KEAM 2026 17th April Question Paper with Solutions

Using $n(A \cup B) = n(A) + n(B) - n(A \cap B)$, we get $99 = n(A) + 61 - 28$, so $n(A) = 66$.

Therefore, $n(A') = n(U) - n(A) = 116 - 66 = 50$.

$(f \circ g)(x) = \sin^{-1}(x-2)$, which requires $-1 \le x-2 \le 1$.

So the largest domain of $g$ is $[1,3]$.

Since $4\sin x + 3\cos x \in [-5,5]$, the denominator $7+4\sin x+3\cos x \in [2,12]$.

Hence $f(x) = \dfrac{10}{\text{denominator}} \in \left[\dfrac{10}{12}, \dfrac{10}{2}\right] = \left[\dfrac{5}{6}, 5\right]$.

$g(2x) = e^{2x}+5$.

$(f \circ g)(x) = g(x)^2 - 10\,g(x) = (e^x+5)^2 - 10(e^x+5) = e^{2x} - 25$

So $g(2x) - (f\circ g)(x) = (e^{2x}+5) - (e^{2x}-25) = 30$

$5z - 2\overline{z} = \dfrac{7-7i}{1+i} = \dfrac{7(1-i)^2}{2} = \dfrac{7(-2i)}{2} = -7i$

With $z=x+iy$: $5(x+iy) - 2(x-iy) = 3x + 7iy = -7i$

So $3x = 0 \Rightarrow x = 0$ and $7y = -7 \Rightarrow y = -1$. Hence $x+y = -1$.

For $z = 1+i$, $z^2 = 2i$ and $\overline{z} = 1-i$.

$\dfrac{24\overline{z}}{z^2} = \dfrac{24(1-i)}{2i} = -12-12i$

$z - \dfrac{24\overline{z}}{z^2} = (1+i)-(-12-12i) = 13+13i = 13z$

Therefore $\lambda = 13$.

Let $z=x+iy$. Then $\dfrac{x}{2+i}+\dfrac{y}{1+2i}=\dfrac{3}{1-2i}$

Rationalising each term: $\dfrac{x(2-i)}{5}+\dfrac{y(1-2i)}{5}=\dfrac{3(1+2i)}{5}$

$(2x+y) - i(x+2y) = 3+6i \Rightarrow 2x+y=3,\ x+2y=-6$

Solving gives $x=4,\ y=-5$, so $z=4-5i$.

$\left(\alpha+\dfrac{i}{\alpha}\right)^2 = \alpha^2 - \dfrac{1}{\alpha^2} + 2i = 2+2i$

So $\left(\alpha+\dfrac{i}{\alpha}\right)^{16} = (2+2i)^8$

Since $2+2i = 2\sqrt{2}\,e^{i\pi/4}$, $(2+2i)^8 = (2\sqrt2)^8 e^{i2\pi} = (2\sqrt2)^8 = 8^4 = 4096$

Let the GP be $a, ar, ar^2,\ldots$. Given $a+ar=12$ and $ar^2=16$.

$\dfrac{ar^2}{a(1+r)} = \dfrac{16}{12} \Rightarrow \dfrac{r^2}{1+r}=\dfrac{4}{3} \Rightarrow 3r^2-4r-4=0$

Solving, $r=2$ or $r=-\dfrac{2}{3}$. Since $r>0$, $r=2$.

Let the terms be $a, ar, ar^2, ar^3$. Given $ar^3 = 8a \Rightarrow r^3=8 \Rightarrow r=2$.

Sum: $a(1+2+4+8)=15a=960 \Rightarrow a=64$

Since $r=2>1$, the smallest term is $a=64$.

Let the G.P. have terms $t_2=\cos\theta,\ t_3=\sqrt2\sin\theta,\ t_4=\sqrt3\tan\theta$.

$r=\dfrac{t_3}{t_2}=\sqrt2\tan\theta$ and $r=\dfrac{t_4}{t_3}=\dfrac{\sqrt3}{\sqrt2\cos\theta}$

Equating: $2\tan\theta\cos\theta=\sqrt3 \Rightarrow \sin\theta=\dfrac{\sqrt3}{2}\Rightarrow \theta=60^\circ$

So $\cos\theta=\tfrac12$, $r=\sqrt2\tan60^\circ=\sqrt6$, and $t_1=\dfrac{t_2}{r}=\dfrac{1/2}{\sqrt6}=\dfrac{1}{2\sqrt6}$

$A=\dfrac{a+b}{2},\ G=\sqrt{ab}$. Given $\dfrac{A}{G}=\dfrac{5}{3}\Rightarrow \dfrac{a+b}{2\sqrt{ab}}=\dfrac53 \Rightarrow 3(a+b)=10\sqrt{ab}$

Squaring: $9(a+b)^2=100ab \Rightarrow 9(a^2+b^2)+18ab=100ab \Rightarrow 9(a^2+b^2)=82ab$

Therefore $a^2+b^2:ab=82:9$

All $3! = 6$ permutations of digits 3, 5, 7 form the 3-digit numbers, and each digit appears exactly twice in each place (hundreds, tens, units).

Sum $= 2(3+5+7)(100+10+1) = 2 \times 15 \times 111 = 3330$

With 4 girls and 3 boys required to alternate strictly, the only possible pattern is G B G B G B G.

The girls can be arranged in $4!$ ways and the boys in $3!$ ways, giving $4! \times 3! = 24 \times 6 = 144$ arrangements.

Putting $x=1$: $(2-1)^9=1=a_0+a_1+\cdots+a_9$

Here $a_0=2^9=512$ and $a_9=(-1)^9=-1$ (coefficient of $x^9$).

So $a_1+\cdots+a_8 = 1-a_0-a_9 = 1-512+1=-510$

BANANA has letters B(1), A(3), N(2). Since B occurs only once, the word cannot start and end with B.

Case A...A: Remaining letters B,A,N,N arranged in the middle 4 spots: $\dfrac{4!}{2!}=12$

Case N...N: Remaining letters B,A,A,A arranged in the middle 4 spots: $\dfrac{4!}{3!}=4$

Total $=12+4=16$

Coefficient of $x^3$ in $(2+x)^n$ is $\binom{n}{3}2^{n-3}=160$. Testing $n=6$: $\binom{6}{3}2^3=20\times8=160$, so $n=6$.

In $(2-x^2)^n=(2-x^2)^6$, the term with $x^6$ needs $k=3$: $\binom{6}{3}2^3(-1)^3=20\times8\times(-1)=-160$

For a non-unique solution, the determinant of the coefficient matrix must be zero:

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

Expanding gives $8\alpha^2-7\alpha-1=0 \Rightarrow (8\alpha+1)(\alpha-1)=0$

So $\alpha=1$ or $\alpha=-\tfrac18$. Since $\alpha$ is an integer, $\alpha=1$

$B^TA^T=(AB)^T$. Computing $$AB=\begin{bmatrix}2&x\\3&1\end{bmatrix}\begin{bmatrix}3&2\\1&4\end{bmatrix}=\begin{bmatrix}6+x&4+4x\\10&10\end{bmatrix}$$

So $$(AB)^T=\begin{bmatrix}6+x&10\\4+4x&10\end{bmatrix}=C=\begin{bmatrix}3x+2&10\\14-x&10\end{bmatrix}$$

Equating: $6+x=3x+2 \Rightarrow x=2$ (also checks: $4+4(2)=12=14-2$)

Expanding the determinant $$\begin{vmatrix}2&-2&4\\-5&x+2&-10\\-1&1&x+1\end{vmatrix}=0$$

gives $2x^2-18=0 \Rightarrow x^2=9 \Rightarrow x=\pm3$

Given $A^2-A=20I \Rightarrow A^2=A+20I$.

Multiplying both sides by $A^{-1}$: $A = I+20A^{-1} \Rightarrow 20A^{-1}=A-I$

So $A^{-1}=\dfrac{1}{20}(A-I)$

$6(2x+3)+x>53-2x \Rightarrow 12x+18+x>53-2x \Rightarrow 13x+18>53-2x$

$15x>35 \Rightarrow x>\dfrac{7}{3}$

So the solution set is $\left(\dfrac73,\infty\right)$

Since $\angle A=\angle D=90^\circ$, sides $AB \parallel CD$ and $AD$ is perpendicular to both, so $BC=\sqrt{(AB-CD)^2+AD^2}$

$BC=\sqrt{(7a+1-3a-1)^2+(3a)^2}=\sqrt{16a^2+9a^2}=5a$

Perimeter $=(7a+1)+5a+(3a+1)+3a=18a+2$

Given $56<18a+2<92 \Rightarrow 54<18a<90 \Rightarrow 3<a<5$

Pairing terms: $\cot10^\circ\cot80^\circ=\cot10^\circ\tan10^\circ=1$ and $\cot20^\circ\cot70^\circ=\cot20^\circ\tan20^\circ=1$

Remaining: $\cot30^\circ\cot45^\circ\cot60^\circ=\sqrt3\times1\times\dfrac{1}{\sqrt3}=1$

Total product $=1\times1\times1=1$

$\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2(1/2)}{1-1/4}=\dfrac{1}{3/4}=\dfrac43$

$\sec^2 2\alpha=1+\tan^2 2\alpha=1+\dfrac{16}{9}=\dfrac{25}{9}$

$\tan^2(2\alpha)\sec^2(2\alpha)=\dfrac{16}{9}\times\dfrac{25}{9}=\dfrac{400}{81}$

$(3\cos x-2\sec x)^2=9\cos^2x-12+4\sec^2x$

Since $4\sec^2x=4+4\tan^2x$, this becomes $9\cos^2x-12+4+4\tan^2x=9\cos^2x+4\tan^2x-8$

Comparing with $9\cos^2x+4\tan^2x+k$, we get $k=-8$

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

Since $\alpha-\beta=\dfrac{\pi}{4}$, this equals $2+2\cos\dfrac{\pi}{4}=2+2\cdot\dfrac{\sqrt2}{2}=2+\sqrt2$

Using $\cos^{-1}(a)=\dfrac{\pi}{2}-\sin^{-1}(a)$, the equation becomes $\sin^{-1}(x-2)+\sin^{-1}(y+1)=\dfrac{\pi}{2}$

Whenever $\sin^{-1}p+\sin^{-1}q=\dfrac{\pi}{2}$, we have $p^2+q^2=1$.

So $(x-2)^2+(y+1)^2=1 \Rightarrow x^2-4x+4+y^2+2y+1=1 \Rightarrow x^2+y^2-4x+2y+4=0$

Let $\theta=\cot^{-1}\left(\dfrac43\right)$, so $\cos\theta=\dfrac45,\ \sin\theta=\dfrac35$

$\cos2\theta=2\cos^2\theta-1=2\cdot\dfrac{16}{25}-1=\dfrac{7}{25}$

Given $2\theta=\cos^{-1}\left(\dfrac{x}{5}\right)$, so $\dfrac{x}{5}=\dfrac{7}{25}\Rightarrow x=\dfrac75$

At $x=\dfrac16$: $\sin^{-1}(3x)=\sin^{-1}\left(\dfrac12\right)=\dfrac{\pi}{6}$

So $2\sin^{-1}(\alpha)=\dfrac{\pi}{2}-\dfrac{\pi}{6}=\dfrac{\pi}{3} \Rightarrow \sin^{-1}(\alpha)=\dfrac{\pi}{6}$

Hence $\alpha=\sin\dfrac{\pi}{6}=\dfrac12$

Since $OP$ is the perpendicular from the origin to the line, the line's normal form is $x\cos\alpha+y\sin\alpha=OP$

With $\sin\alpha=\dfrac35$ and $\alpha$ obtuse, $\cos\alpha=-\dfrac45$

$x\left(-\dfrac45\right)+y\left(\dfrac35\right)=5 \Rightarrow -4x+3y=25 \Rightarrow 4x-3y+25=0$

Since the line $x-y=7$ passes through $(4,a)$, substituting gives $4-a=7 \Rightarrow a=-3$

Slope through $(1,5)$ and $(3,-5)$ is $\dfrac{-5-5}{3-1}=-5$. Line: $y-5=-5(x-1) \Rightarrow 5x+y=10$

$x$-intercept: $A(2,0)$; $y$-intercept: $B(0,10)$

Area of $\triangle OAB = \dfrac12|OA \times OB| = \dfrac12(2)(10)=10$

Rewriting $x^2+y^2+8x-6y+c=0$ as $(x+4)^2+(y-3)^2=25-c$, the radius squared is $25-c$

Given area $=\pi r^2=75\pi \Rightarrow r^2=75 \Rightarrow 25-c=75 \Rightarrow c=-50$