86847
Area bounded by the curve \(x y-3 x-2 y-10=0\), \(X\)-axis and the lines \(x=3, x=4\) is
1 \(16 \log 2-13\) sq unit
2 \(16 \log 2-3\) sq unit
3 \(16 \log 2+3\) sq unit
4 None of the above
Explanation:
(C) : Given curve \(x y-3 x-2 y-10=0\) \(x y-2 y=3 x+10 \Rightarrow(x-2) y=10+3 x\) \(y=\frac{10+3 x}{x-2}\) \(\therefore\) The required area, \(A=\int_{3}^{4}(y-0) d x=\int_{3}^{4} y d x=\int_{3}^{4} \frac{10+3 x}{x-2}=\int_{3}^{4}\left(3+\frac{16}{x-2}\right) d x\) \(=[3 \mathrm{x}+16 \log (\mathrm{x}-2)]_{3}^{4}=12+16 \log 2-9-16 \log 1\) \(=16 \log 2+3\) sq unit
CG PET-2006
Application of the Integrals
86848
\(y=\cos ^{2} x, 0 \leq x \leq \frac{\pi}{2}\) and the axes.
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{2 \pi}{3}\)
4 \(\frac{\pi}{4}\)
Explanation:
(D) : Area included between the curve \(y=\cos ^{2} x, 0 \leq x \leq \pi / 2\) and the axes. Required area, \(A=\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x=\int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x\) \(=\left[\frac{1}{2} x+\frac{1}{2} \sin 2 x \cdot \frac{1}{2}\right]_0^{\frac{\pi}{2}}=\frac{1}{2} \cdot \frac{\pi}{2}+0+0-0=\frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi}{4}\)
CG PET-2011
Application of the Integrals
86849
The area bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{32}{3}\) sq unit
3 \(\frac{9}{8}\) squnit
4 \(\frac{8}{3}\) squnit
Explanation:
(C) : Given, \(x^{2}=4 y\) And, \(\quad x=4 y-2\) By solving equation (i) and (ii), intersection point is \(\mathrm{A}(2,1)\) and \(\mathrm{B}\left(-1, \frac{1}{4}\right)\) \(\therefore\) Required area \(=\) shaded region \(=\left[\int_{-1}^{2} \mathrm{y} \cdot \mathrm{dx}_{(\mathrm{x}=4 \mathrm{y}-2)}\right]-\left[\int_{-1}^{2} \mathrm{y} \cdot \mathrm{dx}_{\left(\mathrm{x}^{2}=4 \mathrm{y}\right)}\right]\) \(=\int_{-1}^{2} \frac{1}{4}(\mathrm{x}+2) \mathrm{dx}-\int_{-1}^{2} \frac{1}{4} \mathrm{x}^{2} \mathrm{dx}\) \(=\frac{1}{4}\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2}-\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\) \(\frac{1}{4}\left\{\frac{2^{2}}{2}+2 \times 2-\left(\frac{1}{2}-2\right)\right\}-\frac{1}{12}\left[2^{3}-(-1)^{3}\right]\) \(=\frac{1}{4}\left(6+\frac{3}{2}\right)-\frac{1}{12} \times 9=\frac{15}{8}-\frac{3}{4}=\frac{9}{8}\) square unit
CG PET-2018
Application of the Integrals
86851
The area bounded by the curve \(y=f(x), x\) - axis and the ordinates \(x=l\) and \(x=b\) is \((b-1)\) sin \((3 b\) \(+4)\). Then \(f(x)\) is
1 \((x-1) \cos (3 x+4)\)
2 \(\sin (3 x+4)\)
3 \(\sin (3 x+4)+3(x-1) \cos (3 x+4)\)
4 None of these
Explanation:
(C) : The area bounded by curve \(y=f(x)\) with ordinates \(\mathrm{x}=l\) and \(\mathrm{x}=\mathrm{b}\) \(\int_{l}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{b}-1) \sin (3 \mathrm{~b}+4)\) Differentiating w.r.t \(b\) we get \(f(x)=\frac{d}{d x}(x-1) \sin (3 x+4)\) \(f(b)=\sin (3 b+4)+(b-1) \cos (3 b+4) \times 3\) \(f(x)=\sin (3 x+4)+3(x-1) \cos (3 x+4)\)
86847
Area bounded by the curve \(x y-3 x-2 y-10=0\), \(X\)-axis and the lines \(x=3, x=4\) is
1 \(16 \log 2-13\) sq unit
2 \(16 \log 2-3\) sq unit
3 \(16 \log 2+3\) sq unit
4 None of the above
Explanation:
(C) : Given curve \(x y-3 x-2 y-10=0\) \(x y-2 y=3 x+10 \Rightarrow(x-2) y=10+3 x\) \(y=\frac{10+3 x}{x-2}\) \(\therefore\) The required area, \(A=\int_{3}^{4}(y-0) d x=\int_{3}^{4} y d x=\int_{3}^{4} \frac{10+3 x}{x-2}=\int_{3}^{4}\left(3+\frac{16}{x-2}\right) d x\) \(=[3 \mathrm{x}+16 \log (\mathrm{x}-2)]_{3}^{4}=12+16 \log 2-9-16 \log 1\) \(=16 \log 2+3\) sq unit
CG PET-2006
Application of the Integrals
86848
\(y=\cos ^{2} x, 0 \leq x \leq \frac{\pi}{2}\) and the axes.
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{2 \pi}{3}\)
4 \(\frac{\pi}{4}\)
Explanation:
(D) : Area included between the curve \(y=\cos ^{2} x, 0 \leq x \leq \pi / 2\) and the axes. Required area, \(A=\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x=\int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x\) \(=\left[\frac{1}{2} x+\frac{1}{2} \sin 2 x \cdot \frac{1}{2}\right]_0^{\frac{\pi}{2}}=\frac{1}{2} \cdot \frac{\pi}{2}+0+0-0=\frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi}{4}\)
CG PET-2011
Application of the Integrals
86849
The area bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{32}{3}\) sq unit
3 \(\frac{9}{8}\) squnit
4 \(\frac{8}{3}\) squnit
Explanation:
(C) : Given, \(x^{2}=4 y\) And, \(\quad x=4 y-2\) By solving equation (i) and (ii), intersection point is \(\mathrm{A}(2,1)\) and \(\mathrm{B}\left(-1, \frac{1}{4}\right)\) \(\therefore\) Required area \(=\) shaded region \(=\left[\int_{-1}^{2} \mathrm{y} \cdot \mathrm{dx}_{(\mathrm{x}=4 \mathrm{y}-2)}\right]-\left[\int_{-1}^{2} \mathrm{y} \cdot \mathrm{dx}_{\left(\mathrm{x}^{2}=4 \mathrm{y}\right)}\right]\) \(=\int_{-1}^{2} \frac{1}{4}(\mathrm{x}+2) \mathrm{dx}-\int_{-1}^{2} \frac{1}{4} \mathrm{x}^{2} \mathrm{dx}\) \(=\frac{1}{4}\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2}-\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\) \(\frac{1}{4}\left\{\frac{2^{2}}{2}+2 \times 2-\left(\frac{1}{2}-2\right)\right\}-\frac{1}{12}\left[2^{3}-(-1)^{3}\right]\) \(=\frac{1}{4}\left(6+\frac{3}{2}\right)-\frac{1}{12} \times 9=\frac{15}{8}-\frac{3}{4}=\frac{9}{8}\) square unit
CG PET-2018
Application of the Integrals
86851
The area bounded by the curve \(y=f(x), x\) - axis and the ordinates \(x=l\) and \(x=b\) is \((b-1)\) sin \((3 b\) \(+4)\). Then \(f(x)\) is
1 \((x-1) \cos (3 x+4)\)
2 \(\sin (3 x+4)\)
3 \(\sin (3 x+4)+3(x-1) \cos (3 x+4)\)
4 None of these
Explanation:
(C) : The area bounded by curve \(y=f(x)\) with ordinates \(\mathrm{x}=l\) and \(\mathrm{x}=\mathrm{b}\) \(\int_{l}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{b}-1) \sin (3 \mathrm{~b}+4)\) Differentiating w.r.t \(b\) we get \(f(x)=\frac{d}{d x}(x-1) \sin (3 x+4)\) \(f(b)=\sin (3 b+4)+(b-1) \cos (3 b+4) \times 3\) \(f(x)=\sin (3 x+4)+3(x-1) \cos (3 x+4)\)
86847
Area bounded by the curve \(x y-3 x-2 y-10=0\), \(X\)-axis and the lines \(x=3, x=4\) is
1 \(16 \log 2-13\) sq unit
2 \(16 \log 2-3\) sq unit
3 \(16 \log 2+3\) sq unit
4 None of the above
Explanation:
(C) : Given curve \(x y-3 x-2 y-10=0\) \(x y-2 y=3 x+10 \Rightarrow(x-2) y=10+3 x\) \(y=\frac{10+3 x}{x-2}\) \(\therefore\) The required area, \(A=\int_{3}^{4}(y-0) d x=\int_{3}^{4} y d x=\int_{3}^{4} \frac{10+3 x}{x-2}=\int_{3}^{4}\left(3+\frac{16}{x-2}\right) d x\) \(=[3 \mathrm{x}+16 \log (\mathrm{x}-2)]_{3}^{4}=12+16 \log 2-9-16 \log 1\) \(=16 \log 2+3\) sq unit
CG PET-2006
Application of the Integrals
86848
\(y=\cos ^{2} x, 0 \leq x \leq \frac{\pi}{2}\) and the axes.
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{2 \pi}{3}\)
4 \(\frac{\pi}{4}\)
Explanation:
(D) : Area included between the curve \(y=\cos ^{2} x, 0 \leq x \leq \pi / 2\) and the axes. Required area, \(A=\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x=\int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x\) \(=\left[\frac{1}{2} x+\frac{1}{2} \sin 2 x \cdot \frac{1}{2}\right]_0^{\frac{\pi}{2}}=\frac{1}{2} \cdot \frac{\pi}{2}+0+0-0=\frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi}{4}\)
CG PET-2011
Application of the Integrals
86849
The area bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{32}{3}\) sq unit
3 \(\frac{9}{8}\) squnit
4 \(\frac{8}{3}\) squnit
Explanation:
(C) : Given, \(x^{2}=4 y\) And, \(\quad x=4 y-2\) By solving equation (i) and (ii), intersection point is \(\mathrm{A}(2,1)\) and \(\mathrm{B}\left(-1, \frac{1}{4}\right)\) \(\therefore\) Required area \(=\) shaded region \(=\left[\int_{-1}^{2} \mathrm{y} \cdot \mathrm{dx}_{(\mathrm{x}=4 \mathrm{y}-2)}\right]-\left[\int_{-1}^{2} \mathrm{y} \cdot \mathrm{dx}_{\left(\mathrm{x}^{2}=4 \mathrm{y}\right)}\right]\) \(=\int_{-1}^{2} \frac{1}{4}(\mathrm{x}+2) \mathrm{dx}-\int_{-1}^{2} \frac{1}{4} \mathrm{x}^{2} \mathrm{dx}\) \(=\frac{1}{4}\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2}-\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\) \(\frac{1}{4}\left\{\frac{2^{2}}{2}+2 \times 2-\left(\frac{1}{2}-2\right)\right\}-\frac{1}{12}\left[2^{3}-(-1)^{3}\right]\) \(=\frac{1}{4}\left(6+\frac{3}{2}\right)-\frac{1}{12} \times 9=\frac{15}{8}-\frac{3}{4}=\frac{9}{8}\) square unit
CG PET-2018
Application of the Integrals
86851
The area bounded by the curve \(y=f(x), x\) - axis and the ordinates \(x=l\) and \(x=b\) is \((b-1)\) sin \((3 b\) \(+4)\). Then \(f(x)\) is
1 \((x-1) \cos (3 x+4)\)
2 \(\sin (3 x+4)\)
3 \(\sin (3 x+4)+3(x-1) \cos (3 x+4)\)
4 None of these
Explanation:
(C) : The area bounded by curve \(y=f(x)\) with ordinates \(\mathrm{x}=l\) and \(\mathrm{x}=\mathrm{b}\) \(\int_{l}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{b}-1) \sin (3 \mathrm{~b}+4)\) Differentiating w.r.t \(b\) we get \(f(x)=\frac{d}{d x}(x-1) \sin (3 x+4)\) \(f(b)=\sin (3 b+4)+(b-1) \cos (3 b+4) \times 3\) \(f(x)=\sin (3 x+4)+3(x-1) \cos (3 x+4)\)
86847
Area bounded by the curve \(x y-3 x-2 y-10=0\), \(X\)-axis and the lines \(x=3, x=4\) is
1 \(16 \log 2-13\) sq unit
2 \(16 \log 2-3\) sq unit
3 \(16 \log 2+3\) sq unit
4 None of the above
Explanation:
(C) : Given curve \(x y-3 x-2 y-10=0\) \(x y-2 y=3 x+10 \Rightarrow(x-2) y=10+3 x\) \(y=\frac{10+3 x}{x-2}\) \(\therefore\) The required area, \(A=\int_{3}^{4}(y-0) d x=\int_{3}^{4} y d x=\int_{3}^{4} \frac{10+3 x}{x-2}=\int_{3}^{4}\left(3+\frac{16}{x-2}\right) d x\) \(=[3 \mathrm{x}+16 \log (\mathrm{x}-2)]_{3}^{4}=12+16 \log 2-9-16 \log 1\) \(=16 \log 2+3\) sq unit
CG PET-2006
Application of the Integrals
86848
\(y=\cos ^{2} x, 0 \leq x \leq \frac{\pi}{2}\) and the axes.
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{2 \pi}{3}\)
4 \(\frac{\pi}{4}\)
Explanation:
(D) : Area included between the curve \(y=\cos ^{2} x, 0 \leq x \leq \pi / 2\) and the axes. Required area, \(A=\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x=\int_{0}^{\frac{\pi}{2}} \frac{1+\cos 2 x}{2} d x\) \(=\left[\frac{1}{2} x+\frac{1}{2} \sin 2 x \cdot \frac{1}{2}\right]_0^{\frac{\pi}{2}}=\frac{1}{2} \cdot \frac{\pi}{2}+0+0-0=\frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi}{4}\)
CG PET-2011
Application of the Integrals
86849
The area bounded by the curve \(x^{2}=4 y\) and the straight line \(x=4 y-2\) is
1 \(\frac{4}{3}\) squnit
2 \(\frac{32}{3}\) sq unit
3 \(\frac{9}{8}\) squnit
4 \(\frac{8}{3}\) squnit
Explanation:
(C) : Given, \(x^{2}=4 y\) And, \(\quad x=4 y-2\) By solving equation (i) and (ii), intersection point is \(\mathrm{A}(2,1)\) and \(\mathrm{B}\left(-1, \frac{1}{4}\right)\) \(\therefore\) Required area \(=\) shaded region \(=\left[\int_{-1}^{2} \mathrm{y} \cdot \mathrm{dx}_{(\mathrm{x}=4 \mathrm{y}-2)}\right]-\left[\int_{-1}^{2} \mathrm{y} \cdot \mathrm{dx}_{\left(\mathrm{x}^{2}=4 \mathrm{y}\right)}\right]\) \(=\int_{-1}^{2} \frac{1}{4}(\mathrm{x}+2) \mathrm{dx}-\int_{-1}^{2} \frac{1}{4} \mathrm{x}^{2} \mathrm{dx}\) \(=\frac{1}{4}\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2}-\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\) \(\frac{1}{4}\left\{\frac{2^{2}}{2}+2 \times 2-\left(\frac{1}{2}-2\right)\right\}-\frac{1}{12}\left[2^{3}-(-1)^{3}\right]\) \(=\frac{1}{4}\left(6+\frac{3}{2}\right)-\frac{1}{12} \times 9=\frac{15}{8}-\frac{3}{4}=\frac{9}{8}\) square unit
CG PET-2018
Application of the Integrals
86851
The area bounded by the curve \(y=f(x), x\) - axis and the ordinates \(x=l\) and \(x=b\) is \((b-1)\) sin \((3 b\) \(+4)\). Then \(f(x)\) is
1 \((x-1) \cos (3 x+4)\)
2 \(\sin (3 x+4)\)
3 \(\sin (3 x+4)+3(x-1) \cos (3 x+4)\)
4 None of these
Explanation:
(C) : The area bounded by curve \(y=f(x)\) with ordinates \(\mathrm{x}=l\) and \(\mathrm{x}=\mathrm{b}\) \(\int_{l}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=(\mathrm{b}-1) \sin (3 \mathrm{~b}+4)\) Differentiating w.r.t \(b\) we get \(f(x)=\frac{d}{d x}(x-1) \sin (3 x+4)\) \(f(b)=\sin (3 b+4)+(b-1) \cos (3 b+4) \times 3\) \(f(x)=\sin (3 x+4)+3(x-1) \cos (3 x+4)\)
