1
GATE EE 2011
+1
-0.3
The two vectors $$\left[ {\matrix{ {1,} & {1,} & {1} \cr } } \right]$$ and $$\left[ {\matrix{ {1,} & {a,} & {{a^2}} \cr } } \right]$$ where $$a = {{ - 1} \over 2} + j{{\sqrt 3 } \over 2}$$ are
A
Orthonormal
B
Orthogonal
C
Parallel
D
Collinear
2
GATE EE 2011
+1
-0.3
With $$K$$ as constant, the possible solution for the first order differential equation $${{dy} \over {dx}} = {e^{ - 3x}}$$ is
A
$${{ - 1} \over 3}{e^{ - 3x}} + K$$
B
$${1 \over 3}\left( { - 1} \right){e^{ 3x}} + K$$
C
$$- 3{e^{ - 3x}} + K$$
D
$$- 3{e^{ - x}} + K$$
3
GATE EE 2011
+2
-0.6
Solution, the variable $${x_1}$$ and $${x_2}$$ for the following equations is to be obtained by employing the Newton $$-$$ Raphson iteration method
equation (i) $$10\,{x_2}\,\sin \,{x_1} - 0.8 = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$10\,x_2^2\, - 10\,{x_2}\cos \,{x_1} - 0.6 = 0$$
Assuming the initial values $${x_1} = 0.0$$ and $${x_2} = 1.0$$ the Jacobian matrix is
A
$$\left[ {\matrix{ {10} & { - 0.8} \cr 0 & { - 0.6} \cr } } \right]$$
B
$$\left[ {\matrix{ {10} & 0 \cr 0 & {10} \cr } } \right]$$
C
$$\left[ {\matrix{ 0 & { - 0.8} \cr {10} & { - 0.6} \cr } } \right]$$
D
$$\left[ {\matrix{ {10} & 0 \cr {10} & { - 10} \cr } } \right]$$
4
GATE EE 2011
+2
-0.6
Given $$f(t)$$ and $$g(t)$$ as shown below

$$g(t)$$ can be expressed as

A
$$g(t)=f(2t-3)$$
B
$$g\left( t \right) = f\left( {{t \over 2} - 3} \right)$$
C
$$g\left( t \right) = f\left( {2t - {3 \over 2}} \right)$$
D
$$g\left( t \right) = f\left( {{t \over 2} - {3 \over 2}} \right)$$
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