AP EAPCET 2025 22nd May Morning Shift Question Paper with Solutions

$𝑓(𝑥+𝑦)=𝑓\left(𝑥\right)+𝑓\left(𝑦\right)+𝑥𝑦...\left(𝑖\right)$

$𝑥,𝑦\in 𝑁$

From Eq. (i),

$\begin{array}{cc}& 𝑓\left(1\right)=2=\frac{1(1+3)}{2}\\ & 𝑥=1,𝑦=1\\ & 𝑓\left(2\right)=𝑓\left(1\right)+𝑓\left(1\right)+1\\ & 𝑓\left(2\right)=5=\frac{2(2+3)}{2}\end{array}$

From Eq. (i), $𝑥=1,𝑦=2$

$\begin{array}{cc}𝑓\left(3\right)& =𝑓\left(1\right)+𝑓\left(2\right)+2\\ & =2+5+2=9=\frac{3(3+3)}{2}\\ 𝑓\left(4\right)& =𝑓\left(2\right)+𝑓\left(2\right)+4\\ & =14=\frac{4(4+3)}{2}\\ \Rightarrow 𝑓\left(𝑥\right)=& \frac{𝑛(𝑛+3)}{2}=\frac{{𝑛}^2}{2}+\frac{3𝑛}{2}\\ \sum _{𝑘=0}^{10}𝑓\left(𝑘\right)=& \sum _{𝑘=1}^{10}\frac{𝑘(𝑘+3)}{2}=\frac{1}{2}\sum _{𝑘=1}^{10}{𝑘}^2+\frac{3}{2}\sum _{𝑘=1}^{10}𝑘\\ =& \frac{1}{2}(385+165)=275\end{array}$

$\begin{array}{cc}𝑓\left(𝑥\right)& =\{\begin{array}{cc}1-𝑥,& 𝑥\in [-1,1]\\ 𝑥-1,& 𝑥\in (1,2]\end{array}\\ & -1\le 𝑥\le 1\\ & +1\ge -𝑥\ge -1\\ & +2\ge 1-𝑥\ge +0\\ 𝑥& \in [-1,1]\\ 𝑓\left(𝑥\right)& \in [0,2]...\left(𝑖\right)\end{array}$

Now $𝑥\in (1,2],𝑓\left(𝑥\right)\in (0,1]...\left(𝑖𝑖\right)$

From Eqs. (i) and (ii) range $\equiv [0,2]$

$\begin{array}{cc}& \mathrm{AM}\ge \mathrm{GM}\\ & \begin{array}{cc}\therefore & \frac{1+3+3^2+3^3+3^{𝑛-1}}{𝑛}\\ & \ge {(1\cdot 3\cdot 3^2\cdot 3^3\cdot \dots \cdot 3^{𝑛-1})}^{\frac{1}{𝑛}}\\ \Rightarrow & \frac{3^{𝑛}-1}{2𝑛}\ge {\left(3^{1+2+3+\dots 𝑛-1}\right)}^{\frac{1}{𝑛}}\\ \Rightarrow & \frac{3^{𝑛}-1}{2}\ge 𝑛{\left(3^{𝑛}\frac{(𝑛-1)}{2}\right)}^{\frac{1}{𝑛}}\\ \Rightarrow & \frac{3^{𝑛}-1}{2}\ge 𝑛3^{\frac{𝑛-1}{2}}\end{array}\end{array}$

$2𝑥+𝑝𝑦+6𝑧=8,𝑥+2𝑦+𝑞𝑧=5$ and $𝑥+𝑦+3𝑧=4$ may have no solution.

$\begin{array}{cc}& 𝑥=\frac{\Delta 𝑥}{\Delta },𝑦=\frac{\Delta 𝑦}{\Delta },𝑧=\frac{\Delta 𝑧}{\Delta }\\ & \Delta =\mid \begin{array}{ccc}2& 𝑝& 6\\ 1& 2& 𝑞\\ 1& 1& 3\end{array}\mid =(𝑝-2)(𝑞-3)\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& {\Delta }_{𝑥}=\mid \begin{array}{ccc}8& 𝑝& 6\\ 5& 2& 𝑞\\ 4& 1& 3\end{array}\mid =(15-4𝑞)(2-𝑝)\\ & {\Delta }_{𝑦}=\mid \begin{array}{ccc}2& 8& 6\\ 1& 5& 𝑞\\ 1& 4& 3\end{array}\mid =0\\ & {\Delta }_{𝑧}=\mid \begin{array}{ccc}2& 𝑝& 8\\ 1& 2& 5\\ 1& 1& 4\end{array}\mid =𝑝-2\end{array}\\ & \text{ For no solution }\\ & \begin{array}{cc}& (𝑝-2)(𝑞-3)=0\\ & 𝑝-2\ne 0\\ & 𝑞=3\\ & 𝑝\ne 2\end{array}\end{array}$

A is det of order 3 with entries 0 or 1 .

$\therefore 𝐵$ and $𝐶$ has same number of elements.

(c) $𝐴=⎡⎣⎢\begin{array}{ccc}𝑥& 𝑦& 0\\ -3& 1& 2\\ 1& -2& 𝑧\end{array}⎤⎦⎥$ and

$𝐵=⎡⎣⎢\begin{array}{ccc}1& -2& -2\\ 2& 0& 1\\ 2& 1& 0\end{array}⎤⎦⎥$

Cofactor of $𝑧=9$, cofactor of 1 in $3^{\text{rd }}$ row of $𝐴=4$

$\begin{array}{cccc}& & 𝑥+3𝑦& =9...\left(𝑖\right)\\ \Rightarrow & & 𝑥& =3\\ & & 2𝑦& =4...\left(𝑖𝑖\right)\\ \Rightarrow & & 𝑦& =2\end{array}$

Cofactor of $𝑥=3$

$𝑧+4=3\Rightarrow 𝑧=-1...\left(𝑖𝑖𝑖\right)$

$\begin{array}{cc}\therefore 𝐴𝐵& =⎡⎣⎢\begin{array}{ccc}3& 2& 0\\ -3& 1& 2\\ 1& -2& -1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}1& -2& -2\\ 2& 0& 1\\ 2& 1& 0\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}7& -6& -4\\ 3& 8& 7\\ -5& -3& -4\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& |𝑧-1|+|𝑧-5|\ge |𝑧-1-𝑧+5|\\ & \ge 4[\because \left|{𝑧}_1\right|+\left|{𝑧}_2\right|\ge |{𝑧}_1-{𝑧}_2|]\\ & \text{ Minimum value }=4\end{array}$

$\begin{array}{cc}& 𝑧=𝑥+𝑖𝑦\\ & \begin{array}{cc}2\mid 𝑧& -2-3𝑖|=3|𝑧+𝑖-2\mid \\ \text{ Let }𝑧& =𝑥+𝑖𝑦\\ 2\mid (𝑥& -2)+𝑖(𝑦-3)\mid \\ & =3\left|\right(𝑥-2)+𝑖(𝑦+1\left)\right|\end{array}\end{array}$

$\begin{array}{cc}& \text{ Squaring on both sides, we get }\\ & \begin{array}{cc}& \begin{array}{cc}& 4{𝑥}^2-16𝑥+16+4{𝑦}^2-24𝑦+36\\ & =9{𝑥}^2-36𝑥+36+9{𝑦}^2+18𝑦+9\\ \Rightarrow & 5{𝑥}^2+5{𝑦}^2-20𝑥+42𝑦-7=0\end{array}\\ & \text{ Centre }\equiv (2,\frac{-21}{5})\end{array}\end{array}$

$\begin{array}{cc}& {𝑥}^7=1\\ & \Rightarrow {𝑥}^7-1=0\\ & \therefore {𝑧}^7-1=0\\ & (𝑧-1)(1+𝑧+{𝑧}^2+{𝑧}^3+{𝑧}^4+{𝑧}^5+{𝑧}^6)=0\\ & 𝑧\ne 1\\ & 1+𝑧+{𝑧}^2+{𝑧}^3+{𝑧}^4+{𝑧}^5+{𝑧}^6=0...\left(𝑖\right)\end{array}$

Now,

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& (2𝑘-1){𝑥}^2-2(3𝑘-2)𝑥+4𝑘>0,\forall 𝑥\in ℝ\\ & \Rightarrow 4(3𝑘-2{)}^2-4(2𝑘-1)4𝑘<0\\ & \Rightarrow 9{𝑘}^2-12𝑘+4-8{𝑘}^2+4𝑘<0\\ & \Rightarrow {𝑘}^2-8𝑘+4<0,\\ & 𝑘=\frac{8\pm \sqrt{\surd 64-16}}{2}=\frac{8\pm 4\sqrt{\surd 3}}{2}\\ & \Rightarrow 𝑘\in (4-2\sqrt{\surd 3},4+2\sqrt{\surd 3})\\ & 𝑘=1,2,3,\dots ,7\\ & \therefore \text{ Sum of integral value }=28\end{array}\end{array}$

Given, $𝑓\left(𝑥\right)=\frac{{𝑥}^2-5𝑥+7}{{𝑥}^2-5𝑥-7}=𝑦$

$\Rightarrow (𝑦-1){𝑥}^2-5(𝑦-1)𝑥-(7𝑦+7)=0$

for $𝑥\in 𝑅$

$\begin{array}{cc}& 𝐷\ge 0\\ & \Rightarrow 25(𝑦-1{)}^2+4(𝑦-1\left)\right(7𝑦+𝜋\ge 0\\ & \Rightarrow (𝑦-1)[25𝑦-25+28𝑦+28]\ge 0\\ & (𝑦-1)(53𝑦+3)\ge 0\\ & 𝑦\in (-\infty ,\frac{-3}{53}]\cup (1,\infty )\text{ at }𝑦=1,𝑥\text{ is not }\end{array}$

possible

$\therefore$ Required sum $=2+(-1)=1$

$f(x)=18 x^3-33 x^2+20 x-4=0

${𝑓}^{\prime }\left(𝑥\right)=54{𝑥}^2-66𝑥+20$$𝑥=𝛼$ is repeated roots, ${𝑓}^{\prime }\left(𝛼\right)=0$ $\begin{array}{cc}\text{ and }& 54{𝛼}^2-66𝛼+20=0\\ \Rightarrow & 27{𝛼}^2-33𝛼+10=0\\ \Rightarrow & 27{𝛼}^2-18𝛼-15𝛼+10=0\\ \Rightarrow & (9𝛼-5)(3𝛼-2)=0\Rightarrow 𝛼=\frac{2}{3}\text{ or }\frac{5}{9}\end{array}$

We start with the equation: $6{𝑥}^4-5{𝑥}^3+13{𝑥}^2-5𝑥+6=0$.

Divide both sides by ${𝑥}^2$ (where $𝑥\ne 0$) to get:

$6{𝑥}^2-5𝑥+13-\frac{5}{𝑥}+\frac{6}{{𝑥}^2}=0$

Rewrite the expression by grouping terms: $6({𝑥}^2+\frac{1}{{𝑥}^2})-5(𝑥+\frac{1}{𝑥})+13=0$

Let $𝑡=𝑥+\frac{1}{𝑥}$. Then ${𝑥}^2+\frac{1}{{𝑥}^2}={𝑡}^2-2$.

Substitute these into the equation: $6({𝑡}^2-2)-5𝑡+13=0$

Simplify: $6{𝑡}^2-12-5𝑡+13=06{𝑡}^2-5𝑡+1=0$

This is a quadratic equation in $𝑡$. The discriminant ($𝐷$) is:

$𝐷=(-5{)}^2-4\cdot 6\cdot 1=25-24=1$ (which is greater than 0).

This means there are two real solutions for $𝑡$: $𝑡=\frac{5\pm 1}{12}=\frac{6}{12}\text{ or }\frac{4}{12}𝑡=\frac{1}{2}\text{or}\frac{1}{3}$

So, $𝑥+\frac{1}{𝑥}=\frac{1}{2}$ or $𝑥+\frac{1}{𝑥}=\frac{1}{3}$.

Now solve for $𝑥$ in each case:

If $𝑥+\frac{1}{𝑥}=\frac{1}{2}$, multiply both sides by $𝑥$: ${𝑥}^2+1=\frac{1}{2}𝑥$. Rearranged: $2{𝑥}^2-𝑥+2=0$

If $𝑥+\frac{1}{𝑥}=\frac{1}{3}$, multiply both sides by $𝑥$: ${𝑥}^2+1=\frac{1}{3}𝑥$. Rearranged: $3{𝑥}^2-𝑥+3=0$

For both equations, the discriminant is less than 0. This means there are no real solutions for $𝑥$.

Therefore, all the solutions for $𝑥$ are complex numbers.

We have, $E,E,L,R,T,T$

Words starting with $E,L,R$, where first letter is T

(i) Starting with $E:\frac{5!}{2!}=60$

(ii) Starting with $L:\frac{5!}{2!2!}=30$

(iii) Starting with $R:\frac{5!}{2!2!}=30$

Total $=60+30+30=120$

Fixed TE next is T

Letters before T is E, L, R

Each gives $3!=6$

$\Rightarrow 3\times 6=18$

Now, fixed TETL, next is E one word before TETLE

$𝑘$ is remaining letter which is before 2 letters.

Final letter is TETLER =1

Rank $=120+18+2+1=141$.

Given digits $0,1,2,3,5,7$

To find rank of 70513

Total number start with $1=5!=120$

Number start with $2=5!=120$

Number start with $3=5!=120$

Number start with $5=5!=120$

Number start with $701=6$

Number start with $702=6$

Number start with $703=6$

$\begin{array}{cc}& 70512=1\\ & 70513=1\end{array}$

Rank of 70513

$\begin{array}{cc}& =4\times 120+3\times 6+2\\ & =500\end{array}$

Number of divisor of 7!.

$\begin{array}{cc}7!& =7\times 6\times 5\times 4\times 3\times 2\times 1\\ & =2^4\times 3^2\times 5^1\times 7^1\end{array}$

Number of divisor

$\begin{array}{cc}& =(4+1)(2+1)(1+1)(1+1)\\ & =60\end{array}$

$9^{11}+{11}^9=(10-1{)}^{11}+(10+1{)}^9$

$\begin{array}{cc}={}^{11}{𝐶}_0{10}^{11}& +{}^{11}{𝐶}_1{10}^9+\dots {}^{11}{𝐶}_{11}\\ & +{}^9{𝐶}_0{10}^{10}+{}^9{𝐶}_1{10}^8+\dots +{}^9{𝐶}_9\end{array}$

Sum of last two terms of both expansion as other term have at least ${10}^2$ as a factor

$(110-1)+90+1=200$

$\therefore 9^{11}+{11}^9=$ Multiple of

$100+200=\text{ Multiple of }100$

Since $200=2\times {10}^2$

So, $9^{11}+{11}^9$ is divisible by ${10}^2$

$\therefore 𝑘=2$

Given expansion ${({𝑥}^{\frac{1}{2}}+{𝑦}^{\frac{1}{𝑘}})}^{10}$

General term $={}^{10}{𝐶}_{𝑟}{𝑥}^{\frac{10-𝑟}{2}}{𝑦}^{\frac{𝑟}{𝑘}}$

Given expansion have only 9 irrational term

$\therefore$ Expansion has only two rational term Two rational term is possible

LCM of 2 and $𝑘$ is 10

$\therefore 𝑘=5$

$\begin{array}{cc}& \frac{𝑥+1}{(𝑥-1{)}^2({𝑥}^2+1)}=\frac{𝐴}{𝑥-1}+\frac{𝐵}{(𝑥-1{)}^2}+\frac{𝐶𝑥+𝐷}{{𝑥}^2+1}\\ & \begin{array}{c}𝑥+1=𝐴(𝑥-1)({𝑥}^2+1)+𝐵({𝑥}^2+1)+(𝐶𝑥+𝐷)(𝑥-1{)}^2...(𝑖)\end{array}\end{array}$

Put $𝑥=1$

$2=2𝐵\Rightarrow 𝐵=1$

In Eq. (i) put $𝑥=0$

$\begin{array}{cc}& 1=-𝐴+𝐵+𝐷...\left(𝑖𝑖\right)\\ \Rightarrow & 𝐴=𝐷...\left(𝑖𝑖\right)\end{array}$

In Eq. (i) put $𝑥=-1$

$\begin{array}{cc}0& =-4𝐴+2-4𝐶+4𝐷\\ 4𝐶& =2\Rightarrow 𝐶=\frac{1}{2}...\left(𝑖𝑖𝑖\right)\end{array}$

In Eq. (i) put $𝑥=2$

$3=5𝐴+5+1+𝐷\Rightarrow 6𝐴=-3$

$\begin{array}{c}𝐴=-\frac{1}{2}=𝐷\\ \sqrt{\surd 3{𝐴}^2+4{𝐷}^2+5{𝐶}^2+{𝐵}^2}\end{array}$

$\begin{array}{cc}& =\sqrt{\surd \frac{3}{4}+\frac{4}{4}+\frac{5}{4}+1}\\ & =\sqrt{\surd \frac{16}{4}}=2\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}𝐴+𝐵=& \frac{𝜋}{4}\\ \Rightarrow \frac{\cos 𝐵-\sin 𝐵}{\cos 𝐵+\sin 𝐵}& =\frac{\sin (\frac{𝜋}{4}+𝐴)-\sin (\frac{𝜋}{4}-𝐴)}{\sin (\frac{𝜋}{4}+𝐴)+\sin (\frac{𝜋}{4}-𝐴)}\\ & =\frac{2\cos \frac{𝜋}{4}\sin 𝐴}{2\sin \frac{𝜋}{4}\cos 𝐴}=\tan 𝐴\end{array}\end{array}$

$7\cos 𝜃-\sin 𝜃=5,\tan 𝜃>0$

$7-\tan 𝜃=5\sec 𝜃$

Squaring both sides, we get

$\begin{array}{cc}& \Rightarrow 49+{\tan }^2𝜃-14\tan 𝜃=25(1+{\tan }^2𝜃)\\ & \Rightarrow 24{\tan }^2𝜃+14\tan 𝜃-24=0\\ & \Rightarrow 12{\tan }^2𝜃+7\tan 𝜃-12=0\\ & \Rightarrow 12{\tan }^2𝜃+16\tan 𝜃-9\cdot \tan 𝜃-12=0\\ & \Rightarrow (3\tan 𝜃+4)(4\tan 𝜃-3)=0\\ & \tan 𝜃=\frac{3}{4}\end{array}$

$\begin{array}{cc}& {\sin }^3{10}^{\circ }+{\sin }^3{50}^{\circ }-{\sin }^3{70}^{\circ }\\ & \begin{array}{cc}& \Rightarrow \sin 3𝑥=3\sin 𝑥-4{\sin }^3𝑥\\ & \Rightarrow {\sin }^3𝑥=\frac{1}{4}(3\sin 𝑥-\sin 3𝑥)\\ & \Rightarrow \frac{1}{4}[3\sin {10}^{\circ }-\sin {30}^{\circ }+3\sin {50}^{\circ }-\sin {150}^{\circ }-3\sin {70}^{\circ }+\sin {210}^{\circ }]\\ & \Rightarrow \frac{1}{4}[3\sin {10}^{\circ }+3\sin {50}^{\circ }-3\sin {70}^{\circ }-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}]\\ & \Rightarrow \frac{3}{4}[\sin {10}^{\circ }+\sin {50}^{\circ }-\sin {70}^{\circ }-\frac{1}{2}]\\ & \Rightarrow \frac{3}{4}[2\sin {30}^{\circ }\cos {20}^{\circ }-\sin {70}^{\circ }-\frac{1}{2}]\\ & \Rightarrow \frac{3}{4}[\cos {20}^{\circ }-\sin {70}^{\circ }-\frac{1}{2}]\\ & \Rightarrow \frac{3}{4}[\sin {70}^{\circ }-\sin {70}^{\circ }-\frac{1}{2}]=-\frac{3}{8}\end{array}\end{array}$

$\sin 2𝑥+\cos 4𝑥=2𝑥\in [-𝜋,𝜋]$

$\begin{array}{cc}& \Rightarrow \sin 2𝑥+1-2{\sin }^{2}2𝑥=2\\ & \Rightarrow 2{\sin }^{2}2𝑥-\sin 2𝑥+1=0\\ & \Rightarrow \sin 2𝑥=\frac{1\pm \sqrt{\surd 1-8}}{2}\end{array}$

Not possible

Number of solution $=0$