1
GATE CE 2025 Set 1
MCQ (Single Correct Answer)
+1
-0.33

Let $A=\left[\begin{array}{cc}1 & 1 \\ 1 & 3 \\ -2 & -3\end{array}\right]$ and $b=\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$. For $\mathrm{Ax}=\mathrm{b}$ to be solvable, which one of the following options is the correct condition on $b_1, b_2$ and $b_3$ :

A
$b_1+b_2+b_3=1$
B
$3 b_1+b_2+2 b_3=0$
C
$b_1+3 b_2+b_3=2$
D
$b_1+b_2+b_3=2$
2
GATE CE 2025 Set 1
MCQ (Single Correct Answer)
+1
-0.33

Which one of the following options is the correct Fourier series of the periodic function $f(x)$ described below:

$$ f(x)=\left\{\begin{array}{cl} 0 & \text { if }-2 < x < -1 \\ 2 k & \text { if }-1 < x < 1 \text {; period }=4 \\ 0 & \text { if }-1 < x < 2 \end{array}\right. $$

A
$f(x)=\frac{k}{2}+\frac{2 k}{\pi}\left(\cos \frac{\pi}{2} x-\frac{1}{3} \cos \frac{3 \pi}{2} x+\frac{1}{5} \cos \frac{5 \pi}{2} x-+\ldots\right)$
B
$f(x)=\frac{k}{2}+\frac{2 k}{\pi}\left(\sin \frac{\pi}{2} x-\frac{1}{3} \sin \frac{3 \pi}{2} x+\frac{1}{5} \sin \frac{5 \pi}{2} x-+\ldots\right)$
C
$f(x)=k+\frac{4 k}{\pi}\left(\cos \frac{\pi}{2} x-\frac{1}{3} \cos \frac{3 \pi}{2} x+\frac{1}{5} \cos \frac{5 \pi}{2} x-+\ldots\right)$
D
$f(x)=k+\frac{4 k}{\pi}\left(\sin \frac{\pi}{2} x-\frac{1}{3} \sin \frac{3 \pi}{2} x+\frac{1}{5} \sin \frac{5 \pi}{2} x-+\ldots\right)$
3
GATE CE 2025 Set 1
MCQ (Single Correct Answer)
+1
-0.33

$X$ is the random variable that can take any one of the values, $0,1,7,11$ and 12 . The probability mass function for $X$ is

$$ \begin{aligned} & \mathrm{P}(X=0)=0.4 ; \mathrm{P}(X=1)=0.3 ; \mathrm{P}(X=7)=0.1 ; \\ & \mathrm{P}(X=11)=0.1 ; \mathrm{P}(X=12)=0.1 \end{aligned} $$

Then, the variance of $X$ is

A
20.81
B
28.40
C
31.70
D
10.89
4
GATE CE 2025 Set 1
MCQ (Single Correct Answer)
+1
-0.33
Which of the following equations belong/belongs to the class of second-order, linear, homogeneous partial differential equations:
A
$\frac{\partial^2 u}{\partial t^2}=c^2\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)+x y$
B
$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=0$
C
$\frac{\partial u}{\partial t}=c \frac{\partial u}{\partial x}$
D
$\left(\frac{\partial^2 u}{\partial t^2}\right)^2=c^2 \frac{\partial^2 u}{\partial x^2}$
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