KEAM 2025 23rd April Question Paper with Solutions

Let $n(A)=a, n(B)=b, n(C)=c$.

$ab=160,\quad bc=80,\quad ca=200$

Multiplying all three: $(abc)^2 = 160 \times 80 \times 200 = 2560000 \Rightarrow abc = 1600$

$a = \frac{abc}{bc} = \frac{1600}{80} = 20$

$f(x) = x^2 - 10x - 19 = 5$

$\Rightarrow x^2 - 10x - 24 = 0 \Rightarrow (x-12)(x+2) = 0$

$\Rightarrow x = 12, -2$

So $f^{-1}(5) = \{-2, 12\}$

$f(x+y) + f(y-x) = \cos(x+y) + \cos(y-x) = 2\cos x \cos y$

So $\frac{1}{2}[f(x+y)+f(y-x)] = \cos x \cos y = f(x)f(y)$

Hence the expression equals $\cos x \cos y - \cos x \cos y = 0$

For $g(f(x))$ to be defined, we need $x$ in the domain of $f$ (i.e. $x>0$) and $f(x)$ in the domain of $g$, i.e. $-1 \le f(x) \le 1$.

$-1 \le \log_5 x \le 1 \Rightarrow 5^{-1} \le x \le 5^1$

So the domain is $\left[\frac{1}{5}, 5\right]$

$z = 1 + \frac{1}{i} = 1 - i$ (since $\frac{1}{i} = -i$)

$z^2 = (1-i)^2 = 1 - 2i + i^2 = -2i$

$z^4 = (z^2)^2 = (-2i)^2 = 4i^2 = -4$

Let $w = 2\sqrt{2} + 2\sqrt{2}i$

$|w| = \sqrt{(2\sqrt2)^2 + (2\sqrt2)^2} = \sqrt{8+8} = 4$

$|w^2| = |w|^2 = 16$

$z + \bar z = 6 \Rightarrow 2\,\text{Re}(z) = 6 \Rightarrow \text{Re}(z) = 3$

$z - \bar z = 4i \Rightarrow 2i\,\text{Im}(z) = 4i \Rightarrow \text{Im}(z) = 2$

$|z|^2 = 3^2 + 2^2 = 13$

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

So $\text{Re}(z) = \frac{2\alpha-1}{\alpha^2+1}$, and $\text{Im}(\bar z) = \frac{\alpha+2}{\alpha^2+1}$

Given $4\,\text{Re}(z) = 3\,\text{Im}(\bar z)$:

$4(2\alpha-1) = 3(\alpha+2) \Rightarrow 8\alpha - 4 = 3\alpha + 6 \Rightarrow 5\alpha = 10 \Rightarrow \alpha = 2$

Let the first term be $a=4$ and common ratio $r$. Third term: $ar^2 = 8 \Rightarrow r^2 = 2$

$21^{st}\text{ term} = ar^{20} = a(r^2)^{10} = 4 \times 2^{10} = 4 \times 1024 = 4096$

For a G.P., $a_1 a_2 a_3 = a_2^3 = 64 \Rightarrow a_2 = 4$

$a_1 a_2 a_3 a_4 a_5 = a_3^5 = 32 \Rightarrow a_3 = 2$

Common ratio $r = \dfrac{a_3}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$

$t_n = \frac{n(n+6)}{n+4} = 5$

$n^2+6n = 5n+20 \Rightarrow n^2+n-20=0 \Rightarrow (n+5)(n-4)=0$

Since $n$ is a positive integer, $n=4$

Let the G.P. have first term $a$ and common ratio $r$. Product of first 5 terms:

$a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 = a^5 r^{10} = (ar^2)^5 = t_3^5$

Given this product is 32: $t_3^5 = 32 \Rightarrow t_3 = 2$

$\alpha = {}^{10}C_0 + {}^{10}C_2 + {}^{10}C_4 + {}^{10}C_6 + {}^{10}C_8 + {}^{10}C_{10}$ (sum of coefficients at even positions) $= 2^{9} = 512$

$\beta = {}^{10}C_1 + {}^{10}C_3 + {}^{10}C_5 + {}^{10}C_7 + {}^{10}C_9$ (sum of coefficients at odd positions) $= 2^{9} = 512$

Since the sum of even-indexed and odd-indexed binomial coefficients are each equal to $2^{n-1}$, we get $\alpha - \beta = 0$

$\frac{\alpha}{\beta} = \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$

$1 + \frac{\alpha}{\beta} = 1 + \frac{n-r+1}{r} = \frac{r+n-r+1}{r} = \frac{n+1}{r}$

${}^{11}P_r = \frac{11!}{(11-r)!} = 7920$

Checking $r=4$: $\frac{11!}{7!} = 11\times10\times9\times8 = 7920$

So $r=4$

General term of $(2x+\alpha)^8$: $T_{k+1} = {}^8C_k (2x)^{8-k}\alpha^k$

Coefficient of $x^2$ ($k=6$): ${}^8C_6 \cdot 2^2 \cdot \alpha^6 = 28 \times 4 \times \alpha^6$

Coefficient of $x^3$ ($k=5$): ${}^8C_5 \cdot 2^3 \cdot \alpha^5 = 56 \times 8 \times \alpha^5$

Equating: $112\alpha^6 = 448\alpha^5 \Rightarrow \alpha = 4$

Total arrangements of all 5 digits $\{0,2,4,6,8\}$ = $5! = 120$

Of these, numbers starting with 0 (not valid 5-digit numbers) = $4! = 24$

Valid 5-digit numbers $= 120 - 24 = 96$

$B = 3A$ where $A$ is $3\times3$, so $|B| = 3^3|A| = 27\times5 = 135$

For a $3\times3$ matrix, $|\text{adj }B| = |B|^{2} = 135^2$

$\frac{|\text{adj }B|}{|3A|} = \frac{135^2}{135} = 135$

$$\begin{pmatrix}-1 & 2\\ 3 & -4\\ -5 & 6\end{pmatrix}\begin{pmatrix}7\\8\end{pmatrix} = \begin{pmatrix}-7+16\\21-32\\-35+48\end{pmatrix} = \begin{pmatrix}9\\-11\\13\end{pmatrix}$$

So $\alpha = 9, \beta = -11$, and $\alpha+\beta = -2$

For the matrix to be singular, its determinant must be zero:

$(8-k)(4-k) - (2)(-2) = 0$

$32 - 12k + k^2 + 4 = 0 \Rightarrow k^2 - 12k + 36 = 0 \Rightarrow (k-6)^2 = 0$

$\Rightarrow k = 6$

For the system to not have a unique solution, the coefficient determinant must be zero:

$$\begin{vmatrix}1 & 1 & 1\\2 & 3 & -m\\3 & 5 & 3\end{vmatrix} = 0$$

Expanding: $1(9+5m) - 1(6+3m) + 1(10-9) = 0 \Rightarrow 9+5m-6-3m+1=0 \Rightarrow 2m+4=0$

$\Rightarrow m = -2$

$-4 \le 2-3x < 7$

Subtract 2: $-6 \le -3x < 5$

Divide by $-3$ (inequality signs flip and reverse order): $2 \ge x > -\frac{5}{3}$

So the solution set is $\left(-\frac{5}{3}, 2\right]$

$-5 < x \le -1$

Multiply by 5: $-25 < 5x \le -5$

Add 4: $-21 < 5x+4 \le -1$

Comparing with $-21 < 5x+4 \le b$, the least value of $b$ satisfying the inequality is $b=-1$

Since $75^\circ = 90^\circ - 15^\circ$, $\tan 75^\circ = \cot 15^\circ$

$\tan 15^\circ + \cot 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} + \frac{\cos 15^\circ}{\sin 15^\circ} = \frac{1}{\sin 15^\circ \cos 15^\circ} = \frac{2}{\sin 30^\circ}$

$= \frac{2}{1/2} = 4$

Given $x+z=2y$ and $y=\frac{\pi}{4}$, so $x+z=\frac{\pi}{2}$, i.e. $z = \frac{\pi}{2}-x$

$\tan x \tan y \tan z = \tan x \cdot 1 \cdot \tan\left(\frac{\pi}{2}-x\right) = \tan x \cdot \cot x = 1$

$a = \sin x+\sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$

$b = \cos x+\cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$

$\frac{a}{b} = \tan\left(\frac{x+y}{2}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt3$

Since $\frac{\pi}{2}<\alpha<\frac{3\pi}{2}$ and $\sin\alpha = \frac{12}{13} > 0$, $\alpha$ must lie in the second quadrant (where sine is positive), so $\cos\alpha < 0$.

$\cos\alpha = -\sqrt{1-\sin^2\alpha} = -\sqrt{1-\frac{144}{169}} = -\frac{5}{13}$

$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{12/13}{-5/13} = -\frac{12}{5}$

Using the identity $\tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}x$ for $|x|<1$:

$f\left(\frac{1}{\sqrt3}\right) = 2\tan^{-1}\left(\frac{1}{\sqrt3}\right) = 2\times\frac{\pi}{6} = \frac{\pi}{3}$

Using $\sin^{-1}\alpha + \cos^{-1}\alpha = \frac{\pi}{2}$:

$5\sin^{-1}\alpha + 3\cos^{-1}\alpha = 3(\sin^{-1}\alpha+\cos^{-1}\alpha) + 2\sin^{-1}\alpha = \pi$

$3\cdot\frac{\pi}{2} + 2\sin^{-1}\alpha = \pi \Rightarrow 2\sin^{-1}\alpha = -\frac{\pi}{2} \Rightarrow \sin^{-1}\alpha = -\frac{\pi}{4}$

$\Rightarrow \alpha = -\frac{1}{\sqrt2}$

Let $\cot\theta = \sqrt{\frac{1-x}{1+x}}$, so $\cot^2\theta = \frac{1-x}{1+x}$

$\sec^2\theta = 1+\tan^2\theta = 1+\frac{1}{\cot^2\theta} = 1+\frac{1+x}{1-x} = \frac{(1-x)+(1+x)}{1-x} = \frac{2}{1-x}$

The perpendicular distance from $A(-2,2)$ to the line $x+y=6$ gives the height of the equilateral triangle (from vertex A to side BC):

$h = \frac{|-2+2-6|}{\sqrt2} = \frac{6}{\sqrt2} = 3\sqrt2$

For an equilateral triangle, $h = \frac{\sqrt3}{2}a$, so $a = \frac{2h}{\sqrt3} = \frac{2(3\sqrt2)}{\sqrt3} = \frac{6\sqrt2}{\sqrt3} = 2\sqrt6$

$x^2-4x+8y+4=0 \Rightarrow (x-2)^2 - 4 + 8y+4=0 \Rightarrow (x-2)^2 = -8y$

This is of the form $(x-h)^2 = 4p(y-k)$ with vertex $(2,0)$ and $4p=-8 \Rightarrow p=-2$

Focus $= (h, k+p) = (2, -2)$

Since the circle touches the x-axis at $(9,0)$, its center has x-coordinate 9 and its y-coordinate equals the radius $r$, i.e. center $=(9,r)$.

Since it also touches the line $y=14$ (and the center lies below this line), $14-r=r \Rightarrow r=7$

Equation: $(x-9)^2+(y-7)^2=49$

Let $a=m/2, b=n/2$ be semi-axes. Then $m^2-n^2=4(a^2-b^2)=45 \Rightarrow a^2-b^2=\frac{45}{4}$

Eccentricity: $e^2=\frac{5}{9}=1-\frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2}=\frac{4}{9}$

Substituting: $a^2-\frac{4}{9}a^2=\frac{45}{4} \Rightarrow \frac{5}{9}a^2=\frac{45}{4} \Rightarrow a^2=\frac{81}{4} \Rightarrow a=\frac{9}{2}$

Major axis $m=2a=9$

$4y = x^2-6x+17 = (x-3)^2+8$

$\Rightarrow (x-3)^2 = 4y-8 = 4(y-2)$

This is of the form $(x-h)^2=4p(y-k)$, so the vertex is $(3,2)$

$\frac{(2x-6)^2}{2} - \frac{(4y+7)^2}{16} = 1 \Rightarrow \frac{4(x-3)^2}{2} - \frac{16(y+7/4)^2}{16} = 1$

$\Rightarrow \frac{(x-3)^2}{1/2} - (y+7/4)^2 = 1$

Here $a^2=\frac12, b^2=1$

$e^2 = 1+\frac{b^2}{a^2} = 1+2 = 3 \Rightarrow e=\sqrt3$