MAH CET MCA 2025 Previous Year Questions (PYQs) – Page 2 of 10
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The population of a town grows at $10%$ per year. How long will it take for the population to triple?
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Solution
$(1.1)^t = 3 \Rightarrow t=\dfrac{\log 3}{\log 1.1}\approx 11.53\ \text{years}$.
$\boxed{t=\dfrac{\log 3}{\log 1.1}\approx 11.53\ \text{years}}$
A company models the rate of change of concentration by
$(y^2+3xy),dx+(x^2+xy),dy=0$, with $y=1$ when $x=1$. Find $y$ as a function of $x$ (choose the correct option).
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Solution
Put $y=vx$. Then $v+x,dv/dx=-\dfrac{v(v+3)}{1+v}$.
$\displaystyle \frac{1+v}{v(v+2)},dv=-\frac{2,dx}{x} \Rightarrow \ln!\big(v(v+2)\big)=-4\ln x+C$.
$\Rightarrow v(v+2)=\dfrac{C}{x^{4}}$. Using $(x,y)=(1,1)$ gives $C=3$.
$\displaystyle \therefore \frac{y}{x}!\left(\frac{y}{x}+2\right)=\frac{3}{x^{4}} \Rightarrow y=y(x)=x!\left(-1+\sqrt{1+\frac{3}{x^{4}}}\right)$ (positive branch by IC).$\boxed{y=x!\left(-1+\sqrt{1+\dfrac{3}{x^{4}}}\right)}$
A family of parabolas $y=ax^{2}$ $(a>0)$ has vertices on the $x$–axis and axis parallel to $y$–axis. The differential equation of this family is:
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Solution
$y=ax^{2}$, $a=\dfrac{y}{x^{2}}$. Differentiate: $y'=2ax=\dfrac{2y}{x}$.
$\Rightarrow x,y'-2y=0$.
The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ and $(1+x)^{2n-1}$ are in the ratio
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Solution
$\text{Coeff}{x^n},(1+x)^{2n}=\binom{2n}{n}$ and $\text{Coeff}{x^n},(1+x)^{2n-1}=\binom{2n-1}{n}$.
$\displaystyle \binom{2n}{n}=\frac{(2n)}{n,n}=\frac{2n}{n}\cdot\frac{(2n-1)}{n,(n-1)}=2\binom{2n-1}{n}$.
So ratio $=2:1$.
The equation of a parabola is $y^2=4x$. Which of the following points lies on the parabola?
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Solution
For $y^2=4x$, we need $x=\dfrac{y^2}{4}$.
For $y=\pm4$, $x=\dfrac{16}{4}=4$ $\Rightarrow$ $(4,4)$ and $(4,-4)$ satisfy.
Others give $x\neq\dfrac{y^2}{4}$.
In $\triangle ABC$, sides are $a=9$, $b=7$, $c=12$. Find angle $C$ (opposite side $c$) using the Law of Cosines.
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Solution
$\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{81+49-144}{2\cdot9\cdot7}=\dfrac{-14}{126}=-\dfrac{1}{9}$.
$C=\cos^{-1}\left(-\dfrac{1}{9}\right)\approx97.57^\circ$.
The number of real solutions of $\sqrt{1+\cos 2x}=\sqrt{2},\cos^{-1}(\cos x)$ in $\left[\frac{\pi}{2},\pi\right]$ is
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Solution
$\sqrt{1+\cos 2x}=\sqrt{2},|\cos x|$. On $\left[\frac{\pi}{2},\pi\right]$, $\cos x\le 0$ so $|\cos x|=-\cos x$.
LHS $=\sqrt{2},|\cos x|$, RHS $=\sqrt{2},|\cos x|$ only if $\cos^{-1}(\cos x)=|\cos x|$, which never holds for $x\in\left[\frac{\pi}{2},\pi\right]$.
Hence no real solution.
Given $,\dfrac{dy}{dx}+y=1,$ with $y(0)=2$, find $y(1)$.
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Solution
Solve linear ODE: $y=Ce^{-x}+1$. Using $y(0)=2\Rightarrow C=1$.
So $y(1)=e^{-1}+1$.
The valid range of $x$ for which $\cos^{-1}(2x-3)$ is defined is
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Solution
For $\cos^{-1}(,\cdot,)$ to be defined, $-1\le 2x-3\le 1$.
This gives $2\le 2x\le 4\Rightarrow 1\le x\le 2$.
The perpendicular distance of the point $P(1,2,3)$ from the line $\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}$ is
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Solution
Line passes through $A(6,7,7)$ with direction $\vec v=\langle3,2,-2\rangle$.
$\vec{AP}=\langle-5,-5,-4\rangle$. Distance $=\dfrac{\lVert \vec{AP}\times\vec v\rVert}{\lVert\vec v\rVert}=\dfrac{\sqrt{833}}{\sqrt{17}}=\sqrt{49}=7$.
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