Question 1 |
In matrix equation [A]{X}={R},
[A]=\begin{bmatrix} 4 & 8 & 4\\ 8& 16 & -4\\ 4& -4 & 15 \end{bmatrix}, \{X\}=\begin{Bmatrix} 2\\ 1\\ 4 \end{Bmatrix} and \{R\}=\begin{Bmatrix} 32\\ 16\\ 64 \end{Bmatrix}.
One of the eigenvalues of matrix [A] is
[A]=\begin{bmatrix} 4 & 8 & 4\\ 8& 16 & -4\\ 4& -4 & 15 \end{bmatrix}, \{X\}=\begin{Bmatrix} 2\\ 1\\ 4 \end{Bmatrix} and \{R\}=\begin{Bmatrix} 32\\ 16\\ 64 \end{Bmatrix}.
One of the eigenvalues of matrix [A] is
4 | |
8 | |
15 | |
16 |
Question 1 Explanation:
Given that AX=R
\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right]=\left[\begin{array}{l} 32 \\ 16 \\ 64 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right]=16\left[\begin{array}{c} 32 \\ 4 \end{array}\right](\because A X=\lambda X) \end{array}
\therefore One of eigen value of the given matrix A is given by \lambda=16
\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right]=\left[\begin{array}{l} 32 \\ 16 \\ 64 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{array}\right]\left[\begin{array}{l} 2 \\ 1 \\ 4 \end{array}\right]=16\left[\begin{array}{c} 32 \\ 4 \end{array}\right](\because A X=\lambda X) \end{array}
\therefore One of eigen value of the given matrix A is given by \lambda=16
Question 2 |
The directional derivative of the function f(x,y)=x^2+y^2 along aline directed from (0,0) to (1,1), evaluated at the point x=1, y=1 is
\sqrt{2} | |
2 | |
2 \sqrt{2} | |
4 \sqrt{2} |
Question 2 Explanation:
Given function is f(x, y)=x^{2}+y^{2}
The direction vector a is given by
\overline{\mathrm{a}}=(1,1)-(0,0)=\overline{\mathrm{i}}+\overline{\mathrm{j}}
Let the given point be P=(x, y)=(1,1)
Now, the directional directive of f(x, y) in the direction of vector a at point is given by
\mathrm{D} . \mathrm{D}=(\nabla \mathrm{f}) \cdot \cdot \frac{\overline{\mathrm{a}}}{|\mathrm{a}|}
\Rightarrow D . D=(2 x \bar{i}+2 x \bar{j}) \cdot \frac{(\bar{i}+\bar{j})}{\sqrt{1+1}}
\therefore D . D=\frac{2+2}{\sqrt{2}}=2 \sqrt{2}
The direction vector a is given by
\overline{\mathrm{a}}=(1,1)-(0,0)=\overline{\mathrm{i}}+\overline{\mathrm{j}}
Let the given point be P=(x, y)=(1,1)
Now, the directional directive of f(x, y) in the direction of vector a at point is given by
\mathrm{D} . \mathrm{D}=(\nabla \mathrm{f}) \cdot \cdot \frac{\overline{\mathrm{a}}}{|\mathrm{a}|}
\Rightarrow D . D=(2 x \bar{i}+2 x \bar{j}) \cdot \frac{(\bar{i}+\bar{j})}{\sqrt{1+1}}
\therefore D . D=\frac{2+2}{\sqrt{2}}=2 \sqrt{2}
Question 3 |
The differential equation \frac{dy}{dx}+4y=5 is valid in the domain 0\leq x\leq 1 with y(0)=2.25. The solution of the differential equation is
y=e^{-4x}+5 | |
y=e^{-4x}+1.25 | |
y=e^{4x}+5 | |
y=e^{4x}+1.25 |
Question 3 Explanation:
\begin{array}{l} \text { Given } \frac{\mathrm{dy}}{\mathrm{dx}}+4 \mathrm{y}=5,0 \leq \mathrm{x} \leq 1 \; \; \ldots(1) \\ \because \frac{d y}{d x}+P(x, y)=Q(x) \\ \text { With } \mathrm{y}(0)=2.25\;\;\; \ldots(2) \\ \text { Here, I.F }=\mathrm{e}^{\int 4 \mathrm{dx}}=\mathrm{e}^{4 \mathrm{x}} \end{array}
The general solution of (1) is given by
y \cdot e^{4 x}=\int(5)\left(e^{4 x}\right) d x+c
\Rightarrow \mathrm{y} \cdot \mathrm{e}^{4 \mathrm{x}}=\frac{5}{4} \mathrm{e}^{4 \mathrm{x}}+\mathrm{c}\cdots(3)
Using (2) and (3)
(2.25)(1)=\left(\frac{5}{4}\right)(1)+c
c=1 \cdots(4)
The solution of (1) from (3) and (4) is
y \cdot e^{4 x}=\frac{5}{4} e^{4 x}+1
or y=\frac{5}{4}+e^{4 x}=e^{-4 x}+1.25
The general solution of (1) is given by
y \cdot e^{4 x}=\int(5)\left(e^{4 x}\right) d x+c
\Rightarrow \mathrm{y} \cdot \mathrm{e}^{4 \mathrm{x}}=\frac{5}{4} \mathrm{e}^{4 \mathrm{x}}+\mathrm{c}\cdots(3)
Using (2) and (3)
(2.25)(1)=\left(\frac{5}{4}\right)(1)+c
c=1 \cdots(4)
The solution of (1) from (3) and (4) is
y \cdot e^{4 x}=\frac{5}{4} e^{4 x}+1
or y=\frac{5}{4}+e^{4 x}=e^{-4 x}+1.25
Question 4 |
An analytic function f(z) of complex variable z=x+iy may be written as f(z)=u(x,y)+iv(x,y). Then u(x,y) and v(x,y) must satisfy
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=\frac{\partial v}{\partial x} | |
\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y} and \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} |
Question 4 Explanation:
Given that the complex function f(z)=u(x,y)+ i v(x,y) is an analytic function.
\Rightarrow the Cauchy-Riemann equation will satisfy for u(x,y) & v(x,y)
\therefore \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}} \text{ and } \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=-\frac{\partial \mathrm{v}}{\partial \mathrm{y}}
\Rightarrow the Cauchy-Riemann equation will satisfy for u(x,y) & v(x,y)
\therefore \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=\frac{\partial \mathrm{v}}{\partial \mathrm{y}} \text{ and } \frac{\partial \mathrm{u}}{\partial \mathrm{x}}=-\frac{\partial \mathrm{v}}{\partial \mathrm{y}}
Question 5 |
A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plane. At the instant shown, edge QR is parallel to the x-axis, and the body moves such that velocities of points P and R are V_P \; and \; V_R, in the x and y directions, respectively. The magnitude of the angular velocity of the body is
2V_R | |
2V_P | |
V_R/\sqrt{3} | |
V_P/\sqrt{3} |
Question 5 Explanation:
\begin{array}{l} \Rightarrow \mathrm{V}_{\mathrm{R}}=(\mathrm{IR}) \omega \\ \Rightarrow \omega=\frac{\mathrm{V}_{\mathrm{R}}}{(\mathrm{IR})} \\ \Rightarrow \omega \times \frac{\mathrm{V}_{\mathrm{R}}}{\frac{1}{2}} \\ \Rightarrow \omega=2 \mathrm{V}_{\mathrm{R}} \end{array}
There are 5 questions to complete.