KEAM 2025 28th April Question Paper with Solutions

$A=\{2,4,6,\ldots,34\}$ (multiples of 2 less than 36).

$B=\{18,21,24,27,30,33\}$ (multiples of 3 greater than 16).

$C=\{4,8,12,\ldots,40\}$ (multiples of 4 less than 42).

$A\cap B=\{18,24,30\}$, so $(A\cap B)\cap C=\{24\}$.

Number of subsets with exactly 2 or 6 elements $= {}^8C_2 + {}^8C_6 = 28+28=56$.

$f(x)=[2x]$.

$f(-2.3)=[-4.6]=-5$

$f(2.9)=[5.8]=5$

So the image of $\{-2.3,2.9\}$ is $\{-5,5\}$.

$f(f(x))=a(ax+bx^2)+b(ax+bx^2)^2$

$=a^2x+abx^2+b(a^2x^2+2abx^3+b^2x^4)$

Coefficient of $x^3$ comes only from $b\cdot 2abx^3=2ab^2x^3$.

$|z|=\sqrt{1+\tan^2\theta}=|\sec\theta|$

For $\pi<\theta<\dfrac{3\pi}{2}$ (third quadrant), $\cos\theta<0$, so $\sec\theta<0$.

Hence $|z|=-\sec\theta$.

$z=\dfrac{3+i}{2-i}=\dfrac{(3+i)(2+i)}{(2-i)(2+i)}=\dfrac{6+3i+2i+i^2}{5}=\dfrac{5+5i}{5}=1+i$

$z^{-1}=\dfrac{1}{1+i}=\dfrac{1-i}{(1+i)(1-i)}=\dfrac{1-i}{2}$

$|z-2|+|z-4|$ represents the sum of distances of $z$ from the points $2$ and $4$ on the real axis.

By the triangle inequality, this sum is minimized (and equals the distance between the two fixed points) when $z$ lies on the segment joining them.

Minimum value $=|4-2|=2$.

$z=\dfrac{1}{\sqrt2}(1+i)=\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}=e^{i\pi/4}$

Rotating clockwise by $\dfrac{\pi}{4}$ multiplies by $e^{-i\pi/4}$:

$z\cdot e^{-i\pi/4}=e^{i\pi/4}e^{-i\pi/4}=e^0=1$

Let the common ratio be $r$. Since $54$ and $162$ are consecutive terms, $r=\dfrac{162}{54}=3$.

Then $6\cdot r=18=y$ and $x\cdot r=6\Rightarrow x=2$.

$\dfrac{y}{x}=\dfrac{18}{2}=9$

Since $1,a,b,c,16$ are in GP, $16=1\cdot r^4\Rightarrow r=2$.

So $a=2, b=4, c=8$, giving $abc=64$.

$\sqrt[3]{abc}=\sqrt[3]{64}=4$

The series $\sqrt3+\sqrt{12}+\sqrt{48}+\ldots$ has first term $a=\sqrt3$ and common ratio $r=\dfrac{\sqrt{12}}{\sqrt3}=2$.

Sum of 10 terms $=\dfrac{a(r^{10}-1)}{r-1}=\sqrt3(2^{10}-1)=1023\sqrt3$

Let AM $=\dfrac{a+b}{2}$ and GM $=\sqrt{ab}$.

Given AM+GM$=18$ and AM-GM$=8$, so AM$=13$, GM$=5$.

Then $a+b=26$, $ab=25$, giving $t^2-26t+25=0\Rightarrow t=1,25$.

General term of $\left(\sqrt{x}+\dfrac1x\right)^{10}$: $T_{k+1}={}^{10}C_k\,x^{(10-k)/2}\,x^{-k}={}^{10}C_k\,x^{(10-3k)/2}$

For coefficient of $x^5$: $\dfrac{10-3k}{2}=5\Rightarrow k=0$

Coefficient $={}^{10}C_0=1$

Total 4-digit arrangements (ignoring leading-zero restriction) from 6 digits $={}^6P_4=360$.

Arrangements with 0 as the leading digit: fix 0 first, arrange remaining 3 digits from 5 in the other places $={}^5P_3=60$.

Valid 4-digit numbers $=360-60=300$

Total ways to select 3 balls from 12 $={}^{12}C_3=220$

Ways with no white ball (from 9 red+black) $={}^9C_3=84$

Ways with at least one white ball $=220-84=136$

We need the number of solutions of $x_1+x_2+x_3+x_4=6$ with $1\le x_i\le6$.

Substituting $y_i=x_i-1\ge0$, we need $y_1+y_2+y_3+y_4=2$, whose number of non-negative solutions is ${}^{2+3}C_3={}^5C_3=10$ (the upper bound of 6 is never violated here).

$2+{}^{15}C_1+{}^{15}C_2+\ldots+{}^{15}C_{14}=({}^{15}C_0+{}^{15}C_1+\ldots+{}^{15}C_{15})-{}^{15}C_{15}+1$

$=2^{15}-1+1=2^{15}$

Expanding the determinant: $a(bc-1)-1(c-1)+1(1-b)=abc-a-b-c+2$

Given this equals 2, we get $abc=a+b+c$.

For positive integers, $a=1,b=2,c=3$ satisfies $1\cdot2\cdot3=6=1+2+3$.

So $a+b+c=6$

With $x+y=-1$, row 2 becomes $(-1, y+1, x+1)$.

Substituting any pair satisfying $x+y=-1$ (e.g. $x=2,y=-3$) and expanding the determinant gives $\Delta=0$ identically.

This is because row 2 becomes a linear combination of rows 1 and 3 when $x+y=-1$.

$A$ is upper triangular, so $|A|=1\cdot2\cdot(-2)=-4$

For a $3\times3$ matrix, $|adj(adjA)|=|A|^{(n-1)^2}=|A|^4$

$|adj(adjA)|=(-4)^4=256$

The system has a unique solution when the determinant of coefficients is non-zero.

$$\Delta=\begin{vmatrix}1&-2&1\\2&-1&2\\1&2&\lambda\end{vmatrix}$$

Expanding gives $\Delta=3(\lambda-1)\cdot(\ldots)$, which vanishes only when $\lambda=1$. So for a unique solution, $\lambda\neq1$.

$\dfrac{x+1}{x-1}-2<0 \Rightarrow \dfrac{x+1-2(x-1)}{x-1}<0 \Rightarrow \dfrac{3-x}{x-1}<0$

$\Rightarrow \dfrac{x-3}{x-1}>0 \Rightarrow x<1 \text{ or } x>3$

$|x+2|<3 \Rightarrow -3<x+2<3 \Rightarrow -5<x<1$

$\tan 2x=\tan\big((x+y)+(x-y)\big)=\dfrac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)}$

$=\dfrac{\frac65+\frac45}{1-\frac65\cdot\frac45}=\dfrac{2}{1-\frac{24}{25}}=\dfrac{2}{\frac1{25}}=50$

Since $\dfrac{\pi}{12}+\dfrac{\pi}{6}=\dfrac{\pi}{4}$, using $\tan(A+B)$:

$1=\tan\dfrac{\pi}{4}=\dfrac{\tan\frac{\pi}{12}+\tan\frac{\pi}{6}}{1-\tan\frac{\pi}{12}\tan\frac{\pi}{6}}$

$\Rightarrow \tan\dfrac{\pi}{12}+\tan\dfrac{\pi}{6}=1-\tan\dfrac{\pi}{12}\tan\dfrac{\pi}{6}$

So the given expression $=\left(1-\tan\frac{\pi}{12}\tan\frac{\pi}{6}\right)+\tan\frac{\pi}{12}\tan\frac{\pi}{6}=1$

Period of $2\sin4x$ is $\dfrac{2\pi}{4}=\dfrac{\pi}{2}$

Period of $3\cos2x$ is $\dfrac{2\pi}{2}=\pi$

Period of $f(x)$ is LCM of $\dfrac{\pi}{2}$ and $\pi$, which is $\pi$

$\dfrac{1-\cos2x}{1+\cos2x}=\dfrac{2\sin^2x}{2\cos^2x}=\tan^2x$

So the expression becomes $\tan^2x-\sec^2x=-1$ (using $\sec^2x-\tan^2x=1$)

For $\theta\in\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$, the standard result is $\sin^{-1}(\sin\theta)=\pi-\theta$

This ensures the result lies within the principal range $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ of $\sin^{-1}$.

Since $\cos^{-1}x,\cos^{-1}y,\cos^{-1}z\in[0,\pi]$ each, their sum can equal $3\pi$ only if each term equals its maximum value $\pi$.

So $\cos^{-1}x=\cos^{-1}y=\cos^{-1}z=\pi \Rightarrow x=y=z=-1$

$x+y+z=-3$

Using $\cos^{-1}x=\dfrac{\pi}{2}-\sin^{-1}x$ and similarly for $y$:

$\cos^{-1}x+\cos^{-1}y=\pi-(\sin^{-1}x+\sin^{-1}y)=\pi-\dfrac{2\pi}{3}=\dfrac{\pi}{3}$

The line through $(1,0)$ and $(4,3)$ has slope 1, i.e. inclination $45^\circ$.

Rotating counter-clockwise by $15^\circ$ gives a new inclination of $60^\circ$, so the new slope is $\tan60^\circ=\sqrt3$.

Equation through $(1,0)$: $y-0=\sqrt3(x-1)\Rightarrow \sqrt3x-y-\sqrt3=0$

Let the point be $(t,1-t)$ on $x+y=1$.

Distance from $5x+12y=0$: $\dfrac{|5t+12(1-t)|}{13}=\dfrac{|12-7t|}{13}=2$

$\Rightarrow |12-7t|=26 \Rightarrow t=-2$ (integer solution)

Point: $(-2,3)$

$x+\sqrt3y=2\sqrt3$; dividing by $\sqrt{1^2+(\sqrt3)^2}=2$:

$\dfrac12x+\dfrac{\sqrt3}{2}y=\sqrt3$

Comparing with $x\cos\alpha+y\sin\alpha=p$: $\cos\alpha=\dfrac12,\sin\alpha=\dfrac{\sqrt3}{2}\Rightarrow\alpha=\dfrac{\pi}{3}$, and $p=\sqrt3$

Centers are $(2,3)$ and $(2,-3)$, distance between centers $d=6$, radii $3$ and $a$.

Two distinct intersection points require $|a-3|<d<a+3$, i.e. $|a-3|<6$ and $6<a+3$.

This gives $3<a<9$

For $y^2=16x$, $4a=16\Rightarrow a=4$, so the latus rectum is the chord $x=4$.

Its endpoints are $(4,8)$ and $(4,-8)$.

Given one end is $(4,8)$, the other end is $(4,-8)$

Length of latus rectum $=\dfrac{2b^2}{a}$, and major axis $=2a$.

Given $\dfrac{2b^2}{a}=\dfrac{1}{4}(2a)=\dfrac{a}{2}\Rightarrow \dfrac{b^2}{a^2}=\dfrac14$

$e=\sqrt{1-\dfrac{b^2}{a^2}}=\sqrt{1-\dfrac14}=\dfrac{\sqrt3}{2}$

$x^2-4y^2=16 \Rightarrow \dfrac{x^2}{16}-\dfrac{y^2}{4}=1$, so $a^2=16, b^2=4$

$c^2=a^2+b^2=20 \Rightarrow c=2\sqrt5$

Distance between foci $=2c=4\sqrt5$

Both vectors $2i+j+2k$ and $i+2j-2k$ have magnitude 3.

The bisector direction is along the sum of their unit vectors: $\dfrac{2i+j+2k}{3}+\dfrac{i+2j-2k}{3}=\dfrac{3i+3j}{3}=i+j$

Unit vector along bisector $=\dfrac{i+j}{\sqrt2}$

$|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b=2-2\cos\theta$

Given $|\vec a-\vec b|=1$: $1=2-2\cos\theta\Rightarrow\cos\theta=\dfrac12\Rightarrow\theta=\dfrac{\pi}{3}$