1
GATE EE 2013
+1
-0.3
When the Newton-Raphson method is applied to solve the equation $$\,\,f\left( x \right) = {x^3} + 2x - 1 = 0,\,\,$$ the solution at the end of the first iteration with the initial value as $${x_0} = 1.2$$ is
A
$$-0.82$$
B
$$0.49$$
C
$$0.705$$
D
$$1.69$$
2
GATE EE 2013
+1
-0.3
Square roots of $$-i,$$ where $$i = \sqrt { - 1}$$ are
A
$$i, -i$$
B
\eqalign{ & \cos \left( { - {\pi \over 4}} \right) + \sin \left( { - {\pi \over 4}} \right), \cr & \cos \left( {{{3\pi } \over 4}} \right) + i\,\sin \left( {{{3\pi } \over 4}} \right) \cr}
C
\eqalign{ & \cos \left( {{\pi \over 4}} \right) + i\sin \left( {{{3\pi } \over 4}} \right), \cr & \cos \left( {{{3\pi } \over 4}} \right) + i\sin \left( {{\pi \over 4}} \right) \cr}
D
\eqalign{ & \cos \left( {{{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right), \cr & \cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( {{{3\pi } \over 4}} \right) \cr}
3
GATE EE 2013
+2
-0.6
$$\oint {{{{z^2} - 4} \over {{z^2} + 4}}} dz\,\,$$ evaluated anticlockwise around the circular $$\left| {z - i} \right| = 2,$$ where $$i = \sqrt { - 1}$$, is
A
$$- 4\pi$$
B
$$0$$
C
$$2 + \pi$$
D
$$2+2$$
4
GATE EE 2013
+2
-0.6
Thyristor $$T$$ in the figure below is initially off and is triggered with a single pulse of width $$10\mu s.$$ It is given that $$L = \left( {{{100} \over \pi }} \right)\mu H$$ and $$C = \left( {{{100} \over \pi }} \right)\mu F.$$ Assuming latching and holding currents of the thyristor are both zero and the initial charge on $$C$$ is zero, $$T$$ conducts for
A
$$10\mu s$$
B
$$50$$ $$s$$
C
$$100\mu s$$
D
$$200\mu s$$
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