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GATE - EC 2025 Answers with Explanations
Q.1
Current population 11,02,500
increasing 5% per annum
Let assume 2 years age, the population were, x
then, x*(1+5/100)^2 = 11,02,500
Or, x = 1000000
Option C
Q.2
p/q + q/p = 3
Or, (p/q)^2 + (q/p)^2 = (p/q + q/p)^2 - 2
Or, (p/q)^2 + (q/p)^2 = 9-2=7
Option B
Q.3
Option C
Q.4
Option C
Q.5
Option B
Q.6
Option C
Q.7
Option D
Q.8
Total Volume = 1/3*pi*r^2*l = 1/3*pi*1*1 = pi/3
Option A
Q.9
Option C
Q.10
The area of the triangle is 1/2*3x*3x = 9/2*x^2
The area of the regular convex hexagon is 9/2*x^2 - 3*1/2*x*x = 6/2*x^2
Or, area of hexagon: triangle = 2:3
Option D
Electronics and Communication (EC)
Q.1
$ \mathbf{F}(x,y)=-x\hat{i}+y\hat{j} $
$$
\int_C \mathbf{F}\cdot d\mathbf{r}
$$
$$
=\int_C(-x\,dx+y\,dy)
$$
Parameterization:
$$
x=\cos t,\qquad
y=\sin t
$$
$$
dx=-\sin t\,dt,\qquad
dy=\cos t\,dt
$$
$$
\int_{0}^{\pi/4}
\left[
(-\cos t)(-\sin t)
+
(\sin t)(\cos t)
\right]dt
$$
$$
=
\int_{0}^{\pi/4}
2\sin t\cos t\,dt
$$
where
$$
0\le t\le\frac{\pi}{4}
$$
$$
2\sin t\cos t=\sin2t
$$
$$
=\int_{0}^{\pi/4}\sin2t\,dt
$$
$$
=\left[-\frac12\cos2t\right]_{0}^{\pi/4}
$$
$$
=-\frac12\cos\frac{\pi}{2}
+\frac12\cos0
$$
$$
=0+\frac12
=\frac12
$$
$$
\boxed{\int_C \mathbf{F}\cdot d\mathbf{r}=\frac12}
$$
Option A
Q.2
Find the integrating factor of the differential equation
$$
\frac{dy}{dx}
+
\frac{x}{1-x^2}y
=
x\sqrt{y}
$$
Solution
The given differential equation is
$$
\frac{dy}{dx}
+
\frac{x}{1-x^2}y
=
x\sqrt{y}
$$
Since the equation contains the term $\sqrt{y}$, it is not linear.
Let
$$
v=\sqrt{y}
$$
Then,
$$
y=v^2
$$
Differentiating,
$$
\frac{dy}{dx}
=
2v\frac{dv}{dx}
$$
Substitute these values into the given equation:
$$
2v\frac{dv}{dx}
+
\frac{x}{1-x^2}v^2
=
xv
$$
Divide both sides by $2v$:
$$
\frac{dv}{dx}
+
\frac{x}{2(1-x^2)}v
=
\frac{x}{2}
$$
This is now a linear differential equation of the form
$$
\frac{dv}{dx}+P(x)v=Q(x)
$$
where
$$
P(x)=
\frac{x}{2(1-x^2)}.
$$
The integrating factor is
$$
\text{I.F.}
=
e^{\int P(x)\,dx}
=
e^{\int
\frac{x}{2(1-x^2)}dx}
$$
Let
$$
u=1-x^2
$$
Then,
$$
du=-2x\,dx
$$
Therefore,
$$
\int
\frac{x}{2(1-x^2)}dx
=
-\frac14
\int
\frac{du}{u}
$$
Hence,
$$
=
-\frac14\ln|1-x^2|
$$
Therefore,
$$
\text{I.F.}
=
e^{-\frac14\ln(1-x^2)}
$$
Using the identity
$$
e^{\ln a}=a,
$$
we obtain
$$
\boxed{
\text{I.F.}
=
(1-x^2)^{-1/4}
}
$$
Hence, the correct answer is
$$
\boxed{(1-x^2)^{-1/4}}
$$
which corresponds to Option (B).
Q.3
Var(Y) = E(Y^2) - (E(Y))^2
= E((2X+3)^2) - (E(2X+3))^2
=4* {E(X^2) - (E(X)^2)}
or, Var(Y) = 4*Var(X)
or, σY^2 = 4*σx^2
Option B
Q.4
Option