1
GATE CE 2005
MCQ (Single Correct Answer)
+2
-0.6
The solution $${{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} + 17y = 0;$$ $$y\left( 0 \right) = 1,{\left( {{{d\,y} \over {d\,x}}} \right)_{x = {\raise0.5ex\hbox{$\scriptstyle \pi $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}}} = 0\,\,$$ in the range $$0 < x < {\pi \over 4}$$ is given by
A
$${e^{ - x}}\left[ {\cos \,4x + {1 \over 4}\sin \,4x} \right]$$
B
$${e^x}\left[ {\cos \,4x - {1 \over 4}\sin \,4x} \right]$$
C
$${e^{ - 4x}}\left[ {\cos \,4x - {1 \over 4}\sin \,x} \right]$$
D
$${e^{ - 4x}}\left[ {\cos \,4x - {1 \over 4}\sin \,4x} \right]$$
2
GATE CE 2005
MCQ (Single Correct Answer)
+1
-0.3
Which one of the following is not true for the complex number z1 and z2 ?
A
$${{{z_1}} \over {{z_2}}} = {{{z_1}\overline {{z_2}} } \over {{{\left| {{z_2}} \right|}^2}}}$$
B
$$\left| {{z_1}\, + \,\,{z_2}} \right| \le \,\left| {{z_1}} \right|\, + \,\left| {{z_2}} \right|$$
C
$$\left| {{z_1}\, + \,\,{z_2}} \right| \le \,\left| {\left| {{z_1}} \right|\, - \,\left| {{z_2}} \right|} \right|$$
D
$${\left| {{z_1}\, + \,\,{z_2}} \right|^2}\, + \,{\left| {{z_1}\, - \,\,{z_2}} \right|^2} = \,\,2{\left| {{z_1}} \right|^2}\, + \,2{\left| {{z_2}} \right|^2}$$
3
GATE CE 2005
MCQ (Single Correct Answer)
+2
-0.6
Consider likely applicability of Cauchy's Integral theorem to evaluate the following integral counterclockwise around the unit circle C.

$$I\, = \,\oint\limits_C {\sec z\,dz} $$, z being a complex variable. The value of I will be
A
I = 0 ; Singularities set = $$\phi $$
B
I = 0 ; Singularities set = $$\left\{ { \pm {{\left( {2n + 1} \right)} \over 2}\pi \,\,;\,n = 0,1,2,.....} \right\}$$
C
I = 0 ; Singularities set = $$\left\{ { \pm \,n\pi \,\,;\,n = 0,1,2,.....} \right\}$$
D
None of the above
4
GATE CE 2005
MCQ (Single Correct Answer)
+1
-0.3
Given $$a>0,$$ we wish to calculate it reciprocal value $${1 \over a}$$ by using Newton - Raphson method for $$f(x)=0.$$ The Newton - Raphson algorithm for the function will be
A
$${X_{k + 1}} = {1 \over 2}\left( {{X_k} + {a \over {{X_k}}}} \right)$$
B
$${X_{k + 1}} = {X_k} + {a \over 2}X_k^2$$
C
$${X_{k + 1}} = 2{X_k} - aX_k^2$$
D
$${X_{k + 1}} = 2{X_k} - {a \over 2}X_k^2$$
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