KEAM 2026 20th April Question Paper with Solutions

We need $A\setminus B=\{d\}$, i.e. $d\in A, d\notin B$, and no other element of $A$ is missing from $B$.

So $a,b,c$ must all be in $B$ (otherwise $A\setminus B$ would contain them too), $d$ must not be in $B$, and $e$ can be either in $B$ or not (it doesn't affect $A\setminus B$ since $e\notin A$).

So $B=\{a,b,c\}$ or $B=\{a,b,c,e\}$. Hence $|P|=2$.

For $R$ to be reflexive on $X=\{1,2,\ldots,7\}$, every pair $(x,x)$ must belong to $R$.

$R$ already contains $(1,1)$. The missing pairs are $(2,2),(3,3),(4,4),(5,5),(6,6),(7,7)$.

Adding exactly these makes $R$ reflexive, which matches option E.

$f(x)=\dfrac{2025x+2026}{2027x-2025}$ corresponds to the matrix $$\begin{pmatrix}2025 & 2026\\ 2027 & -2025\end{pmatrix}$$, whose trace is $0$.

For a M\"obius transformation with $a+d=0$, we have $f(f(x))=x$, i.e. $f$ is an involution ($f^2=\text{identity}$).

Since $1000$ is even, $f^{1000}(x)=x$ for all $x$ in the domain. So $f^{1000}(100)=100$.

For $x\le -1$, $f(x)=x^2+2x+3=(x+1)^2+2$.

Since $x\le-1 \Rightarrow (x+1)\le 0 \Rightarrow (x+1)^2\ge 0$, we get $f(x)\ge 2$, and as $x\to-\infty$, $f(x)\to\infty$.

So the range of $f$ is $[2,\infty)$, which is the domain of $f^{-1}$.

$i^1\cdot i^2\cdots i^{2026}=i^{1+2+\cdots+2026}=i^{\frac{2026\cdot 2027}{2}}=i^{2053351}$.

Since $2053351 = 4\times 513337 + 3$, we get $i^{2053351}=i^3=-i$.

$z_1=(7+i\sqrt5)^2+(7-i\sqrt5)^2=2\left[7^2+(i\sqrt5)^2\right]=2(49-5)=88$, which is purely real.

$z_2=(3+2i)^3-(3-2i)^3$. Using $a^3-b^3$ with $a=3+2i,\ b=3-2i$: this equals $2i\,\text{Im}\big[(3+2i)^3\big]\times$ structure that gives a purely imaginary result. Computing directly, $(3+2i)^3=-9+46i$ and $(3-2i)^3=-9-46i$, so $z_2=92i$, purely imaginary.

Hence $z_1$ is purely real and $z_2$ is purely imaginary.

Let $w=z_1\overline{z_2}+z_2\overline{z_1}$. Since $w=z_1\overline{z_2}+\overline{z_1\overline{z_2}}=2\,\text{Re}(z_1\overline{z_2})$, $w$ is already real, so $\overline{w}=w$.

Thus the expression equals $w+\overline w=2w=4\,\text{Re}(z_1\overline{z_2})$.

With $z_1=1+3i,\ z_2=5-3i$: $\overline{z_2}=5+3i$, so $z_1\overline{z_2}=(1+3i)(5+3i)=5+3i+15i-9=-4+18i$, and $2\,\text{Re}(z_1\overline{z_2})=-8$.

So the required value is $2\times(-8)=-16$.

$|\cos\alpha+i\sin\alpha|=1$ and $|\sin\alpha-i\cos\alpha|=1$, so the first term is $1^{1000}=1$.

Similarly $|\sin\alpha+i\cos\alpha|=1$ and $|\cos\alpha-i\sin\alpha|=1$, so the second term is $1^{2000}=1$.

Sum $=1+1=2$.

Since $t_1,t_2,\ldots$ are in G.P. with common ratio $r$, $\dfrac{t_4}{t_2}=r^2$, so $\left(\dfrac{t_4}{t_2}\right)^3=r^6$.

Checking option B: $\dfrac{t_7}{t_4}=r^3$, so $\left(\dfrac{t_7}{t_4}\right)^2=r^6$, which matches.

Given $S_n=100-100^{1-n}$.

$a_1=S_1=100-100^0=99$, and $a_2=S_2-S_1=(100-100^{-1})-99=1-\dfrac{1}{100}=\dfrac{99}{100}$.

Common ratio $r=\dfrac{a_2}{a_1}=\dfrac{99/100}{99}=\dfrac{1}{100}$.

Since $a_1,a_2,6,a_4$ are in G.P. with common ratio $r$: $a_2=a_1r,\ 6=a_1r^2,\ a_4=a_1r^3$.

Then $6a_1+6a_2+a_1a_4+a_2a_4=(a_1+a_2)(6+a_4)$. Substituting $a_1=\dfrac{6}{r^2}$ gives $a_1+a_2=\dfrac{6(1+r)}{r^2}$ and $6+a_4=6(1+r)$.

So $(a_1+a_2)(6+a_4)=\dfrac{36(1+r)^2}{r^2}=\left(\dfrac{6(1+r)}{r}\right)^2=(6+a_2)^2$ since $a_2=6/r$.

Hence the required square is $(6+a_2)^4$.

If $g_1,g_2$ are two geometric means between $2$ and $8$, then $2,g_1,g_2,8$ are in G.P.

Common ratio: $r^3=\dfrac{8}{2}=4 \Rightarrow r=4^{1/3}$.

So $g_1=2r,\ g_2=2r^2$, and $g_1^3+g_2^3=8r^3+8r^6=8(4)+8(16)=32+128=160$.

$\displaystyle \binom{n}{4}=\frac{n(n-1)(n-2)(n-3)}{24}=1365$.

Trying $n=15$: $\binom{15}{4}=\dfrac{15\times14\times13\times12}{24}=1365$, which matches.

Number of diagonals of an $n$-sided polygon $=\dfrac{n(n-3)}{2}=170 \Rightarrow n(n-3)=340$.

Trying $n=20$: $20\times17=340$. So the polygon has $20$ sides.

Expanding $(x^2+\sqrt{1-x^2})^5+(x^2-\sqrt{1-x^2})^5$, the odd-power terms in $\sqrt{1-x^2}$ cancel, leaving twice the even-$k$ terms:

$2\sum_{k=0,2,4}\binom{5}{k}(x^2)^{5-k}(1-x^2)^{k/2}$.

Only $k=2$ can contribute an $x^8$ term: $2\binom{5}{2}x^6(1-x^2)=20x^6-20x^8$, giving coefficient $-20$ for $x^8$ ($k=0$ gives only $x^{10}$, $k=4$ gives only up to $x^6$).

$^{n}P_5=n(n-1)(n-2)(n-3)(n-4)=6720$ and $^{(n-1)}P_4=(n-1)(n-2)(n-3)(n-4)=840$.

Dividing, $n=\dfrac{6720}{840}=8$.

So $^{n}P_3={}^{8}P_3=8\times7\times6=336$.

For any square matrix $A$: $(A+A^T)^T=A^T+A=A+A^T$, so $A+A^T$ is symmetric.

Also $(A-A^T)^T=A^T-A=-(A-A^T)$, so $A-A^T$ is skew-symmetric.

For any nonzero $a$, $(a+a^{-1})^2-(a-a^{-1})^2=4$, a constant independent of $a$.

So for every row, $C_1-C_2=4$, i.e. $C_1=C_2+4C_3$ (since $C_3$ is all $1$'s). This is a linear relation among the columns.

Since the columns are linearly dependent, the determinant is $0$.

$$A=\begin{pmatrix}1 & \tan x\\-\tan x & 1\end{pmatrix}$$, so $$A^T=\begin{pmatrix}1 & -\tan x\\ \tan x & 1\end{pmatrix}$$.

$$AA^T=\begin{pmatrix}1+\tan^2x & -\tan x+\tan x\\ -\tan x+\tan x & \tan^2x+1\end{pmatrix}=\begin{pmatrix}\sec^2x & 0\\0 & \sec^2x\end{pmatrix}$$.

Adding $C_2$ and $C_3$ to $C_1$, each row sum becomes $x+19$ (e.g. row 1: $(x+2)+8+9=x+19$), so $C_1\to(x+19,x+19,x+19)^T$, giving a common factor of $(x+19)$ from that column.

The remaining determinant, after row-reducing $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, produces a factor $(x-2)^2$.

So the determinant is $(x-2)^2(x+19)$, matching $(x-2)^2(ax+b)$ with $a=1,\ b=19$.

$\dfrac{2x-10}{x+2}\ge x-5 \ (x\neq -2)$

$\dfrac{2x-10-(x-5)(x+2)}{x+2}\ge 0 \Rightarrow \dfrac{2x-10-(x^2-3x-10)}{x+2}\ge 0 \Rightarrow \dfrac{-x^2+5x}{x+2}\ge 0 \Rightarrow \dfrac{x(x-5)}{x+2}\le 0$

Using the sign chart with critical points $-2,0,5$: the expression is $\le 0$ for $x\in(-\infty,-2)\cup[0,5]$.

$10<|x+10|\le 25$.

Case $x+10>0$: $10<x+10\le25 \Rightarrow 0<x\le15$.

Case $x+10<0$: $10<-(x+10)\le25 \Rightarrow 20<-x\le35 \Rightarrow -35\le x<-20$.

So $x\in[-35,-20)\cup(0,15]$.

Reduce each angle using periodicity: $\cot(-405^\circ)=\cot(-405^\circ+360^\circ)=\cot(-45^\circ)=-1$, $\tan315^\circ=\tan(-45^\circ)=-1$.

$\cot495^\circ=\cot(495^\circ-360^\circ)=\cot135^\circ=-1$, $\tan(-585^\circ)=\tan(-585^\circ+720^\circ)=\tan135^\circ=-1$.

So the expression $=3(-1)(-1)-5(-1)(-1)=3-5=-2$.

Using $\sin(A+x)-\sin(A-x)=2\cos A\sin x$ and $\cos(A+x)-\cos(A-x)=-2\sin A\sin x$ with $A=\dfrac{3\pi}{4}$:

Expression $=2\cos\dfrac{3\pi}{4}\sin x-2\sin\dfrac{3\pi}{4}\sin x=2\sin x\left(\cos\dfrac{3\pi}{4}-\sin\dfrac{3\pi}{4}\right)$

Since $\cos\dfrac{3\pi}{4}=-\dfrac{1}{\sqrt2},\ \sin\dfrac{3\pi}{4}=\dfrac{1}{\sqrt2}$: $=2\sin x\left(-\dfrac{2}{\sqrt2}\right)=-2\sqrt2\sin x$.

Use the identity $\sin\phi=2\sin\dfrac{\phi}{2}\cos\dfrac{\phi}{2}$ repeatedly starting from $\phi=\dfrac{x}{2^9}$.

$2\sin\left(\dfrac{x}{2^{10}}\right)\cos\left(\dfrac{x}{2^{10}}\right)=\sin\left(\dfrac{x}{2^9}\right)$, then multiplying by $2\cos\left(\dfrac{x}{2^9}\right)$ gives $\sin\left(\dfrac{x}{2^8}\right)$, and so on.

After absorbing all 10 cosine factors ($\cos(x/2)$ through $\cos(x/2^{10})$), we reach $2^{10}\sin\left(\dfrac{x}{2^{10}}\right)\cos\left(\dfrac{x}{2}\right)\cdots\cos\left(\dfrac{x}{2^{10}}\right)=\sin x$.

So $2^{11}\sin\left(\dfrac{x}{2^{10}}\right)\cos\left(\dfrac{x}{2}\right)\cdots\cos\left(\dfrac{x}{2^{10}}\right)=2\sin x$.