KEAM 2025 26th April Question Paper with Solutions

Given $R=\{(1,3),(2,3),(2,4),(3,1),(4,4),(4,1)\}$ on $X=\{1,2,3,4\}$.

Not reflexive since $(1,1) \notin R$. Not transitive since $(2,3), (3,1) \in R$ but $(2,1) \notin R$. Not a function since 2 and 4 each map to two elements, so it is neither 1-1 nor onto.

Not symmetric: $(1,3) \in R$ but $(3,1)$ is present while $(2,3) \in R$ but $(3,2) \notin R$, so symmetry fails.

Hence the relation is not symmetric.

$A$ and $B$ have 11 common elements, say the set $C$ with $|C|=11$.

An element common to $A\times B$ and $B\times A$ must be an ordered pair $(x,y)$ with $x,y\in A\cap B$, i.e. $x,y \in C$.

So the number of common elements $= |C|\times|C| = 11\times11 = 121$.

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

$(x+2)(x-1)\geq 0 \Rightarrow x\leq -2 \text{ or } x\geq 1$

So the domain is $(-\infty,-2]\cup[1,\infty)$.

$f(x)=\sqrt{x^2+4x+4}=\sqrt{(x+2)^2}=|x+2|$

Since $|x+2|\geq 0$ for all real $x$, and takes every non-negative value, the range is $[0,\infty)$.

Given $sz+t\bar z+r=0$ with $z=x+iy$, so $\bar z = x-iy$.

$s(x+iy)+t(x-iy)+r=0 \Rightarrow (s+t)x+r+i(s-t)y=0$

Equating real and imaginary parts: $(s+t)x+r=0$ and $(s-t)y=0$.

Since $s,t$ are distinct positive reals, $s-t\neq 0$, so $y=0$. Thus $z$ lies on the real axis.

$|z-1+i|=5 \Rightarrow |z-(1-i)|=5$

This represents a circle with centre $(1,-1)$ and radius 5.

$z^3+iz^2-iz+1=0$

Group: $z^2(z+i) - i(z+i)\cdot 1$... instead try $z=i$: $i^3+i\cdot i^2 - i\cdot i+1 = -i-i+1+1=2-2i \neq 0$.

Rewrite as $z^3+1+iz(z-1)=0$. Try $z=-i$: $(-i)^3+i(-i)^2-i(-i)+1 = i - i -1+1=0$. So $z=-i$ is a root, $|z|=1$.

Factor out $(z+i)$: $z^3+iz^2-iz+1=(z+i)(z^2+1)$. Roots of $z^2+1=0$ are $z=\pm i$, also giving $|z|=1$.

Hence $|z|=1$ for every root.

Let $a=1+\sin\theta$, $b=\cos\theta$ where $\theta=\frac{2\pi}{27}$. The expression is $\dfrac{a-ib}{a+ib}$.

Multiply numerator and denominator by conjugate: $\dfrac{(a-ib)^2}{a^2+b^2} = \dfrac{a^2-b^2-2iab}{a^2+b^2}$

Real part $=\dfrac{a^2-b^2}{a^2+b^2}=\dfrac{(1+\sin\theta)^2-\cos^2\theta}{(1+\sin\theta)^2+\cos^2\theta}$

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

Denominator $=1+2\sin\theta+\sin^2\theta+\cos^2\theta=2+2\sin\theta=2(1+\sin\theta)$

So real part $=\dfrac{2\sin\theta(1+\sin\theta)}{2(1+\sin\theta)}=\sin\theta=\sin\dfrac{2\pi}{27}$

The G.P. is $9,3,1,\frac13,\frac19,...$ with first term $a=9$ and common ratio $r=\frac13$.

$T_{25}=ar^{24}=9\times\left(\frac13\right)^{24}=3^2\times 3^{-24}=3^{-22}=\dfrac{1}{3^{22}}$

In a G.P. with terms $a,b,c$, the common ratio is $r=\dfrac{b}{a}=\dfrac{c}{b}$, so $b^2=ac$, giving $r=\dfrac{c}{b}$ and also $r=\dfrac{b}{a}$.

Since $r=\dfrac{c}{b}$ and $b=ar$, we get $r=\dfrac{c}{ar}\Rightarrow r^2=\dfrac{c}{a}$.

Fifth term $=ar^4=a(r^2)^2=a\left(\dfrac{c}{a}\right)^2=\dfrac{c^2}{a}$

Sum of first $n$ terms of G.P.: $S_n=\dfrac{a(r^n-1)}{r-1}$, with $a=1,r=2$.

$1023=\dfrac{1(2^n-1)}{2-1}=2^n-1 \Rightarrow 2^n=1024=2^{10}$

So $n=10$.

With $l, G_1, G_2, G_3, n$ in G.P. (5 terms), $n=l\cdot r^4$ where $r$ is the common ratio.

$r^4=\dfrac{n}{l} \Rightarrow r=\left(\dfrac{n}{l}\right)^{1/4}$

To divide 25 distinct objects into 5 unordered groups of 5 each, first divide into 5 labelled groups of 5: $\dfrac{25!}{(5!)^5}$ ways.

Since the 5 groups are identical (unordered), divide by $5!$ for the ways of arranging the groups: $\dfrac{25!}{(5!)^5\cdot 5!}=\dfrac{25!}{(5!)^6}$

$\sum_{k=0}^{100}{}^{100}C_k = 2^{100}$, i.e. $1+{}^{100}C_1+{}^{100}C_2+...+{}^{100}C_{99}+1=2^{100}$

since ${}^{100}C_0={}^{100}C_{100}=1$.

Coefficient of $x^{10}$ in $(1-x^2)(1-x^3)^9 = (1-x^3)^9 - x^2(1-x^3)^9$.

In $(1-x^3)^9$, powers of $x$ are multiples of 3 (0,3,6,9,...). So $x^{10}$ never appears in $(1-x^3)^9$.

In $x^2(1-x^3)^9$, powers are $2+3k$; for this to equal 10, $3k=8$, not an integer, so no $x^{10}$ term here either.

Hence the coefficient of $x^{10}$ is $0$.

$\sum_{k=0}^{21}{}^{21}C_k = 2^{21}$

By symmetry ${}^{21}C_k = {}^{21}C_{21-k}$, so ${}^{21}C_1+...+{}^{21}C_{10} = {}^{21}C_{11}+...+{}^{21}C_{20}$.

Since 21 is odd, all terms $k=0$ to $21$ pair up evenly (no middle term): $2\left({}^{21}C_1+...+{}^{21}C_{10}\right) + 2\cdot{}^{21}C_0 = 2^{21}$

So ${}^{21}C_1+...+{}^{21}C_{10} = \dfrac{2^{21}}{2}-{}^{21}C_0 = 2^{20}-1$

General term: ${}^{10}C_r\left(\dfrac{\sqrt x}{2}\right)^{10-r}\left(\dfrac{1}{3x^2}\right)^r = {}^{10}C_r\dfrac{1}{2^{10-r}3^r}x^{\frac{10-r}{2}-2r}$

For constant term, exponent of $x$ is zero: $\dfrac{10-r}{2}-2r=0 \Rightarrow 10-r=4r \Rightarrow r=2$

Constant term $={}^{10}C_2\dfrac{1}{2^8\cdot3^2}=45\times\dfrac{1}{256\times9}=\dfrac{45}{2304}=\dfrac{5}{256}$

$B$ is $3\times2$ and $C=BA$ is $3\times3$.

For $BA$ to be defined, $A$ must have 2 rows (matching $B$'s columns), and the result has as many columns as $A$.

Since $BA$ is $3\times3$, $A$ must be $2\times3$.

$\det(QPQ^{-1}) = \det(Q)\det(P)\det(Q^{-1}) = \det(P)$ since $\det(Q)\det(Q^{-1})=1$.

$P$ is upper triangular with diagonal entries $1,2,3$, so $\det(P)=1\times2\times3=6$.

Hence $\det(QPQ^{-1})=6$.

$P=\dfrac12(A+A^T)$ is always symmetric, since $P^T=\dfrac12(A^T+A)=P$.

This holds for any square matrix $A$, in particular this one, so $P^T=P$.

$\sec^2x+\csc^2x = \dfrac{1}{\cos^2x}+\dfrac{1}{\sin^2x}=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\dfrac{1}{\sin^2x\cos^2x}=\sec^2x\csc^2x$

So $\sec^2x+\csc^2x-\sec^2x\csc^2x=0$

$7x+4<9x+8 \Rightarrow -4<2x \Rightarrow x>-2$

Solution set: $(-2,\infty)$

$\dfrac{3(x+3)}{7}\leq\dfrac{6(x-1)}{5}$

Cross multiply (both denominators positive): $15(x+3)\leq42(x-1)$

$15x+45\leq42x-42 \Rightarrow 87\leq27x \Rightarrow x\geq\dfrac{87}{27}=\dfrac{29}{9}$

Solution set: $\left[\dfrac{29}{9},\infty\right)$

$\sin15^\circ\sin75^\circ = \sin15^\circ\cos15^\circ = \dfrac12\sin30^\circ=\dfrac14$

So $\sin15^\circ\sin45^\circ\sin75^\circ = \dfrac14\times\dfrac{1}{\sqrt2}=\dfrac{1}{4\sqrt2}$

$\sin\theta=\dfrac15$, $\theta$ in 2nd quadrant, so $\cos\theta<0$.

$\cos\theta = -\sqrt{1-\sin^2\theta}=-\sqrt{1-\dfrac{1}{25}}=-\sqrt{\dfrac{24}{25}}=\dfrac{-2\sqrt6}{5}$

$\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{-5}{2\sqrt6}$

$4\sin\left(\dfrac{x}{4}\right)\cos\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{4}\right)$

$=2\left[2\sin\left(\dfrac{x}{4}\right)\cos\left(\dfrac{x}{4}\right)\right]\cos\left(\dfrac{x}{2}\right)=2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)=\sin x$

Using sum-to-product: $\cos75^\circ-\cos15^\circ = -2\sin45^\circ\sin30^\circ$

$\cos75^\circ+\cos15^\circ = 2\cos45^\circ\cos30^\circ$

Ratio $=\dfrac{-2\sin45^\circ\sin30^\circ}{2\cos45^\circ\cos30^\circ}=-\tan45^\circ\tan30^\circ=-1\times\dfrac{1}{\sqrt3}=\dfrac{-1}{\sqrt3}$

Denominator: $(1+\sin\alpha+\cos\alpha)(1+\sin\alpha-\cos\alpha) = (1+\sin\alpha)^2-\cos^2\alpha$

$=1+2\sin\alpha+\sin^2\alpha-\cos^2\alpha = 2\sin\alpha+2\sin^2\alpha = 2\sin\alpha(1+\sin\alpha)$

So expression $=\dfrac{2\sin\alpha(1+\sin\alpha)}{2\sin\alpha(1+\sin\alpha)}=1$

$\dfrac{\pi}{2}-\cos^{-1}\left(\dfrac12\right) = \sin^{-1}\left(\dfrac12\right) = \dfrac{\pi}{6}$

So $\sin^{-1}\left(\dfrac{x}{1+x}\right)=\dfrac{\pi}{6} \Rightarrow \dfrac{x}{1+x}=\sin\dfrac{\pi}{6}=\dfrac12$

$2x = 1+x \Rightarrow x=1$

$\tan^{-1}x = \tan^{-1}3-\dfrac{\pi}{4} = \tan^{-1}3-\tan^{-1}1$

Using $\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left(\dfrac{A-B}{1+AB}\right)$: $\tan^{-1}x=\tan^{-1}\left(\dfrac{3-1}{1+3}\right)=\tan^{-1}\left(\dfrac12\right)$

So $x=\dfrac12$

Distance of $4x-3y+k=0$ from $(1,2)$: $\dfrac{|4(1)-3(2)+k|}{\sqrt{16+9}}=\dfrac{|k-2|}{5}=5$

$|k-2|=25 \Rightarrow k-2=25 \text{ or } k-2=-25$

$k=27 \text{ or } k=-23$

The two given sides $x+y=5$ and $x-y=-5$ meet at a vertex: solving, $x=0,y=5$, so one vertex is $(0,5)$.

Since diagonals of a parallelogram bisect each other, the vertex opposite to $(0,5)$ is $2(3,6)-(0,5) = (6,7)$.

So $(6,7)$ is a vertex of the parallelogram.