87127
The degree of the differential equation \(x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots \ldots\)
1 3
2 2
3 1
4 Not defined
Explanation:
(C) : Given that equation, \(x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots .\) We know that, \(\mathrm{e}^{\mathrm{x}}=1+\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+\frac{\mathrm{x}^{4}}{4 !}+\ldots\) Then, \(\quad x=1+y^{\prime}++\frac{1}{2 !}\left(y^{\prime}\right)^{2}+\frac{1}{3 !}\left(y^{\prime}\right)^{3}+\ldots\). \(\mathrm{x}=\mathrm{e}^{\mathrm{y}^{\prime}}\) Taking \(\log\) both side, we get- \(\mathrm{y}^{\prime}=\log \mathrm{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\log \mathrm{x}\) Hence, degree of differential equation is 1 .
Manipal UGET-2016
Differential Equation
87128
If \(1, \omega, \omega^{2}\) are cube roots of unity, then \(\left|\begin{array}{ccc}1& \omega^{n}& \omega^{2 n}\\ \omega^{2 n} &1& \omega^{n}\\ \omega^{n} &\omega^{2 n}& 1\end{array}\right|\) has value
87130
The differential equation of all parabolas whose axis of symmetry is parallel to \(\mathrm{x}\)-axis is of order
1 2
2 3
3 1
4 4
Explanation:
(B) : Since, the parabola is symmetric about a line parallel to \(\mathrm{x}\)-axis. So, parabola may be leftward or rightward. Let us consider the parabola is rightward. The family of parabola is rightward whose axis of symmetry is parabola to \(\mathrm{x}\)-axis is represented by the equation. \((y-h)^{2}=4 a(x-k)\) Where \(\mathrm{h}, \mathrm{k}\) and a are arbitrary constant since, there are three arbitrary constant in the equation. So, order of differential equation is 3 .
AMU-2016
Differential Equation
87131
The degree of the differential equation satisfying \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=\mathbf{a}(\mathbf{x}-\mathbf{y})\) is
1 1
2 3
3 2
4 none of these
Explanation:
(A) : The differential equation is, \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)\) Put \(\mathrm{x}=\sin \alpha\) and \(\mathrm{y}=\sin \beta\) \(\because \quad \sqrt{1-\sin ^{2} \alpha}+\sqrt{1-\sin ^{2} \beta}=\mathrm{a}(\sin \alpha-\sin \beta)\) \(\cos \alpha+\cos \beta=\mathrm{a}(\sin \alpha-\sin \beta)\) \(2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=\mathrm{a}\left(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)\) \(\cot \left(\frac{\alpha-\beta}{2}\right)=a \sin \left(\frac{\alpha-\beta}{2}\right)\) \(\frac{\alpha-\beta}{2}=\cot ^{-1} \mathrm{a} \Rightarrow \alpha-\beta=2 \cot ^{-1} \mathrm{a}\) \(\sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a \Rightarrow \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=0\) Thus, degree of differential equation is 1 .
87127
The degree of the differential equation \(x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots \ldots\)
1 3
2 2
3 1
4 Not defined
Explanation:
(C) : Given that equation, \(x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots .\) We know that, \(\mathrm{e}^{\mathrm{x}}=1+\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+\frac{\mathrm{x}^{4}}{4 !}+\ldots\) Then, \(\quad x=1+y^{\prime}++\frac{1}{2 !}\left(y^{\prime}\right)^{2}+\frac{1}{3 !}\left(y^{\prime}\right)^{3}+\ldots\). \(\mathrm{x}=\mathrm{e}^{\mathrm{y}^{\prime}}\) Taking \(\log\) both side, we get- \(\mathrm{y}^{\prime}=\log \mathrm{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\log \mathrm{x}\) Hence, degree of differential equation is 1 .
Manipal UGET-2016
Differential Equation
87128
If \(1, \omega, \omega^{2}\) are cube roots of unity, then \(\left|\begin{array}{ccc}1& \omega^{n}& \omega^{2 n}\\ \omega^{2 n} &1& \omega^{n}\\ \omega^{n} &\omega^{2 n}& 1\end{array}\right|\) has value
87130
The differential equation of all parabolas whose axis of symmetry is parallel to \(\mathrm{x}\)-axis is of order
1 2
2 3
3 1
4 4
Explanation:
(B) : Since, the parabola is symmetric about a line parallel to \(\mathrm{x}\)-axis. So, parabola may be leftward or rightward. Let us consider the parabola is rightward. The family of parabola is rightward whose axis of symmetry is parabola to \(\mathrm{x}\)-axis is represented by the equation. \((y-h)^{2}=4 a(x-k)\) Where \(\mathrm{h}, \mathrm{k}\) and a are arbitrary constant since, there are three arbitrary constant in the equation. So, order of differential equation is 3 .
AMU-2016
Differential Equation
87131
The degree of the differential equation satisfying \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=\mathbf{a}(\mathbf{x}-\mathbf{y})\) is
1 1
2 3
3 2
4 none of these
Explanation:
(A) : The differential equation is, \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)\) Put \(\mathrm{x}=\sin \alpha\) and \(\mathrm{y}=\sin \beta\) \(\because \quad \sqrt{1-\sin ^{2} \alpha}+\sqrt{1-\sin ^{2} \beta}=\mathrm{a}(\sin \alpha-\sin \beta)\) \(\cos \alpha+\cos \beta=\mathrm{a}(\sin \alpha-\sin \beta)\) \(2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=\mathrm{a}\left(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)\) \(\cot \left(\frac{\alpha-\beta}{2}\right)=a \sin \left(\frac{\alpha-\beta}{2}\right)\) \(\frac{\alpha-\beta}{2}=\cot ^{-1} \mathrm{a} \Rightarrow \alpha-\beta=2 \cot ^{-1} \mathrm{a}\) \(\sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a \Rightarrow \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=0\) Thus, degree of differential equation is 1 .
87127
The degree of the differential equation \(x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots \ldots\)
1 3
2 2
3 1
4 Not defined
Explanation:
(C) : Given that equation, \(x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots .\) We know that, \(\mathrm{e}^{\mathrm{x}}=1+\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+\frac{\mathrm{x}^{4}}{4 !}+\ldots\) Then, \(\quad x=1+y^{\prime}++\frac{1}{2 !}\left(y^{\prime}\right)^{2}+\frac{1}{3 !}\left(y^{\prime}\right)^{3}+\ldots\). \(\mathrm{x}=\mathrm{e}^{\mathrm{y}^{\prime}}\) Taking \(\log\) both side, we get- \(\mathrm{y}^{\prime}=\log \mathrm{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\log \mathrm{x}\) Hence, degree of differential equation is 1 .
Manipal UGET-2016
Differential Equation
87128
If \(1, \omega, \omega^{2}\) are cube roots of unity, then \(\left|\begin{array}{ccc}1& \omega^{n}& \omega^{2 n}\\ \omega^{2 n} &1& \omega^{n}\\ \omega^{n} &\omega^{2 n}& 1\end{array}\right|\) has value
87130
The differential equation of all parabolas whose axis of symmetry is parallel to \(\mathrm{x}\)-axis is of order
1 2
2 3
3 1
4 4
Explanation:
(B) : Since, the parabola is symmetric about a line parallel to \(\mathrm{x}\)-axis. So, parabola may be leftward or rightward. Let us consider the parabola is rightward. The family of parabola is rightward whose axis of symmetry is parabola to \(\mathrm{x}\)-axis is represented by the equation. \((y-h)^{2}=4 a(x-k)\) Where \(\mathrm{h}, \mathrm{k}\) and a are arbitrary constant since, there are three arbitrary constant in the equation. So, order of differential equation is 3 .
AMU-2016
Differential Equation
87131
The degree of the differential equation satisfying \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=\mathbf{a}(\mathbf{x}-\mathbf{y})\) is
1 1
2 3
3 2
4 none of these
Explanation:
(A) : The differential equation is, \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)\) Put \(\mathrm{x}=\sin \alpha\) and \(\mathrm{y}=\sin \beta\) \(\because \quad \sqrt{1-\sin ^{2} \alpha}+\sqrt{1-\sin ^{2} \beta}=\mathrm{a}(\sin \alpha-\sin \beta)\) \(\cos \alpha+\cos \beta=\mathrm{a}(\sin \alpha-\sin \beta)\) \(2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=\mathrm{a}\left(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)\) \(\cot \left(\frac{\alpha-\beta}{2}\right)=a \sin \left(\frac{\alpha-\beta}{2}\right)\) \(\frac{\alpha-\beta}{2}=\cot ^{-1} \mathrm{a} \Rightarrow \alpha-\beta=2 \cot ^{-1} \mathrm{a}\) \(\sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a \Rightarrow \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=0\) Thus, degree of differential equation is 1 .
87127
The degree of the differential equation \(x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots \ldots\)
1 3
2 2
3 1
4 Not defined
Explanation:
(C) : Given that equation, \(x=1+\left(\frac{d y}{d x}\right)+\frac{1}{2 !}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{3 !}\left(\frac{d y}{d x}\right)^{3}+\ldots .\) We know that, \(\mathrm{e}^{\mathrm{x}}=1+\mathrm{x}+\frac{\mathrm{x}^{2}}{2 !}+\frac{\mathrm{x}^{3}}{3 !}+\frac{\mathrm{x}^{4}}{4 !}+\ldots\) Then, \(\quad x=1+y^{\prime}++\frac{1}{2 !}\left(y^{\prime}\right)^{2}+\frac{1}{3 !}\left(y^{\prime}\right)^{3}+\ldots\). \(\mathrm{x}=\mathrm{e}^{\mathrm{y}^{\prime}}\) Taking \(\log\) both side, we get- \(\mathrm{y}^{\prime}=\log \mathrm{x}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\log \mathrm{x}\) Hence, degree of differential equation is 1 .
Manipal UGET-2016
Differential Equation
87128
If \(1, \omega, \omega^{2}\) are cube roots of unity, then \(\left|\begin{array}{ccc}1& \omega^{n}& \omega^{2 n}\\ \omega^{2 n} &1& \omega^{n}\\ \omega^{n} &\omega^{2 n}& 1\end{array}\right|\) has value
87130
The differential equation of all parabolas whose axis of symmetry is parallel to \(\mathrm{x}\)-axis is of order
1 2
2 3
3 1
4 4
Explanation:
(B) : Since, the parabola is symmetric about a line parallel to \(\mathrm{x}\)-axis. So, parabola may be leftward or rightward. Let us consider the parabola is rightward. The family of parabola is rightward whose axis of symmetry is parabola to \(\mathrm{x}\)-axis is represented by the equation. \((y-h)^{2}=4 a(x-k)\) Where \(\mathrm{h}, \mathrm{k}\) and a are arbitrary constant since, there are three arbitrary constant in the equation. So, order of differential equation is 3 .
AMU-2016
Differential Equation
87131
The degree of the differential equation satisfying \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=\mathbf{a}(\mathbf{x}-\mathbf{y})\) is
1 1
2 3
3 2
4 none of these
Explanation:
(A) : The differential equation is, \(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)\) Put \(\mathrm{x}=\sin \alpha\) and \(\mathrm{y}=\sin \beta\) \(\because \quad \sqrt{1-\sin ^{2} \alpha}+\sqrt{1-\sin ^{2} \beta}=\mathrm{a}(\sin \alpha-\sin \beta)\) \(\cos \alpha+\cos \beta=\mathrm{a}(\sin \alpha-\sin \beta)\) \(2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=\mathrm{a}\left(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}\right)\) \(\cot \left(\frac{\alpha-\beta}{2}\right)=a \sin \left(\frac{\alpha-\beta}{2}\right)\) \(\frac{\alpha-\beta}{2}=\cot ^{-1} \mathrm{a} \Rightarrow \alpha-\beta=2 \cot ^{-1} \mathrm{a}\) \(\sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a \Rightarrow \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=0\) Thus, degree of differential equation is 1 .
