KEAM 2025 29th April Question Paper with Solutions

Number of subsets of $A$ is $2^m$ and of $B$ is $2^n$.

Given $2^m - 2^n = 48$.

Trying $m=6, n=4$: $2^6 - 2^4 = 64 - 16 = 48$, which satisfies the condition.

$f(x) = \dfrac{x}{x-1}$

$f(a) = \dfrac{a}{a-1}, \quad f(a+1) = \dfrac{a+1}{a}$

$\dfrac{f(a)}{f(a+1)} = \dfrac{a}{a-1} \times \dfrac{a}{a+1} = \dfrac{a^2}{a^2-1} = f(a^2)$

Since $f(x+y)=f(x)+f(y)$, this is Cauchy's functional equation, so $f(x) = kx$ for some constant $k$.

Given $f(1) = 3 \Rightarrow k = 3$, so $f(x) = 3x$.

$f(0)+f(1)+f(2)+f(3) = 0+3+6+9 = 18$

$z = 2+i$

$z^2 - 4z + 15 = z^2 - 4z + 4 + 11 = (z-2)^2 + 11$

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

$\therefore z^2-4z+15 = -1+11 = 10$

$\dfrac{i}{2} - \dfrac{2}{i} = \dfrac{i}{2} - \dfrac{2}{i}\times\dfrac{-i}{-i} = \dfrac{i}{2} + 2i = \dfrac{i+4i}{2} = \dfrac{5i}{2}$

Modulus $= \left|\dfrac{5i}{2}\right| = \dfrac{5}{2}$

Let $z = x+iy$. Then $z - 2 - 3i = (x-2) + i(y-3)$.

Real part $= x-2$, Imaginary part $= y-3$.

Given they are equal: $x - 2 = y - 3 \Rightarrow y = x+1$

This is a straight line.

$k = 4n+3$

$i^k = i^{4n+3} = (i^4)^n \cdot i^3 = 1 \cdot i^3 = -i$

Let the G.P. have first term $a$ and common ratio $r$.

Sum of first three terms: $a(1+r+r^2) = 14$

Sum of next three terms: $ar^3(1+r+r^2) = 112$

Dividing: $r^3 = \dfrac{112}{14} = 8 \Rightarrow r = 2$

So $a(1+2+4)=14 \Rightarrow a = 2$

$T_{100} = ar^{99} = 2 \cdot 2^{99} = 2^{100}$

Let first term $= a$, common ratio $= r$.

Product of first 3 terms: $a^3r^3 = 216 \Rightarrow ar = 6$

Product of first 4 terms: $a^4r^6 = 324$

Since $(ar)^3 \cdot ar^3 = a^4r^6$, we get $216 \cdot ar^3 = 324 \Rightarrow ar^3 = 1.5$

So $r^2 = \dfrac{ar^3}{ar} = \dfrac{1.5}{6} = 0.25 \Rightarrow r = 0.5$

$a = \dfrac{6}{r} = 12$

Let first term $=a$, common ratio $=r$.

Product of first four terms: $a \cdot ar \cdot ar^2 \cdot ar^3 = a^4r^6 = \dfrac{1}{1024}$

Product of 2nd and 3rd terms $= ar \cdot ar^2 = a^2r^3 = \sqrt{a^4r^6} = \sqrt{\dfrac{1}{1024}} = \dfrac{1}{32}$

A.M. of $a$ and $c$ is 16: $\dfrac{a+c}{2}=16$

Given $a=8 \Rightarrow c = 32-8 = 24$

G.M. of $a$ and $c$ $= \sqrt{ac} = \sqrt{8 \times 24} = \sqrt{192} = 8\sqrt{3}$

$^{n}P_5 = 42\, {}^{n}P_3$

$\dfrac{n!}{(n-5)!} = 42 \cdot \dfrac{n!}{(n-3)!}$

$(n-3)(n-4) = 42$

$n^2 - 7n - 30 = 0 \Rightarrow (n-10)(n+3) = 0$

Since $n$ must be positive, $n = 10$

INDEPENDENCE has 12 letters: I(1), N(3), D(2), E(4), P(1), C(1).

Fix I at the first position and P at the last position. The remaining 10 letters to arrange are N(3), D(2), E(4), C(1).

Number of arrangements $= \dfrac{10!}{3!\,2!\,4!} = \dfrac{3628800}{6 \times 2 \times 24} = 12600$

Number of ways of getting exactly 2 heads $= {}^4C_2 = 6$

Number of ways of getting exactly 3 heads $= {}^4C_3 = 4$

Total ways of getting 2 or 3 heads $= 6+4 = 10$

$^{5}C_r$ takes its greatest value at $r=2$ (or 3), giving $^{5}C_2 = 10$.

$^{6}C_r$ takes its greatest value at $r=3$, giving $^{6}C_3 = 20$.

$\dfrac{^{5}C_r}{^{6}C_r} = \dfrac{10}{20} = \dfrac{1}{2}$

Let $f(x)=(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$

Put $x=1$: $f(1)=(1+1-2)^6=0 = 1+a_1+a_2+\ldots+a_{12}$

Put $x=-1$: $f(-1)=(1-1-2)^6=64 = 1-a_1+a_2-a_3+\ldots+a_{12}$

Adding: $2(1+a_2+a_4+\ldots+a_{12}) = 64 \Rightarrow a_2+a_4+\ldots+a_{12} = 31$

For a symmetric matrix, $A = A^T$, so corresponding off-diagonal elements are equal.

$A_{13}=A_{31} \Rightarrow x = 4$

$A_{12}=A_{21} \Rightarrow y = 2$

$A_{23}=A_{32} \Rightarrow t = -3$

$$A^2 = \begin{bmatrix} X & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} X & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} X^2 & 0 \\ X+1 & 1 \end{bmatrix}$$

Given $$A^2=B=\begin{bmatrix} 16 & 0 \\ 5 & 1 \end{bmatrix}$$

So $X^2=16$ and $X+1=5 \Rightarrow X=4$, which is consistent with both equations.

Given $\alpha+\beta+\gamma=0$.

Splitting the third column: the determinant equals the determinant with column $(e^{3\alpha}, e^{3\beta}, e^{3\gamma})$ minus the determinant with column $(1,1,1)$.

Using the identity for such symmetric exponential determinants under the constraint $\alpha+\beta+\gamma=0$, both resulting determinants simplify identically, making the overall determinant equal to $0$.

Points $(2,-3)$, $(x,1)$, $(0,5)$ are collinear, so the slope between any two pairs must be equal.

Slope of line through $(2,-3)$ and $(0,5)$: $\dfrac{5-(-3)}{0-2} = -4$

Equation of line: $y - 5 = -4(x-0) \Rightarrow y = -4x+5$

At $y=1$: $1 = -4x+5 \Rightarrow x = 1$

$\dfrac{x-3}{x-5} > 3$

Case $x>5$: $x-3 > 3(x-5) \Rightarrow x-3 > 3x-15 \Rightarrow x < 6$, giving $5<x<6$.

Case $x<5$: multiplying flips the inequality, giving $x>6$, which contradicts $x<5$ (no solution here).

So $x \in (5,6)$

$\left|\dfrac{1}{x}-2\right| < 4 \Rightarrow -4 < \dfrac{1}{x}-2 < 4 \Rightarrow -2 < \dfrac{1}{x} < 6$

Solving $\dfrac{1}{x} > -2$ and $\dfrac{1}{x} < 6$ together (considering sign of $x$), the solution set is

$x \in \left(-\infty, \dfrac{-1}{2}\right) \cup \left(\dfrac{1}{6}, \infty\right)$

$\cos\left(\dfrac{x}{2}\right) = \sqrt{\dfrac{1+\cos x}{2}}$ (positive since $x/2 \in [0,\pi/4]$)

$= \sqrt{\dfrac{1+\frac{4}{5}}{2}} = \sqrt{\dfrac{9/5}{2}} = \sqrt{\dfrac{9}{10}} = \dfrac{3}{\sqrt{10}}$

Using product-to-sum: $\sin A \sin B = \dfrac{1}{2}[\cos(A-B)-\cos(A+B)]$

$\sin\dfrac{5\pi}{12}\sin\dfrac{\pi}{12} = \dfrac{1}{2}\left[\cos\dfrac{\pi}{3} - \cos\dfrac{\pi}{2}\right] = \dfrac{1}{2}\left[\dfrac{1}{2}-0\right] = \dfrac{1}{4}$

$1-\sin^6\theta-\cos^6\theta = 1 - (\sin^2\theta+\cos^2\theta)(\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta)$

$= 1 - \left[(\sin^2\theta+\cos^2\theta)^2 - 3\sin^2\theta\cos^2\theta\right] = 3\sin^2\theta\cos^2\theta = \dfrac{3}{4}\sin^2 2\theta$

So the expression becomes $\dfrac{\frac{3}{4}\sin^2 2\theta}{\cos^2 2\theta} = \dfrac{3}{4}\tan^2 2\theta$

Given $\dfrac{\cos A}{\cos B}=\alpha$, by componendo-dividendo:

$\dfrac{\alpha+1}{\alpha-1} = \dfrac{\cos A+\cos B}{\cos A-\cos B} = \dfrac{2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}{-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)}$

$= -\cot\left(\dfrac{A+B}{2}\right)\cot\left(\dfrac{A-B}{2}\right)$

$\tan^{-1}2x+\tan^{-1}3x = \dfrac{\pi}{4} \Rightarrow \tan\left(\tan^{-1}2x+\tan^{-1}3x\right)=1$

$\dfrac{2x+3x}{1-6x^2}=1 \Rightarrow 5x = 1-6x^2 \Rightarrow 6x^2+5x-1=0$

$(6x-1)(x+1)=0 \Rightarrow x=\dfrac{1}{6} \text{ or } x=-1$

$x=-1$ is extraneous (makes LHS negative), so $x=\dfrac{1}{6}$

For $\cos^{-1}([x])$ to be defined, we need $-1 \le [x] \le 1$.

$[x]=-1$ for $x \in [-1,0)$, $[x]=0$ for $x \in [0,1)$, $[x]=1$ for $x \in [1,2)$.

Combining: domain is $[-1,2)$

$\sin^{-1}\left(\dfrac{3\sin2\alpha}{5+4\cos2\alpha}\right)=\dfrac{\pi}{2} \Rightarrow \dfrac{3\sin2\alpha}{5+4\cos2\alpha}=1$

$3\sin2\alpha = 5+4\cos2\alpha \Rightarrow 3\sin2\alpha - 4\cos2\alpha = 5$

Given, $\tan\theta = \dfrac{\pi}{4} \Rightarrow \tan\theta = 1$, and slope of one line $m_1 = \dfrac{1}{2}$

Using the angle between two lines formula,

$\tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right|$

$\left|\dfrac{m_2 - \frac{1}{2}}{1 + \frac{m_2}{2}}\right| = 1$

Taking $+1$: $m_2 - \dfrac{1}{2} = 1 + \dfrac{m_2}{2} \Rightarrow \dfrac{m_2}{2} = \dfrac{3}{2} \Rightarrow m_2 = 3$

Taking $-1$: $m_2 - \dfrac{1}{2} = -1 - \dfrac{m_2}{2} \Rightarrow \dfrac{3m_2}{2} = -\dfrac{1}{2} \Rightarrow m_2 = \dfrac{-1}{3}$

So the slope of the other line is $3 \text{ or } \dfrac{-1}{3}$

Given points $\left(\dfrac{-1}{2}, 1\right)$ and $(1, 2)$

Slope, $m = \dfrac{2 - 1}{1 - \left(\frac{-1}{2}\right)} = \dfrac{1}{\frac{3}{2}} = \dfrac{2}{3}$

Equation of line: $y - 2 = \dfrac{2}{3}(x - 1)$

$y = \dfrac{2}{3}x - \dfrac{2}{3} + 2 = \dfrac{2}{3}x + \dfrac{4}{3}$

So the y-intercept is $\dfrac{4}{3}$

Base of the equilateral triangle lies along $2x - y = 1$, and the opposite vertex is $(-1, 2)$

Perpendicular distance from $(-1,2)$ to the line (this is the height of the triangle):

$h = \dfrac{|2(-1) - 2 - 1|}{\sqrt{2^2 + (-1)^2}} = \dfrac{5}{\sqrt{5}} = \sqrt{5}$

For an equilateral triangle of side $a$, height $h = \dfrac{\sqrt{3}}{2}a$

$a = \dfrac{2h}{\sqrt{3}} = \dfrac{2\sqrt{5}}{\sqrt{3}} = 2\sqrt{\dfrac{5}{3}} \text{ units}$

Let centre be $(0,k)$ on the y-axis. Since the circle passes through $(4,0)$ and $(0,2)$, distances to centre are equal:

$\sqrt{16+k^2} = |k-2| \Rightarrow 16+k^2 = k^2-4k+4 \Rightarrow 4k=-12 \Rightarrow k=-3$

Radius $= \sqrt{16+9} = \sqrt{25} = 5$

Let semi-major axis $=a$, semi-minor axis $=b$. Given major axis $=2\times$minor axis, so $2a = 2(2b) \Rightarrow a=2b$

Eccentricity $e = \sqrt{1-\dfrac{b^2}{a^2}} = \sqrt{1-\dfrac{1}{4}} = \dfrac{\sqrt{3}}{2}$

For $\dfrac{x^2}{9}-\dfrac{y^2}{25}=1$: $a^2=9 \Rightarrow a=3$, $b^2=25 \Rightarrow b=5$

Length of transverse axis $=2a=6$

Length of conjugate axis $=2b=10$

$(x-1)^2 = 2(y-2)$ is of the form $X^2=4aY$ with $4a=2 \Rightarrow a=\dfrac{1}{2}$

Vertex is $(1,2)$, and the parabola opens upward.

Directrix: $y = 2-a = 2-\dfrac{1}{2} = \dfrac{3}{2} \Rightarrow 2y-3=0$