KEAM 2026 22nd April Question Paper with Solutions

Given $f(x) = \lambda x + 1$ is its own inverse, so $f(f(x)) = x$ for all $x$.

$f(f(x)) = \lambda(\lambda x + 1) + 1 = \lambda^2 x + \lambda + 1$

Comparing with $x$: $\lambda^2 = 1$ and $\lambda + 1 = 0$, which gives $\lambda = -1$.

Given $f(x) = x+1$, so $f^{-1}(x) = x - 1$.

$f^{-1}(g(x)) = g(x) - 1 = x^3 - 1$

Since $\cos x \in [-1,1]$, the greatest integer function $[\cos x]$ takes the value $-1$ when $\cos x \in [-1,0)$, the value $0$ when $\cos x \in [0,1)$, and the value $1$ when $\cos x = 1$.

Hence the range is $\{-1, 0, 1\}$.

Let $z = x + iy$. Then

$(1-i)z = (1-i)(x+iy) = (x+y) + i(y-x)$

Given $\text{Im}[(1-i)z] = 1 \Rightarrow y - x = 1 \Rightarrow y = x+1$

Note that $\frac{\sqrt3}{2}+\frac{i}{2} = \cos30^\circ + i\sin30^\circ$ and $\frac{\sqrt3}{2}-\frac{i}{2} = \cos30^\circ - i\sin30^\circ$.

Using De Moivre's theorem, their cubes are $\cos90^\circ + i\sin90^\circ$ and $\cos90^\circ - i\sin90^\circ$.

Sum $= 2\cos90^\circ = 0$

Let first term be $a$ and common ratio $r$.

$ar^2 = 108, \quad ar^5 = -32$

Dividing: $r^3 = \frac{-32}{108} = -\frac{8}{27} \Rightarrow r = -\frac{2}{3}$

$a = \frac{108}{r^2} = \frac{108}{4/9} = 243$

Let the three geometric means be inserted between 4 and 324, giving 5 terms in G.P. with $a=4$ and 5th term $=324$.

$ar^4 = 324 \Rightarrow r^4 = 81 \Rightarrow r = 3$

The means are $4(3)=12,\ 4(9)=36,\ 4(27)=108$

Let the G.P. have first term $A$ and common ratio $R$. Then

$a = AR^2,\quad b = AR^6,\quad c = AR^{10}$

$ac = A^2R^{12} = (AR^6)^2 = b^2$

$(ac)^4 = b^8$

Since $a, x, y, b$ are in G.P. with common ratio $r$: $x = ar,\ y = ar^2,\ b = ar^3$

$(x+y)^2 = (ar+ar^2)^2 = a^2r^2(1+r)^2$

$(a+x)(y+b) = a(1+r)\cdot ar^2(1+r) = a^2r^2(1+r)^2$

Hence $(x+y)^2 = (a+x)(y+b)$

Digits available: 0,1,2,3,4,5,6 (no repetition assumed for a well-defined count of 3-digit even numbers).

Units digit must be even: 0, 2, 4, or 6.

Case units = 0: hundreds digit has 6 choices, tens digit has 5 choices $\Rightarrow 30$

Case units = 2, 4, or 6 (3 cases): hundreds digit cannot be 0 or the units digit (5 choices), tens digit has 5 remaining choices $\Rightarrow 5\times5=25$ each, total $75$

Total $= 30 + 75 = 105$

Since 3 particular players are always included, we need to choose the remaining $11-3=8$ players from the remaining $16-3=13$ players.

Number of ways $= \binom{13}{8} = 1287$

Committee of 6 from 7 women and 4 men, with at least 3 women:

$3W,3M: \binom{7}{3}\binom{4}{3}=35\times4=140$

$4W,2M: \binom{7}{4}\binom{4}{2}=35\times6=210$

$5W,1M: \binom{7}{5}\binom{4}{1}=21\times4=84$

$6W,0M: \binom{7}{6}=7$

Total $=140+210+84+7=441$

In $\left(2\alpha - \dfrac{y}{2}\right)^8$, the term containing $y^3$ is

$\binom{8}{3}(2\alpha)^5\left(-\frac{1}{2}\right)^3 y^3 = 56\times32\alpha^5\times\left(-\frac18\right)y^3 = -224\alpha^5 y^3$

Given coefficient $=-7$: $-224\alpha^5=-7 \Rightarrow \alpha^5=\frac{1}{32} \Rightarrow \alpha=\frac12$

$(1-2x)^3 = 1 - 6x + 12x^2 - 8x^3$

$(1+ax)(1-6x+12x^2-8x^3)$, coefficient of $x^2$ is $a_2 = 12 + a(-6) = 12-6a$

Given $a_2=0 \Rightarrow 12-6a=0 \Rightarrow a=2$

In each row, the third entry can be expanded as

$\sin(\theta+75^\circ) = \sin\theta\cos75^\circ + \cos\theta\sin75^\circ$

So the third column is a linear combination of the first two columns ($\cos75^\circ\times$ column 1 $+\sin75^\circ\times$ column 2).

Since one column is a linear combination of the others, the determinant $=0$

Given $S = I+A+A^2+\cdots+A^{10}=O$

Multiplying by $A$: $AS = A+A^2+\cdots+A^{11}=O$

Subtracting: $AS - S = A^{11}-I = O \Rightarrow A^{11}=I$

Since $A$ is non-singular, $A^{10}=A^{-1}$

$\frac{21x-6}{4}-9\le0 \Rightarrow 21x-6\le36 \Rightarrow x\le2$

$\frac{x-1}{3}+1\ge0 \Rightarrow x-1\ge-3 \Rightarrow x\ge-2$

Combining, $x\in[-2,2]$

$|2x-3|<5 \Rightarrow -5<2x-3<5 \Rightarrow -2<2x<8 \Rightarrow -1<x<4$

Using sum-to-product: $\sin40^\circ+\sin20^\circ = 2\sin30^\circ\cos10^\circ = \cos10^\circ$

$\cos40^\circ+\cos20^\circ = 2\cos30^\circ\cos10^\circ = \sqrt3\cos10^\circ$

$\frac{\sqrt3(\sin40^\circ+\sin20^\circ)}{\cos40^\circ+\cos20^\circ} = \frac{\sqrt3\cos10^\circ}{\sqrt3\cos10^\circ}=1$

$\sec^2\theta+\tan^2\theta = 1+2\tan^2\theta = 7 \Rightarrow \tan^2\theta=3 \Rightarrow \tan\theta=\sqrt3$

Since $0<\theta<\frac{\pi}{2}$, $\theta=60^\circ$

$\tan2\theta=\tan120^\circ=-\sqrt3$

$\tan75^\circ+\tan15^\circ = \frac{\sin90^\circ}{\cos75^\circ\cos15^\circ} = \frac{1}{\cos75^\circ\cos15^\circ}$

$\tan75^\circ-\tan15^\circ = \frac{\sin60^\circ}{\cos75^\circ\cos15^\circ}$

$\frac{\tan75^\circ+\tan15^\circ}{\tan75^\circ-\tan15^\circ} = \frac{1}{\sin60^\circ} = \frac{2}{\sqrt3}$

Multiply and divide by $2\sin36^\circ$:

$4\cos36^\circ\cos72^\circ = \frac{2\sin36^\circ\cdot2\cos36^\circ\cos72^\circ}{\sin36^\circ} = \frac{2\sin72^\circ\cos72^\circ}{\sin36^\circ} = \frac{\sin144^\circ}{\sin36^\circ}$

Since $\sin144^\circ=\sin36^\circ$, the value $=1$

The maximum value of $8\sin x+15\cos x$ is $\sqrt{8^2+15^2}=\sqrt{289}=17$

So maximum of $(8\sin x+15\cos x+3)^2$ is $(17+3)^2=400$

Maximum of $f(x) = 400-15=385$

Since $\frac{2\pi}{3}\in[0,\pi]$: $\cos^{-1}\left(\cos\frac{2\pi}{3}\right)=\frac{2\pi}{3}$

Since $\sin\frac{2\pi}{3}=\sin\frac{\pi}{3}$ and $\frac{\pi}{3}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$: $\sin^{-1}\left(\sin\frac{2\pi}{3}\right)=\frac{\pi}{3}$

Sum $=\frac{2\pi}{3}+\frac{\pi}{3}=\pi$

Let $\theta=\sec^{-1}(3)$, so $\sec\theta=3$

$\tan^2\theta=\sec^2\theta-1=9-1=8$

Dividing numerator and denominator by $\cos x$:

$\frac{\cos x-\sqrt3\sin x}{\sqrt3\cos x+\sin x} = \frac{1-\sqrt3\tan x}{\sqrt3+\tan x} = \frac{\frac{1}{\sqrt3}-\tan x}{1+\frac{1}{\sqrt3}\tan x} = \tan\left(\frac{\pi}{6}-x\right)$

So the expression equals $\frac{\pi}{6}-x$

Line $3x+2y=6$ i.e. $\frac{x}{2}+\frac{y}{3}=1$ has x-intercept 2 and y-intercept 3.

New line has x-intercept $3\times2=6$ and y-intercept $4\times3=12$

Equation: $\frac{x}{6}+\frac{y}{12}=1 \Rightarrow 2x+y=12$

The line joining origin to the foot of perpendicular $(1,1)$ has slope $1$, and is perpendicular to $y=mx+c$.

$1\times m=-1 \Rightarrow m=-1$

Since $(1,1)$ lies on the line: $1=-1(1)+c \Rightarrow c=2$

Rewrite $6x+8y+18=0$ as $3x+4y+9=0$ (same direction ratios as $3x+4y-6=0$, confirming parallel lines).

Distance $= \frac{|9-(-6)|}{\sqrt{3^2+4^2}} = \frac{15}{5}=3$

Setting $y=0$ in $x^2+y^2-6x-12y-55=0$: $x^2-6x-55=0$

$x=\frac{6\pm\sqrt{36+220}}{2}=\frac{6\pm16}{2} = 11, -5$

$|AB| = 11-(-5)=16$

Vertex is at distance 6 from origin, focus at distance 3 from origin, both on positive x-axis, so vertex is at $(6,0)$ and focus at $(3,0)$, with the parabola opening towards the origin (negative x direction).

Here $a =$ distance between vertex and focus $=3$.

Equation: $y^2=-4a(x-6) = -12(x-6)$

Foci $(\pm2,0) \Rightarrow c=2$; vertices $(\pm3,0)\Rightarrow a=3$

Eccentricity $e=\frac{c}{a}=\frac{2}{3}$

$9x^2-16y^2=144 \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$

$a=4,\ b=3,\ c=\sqrt{16+9}=5$

Vertex $S=(4,0)$, focus $A=(5,0)$

$AS = 5-4=1$

For the angle between $\vec a=x\hat i+3\hat j+\hat k$ and $\vec b=x\hat i-x\hat j+2\hat k$ to be acute, we need $\vec a\cdot\vec b>0$ (and the vectors not identical in direction).

$\vec a\cdot\vec b = x^2-3x+2 = (x-1)(x-2)>0$

This holds for $x<1$ or $x>2$, i.e. $x\in(-\infty,1)\cup(2,\infty)$