{3}}{3}=\frac{1}{4}\) \(\frac{2(\mathrm{~b}-1)^{3}}{3}=\frac{1}{4}-\frac{1}{3}=\frac{-1}{12}\) \((b-1)^{3}=\frac{-1}{8} \quad(1-b)^{3}=\frac{1}{8}\) \(1-\mathrm{b}=\frac{1}{2}\) or \(\mathrm{b}=\frac{1}{2}\)
AMU-2014
Application of the Integrals
87005
Area of the region bounded by the curve \(y^{2}=x\) and the lines \(x=1, x=4\) and \(X\)-axis in the first quadrant is
1 \(\frac{14}{3}\)
2 \(\frac{7}{3}\)
3 \(\frac{28}{3}\)
4 14
Explanation:
(A): According to question, Area of region \(\mathrm{OABC}=\int_{\mathrm{a}}^{b} \mathrm{ydx}\) \(\mathrm{A}=\int_{1}^{4} \sqrt{\mathrm{x}} \mathrm{dx} \Rightarrow \mathrm{A}=\frac{2}{3}\left[\mathrm{x}^{3 / 2}\right]_{1}^{4}=\frac{2}{3}\left[(4)^{3 / 2}-(1)^{3 / 2}\right]\) \(\mathrm{A}=\frac{2}{3}[8-1]=\frac{2 \times 7}{3}\) So, \(\quad \mathrm{A}=\frac{14}{3}\) square unit.
