Question 1 |
The condition for which the eigenvalues of the matrix
A=\begin{bmatrix} 2&1 \\ 1& k \end{bmatrix}
are positive, is
A=\begin{bmatrix} 2&1 \\ 1& k \end{bmatrix}
are positive, is
k\gt1/2 | |
k\gt-2 | |
k\gt0 | |
k\lt-1/2 |
Question 1 Explanation:
All Eigen values of A=\left[\begin{array}{ll}2 & 1 \\ 1 & k\end{array}\right] are positive
2>0
\therefore 2 \times 2 leading minor must be greater than zero
\left|\begin{array}{ll}2 & 1 \\ 1 & k\end{array}\right|>0
2 k-1>0
2 k>1
k>\frac{1}{2}
2>0
\therefore 2 \times 2 leading minor must be greater than zero
\left|\begin{array}{ll}2 & 1 \\ 1 & k\end{array}\right|>0
2 k-1>0
2 k>1
k>\frac{1}{2}
Question 2 |
The values of x for which the function
f( x)=\frac{x^{2}-3x-4}{x^{2}+3x-4}
is NOT continuous are
f( x)=\frac{x^{2}-3x-4}{x^{2}+3x-4}
is NOT continuous are
4 and -1 | |
4 and 1 | |
-4 and 1 | |
-4 and -1 |
Question 2 Explanation:
f(x)=\frac{x^{2}-3 x-4}{x^{2}+3 x-4} is not continous
when
\begin{aligned} x^{2}+3 x-4 &=0 \\ (x+4)(x-1) &=0 \\ x &=-4,1 \end{aligned}
when
\begin{aligned} x^{2}+3 x-4 &=0 \\ (x+4)(x-1) &=0 \\ x &=-4,1 \end{aligned}
Question 3 |
Laplace transform of cos(\omega t) is
\frac{s}{s^{2}+\omega ^{2}} | |
\frac{\omega }{s^{2}+\omega ^{2}} | |
\frac{s}{s^{2}\omega ^{2}} | |
\frac{\omega }{s^{2}-\omega ^{2}} |
Question 3 Explanation:
L(\cos \omega t)=\frac{s}{s^{2}+\omega^{2}}
Question 4 |
A function of the Complex Variables z=x+iy, is given as f(x,y)=u(x,y)+i v(x,y) , where u(x,y)=2kxy and v(x,y)=x^{2}-y^{2}. The value of k, for which the function is analytic, is _____
-5 | |
-1 | |
-8 | |
1 |
Question 4 Explanation:
Given that ( f(z)=u+i v) is analytic
\begin{aligned} u(x, y) &=2 k x y & & v=x^{2}-y^{2} \\ u_{x} &=2 k y & & v_{y}=-2 y \\ u_{x} &=v_{y} & & \\ k &=-1 & & v_{x}=2 x \\ u_{y} &=2 k x & & & \\ u_{y} &=-v_{x} & & & \\ 2 k x &=-2 x & & & \\ k &=-1 & & \end{aligned}
\begin{aligned} u(x, y) &=2 k x y & & v=x^{2}-y^{2} \\ u_{x} &=2 k y & & v_{y}=-2 y \\ u_{x} &=v_{y} & & \\ k &=-1 & & v_{x}=2 x \\ u_{y} &=2 k x & & & \\ u_{y} &=-v_{x} & & & \\ 2 k x &=-2 x & & & \\ k &=-1 & & \end{aligned}
Question 5 |
Numerical integration using trapezoidal rule gives the best result for a single variable function
linear | |
parabolic | |
logarithmic | |
hyperbolic |
Question 5 Explanation:
Trapezoidal rule gives the best result in single variable function when the function is linear (degree 1)
There are 5 questions to complete.