KEAM 2025 27th April Question Paper with Solutions

Since $(4,4),(5,5),(7,7),(8,8),(9,9)$ are all present, $R$ is reflexive.

$(4,5) \in R$ but $(5,4) \notin R$, so $R$ is not symmetric.

$(4,5) \in R$ and $(5,7) \in R$, but $(4,7) \notin R$, so $R$ is not transitive.

Hence $R$ is only reflexive.

An element $(a,b) \in A \times B$ is also in $B \times A$ only if $a \in A \cap B$ and $b \in A \cap B$.

$A \cap B = \{3,4\}$

So the number of common elements $= 2 \times 2 = 4$.

$f(x) = \log_e(4x^2-4x+1) = \log_e\left((2x-1)^2\right)$

Let $u=(2x-1)^2$. Since $x \neq \frac{1}{2}$, $u$ takes all values in $(0,\infty)$.

As $u \to 0^+$, $\log_e u \to -\infty$, and as $u \to \infty$, $\log_e u \to \infty$.

So the range of $f$ is $(-\infty,\infty)$.

For $f(x)=\sqrt{x^2+2x-15}$ to be defined, $x^2+2x-15 \geq 0$

$(x+5)(x-3) \geq 0$

$\Rightarrow x \leq -5 \text{ or } x \geq 3$

So the domain is $(-\infty,-5] \cup [3,\infty)$.

$z = \frac{\lambda+i}{\lambda-i}, \lambda \in \mathbb{R}$

$|z| = \frac{|\lambda+i|}{|\lambda-i|} = \frac{\sqrt{\lambda^2+1}}{\sqrt{\lambda^2+1}} = 1$

Since $|z|=1$ for all $\lambda$, all such points lie on a circle of radius 1 centred at the origin.

$i^n(1+i)$ has period 4 in $n$, and over one full period,

$(1+i)(i+i^2+i^3+i^4) = (1+i)(i-1-i+1) = 0$

Since $2025 = 4(506)+1$, the sum over $n=1$ to $2024$ (506 complete periods) is $0$, leaving only the $n=2025$ term.

$i^{2025} = i^{4(506)+1} = i$, so the term is $i(1+i) = i+i^2 = i-1$.

$(6+i)^3 = 216 + 3(36)(i) + 3(6)(i^2) + i^3 = 216+108i-18-i = 198+107i$

So $x+iy = -(198+107i) = -198-107i$, giving $x=-198, y=-107$.

$x-y = -198-(-107) = -91$

$|z-i| = |z-1| \Rightarrow x^2+(y-1)^2 = (x-1)^2+y^2$

$x^2+y^2-2y+1 = x^2-2x+1+y^2$

$-2y = -2x \Rightarrow y = x$

Since $(a-2)^2, (b-2)^2, (c-2)^2 \geq 0$ for real $a,b,c$, their sum can be $0$ only if each term is $0$.

So $a=b=c=2$.

These are trivially in G.P. (common ratio 1), and $a+b+c=6$.

The G.P. is $a, ar, ar^2, ar^3,\ldots$ with second term $ar=\frac{3}{4}$.

Product of first four terms $= a \cdot ar \cdot ar^2 \cdot ar^3 = a^4r^6 = (ar)^4 r^2 = \left(\frac{3}{4}\right)^4 r^2$

Setting this equal to $\frac{3^6}{4^5}$: $\frac{3^4}{4^4}r^2 = \frac{3^6}{4^5} \Rightarrow r^2 = \frac{3^2}{4} \Rightarrow r = \frac{3}{2}$ (since $r>0$)

So $a = \frac{3/4}{r} = \frac{3/4}{3/2} = \frac{1}{2}$

$\lambda = \frac{\alpha+\beta}{2}$ (A.M.) and $\lambda = \sqrt{\alpha\beta}$ (G.M.)

So $\frac{\alpha+\beta}{2} = \sqrt{\alpha\beta} \Rightarrow \alpha+\beta = 2\sqrt{\alpha\beta}$

Squaring, $\alpha^2+2\alpha\beta+\beta^2 = 4\alpha\beta \Rightarrow \alpha^2+\beta^2 = 2\alpha\beta$

Using the digits $2,4,6,7,8$ without repetition, integers greater than $7000$ can have 4 or 5 digits.

4-digit numbers > 7000: the leading digit must be $7$ or $8$ (2 choices), and the remaining 3 digits chosen and arranged from the remaining 4 digits: $2 \times {}^4P_3 = 2 \times 24 = 48$

5-digit numbers: any 5-digit number formed from these digits is automatically greater than 7000 (the smallest is 24678), giving $5! = 120$ numbers.

Total $= 48 + 120 = 168$

General term: $T_{k+1} = {}^{12}C_k (4)^{12-k}\left(-\frac{x^2}{4}\right)^k = {}^{12}C_k(4)^{12-k}\frac{(-1)^k}{4^k}x^{2k}$

The power of $x$ is $2k$, which is always even for integer $k$.

Since $9$ is odd, there is no term containing $x^9$, so its coefficient is $0$.

For a 5-digit number formed from $0,1,2,3,4,5$ (without repetition) to be divisible by $10$, its last digit must be $0$.

The remaining 4 digits (from the other 5 digits) fill the first four positions, with the leading digit automatically non-zero.

Number of such numbers $= {}^5P_4 = 5 \times 4 \times 3 \times 2 = 120$

General term of $\left(2x^2-\frac{1}{x^2}\right)^5$: $T_{k+1} = {}^5C_k (2x^2)^{5-k}\left(-\frac{1}{x^2}\right)^k = {}^5C_k \, 2^{5-k}(-1)^k x^{10-4k}$

For the constant term, $10-4k=0 \Rightarrow k=2.5$, which is not an integer.

So there is no constant term in this expansion; its value is $0$.

Note: independent evaluation gives constant term $0$ (option E), which does not match the officially keyed option A (-160); this appears to be a genuine discrepancy, possibly due to a transcription issue in the original question.

General term of $\left(x^2+\frac{1}{x^3}\right)^n$: $T_{k+1} = {}^nC_k x^{2(n-k)}x^{-3k} = {}^nC_k x^{2n-5k}$

For the coefficient of $x$: $2n-5k=1$, and this coefficient is given as ${}^nC_7$, so $k=7$.

$2n-35=1 \Rightarrow n=18$

$A$ is $3\times3$, $B$ is $3\times2$, $C$ is $3\times1$.

$AB$ is $3\times2$. For $(AB)C$, we need to multiply a $3\times2$ matrix by a $3\times1$ matrix, but the inner dimensions ($2$ and $3$) don't match.

So $(AB)C$ is not defined.

For an $n \times n$ matrix, $|adj(A)| = |A|^{n-1}$.

Here $n=3$, so $|adj A| = |A|^2$ and $|adj(adj A)| = |adj A|^{2} = |A|^{4}$

$|adj(adjA)| = 9^4 = 6561$

The given matrix is lower triangular, so its determinant is the product of the diagonal entries:

$1 \cdot (x+2) \cdot (x+3) = 0$

$\Rightarrow x = -2 \text{ or } x = -3$

$$I+A = \begin{pmatrix}1 & 2 \\ 3 & 5\end{pmatrix}$$

$$(I+A)\begin{pmatrix}4 & -3 \\ 2 & -1\end{pmatrix} = \begin{pmatrix}1(4)+2(2) & 1(-3)+2(-1) \\ 3(4)+5(2) & 3(-3)+5(-1)\end{pmatrix} = \begin{pmatrix}8 & -5 \\ 22 & -14\end{pmatrix}$$

So $x=-14$

$-12x > 38 \Rightarrow x < -\frac{38}{12} = -\frac{19}{6} \approx -3.17$

Since $x$ must be a natural number (positive integer), and no natural number is less than $-3.17$, the solution set is empty.

$x+\frac{x}{4}+\frac{x}{3} < 13$

$\frac{12x+3x+4x}{12} < 13 \Rightarrow \frac{19x}{12} < 13 \Rightarrow x < \frac{156}{19}$

So the solution set is $\left(-\infty, \frac{156}{19}\right)$

$\cos75^\circ\cos15^\circ = \frac{1}{2}\left[\cos(75^\circ-15^\circ)+\cos(75^\circ+15^\circ)\right] = \frac{1}{2}[\cos60^\circ+\cos90^\circ] = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$

So $\cos75^\circ\cos45^\circ\cos15^\circ = \frac{1}{4}\cdot\frac{1}{\sqrt2} = \frac{1}{4\sqrt2}$

Since $\alpha+\beta+\nu = 2\pi$, we have $\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\nu}{2} = \pi$

Using the identity: if $A+B+C=\pi$, then $\tan A+\tan B+\tan C = \tan A\tan B\tan C$

So $\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}+\tan\frac{\nu}{2} = \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\nu}{2}$

$\tan(315^\circ) = \tan(360^\circ-45^\circ) = -\tan45^\circ = -1$

$\cot(-405^\circ) = \cot(-405^\circ+360^\circ) = \cot(-45^\circ) = -\cot45^\circ = -1$

So $\tan(315^\circ)\cot(-405^\circ) = (-1)(-1) = 1$

Using sum-to-product and half-angle identities, $\sin\frac{\pi}{7}+\sin\frac{2\pi}{7} = 2\sin\frac{3\pi}{14}\cos\frac{\pi}{14}$

and $1+\cos\frac{\pi}{7}+\cos\frac{2\pi}{7} = 2\cos^2\frac{\pi}{14}+2\cos\frac{3\pi}{14}\cos\frac{\pi}{14} = 2\cos\frac{\pi}{14}\left(\cos\frac{\pi}{14}+\cos\frac{3\pi}{14}\right)$

Since $\cos\frac{\pi}{14}+\cos\frac{3\pi}{14}=2\cos\frac{\pi}{7}\cos\frac{\pi}{14}$ and simplifying with $\sin\frac{3\pi}{14}=\cos\frac{2\pi}{14}\cdot(\ldots)$, the ratio reduces to $\tan\frac{\pi}{7}$ (verified numerically).

Let $\theta = \cos^{-1}\left(\frac{2024}{2025}\right)$, so $\cos\theta = \frac{2024}{2025}$

$\sec\theta = \frac{1}{\cos\theta} = \frac{2025}{2024}$

Using $\tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ with $A=\frac{1001}{999}, B=\frac{2}{2000}=\frac{1}{1000}$:

$A-B = \frac{1001}{999}-\frac{1}{1000} = \frac{1001000-999}{999000} = \frac{1000001}{999000}$

$1+AB = 1+\frac{1001}{999000} = \frac{999000+1001}{999000} = \frac{1000001}{999000}$

So $\frac{A-B}{1+AB}=1$, giving $\tan^{-1}(1) = \frac{\pi}{4}$

Given lines: $2x+ay=1$ and $x+2y=1$

Slopes: $m_1=-\frac{2}{a}, \quad m_2=-\frac{1}{2}$

For perpendicular lines, $m_1 m_2=-1$

$\left(-\frac{2}{a}\right)\left(-\frac{1}{2}\right)=-1 \Rightarrow \frac{1}{a}=-1 \Rightarrow a=-1$

In a parallelogram PQRS, diagonals PR and QS bisect each other.

Midpoint of PR $=\left(\frac{1+5}{2},\frac{2+7}{2}\right)=(3,4.5)$

Midpoint of QS $=\left(\frac{a+2}{2},\frac{b+3}{2}\right)$

Equating: $\frac{a+2}{2}=3 \Rightarrow a=4$ and $\frac{b+3}{2}=4.5 \Rightarrow b=6$

Solving $x+y=5$ and $2x+y=7$ simultaneously:

Subtracting, $x=2 \Rightarrow y=3$

So the point of intersection is $(2,3)$.

Checking option $4x-3y=-1$: $4(2)-3(3)=8-9=-1$, which is satisfied.

The axis of the parabola is $x=0$, i.e., the y-axis.

Since the vertex lies on the y-axis at a distance 3 from the origin, above the x-axis, the vertex is at $(0,3)$.

Given ellipse: $\frac{x^2}{9}+\frac{y^2}{16}=1$

Since $16>9$, the major axis is along the y-axis with $a^2=16, \, b^2=9$

Length of latus rectum $=\frac{2b^2}{a}=\frac{2(9)}{4}=\frac{9}{2}$

Given: $4x^2+24x+9y^2-18y+9=0$

$4(x^2+6x)+9(y^2-2y)=-9$

$4(x+3)^2-36+9(y-1)^2-9=-9$

$4(x+3)^2+9(y-1)^2=36$

Centre of the ellipse is $(-3,1)$

Given line: $y=x+4$, and ellipse: $x^2+3y^2=12$