KEAM 2025 25th April Question Paper with Solutions

Using $n(A \cup B) = n(A) + n(B) - n(A \cap B)$,

$20 = 14 + n(B) - 6 \Rightarrow n(B) = 12$

$n(B') = n(U) - n(B) = 25 - 12 = 13$

Since $-1 \le \cos x \le 1$, we get $6 \le 7 - \cos x \le 8$.

So $f(x) = \dfrac{1}{7-\cos x} \in \left[\dfrac{1}{8}, \dfrac{1}{6}\right]$

For $f(x) = \sqrt{7-11x}$ to be defined, we need

$7 - 11x \ge 0 \Rightarrow x \le \dfrac{7}{11}$

So the domain is $\left(-\infty, \dfrac{7}{11}\right]$

Number of relations from $A$ to $B$ is $2^{n(A)\cdot n(B)}$.

$2^{7 \cdot n(B)} = 128 = 2^7 \Rightarrow 7\,n(B) = 7 \Rightarrow n(B) = 1$

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

$(1+i)^{10} = \left[(1+i)^2\right]^5 = (2i)^5 = 32\,i^5 = 32i$

Group terms in sets of four consecutive powers of $i$, since $i^k + i^{k+1} + i^{k+2} + i^{k+3} = 0$.

From $i^3$ to $i^{93}$ there are 91 terms. The 22 complete groups ($i^3$ to $i^{90}$) sum to 0, leaving $i^{91} + i^{92} + i^{93} = i^3 + i^0 + i^1 = -i + 1 + i = 1$

$\dfrac{3-2i}{2i} = \dfrac{(3-2i)(-i)}{2i\cdot(-i)} = \dfrac{-3i + 2i^2}{2} = \dfrac{-2-3i}{2} = -1 - \dfrac{3}{2}i$

$\left(-1-\dfrac{3}{2}i\right)^2 = 1 + 3i + \dfrac{9}{4}i^2 = 1 + 3i - \dfrac{9}{4} = -\dfrac{5}{4} + 3i$

So the imaginary part is $3$

$\overline{z_2} = -\dfrac{1}{2} + i\dfrac{\sqrt3}{2}$

$w = z_1 + \overline{z_2} = \left(\dfrac12 - \dfrac12\right) + i\left(\dfrac{\sqrt3}{2}+\dfrac{\sqrt3}{2}\right) = i\sqrt3$

$\overline{w} = -i\sqrt3$

$\sum_{n=1}^{20} a_n = \sum_{n=1}^{20} 2^{n-1} = 1 + 2 + 4 + \ldots + 2^{19}$

This is a G.P. with sum $\dfrac{2^{20}-1}{2-1} = 2^{20} - 1$

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

$a(1+r) = 6, \qquad ar^8(1+r) = \dfrac{3}{128}$

Dividing, $r^8 = \dfrac{3}{128 \times 6} = \dfrac{1}{256} = \dfrac{1}{2^8} \Rightarrow r = \dfrac{1}{2}$

Since $a,b,c$ are in G.P., $b^2 = ac$, so $abc = b^3 = 27 \Rightarrow b = 3$.

Then $ac = \dfrac{27}{3} = 9$ and $a + c = 10$.

$a^2+c^2 = (a+c)^2 - 2ac = 100 - 18 = 82$

$a^2+b^2+c^2 = 82 + 9 = 91$

Geometric mean of $\alpha,\beta$ is 6, so $\alpha\beta = 36$.

Since $\alpha,\beta$ are roots of $2x^2-25x+\lambda=0$, product of roots $= \dfrac{\lambda}{2} = 36 \Rightarrow \lambda = 72$

A 5-digit number cannot start with 0, so the first digit must be 1.

Each of the remaining 4 digits can independently be 0 or 1, giving $2^4 = 16$ numbers.

General term: $T_{k+1} = \binom{6}{k}(2x)^{6-k}\left(\dfrac{-5}{x^2}\right)^{k} = \binom{6}{k}2^{6-k}(-5)^k x^{6-3k}$

For the constant term, $6-3k=0 \Rightarrow k=2$

$\binom{6}{2}2^4(-5)^2 = 15 \times 16 \times 25 = 6000$

For a set with $n$ elements, the number of subsets with an odd number of elements is $2^{n-1}$.

Here $n=6$, so the required number is $2^5 = 32$

$\sum_{k=0}^{n+1} {}^{n+1}C_k = 2^{n+1} = 512 = 2^9 \Rightarrow n+1 = 9 \Rightarrow n = 8$

$\sum_{k=0}^{n} {}^{n}C_k = 2^n = 2^8 = 256$

$\dfrac{8}{4}\left({}^7P_4\right) = 2 \times \dfrac{7!}{3!} = 2 \times 840 = 1680$

Also, ${}^8P_4 = \dfrac{8!}{4!} = 1680$

So the expression equals ${}^8P_4$

$|A| = 1(1) - (-1)(2) = 3$, so $$A^{-1} = \dfrac{1}{3}\begin{bmatrix}1 & 1\\ -2 & 1\end{bmatrix}$$

$$X = A^{-1}B = \dfrac{1}{3}\begin{bmatrix}1 & 1\\ -2 & 1\end{bmatrix}\begin{bmatrix}3\\6\end{bmatrix} = \dfrac{1}{3}\begin{bmatrix}9\\0\end{bmatrix} = \begin{bmatrix}3\\0\end{bmatrix}$$

So $x_1+x_2 = 3$

With $a_1=2, r=\tfrac12$: $a_1=2, a_2=1, a_3=\tfrac12, a_4=\tfrac14, a_5=\tfrac18, a_6=\tfrac{1}{16}$

Each row of the determinant is $r^2$ times the row above it (since $a_3=r^2a_1, a_4=r^2a_2$, etc.), except for the constant third column.

Direct evaluation gives $R_2 - r^2R_1$ and $R_3-r^4R_1$ having zero first two entries with the third column making the rows linearly dependent, so the determinant $=0$

For $A$ to be singular, $|A|=0$:

$1(8) - (-1)(4\lambda) = 0 \Rightarrow 8 + 4\lambda = 0 \Rightarrow \lambda = -2$

Expanding along the first row:

$|A| = a(1\cdot4 - (-1)\cdot0) - (-1)(0\cdot4-(-1)\cdot1) + (-a)(0\cdot0-1\cdot1)$

$= 4a + 1 + a = 5a+1$

Given $|A|=26$: $5a+1=26 \Rightarrow a=5$

$|3-4x| \le 11 \Rightarrow -11 \le 3-4x \le 11$

$\Rightarrow -14 \le -4x \le 8 \Rightarrow -2 \le x \le \dfrac{7}{2}$

For $x>11$: $\dfrac{2x-21}{x-11}\ge3 \Rightarrow 2x-21 \ge 3x-33 \Rightarrow x \le 12$, giving $11 < x \le 12$

For $x<11$: the inequality reverses on multiplying by $(x-11)<0$, giving $2x-21 \le 3x-33 \Rightarrow x \ge 12$, which contradicts $x<11$, so no solution here.

Hence $x \in (11, 12]$

Using the triple angle identity $\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta}$ with $\theta = 15^\circ$:

$\dfrac{3\tan15^\circ-\tan^315^\circ}{1-3\tan^215^\circ} = \tan45^\circ = 1$

Since $\sin100^\circ = \sin(180^\circ-100^\circ) = \sin80^\circ$ and $\sin120^\circ=\sin(180^\circ-120^\circ)=\sin60^\circ$,

$\sin60^\circ-\sin80^\circ+\sin100^\circ-\sin120^\circ = \sin60^\circ-\sin80^\circ+\sin80^\circ-\sin60^\circ = 0$

We know $\cos^{-1}x + \sin^{-1}x = \dfrac{\pi}{2}$. Adding this to the given equation $\cos^{-1}x - \sin^{-1}x = \dfrac{\pi}{6}$:

$2\cos^{-1}x = \dfrac{\pi}{2}+\dfrac{\pi}{6} = \dfrac{2\pi}{3} \Rightarrow \cos^{-1}x = \dfrac{\pi}{3}$

$x = \cos\dfrac{\pi}{3} = \dfrac{1}{2}$

$\cot^{-1}(1) = \dfrac{\pi}{4}$

Using $\tan^{-1}\dfrac12+\tan^{-1}\dfrac13 = \tan^{-1}\dfrac{\frac12+\frac13}{1-\frac12\cdot\frac13} = \tan^{-1}(1) = \dfrac{\pi}{4}$, we get $\cot^{-1}(2)+\cot^{-1}(3) = \dfrac{\pi}{4}$

Total $= \dfrac{\pi}{4}+\dfrac{\pi}{4} = \dfrac{\pi}{2}$

Using $\cos(2\sin^{-1}\alpha) = 1-2\alpha^2$:

$1-2\alpha^2 = \dfrac{47}{72} \Rightarrow \alpha^2 = \dfrac12\left(1-\dfrac{47}{72}\right) = \dfrac{25}{144}$

$\alpha = \dfrac{5}{12}$

$\tan\left(\alpha-\dfrac{\pi}{12}\right) = \dfrac{1}{\sqrt3} = \tan\dfrac{\pi}{6}$

$\alpha - \dfrac{\pi}{12} = \dfrac{\pi}{6} \Rightarrow \alpha = \dfrac{\pi}{6}+\dfrac{\pi}{12} = \dfrac{\pi}{4}$

$x = \dfrac{1+\cos2\theta}{\tan\theta-\sec\theta} = \dfrac{2\cos^2\theta}{\frac{\sin\theta-1}{\cos\theta}} = \dfrac{2\cos^3\theta}{\sin\theta-1}$

$y = \dfrac{\tan\theta+\sec\theta}{\sec^2\theta} = \cos\theta(\sin\theta+1)$

$\dfrac{y}{x} = \dfrac{\cos\theta(\sin\theta+1)(\sin\theta-1)}{2\cos^3\theta} = \dfrac{\sin^2\theta-1}{2\cos^2\theta} = \dfrac{-\cos^2\theta}{2\cos^2\theta} = -\dfrac{1}{2}$

$2x^2+2y^2-8y=7 \Rightarrow x^2+(y-2)^2 = \dfrac{15}{2}$

Centre of the circle is $(0,2)$.

The line joining $(-4,2)$ and $(0,2)$ is the horizontal line $y=2$

Slope of $AB = \dfrac{-2-2}{1-(-1)} = -2$

Line through $C(-2,-2)$ with slope $-2$: $y+2 = -2(x+2) \Rightarrow 2x+y+6=0$

The lines $x=4$, $y=-4$, $y=x$ intersect pairwise at $(4,4)$, $(4,-4)$ and $(-4,-4)$.

The base from $(4,-4)$ to $(-4,-4)$ has length 8, and the perpendicular height from $(4,4)$ to the line $y=-4$ is 8.

Area $= \dfrac12\times8\times8 = 32$

Foci at $(\pm3,0)$ give $c=3$. With $e=\dfrac12=\dfrac{c}{a}$, we get $a=6$, so $a^2=36$

$b^2 = a^2(1-e^2) = 36\left(1-\dfrac14\right) = 27$

Equation: $\dfrac{x^2}{36}+\dfrac{y^2}{27}=1$

The axis is horizontal (both points have $y=4$). Distance from vertex $(3,4)$ to focus $(-1,4)$ is $a=4$, and the focus lies to the left, so the parabola opens leftward.

Equation: $(y-4)^2 = -4a(x-3) = -16(x-3)$

The transverse axis joins $(-3,1)$ and $(1,1)$, length 4, so $a=2$ and centre is $(-1,1)$.

Conjugate axis length is 4, so $b=2$, i.e. $b^2=4=a^2$

Equation: $\dfrac{(x+1)^2}{4}-\dfrac{(y-1)^2}{4}=1 \Rightarrow (x+1)^2-(y-1)^2=4$

Radius = perpendicular distance from centre $(-4,0)$ to the line $x+y-2=0$:

$r = \dfrac{|-4+0-2|}{\sqrt{1^2+1^2}} = \dfrac{6}{\sqrt2} = 3\sqrt2$