KEAM 2026 18th April Question Paper with Solutions

We have $n(A)=45,\ n(B)=35,\ n(A\cap B)=13$

$(A\cap B)' \cap (A\cup B) = (A\cup B) - (A\cap B)$

$n(A\cup B) = n(A)+n(B)-n(A\cap B) = 45+35-13 = 67$

$\therefore n((A\cap B)'\cap(A\cup B)) = 67-13 = 54$

For $f(x)=\sqrt{7-x}$ to be defined, $7-x\ge 0 \Rightarrow x\le 7$

For $g(x)=\sqrt{x-5}$ to be defined, $x-5\ge 0 \Rightarrow x\ge 5$

Domain of $h(x)=f(x)g(x)$ is the intersection of both, i.e. $[5,7]$

$A=\{1,3,5,7,9\}$ (odd integers in $[0,10]$), $B=\{2,3,5,7\}$ (primes in $[0,10]$)

Since every element of $A$ is odd, $a+b$ is odd only when $b$ is even, i.e. $b=2$

So pairs are $(1,2),(3,2),(5,2),(7,2),(9,2)$, giving $n(R)=5$

$f(x) = \tan x \sin x$

$f(-x) = \tan(-x)\sin(-x) = (-\tan x)(-\sin x) = \tan x \sin x = f(x)$

Since $f(-x)=f(x)$, $f$ is an even function.

$z = \dfrac{a-\frac{i}{2}}{i-2}$

Multiplying numerator and denominator by the conjugate $(-i-2)$ of $(i-2)$:

$z = \dfrac{\left(a-\frac{i}{2}\right)(-i-2)}{(i-2)(-i-2)} = \dfrac{\left(-2a-\frac12\right) + i(1-a)}{5}$

For $\text{Im}(z)=0$, we need $1-a=0 \Rightarrow a=1$

The two given vertices are $z_1=-3-3i$ and $z_2=3-3i$

Side length: $|z_1-z_2| = |-6| = 6$

Area of equilateral triangle $= \dfrac{\sqrt3}{4}(\text{side})^2 = \dfrac{\sqrt3}{4}(36) = 9\sqrt3$ sq. units

$(3+i)(3-i) = 9-i^2 = 10$

So the base expression is $\dfrac{5i}{10} = \dfrac{i}{2}$

$\left(\dfrac{i}{2}\right)^{2026} = \dfrac{i^{2026}}{2^{2026}}$

Since $2026 = 4(506)+2$, $i^{2026}=i^2=-1$

$\therefore \text{value} = \dfrac{-1}{2^{2026}}$

Let $|z|=r$. Then $z|z| = zr = 24+7i \Rightarrow |z| \cdot r = |24+7i|$

$r \cdot r = \sqrt{24^2+7^2} = \sqrt{576+49} = \sqrt{625}=25$

$r^2 = 25 \Rightarrow r = 5$

For a G.P. with first term $a$ and common ratio $r=\frac13$, sum of first four terms:

$S_4 = a\cdot\dfrac{1-r^4}{1-r} = a\cdot\dfrac{1-\frac{1}{81}}{\frac23} = a\cdot\dfrac{80/81}{2/3} = \dfrac{40a}{27}$

Given $S_4=200$: $\dfrac{40a}{27}=200 \Rightarrow a = \dfrac{200\times27}{40}=135$

Let first term be $a$ and common ratio $r$. Given $ar+ar^2=8$ and $ar^3=4$

$ar(1+r)=8 \quad\text{...(i)}, \qquad ar\cdot r^2=4 \quad\text{...(ii)}$

Dividing (ii) by (i): $\dfrac{r^2}{1+r}=\dfrac{1}{2} \Rightarrow 2r^2-r-1=0 \Rightarrow (2r+1)(r-1)=0$

Since $r\neq1$, we get $r=-\dfrac12$

$\sum_{k=1}^{n}\log_{10}(5^k) = \sum_{k=1}^n k\log_{10}5 = \dfrac{n(n+1)}{2}\log_{10}5$

Given this equals $66\log_{10}5$: $\dfrac{n(n+1)}{2}=66 \Rightarrow n(n+1)=132$

$n=11$ satisfies $11\times12=132$

Let first term $a=512$ and common ratio $r$. The 20th term is $ar^{19}=\dfrac{1}{1024}$

$r^{19} = \dfrac{1}{1024\times512} = \dfrac{1}{2^{19}}$

$\therefore r = \dfrac{1}{2}$

For $k, 6, k+5$ in G.P.: $6^2 = k(k+5)$

$36 = k^2+5k \Rightarrow k^2+5k-36=0 \Rightarrow (k-4)(k+9)=0$

So $k=4$ or $k=-9$

Common ratio $r=\dfrac{6}{k}$: for $k=4$, $r=\dfrac32$; for $k=-9$, $r=-\dfrac23$

Let the initial number of people be $n$. All $n$ people shake hands with each other once, giving $^{n}C_2$ handshakes.

Then 5 people leave, and one new person arrives, shaking hands with the remaining $(n-5)$ people present, adding $(n-5)$ more handshakes.

Total handshakes: $\binom{n}{2}+(n-5) = 100$

Testing $n=14$: $\binom{14}{2}+9 = 91+9=100$ ✓

For a 3-digit number greater than 500 using digits $3,4,5,7$ with repetition, the first digit must be $5$ or $7$ (any such number automatically exceeds 500 since digit 0 is unavailable).

First digit: 2 choices ($5$ or $7$); second and third digits: 4 choices each

Total $= 2\times4\times4 = 32$

$^{n}P_5 = \dfrac{n!}{(n-5)!} = 360360$

Testing $n=15$: $15\times14\times13\times12\times11 = 360360$ ✓

$^{n}C_5 = \dfrac{^{n}P_5}{5!} = \dfrac{360360}{120} = 3003$

In $(4x+a)^7$, the term containing $x^4$ is $\binom{7}{4}(4x)^4a^3 = \binom{7}{4}\cdot256\cdot a^3\cdot x^4$

Coefficient $= 35\times256\times a^3 = 8960\,a^3$

Given $8960\,a^3 = -1120 \Rightarrow a^3 = -\dfrac18 \Rightarrow a = -\dfrac12$

$|A^{-1}| = \dfrac{1}{|A|} \Rightarrow |A| = \dfrac{1}{-7} = -\dfrac17$

$|AB| = |A||B| \Rightarrow 21 = -\dfrac17\,|B|$

$|B| = 21\times(-7) = -147$

For a nontrivial solution of $AX=0$, we need $|A|=0$

$|A| = 9[(-1)(3)-(-3)(1)] - 2[(1)(3)-(-3)(k-1)] + k[(1)(1)-(-1)(k-1)]$

$= 9(0) - 2(3k) + k(k) = k^2-6k = k(k-6)$

Setting $|A|=0$: $k=0$ or $k=6$

Given $BA^2=A$, multiplying both sides by $A^{-2}$ on the right: $B = A\cdot A^{-2} = A^{-1}$

$|A| = (3)(-3)-(5)(-2) = -9+10 = 1$

$$A^{-1} = \dfrac{1}{|A|}\begin{pmatrix}-3 & -5\\ 2 & 3\end{pmatrix} = \begin{pmatrix}-3 & -5\\ 2 & 3\end{pmatrix}$$

At $x=-1$: $$f(-1) = \begin{vmatrix}-1 & -1 & 3\\ 0 & 1 & -2\\ 1 & -1 & 1\end{vmatrix}$$

Expanding along the first row: $-1[(1)(1)-(-2)(-1)] -(-1)[(0)(1)-(-2)(1)] + 3[(0)(-1)-(1)(1)]$

$= -1(-1) +1(2) + 3(-1) = 1+2-3 = 0$

Length $=x+2$, breadth $=x-2$ (with $x>2$)

Area: $A=(x+2)(x-2)=x^2-4 \le 140 \Rightarrow x^2\le144 \Rightarrow x\le12$

Perimeter: $P=2[(x+2)+(x-2)]=4x \ge 20 \Rightarrow x\ge5$

Combining: $x\in[5,12]$

$8+3x > 4(x-3)+2$

$8+3x > 4x-12+2 \Rightarrow 8+3x>4x-10$

$18 > x \Rightarrow x<18$

So the solution set is $(-\infty,18)$

Using the identity $(\text{cosec}\,t-\cot t)(\text{cosec}\,t+\cot t)=1$:

$\text{cosec}\,t-\cot t = \dfrac{1}{5/2}=\dfrac25$

Adding to the given equation: $2\,\text{cosec}\,t = \dfrac52+\dfrac25=\dfrac{29}{10} \Rightarrow \text{cosec}\,t=\dfrac{29}{20}$

Subtracting: $2\cot t = \dfrac52-\dfrac25=\dfrac{21}{10} \Rightarrow \cot t = \dfrac{21}{20}$

$\therefore \tan t = \dfrac{20}{21}$

Given $\sin\theta=\dfrac35$ and $\cos\theta<0$, so $\theta$ lies in the second quadrant.

$\cos\theta = -\sqrt{1-\sin^2\theta} = -\sqrt{1-\frac{9}{25}} = -\dfrac45$

$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac34$

$\sin15^\circ\sin75^\circ = \sin15^\circ\cos15^\circ = \dfrac12\sin30^\circ = \dfrac14$

$\therefore \sin15^\circ\sin45^\circ\sin75^\circ = \dfrac14\times\dfrac{\sqrt2}{2} = \dfrac{\sqrt2}{8}$

Using the sum-to-product formula: $\sin A+\sin B = 2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)$

Here $A=x+\frac{7\pi}{4}, B=x-\frac{7\pi}{4}$, so $\frac{A+B}{2}=x$ and $\frac{A-B}{2}=\frac{7\pi}{4}$

$\sin\left(x+\tfrac{7\pi}{4}\right)+\sin\left(x-\tfrac{7\pi}{4}\right) = 2\sin x\cos\tfrac{7\pi}{4} = 2\sin x\cdot\dfrac{\sqrt2}{2} = \sqrt2\sin x$

$5\pi-6\cos^{-1}(\sqrt3(2x-1))=0 \Rightarrow \cos^{-1}(\sqrt3(2x-1)) = \dfrac{5\pi}{6}$

$\sqrt3(2x-1) = \cos\dfrac{5\pi}{6} = -\dfrac{\sqrt3}{2}$

$2x-1 = -\dfrac12 \Rightarrow 2x=\dfrac12 \Rightarrow x=\dfrac14$

Using $\tan(A+B) = \dfrac{\tan A+\tan B}{1-\tan A\tan B}$ with $\tan A=3, \tan B=7$:

$\tan(\tan^{-1}3+\tan^{-1}7) = \dfrac{3+7}{1-21} = \dfrac{10}{-20} = -\dfrac12$

Let $\sin^{-1}(x)=\theta$, so $x=\sin\theta$. Then $\cos^{-1}\left(\dfrac{3x}{4}\right)=\theta \Rightarrow \dfrac{3x}{4}=\cos\theta=\sqrt{1-x^2}$

Squaring: $\dfrac{9x^2}{16}=1-x^2 \Rightarrow 9x^2+16x^2=16 \Rightarrow 25x^2=16$

$x = \dfrac{4}{5}$ (taking the positive root, consistent with the domain)

The line $x+\sqrt3y=10$ has direction ratios $(\sqrt3,-1)$ and normal direction $(1,\sqrt3)$.

Since $OP$ is perpendicular to the line, $OP$ is along the normal direction $(1,\sqrt3)$.

The angle this direction makes with the $y$-axis (direction $(0,1)$) is $\tan^{-1}\left(\dfrac{1}{\sqrt3}\right)=30^\circ$

Midpoint of $AB$: $M=\left(\dfrac{-4+0}{2},\dfrac{3+5}{2}\right)=(-2,4)$

Line through $C(-1,2)$ and $M(-2,4)$ has slope $\dfrac{4-2}{-2-(-1)}=\dfrac{2}{-1}=-2$

Equation: $y-2=-2(x+1) \Rightarrow y-2=-2x-2 \Rightarrow y+2x=0$

Midpoint of $AB$: $\left(\dfrac{-3+13}{2},-6\right)=(5,-6)$

The perpendicular bisector of the horizontal segment $AB$ is the vertical line $x=5$

$C$ lies on $x=5$ at distance $8$ from $(5,-6)$, so $C=(5,-6+8)=(5,2)$ or $(5,-14)$

Since $C$ is in the first quadrant, $C=(5,2)$

Rewriting $x^2+y^2+6x-8y-24=0$ as $(x+3)^2+(y-4)^2=49$, so radius $r=7$

The chord and the two radii to its endpoints form an isosceles triangle with the two equal sides $=r=7$ and included angle $60^\circ$, making it equilateral.

$\therefore$ chord length $=r=7$ units

Let semi-major axis $=a$. Given $e=\dfrac15$ and latus rectum $\dfrac{2b^2}{a}=\dfrac{48}{5}$