KEAM 2026 19th April Question Paper with Solutions

The relation $R = \{(a_1,a_1), (a_2,a_2), \ldots, (a_n,a_n)\}$ is exactly the identity relation on $X$.

Every element is related to itself, so $R$ is reflexive.

Since each pair is of the form $(a_i,a_i)$, $(a_i,a_i) \in R \Rightarrow (a_i,a_i) \in R$, so it is symmetric.

Similarly, $(a_i,a_i), (a_i,a_i) \in R \Rightarrow (a_i,a_i) \in R$, so it is transitive.

Hence $R$ is reflexive, symmetric and transitive.

Here $X$ has 6 elements and $Y$ has 5 elements.

A one-to-one (injective) function from $X$ to $Y$ requires every element of $X$ to map to a distinct element of $Y$, which needs $|X| \le |Y|$.

Since $6 > 5$, no one-to-one function can be defined from $X$ to $Y$.

Given $f(x) = \dfrac{2x+3}{x-2}$.

$f(f(x)) = \dfrac{2\left(\dfrac{2x+3}{x-2}\right)+3}{\dfrac{2x+3}{x-2}-2} = \dfrac{2(2x+3)+3(x-2)}{2x+3-2(x-2)} = \dfrac{7x}{7} = x$

So $h(x) = f(f(x)) = x$ for all valid $x$, i.e. $h$ is the identity function.

Therefore $h(h(10)) = h(10) = 10$.

Given $f(x) = x^2+4x+4 = (x+2)^2$, for $x \le -2$.

Let $y = (x+2)^2$. Since $x \le -2$, we have $x+2 \le 0$, so $x+2 = -\sqrt{y}$.

Thus $x = -2-\sqrt{y}$.

As $x$ ranges over $(-\infty,-2]$, $y = (x+2)^2$ ranges over $[0,\infty)$.

Hence $f^{-1}(x) = -2-\sqrt{x}, \ x \ge 0$.

The powers of $i$ repeat in cycles of 4: $i,-1,-i,1$, and the sum of any 4 consecutive powers is 0.

The sum $i^{13}+i^{14}+\cdots+i^{2026}$ has $2026-13+1 = 2014$ terms.

$2014 = 4 \times 503 + 2$, so 503 complete cycles (summing to 0) remain, leaving the last 2 terms: $i^{2025}+i^{2026}$.

$2025 = 4(506)+1 \Rightarrow i^{2025}=i$, and $i^{2026}=i^2=-1$.

So the sum $= i - 1$.

Simplify LHS: $(3+i)x+y+(1-i)y+3i-4 = (3x+2y-4) + i(x-y+3)$

Simplify RHS: $(2x+1)i+(x-y+2)i = i(3x-y+3)$, which has real part $0$.

Equating real parts: $3x+2y-4=0$.

Equating imaginary parts: $x-y+3 = 3x-y+3 \Rightarrow x=0$.

Substituting, $2y=4 \Rightarrow y=2$. So $(x,y)=(0,2)$.

For any complex number $z=\dfrac{a+bi}{c+di}$ where $c+di$ is $\pm i$ times the conjugate of $a+bi$ (i.e. $|a+bi|=|c+di|$), we get $|z|=1$.

Check: $|5+7i|=\sqrt{74}=|7-5i|$, so $|z_1|=1$. Similarly $|3+2i|=\sqrt{13}=|3-2i|$, so $|z_2|=1$, and $|1+11i|=\sqrt{122}=|11-i|$, so $|z_3|=1$.

Since $z\bar z=|z|^2$, each term equals 1, so $z_1\overline{z_1}+z_2\overline{z_2}+z_3\overline{z_3}=1+1+1=3$.

Write $1+i=\sqrt2\,e^{i\pi/4}$ and $1-i=\sqrt2\,e^{-i\pi/4}$.

$\dfrac{(1+i)^n}{(1-i)^{n-4}} = \dfrac{(\sqrt2)^n e^{in\pi/4}}{(\sqrt2)^{n-4}e^{-i(n-4)\pi/4}} = (\sqrt2)^4\, e^{i(n\pi/4+(n-4)\pi/4)} = 4\,e^{i(n\pi/2 - \pi)}$

$= 4\left(\cos\left(\dfrac{n\pi}{2}-\pi\right)+i\sin\left(\dfrac{n\pi}{2}-\pi\right)\right) = -4\left(\cos\dfrac{n\pi}{2}+i\sin\dfrac{n\pi}{2}\right) = -4i^n$

The sequence $2,6,18,\ldots,1458$ is a G.P. with first term $a=2$ and common ratio $r=3$.

The $n$-th term is $t_n = a\,r^{n-1} = 2\cdot 3^{n-1}$.

Setting $2\cdot3^{n-1}=1458 \Rightarrow 3^{n-1}=729=3^6 \Rightarrow n-1=6 \Rightarrow n=7$.

So there are 7 terms in the sequence.

Let $t_k = t_1 r^{k-1}$ for the G.P. with common ratio $r$.

Consider the odd-indexed terms $t_1,t_3,t_5,\ldots,t_{2n-1}$. The ratio between consecutive terms is

$\dfrac{t_3}{t_1}=\dfrac{t_1r^2}{t_1}=r^2,\quad \dfrac{t_5}{t_3}=r^2, \ \ldots$

So $t_1,t_3,t_5,\ldots,t_{2n-1}$ form a G.P. with common ratio $r^2$.

The Geometric Mean between 4 and 16 is $\sqrt{4\times16}=\sqrt{64}=8$.

So we need $\dfrac{4^{n+1}+16^{n+1}}{4^n+16^n}=8$.

Substituting $n=-\dfrac12$: $4^n=\tfrac12,\ 16^n=\tfrac14,\ 4^{n+1}=2,\ 16^{n+1}=4$.

$\dfrac{2+4}{\tfrac12+\tfrac14}=\dfrac{6}{0.75}=8$ ✓, which matches the required GM.

So $n=-\dfrac12$.

For a G.P. with first term $a=7$, last term $l=448$, and sum $S=889$, use $S=\dfrac{lr-a}{r-1}$.

$889(r-1) = 448r - 7 \Rightarrow 889r - 889 = 448r - 7 \Rightarrow 441r = 882 \Rightarrow r=2$

Check: with $r=2$, $l=ar^{n-1}=7\cdot2^{n-1}=448 \Rightarrow 2^{n-1}=64 \Rightarrow n=7$, and $S=7(2^7-1)/(2-1)=7\times127=889$. ✓

So the common ratio is 2.

The person can choose any of the 2 entrances, then any of the 3 lifts in that entrance, then the lift can stop at any of the 5 floors.

By the fundamental principle of counting, the total number of ways $= 2\times3\times5 = 30$.

$^9P_5 = \dfrac{9!}{4!} = 9\times8\times7\times6\times5 = 15120$

We need $15120 = 504\times {}^6P_r$, so ${}^6P_r = \dfrac{15120}{504}=30$.

Since ${}^6P_2 = 6\times5 = 30$, we get $r=2$.

The number of 3-digit numbers formed from $\{1,2,3,4\}$ without repetition is ${}^4P_3=24$.

Each digit appears in each place (hundreds, tens, units) equally often: $24/4=6$ times per digit per place.

Sum of digits $=1+2+3+4=10$, so the contribution from each place is $6\times10=60$.

Total sum $=60\times100+60\times10+60\times1=60\times111=6660$.

There are 23 green balls out of 24 (1 is black). We need to choose 3 balls, all green.

Number of ways $= \binom{23}{3} = \dfrac{23\times22\times21}{3\times2\times1} = 1771$

General term in $\left(3x-\dfrac{1}{3x}\right)^4$ is $T_{r+1}=\binom{4}{r}(3x)^{4-r}\left(\dfrac{-1}{3x}\right)^r = \binom{4}{r}(-1)^r 3^{4-2r}x^{4-2r}$

For the coefficient of $\dfrac{1}{x^2}$, set $4-2r=-2 \Rightarrow r=3$.

Coefficient $= \binom{4}{3}(-1)^3 3^{4-6} = 4\times(-1)\times3^{-2} = \dfrac{-4}{9}$

First compute $$\begin{pmatrix}1&1&1\\-1&0&1\\1&0&-1\end{pmatrix}\begin{pmatrix}2\\3\\1\end{pmatrix} = \begin{pmatrix}2+3+1\\-2+0+1\\2+0-1\end{pmatrix}=\begin{pmatrix}6\\-1\\1\end{pmatrix}$$

Then $$(x\ \ 3\ \ -1)\begin{pmatrix}6\\-1\\1\end{pmatrix} = 6x-3-1 = 6x-4$$

Setting this to 0: $6x-4=0 \Rightarrow x=\dfrac{2}{3}$

Compute $P^2$: multiplying $P$ by itself, the third row becomes $(10(1)+100(0)+(-1)(10),\ 10(0)+100(1)+(-1)(100),\ (-1)(-1)) = (0,0,1)$, and rows 1, 2 stay as $(1,0,0)$ and $(0,1,0)$.

So $P^2 = I$, the identity matrix.

Since $4052$ is even, $P^{4052} = (P^2)^{2026} = I^{2026} = I$.

Expanding the determinant along the first row:

$$\begin{vmatrix}11&1&1\\1&21&1\\1&1&31\end{vmatrix} = 11(21\times31-1) - 1(1\times31-1) + 1(1-21)$$

$= 11(650) - 1(30) + 1(-20) = 7150 - 30 - 20 = 7100$

Since $$A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$$, compute $$A^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}=-I$$.

$(\alpha I+\beta A)^2 = \alpha^2 I + 2\alpha\beta A + \beta^2 A^2 = (\alpha^2-\beta^2)I + 2\alpha\beta A$

Setting this equal to $A = 0\cdot I + 1\cdot A$, and comparing coefficients of $I$ and $A$:

$\alpha^2-\beta^2 = 0 \quad \text{and} \quad 2\alpha\beta = 1$

$5 < |x-1| < 15$ means either $5 < x-1 < 15$ or $-15 < x-1 < -5$.

Case 1: $5 < x-1 < 15 \Rightarrow 6 < x < 16$

Case 2: $-15 < x-1 < -5 \Rightarrow -14 < x < -4$

So $-14 < x < -4$ or $6 < x < 16$.

$\dfrac{x-3}{x-2}\ge1 \Rightarrow \dfrac{x-3}{x-2}-1\ge0 \Rightarrow \dfrac{(x-3)-(x-2)}{x-2}\ge0 \Rightarrow \dfrac{-1}{x-2}\ge0$

This holds when $x-2<0$, i.e. $x<2$ (note $x=2$ is excluded as the expression is undefined there).

So the solution set is $(-\infty,2)$.

$\sin\theta\cos\theta>0$ requires $\sin\theta$ and $\cos\theta$ to have the same sign.

In the first quadrant, both $\sin\theta>0$ and $\cos\theta>0$, so the product is positive.

In the third quadrant, both $\sin\theta<0$ and $\cos\theta<0$, so the product is again positive.

In the second and fourth quadrants, $\sin\theta$ and $\cos\theta$ have opposite signs, giving a negative product.

So $\theta$ lies in the first or third quadrant.

Treating this as a quadratic in $\sin x$: $4\sin^2x-2(1+\sqrt3)\sin x+\sqrt3=0$

By the quadratic formula, $\sin x = \dfrac{2(1+\sqrt3)\pm\sqrt{4(1+\sqrt3)^2-16\sqrt3}}{8}$, which simplifies to $\sin x=\dfrac{\sqrt3}{2}$ or $\sin x=\dfrac12$.

For $\sin x=\dfrac{\sqrt3}{2}$: $x=60^\circ$ or $120^\circ$ (both lie in $15^\circ<x<150^\circ$).

For $\sin x=\dfrac12$: $x=30^\circ$ or $150^\circ$; only $30^\circ$ lies strictly within the given range.

So $x=30^\circ,60^\circ,120^\circ$.

Using $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$:

$\tan(\alpha+\beta)=\dfrac{\tfrac{5}{6}+\tfrac{1}{11}}{1-\tfrac{5}{6}\cdot\tfrac{1}{11}}=\dfrac{\tfrac{55+6}{66}}{\tfrac{66-5}{66}}=\dfrac{61}{61}=1$

Since $0<\alpha,\beta<\dfrac{\pi}{2}$, we have $0<\alpha+\beta<\pi$, and $\tan(\alpha+\beta)=1$ gives $\alpha+\beta=\dfrac{\pi}{4}$.

Using the product-to-sum identity $\sin A\sin B=\dfrac12[\cos(A-B)-\cos(A+B)]$:

$\sin6^\circ\sin66^\circ=\dfrac12[\cos60^\circ-\cos72^\circ], \qquad \sin42^\circ\sin18^\circ=\dfrac12[\cos24^\circ-\cos60^\circ]$

Multiplying each by $\cos36^\circ$ and $\cos12^\circ$ respectively and simplifying using known values of $\cos72^\circ,\cos36^\circ,\cos24^\circ,\cos12^\circ$, the whole expression reduces to $\dfrac14(\sin18^\circ+\cos36^\circ)$.

Numerically, LHS $\approx0.2795$ and $\dfrac14(\sin18^\circ+\cos36^\circ)\approx\dfrac14(0.309+0.809)\approx0.2795$, confirming the identity.

For $f(x)=2\sin^{-1}(2x-1)-\dfrac{\pi}{4}$ to be defined, we need $-1\le 2x-1\le1$.

$-1\le2x-1 \Rightarrow x\ge0$, and $2x-1\le1 \Rightarrow x\le1$.

So the domain is $[0,1]$.

Using the sine addition formula, $\sin\dfrac{5\pi}{9}\cos\dfrac{\pi}{9}+\cos\dfrac{5\pi}{9}\sin\dfrac{\pi}{9} = \sin\left(\dfrac{5\pi}{9}+\dfrac{\pi}{9}\right) = \sin\dfrac{6\pi}{9} = \sin\dfrac{2\pi}{3}$

$\sin\dfrac{2\pi}{3} = \sin\left(\pi-\dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{3} = \dfrac{\sqrt3}{2}$

So we need $\sin^{-1}\left(\dfrac{\sqrt3}{2}\right)$. Since the principal range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, the answer is $\dfrac{\pi}{3}$.