GATE CE 2005 | GATE CE Year Wise Previous Years Questions

GATE CE 2005

MCQ (Single Correct Answer)

-0.6

Transformation to linear form by substituting $$v = {y^{1 - n}}$$ of the equation $${{dy} \over {dt}} + p\left( t \right)y = q\left( t \right){y^n},\,\,n > 0$$ will be

$${{dv} \over {dt}} + \left( {1 - n} \right)pv = \left( {1 - n} \right)q$$

$${{dv} \over {dt}} + \left( {1 + n} \right)pv = \left( {1 + n} \right)q$$

$${{dv} \over {dt}} + \left( {1 + n} \right)pv = \left( {1 - n} \right)q$$

$${{dv} \over {dt}} + \left( {1 - n} \right)pv = \left( {1 + n} \right)q$$

GATE CE 2005

MCQ (Single Correct Answer)

-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

$${e^{ - x}}\left[ {\cos \,4x + {1 \over 4}\sin \,4x} \right]$$

$${e^x}\left[ {\cos \,4x - {1 \over 4}\sin \,4x} \right]$$

$${e^{ - 4x}}\left[ {\cos \,4x - {1 \over 4}\sin \,x} \right]$$

$${e^{ - 4x}}\left[ {\cos \,4x - {1 \over 4}\sin \,4x} \right]$$

GATE CE 2005

MCQ (Single Correct Answer)

-0.3

Which one of the following is not true for the complex number z₁ and z₂ ?

$${{{z_1}} \over {{z_2}}} = {{{z_1}\overline {{z_2}} } \over {{{\left| {{z_2}} \right|}^2}}}$$

$$\left| {{z_1}\, + \,\,{z_2}} \right| \le \,\left| {{z_1}} \right|\, + \,\left| {{z_2}} \right|$$

$$\left| {{z_1}\, + \,\,{z_2}} \right| \le \,\left| {\left| {{z_1}} \right|\, - \,\left| {{z_2}} \right|} \right|$$

$${\left| {{z_1}\, + \,\,{z_2}} \right|^2}\, + \,{\left| {{z_1}\, - \,\,{z_2}} \right|^2} = \,\,2{\left| {{z_1}} \right|^2}\, + \,2{\left| {{z_2}} \right|^2}$$

GATE CE 2005

MCQ (Single Correct Answer)

-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

I = 0 ; Singularities set = $$\phi $$

I = 0 ; Singularities set = $$\left\{ { \pm {{\left( {2n + 1} \right)} \over 2}\pi \,\,;\,n = 0,1,2,.....} \right\}$$

I = 0 ; Singularities set = $$\left\{ { \pm \,n\pi \,\,;\,n = 0,1,2,.....} \right\}$$

None of the above

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