86430
\(\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
1 \(\frac{\pi}{2}\)
2 0
3 \(\frac{\pi}{4}\)
4 \(\pi\)
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) We know that, \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\) \(I=\int_{a}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 2}\left(\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\cos x}}\right) d x\) \(2 \mathrm{I}=\int_{0}^{\pi / 2} 1 \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}-0=\frac{\pi}{4}\)
Assam CEE-2021
Integral Calculus
86365
For the curve \(4 x^{5}=5 y^{4}\), the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point \(i\)
1 \(y\left(\frac{5}{4}\right)^{4}\)
2 \(\left(\frac{5}{4}\right)^{4}\)
3 \(\left(\frac{5}{4}\right)^{3}\)
4 \(\left(\frac{4}{5}\right)^{4}\)
Explanation:
(D) : Given,Curve \(4 x^{5}=5 y^{4}\) \(20 x^{4}=20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{x^{4}}{y^{3}}\) We know that, Length of subnormal \((S N)=\left(y \cdot \frac{d x}{d y}\right)=\frac{y^{4}}{x^{4}}\) Length of subtangent \((\mathrm{ST})=\left(\mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\) But given condition, \(\frac{(\mathrm{SN})^{3}}{(\mathrm{ST})^{2}}=\frac{\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3}}{\left(\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\right)^{2}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3} \times\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{4}}\right)^{2}\) \(=\frac{\mathrm{y}^{12}}{\mathrm{x}^{12}} \times \frac{\mathrm{y}^{4}}{\mathrm{x}^{8}}=\frac{\mathrm{y}^{16}}{\mathrm{x}^{20}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{5}}\right)^{4}\) \(=\left(\frac{4}{5}\right)^{4}\)
Karnataka CET-2010
Integral Calculus
86366
The value of \(\int e^{\sin x} \sin 2 x d x\) is
86430
\(\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
1 \(\frac{\pi}{2}\)
2 0
3 \(\frac{\pi}{4}\)
4 \(\pi\)
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) We know that, \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\) \(I=\int_{a}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 2}\left(\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\cos x}}\right) d x\) \(2 \mathrm{I}=\int_{0}^{\pi / 2} 1 \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}-0=\frac{\pi}{4}\)
Assam CEE-2021
Integral Calculus
86365
For the curve \(4 x^{5}=5 y^{4}\), the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point \(i\)
1 \(y\left(\frac{5}{4}\right)^{4}\)
2 \(\left(\frac{5}{4}\right)^{4}\)
3 \(\left(\frac{5}{4}\right)^{3}\)
4 \(\left(\frac{4}{5}\right)^{4}\)
Explanation:
(D) : Given,Curve \(4 x^{5}=5 y^{4}\) \(20 x^{4}=20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{x^{4}}{y^{3}}\) We know that, Length of subnormal \((S N)=\left(y \cdot \frac{d x}{d y}\right)=\frac{y^{4}}{x^{4}}\) Length of subtangent \((\mathrm{ST})=\left(\mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\) But given condition, \(\frac{(\mathrm{SN})^{3}}{(\mathrm{ST})^{2}}=\frac{\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3}}{\left(\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\right)^{2}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3} \times\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{4}}\right)^{2}\) \(=\frac{\mathrm{y}^{12}}{\mathrm{x}^{12}} \times \frac{\mathrm{y}^{4}}{\mathrm{x}^{8}}=\frac{\mathrm{y}^{16}}{\mathrm{x}^{20}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{5}}\right)^{4}\) \(=\left(\frac{4}{5}\right)^{4}\)
Karnataka CET-2010
Integral Calculus
86366
The value of \(\int e^{\sin x} \sin 2 x d x\) is
86430
\(\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
1 \(\frac{\pi}{2}\)
2 0
3 \(\frac{\pi}{4}\)
4 \(\pi\)
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) We know that, \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\) \(I=\int_{a}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 2}\left(\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\cos x}}\right) d x\) \(2 \mathrm{I}=\int_{0}^{\pi / 2} 1 \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}-0=\frac{\pi}{4}\)
Assam CEE-2021
Integral Calculus
86365
For the curve \(4 x^{5}=5 y^{4}\), the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point \(i\)
1 \(y\left(\frac{5}{4}\right)^{4}\)
2 \(\left(\frac{5}{4}\right)^{4}\)
3 \(\left(\frac{5}{4}\right)^{3}\)
4 \(\left(\frac{4}{5}\right)^{4}\)
Explanation:
(D) : Given,Curve \(4 x^{5}=5 y^{4}\) \(20 x^{4}=20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{x^{4}}{y^{3}}\) We know that, Length of subnormal \((S N)=\left(y \cdot \frac{d x}{d y}\right)=\frac{y^{4}}{x^{4}}\) Length of subtangent \((\mathrm{ST})=\left(\mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\) But given condition, \(\frac{(\mathrm{SN})^{3}}{(\mathrm{ST})^{2}}=\frac{\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3}}{\left(\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\right)^{2}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3} \times\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{4}}\right)^{2}\) \(=\frac{\mathrm{y}^{12}}{\mathrm{x}^{12}} \times \frac{\mathrm{y}^{4}}{\mathrm{x}^{8}}=\frac{\mathrm{y}^{16}}{\mathrm{x}^{20}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{5}}\right)^{4}\) \(=\left(\frac{4}{5}\right)^{4}\)
Karnataka CET-2010
Integral Calculus
86366
The value of \(\int e^{\sin x} \sin 2 x d x\) is
86430
\(\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
1 \(\frac{\pi}{2}\)
2 0
3 \(\frac{\pi}{4}\)
4 \(\pi\)
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) We know that, \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\) \(I=\int_{a}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 2}\left(\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\cos x}}\right) d x\) \(2 \mathrm{I}=\int_{0}^{\pi / 2} 1 \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}-0=\frac{\pi}{4}\)
Assam CEE-2021
Integral Calculus
86365
For the curve \(4 x^{5}=5 y^{4}\), the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point \(i\)
1 \(y\left(\frac{5}{4}\right)^{4}\)
2 \(\left(\frac{5}{4}\right)^{4}\)
3 \(\left(\frac{5}{4}\right)^{3}\)
4 \(\left(\frac{4}{5}\right)^{4}\)
Explanation:
(D) : Given,Curve \(4 x^{5}=5 y^{4}\) \(20 x^{4}=20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{x^{4}}{y^{3}}\) We know that, Length of subnormal \((S N)=\left(y \cdot \frac{d x}{d y}\right)=\frac{y^{4}}{x^{4}}\) Length of subtangent \((\mathrm{ST})=\left(\mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\) But given condition, \(\frac{(\mathrm{SN})^{3}}{(\mathrm{ST})^{2}}=\frac{\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3}}{\left(\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\right)^{2}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3} \times\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{4}}\right)^{2}\) \(=\frac{\mathrm{y}^{12}}{\mathrm{x}^{12}} \times \frac{\mathrm{y}^{4}}{\mathrm{x}^{8}}=\frac{\mathrm{y}^{16}}{\mathrm{x}^{20}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{5}}\right)^{4}\) \(=\left(\frac{4}{5}\right)^{4}\)
Karnataka CET-2010
Integral Calculus
86366
The value of \(\int e^{\sin x} \sin 2 x d x\) is
86430
\(\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\)
1 \(\frac{\pi}{2}\)
2 0
3 \(\frac{\pi}{4}\)
4 \(\pi\)
Explanation:
(C) : Let, \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) We know that, \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{\pi / 2} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x\) \(I=\int_{a}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x\) Adding equation (i) and (ii) \(2 I=\int_{0}^{\pi / 2}\left(\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\cos x}}\right) d x\) \(2 \mathrm{I}=\int_{0}^{\pi / 2} 1 \mathrm{dx}\) \(2 \mathrm{I}=[\mathrm{x}]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}-0=\frac{\pi}{4}\)
Assam CEE-2021
Integral Calculus
86365
For the curve \(4 x^{5}=5 y^{4}\), the ratio of the cube of the subtangent at a point on the curve to the square of the subnormal at the same point \(i\)
1 \(y\left(\frac{5}{4}\right)^{4}\)
2 \(\left(\frac{5}{4}\right)^{4}\)
3 \(\left(\frac{5}{4}\right)^{3}\)
4 \(\left(\frac{4}{5}\right)^{4}\)
Explanation:
(D) : Given,Curve \(4 x^{5}=5 y^{4}\) \(20 x^{4}=20 y^{3} \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{x^{4}}{y^{3}}\) We know that, Length of subnormal \((S N)=\left(y \cdot \frac{d x}{d y}\right)=\frac{y^{4}}{x^{4}}\) Length of subtangent \((\mathrm{ST})=\left(\mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\) But given condition, \(\frac{(\mathrm{SN})^{3}}{(\mathrm{ST})^{2}}=\frac{\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3}}{\left(\frac{\mathrm{x}^{4}}{\mathrm{y}^{2}}\right)^{2}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{4}}\right)^{3} \times\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{4}}\right)^{2}\) \(=\frac{\mathrm{y}^{12}}{\mathrm{x}^{12}} \times \frac{\mathrm{y}^{4}}{\mathrm{x}^{8}}=\frac{\mathrm{y}^{16}}{\mathrm{x}^{20}}=\left(\frac{\mathrm{y}^{4}}{\mathrm{x}^{5}}\right)^{4}\) \(=\left(\frac{4}{5}\right)^{4}\)
Karnataka CET-2010
Integral Calculus
86366
The value of \(\int e^{\sin x} \sin 2 x d x\) is
