TG EAPCET 2025 3rd May Morning Shift Question Paper with Solutions

$𝑓\left(𝑥\right)=\frac{{𝑥}^2+𝑥+𝑎}{{𝑥}^2-𝑥+𝑎}$

$\because 𝑓\left(𝑥\right)$ is surjection, then range $=$ Codomain

Range $=𝑅$

Now for range, put $𝑓\left(𝑥\right)=𝑦$.

$\begin{array}{c}𝑦=\frac{{𝑥}^2+𝑥+𝑎}{{𝑥}^2-𝑥+𝑎}\\ {𝑥}^2(𝑦-1)-𝑥(𝑦+1)+𝑎𝑦-𝑎=0\end{array}$

Now $𝑥\in 𝑅\Rightarrow 𝐷\ge 0$

$\begin{array}{cc}& (𝑦+1{)}^2-4(𝑦-1\left)\right(𝑎𝑦-𝑎)\ge 0\\ & \Rightarrow (𝑦+1{)}^2-4𝑎(𝑦-1{)}^2\ge 0,\forall 𝑦\in 𝑅\\ & \Rightarrow {𝑦}^2(1-4𝑎)+𝑦(2+8𝑎)+1-4𝑎\ge 0,\forall 𝑦\in 𝑅\end{array}$

$\begin{array}{cc}& \therefore 1-4𝑎\ge 0\text{ and }𝐷\le 0\\ & 𝑎\le \frac{1}{4}\text{ and }(2+8𝑎{)}^2-4(1-4𝑎{)}^2\le 0\\ & \Rightarrow (1+4𝑎{)}^2-(1-4𝑎{)}^2\le 0\\ & \Rightarrow (1+4𝑎+1-4𝑎)(1+4𝑎-1+4𝑎)\le 0\\ & \Rightarrow 2\left(8𝑎\right)\le 0\\ & \Rightarrow 𝑎\le 0\\ & \therefore 𝑎\in (-\infty ,0).\end{array}$

Given, $𝑓\left(𝑥\right)=\frac{1}{\sqrt{\surd {\log }_{1/3}\left(\frac{𝑥-1}{2-𝑥}\right)}}$

∴ Domain

$\begin{array}{cc}& {\log }_{1/3}\frac{𝑥-1}{2-𝑥}>0\text{ and }\frac{𝑥-1}{2-𝑥}>0\\ & \Rightarrow \frac{𝑥-1}{2-𝑥}<{\left(\frac{1}{3}\right)}^0\Rightarrow \frac{𝑥-1}{2-𝑥}<1\\ & \Rightarrow \frac{𝑥-1}{2-𝑥}-1<0\\ & \Rightarrow \frac{𝑥-1-2+𝑥}{2-𝑥}<0\Rightarrow \frac{2𝑥-3}{2-𝑥}<0\end{array}$

Now,

Now, $𝑥\in (1,\frac{3}{2})$

$\therefore 𝑎=1,𝑏=\frac{3}{2}$

Now, $2𝑏=3=𝑎+2$

$\begin{array}{cc}& \text{ Given, }\\ & \begin{array}{cc}& \frac{1}{2\cdot 7}+\frac{1}{7\cdot 12}+\frac{1}{12\cdot 17}+\dots +10\text{ terms. }\\ & =\frac{1}{5}[\frac{5}{2\cdot 7}+\frac{5}{7\cdot 12}+\frac{5}{12\cdot 17}+\dots +10\text{ terms }]\end{array}\end{array}$

$\begin{array}{cc}& =\frac{1}{5}[\frac{7-2}{2\cdot 7}+\frac{12-7}{7\cdot 12}+\frac{17-12}{12\cdot 17}+\dots +10\text{ terms] }\\ & =\frac{1}{5}\left[\begin{array}{c}\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+(\frac{1}{12}-\frac{1}{17})+\dots \\ +(\frac{1}{2+9\times 5}-\frac{1}{7+9\times 5})\end{array}\right]\\ & =\frac{1}{5}(\frac{1}{2}-\frac{1}{52})=\frac{1}{5}\left(\frac{25}{52}\right)=\frac{5}{52}\end{array}$

Given, $𝑥+𝜆𝑦-2𝑧=1$

$𝑥-𝑦+𝜆𝑧=2$

And $𝑥-2𝑦+3𝑧=3$

∵ Given, system is inconsistent

$\begin{array}{cc}& 𝐷=0\Rightarrow \mid \begin{array}{ccc}1& 𝜆& -2\\ 1& -1& 𝜆\\ 1& -2& 3\end{array}\mid =0\\ & \Rightarrow 1(-3+2𝜆)-𝜆(3-𝜆)-2(-2+1)=0\\ & \Rightarrow 2𝜆-3-3𝜆+{𝜆}^2+2=0\end{array}$

$\therefore {𝜆}_1+{𝜆}_2=1$

Given, $(\sin 𝜃)𝑥+𝑦-2𝑧=0$

$2𝑥-𝑦+(\cos 𝜃)𝑧=0$

And $-3𝑥+(\sec 𝜃)𝑦+3𝑧=0$

∵ system has non-trivial solution.

$\begin{array}{cc}& \Delta =0\\ & \mid \begin{array}{ccc}\sin 𝜃& 1& -2\\ 2& -1& \cos 𝜃\\ -3& \sec 𝜃& 3\end{array}\mid =0\\ & \sin 𝜃(-3-1)-1(6+3\cos 𝜃)-2(2\sec 𝜃-3)=0\\ & \Rightarrow -4\sin 𝜃-6-3\cos 𝜃-4\sec 𝜃+6=0\\ & \Rightarrow 4\sec 𝜃+3\cos 𝜃+4\sin 𝜃=0\\ & \Rightarrow 4+3{\cos }^2𝜃+4\sin 𝜃\cos 𝜃=0\\ & \Rightarrow 4+3\left(\frac{1+\cos 2𝜃}{2}\right)+2\sin 2𝜃=0\\ & \Rightarrow 8+3+3\cos 2𝜃+4\sin 2𝜃=0\\ & \Rightarrow 11+3\cos 2𝜃+4\sin 2𝜃=0\end{array}$

This equation has no solution because the maximum value of $4\sin 2𝜃+3\cos 2𝜃$ is 5 . Since $5+11\ne 0$, there is no value of $𝜃$ that satisfies the equation.

$\begin{array}{cc}& \\ & \begin{array}{cc}& 𝐴=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\\ & \because \mathrm{adj}(\mathrm{adj}𝐴)=|𝐴{|}^{𝑛-2}𝐴\\ & \mathrm{adj}(\mathrm{adj}(\mathrm{adj}𝐴\left)\right)=|\mathrm{adj}𝐴{|}^{2-2}\cdot \mathrm{adj}𝐴\\ & =\mathrm{adj}𝐴=\left|𝐴\right|{𝐴}^{-1}(\because {𝐴}^{-1}=\frac{1}{\left|𝐴\right|}\mathrm{adj}𝐴)\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }\\ & \begin{array}{cc}& \mid \begin{array}{ccc}𝑥& -3& 2\\ -1& -2& 𝑥-1\\ 1& 𝑥-2& 3\end{array}\mid =0\\ & \Rightarrow 𝑥(-6-(𝑥-1\left)\right(𝑥-2\left)\right)+3(-3-(𝑥-1\left)\right)+2(-(𝑥-2)+2)=0\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑥(-6-{𝑥}^2+3𝑥-2)+3(-3-𝑥+1)+2(-𝑥+2+2)=0\\ & \Rightarrow 𝑥(-{𝑥}^2+3𝑥-8)+3(-2-𝑥)+2(4-𝑥)=0\\ & \Rightarrow -{𝑥}^3+3{𝑥}^2-8𝑥-6-3𝑥+8-2𝑥=0\\ & \therefore \text{ Sum of roots }=\frac{-3}{-1}=3.\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}𝑧& =\sqrt{\surd 24-70𝑖}+\sqrt{\surd -24+70𝑖}\\ {𝑧}^2& =24-70𝑖+(-24+70𝑖)\\ & +2\sqrt{\surd 24-70𝑖}\sqrt{\surd -24+70𝑖}\\ & =2𝑖(24-70𝑖)=2(24𝑖+70)\\ & =4(12𝑖+35)\\ 𝑧& =\sqrt{\surd 4}\sqrt{\surd 35+12𝑖}\\ & =2(\pm {\left(\frac{\mathrm{Re}𝑧+\left|𝑧\right|}{2}\right)}^{1/2}\\ & +𝑖{\left(\frac{-\mathrm{Re}𝑧+\left|𝑧\right|}{2}\right)}^{1/2})\\ & =\pm 2({\left(\frac{35+37}{2}\right)}^{1/2}+𝑖{\left(\frac{-35+37}{2}\right)}^{1/2})\\ & =\pm 2(6+𝑖)=\pm (12\pm 2𝑖)\end{array}\end{array}$

$𝑧=\frac{1-𝑖\cos 𝜃}{1+2𝑖\sin 𝜃}$ is purely imaginary

$\begin{array}{cc}& \therefore 𝑧+\overline{𝑧}=0\\ & \Rightarrow \frac{1-𝑖\cos 𝜃}{1+2𝑖\sin 𝜃}+\frac{1+𝑖\cos 𝜃}{1-2𝑖\sin 𝜃}=0\\ & 1-2𝑖\sin 𝜃-𝑖\cos 𝜃-2\sin 𝜃\cos 𝜃+1\\ & \frac{+𝑖\cos 𝜃+2𝑖\sin 𝜃-2\sin 𝜃\cos 𝜃}{(1+2𝑖\sin 𝜃)(1-2𝑖\sin 𝜃)}=0\\ & \Rightarrow (2-4\sin 𝜃\cos 𝜃)=0\\ & \Rightarrow 2\sin 2𝜃=2\Rightarrow \sin 2𝜃=1\\ & \Rightarrow \sin 2𝜃=\sin \frac{𝜋}{2}\\ & \therefore 2𝜃=𝑛𝜋+(-1{)}^{𝑛}\frac{𝜋}{2},𝑛\in 𝑧\\ & \therefore 𝜃=\frac{𝑛𝜋}{2}+(-1{)}^{𝑛}\frac{𝜋}{4},𝑛\in 𝑧.\end{array}$

We have

$\begin{array}{c}\cos 𝛼+\cos 𝛽+\cos 𝛾=0\\ =\sin 𝛼+\sin 𝛽+\sin 𝛾\end{array}$

Let $𝑎=\cos 𝛼+𝑖\sin 𝛼$

$\begin{array}{c}𝑏=\cos 𝛽+𝑖\sin 𝛽\\ 𝑐=\cos 𝛾+𝑖\sin 𝛾\\ \therefore 𝑎+𝑏+𝑐=(\cos 𝛼+\cos 𝛽+\cos 𝛾)\\ +𝑖(\sin 𝛼+\sin 𝛽+\sin 𝛾)\end{array}$

$𝑎+𝑏+𝑐=0$

and ${𝑎}^{-1}+{𝑏}^{-1}+{𝑐}^{-1}=0$

$\begin{array}{cc}& \therefore 𝑎𝑏+𝑏𝑐+𝑐𝑎=0\\ & \Rightarrow \cos (𝛼+𝛽)+\cos (𝛽+𝛾)\\ & +\cos (𝛾+𝛼)=0\end{array}$

Now ${𝑎}^2+{𝑏}^2+{𝑐}^2=(𝑎+𝑏+𝑐{)}^2$

$-2(𝑎𝑏+𝑏𝑐+𝑐𝑎)$

$\therefore \sin 2𝛼+\sin 2𝛽+\sin 2𝛾=0$

$\begin{array}{cc}& {𝑥}^2-𝑥+1=0\to 𝛼\\ & \therefore 𝛼=-𝜔\Rightarrow {𝛼}^3=-{𝜔}^3=-1\end{array}$

Now, ${(𝛼+\frac{1}{𝛼})}^3+{({𝛼}^2+\frac{1}{{𝛼}^2})}^3+\dots$ to 12 terms

$\begin{array}{c}={\left(\frac{{𝛼}^2+1}{𝛼}\right)}^3+{\left(\frac{{𝛼}^4+1}{{𝛼}^2}\right)}^3+(-1-1{)}^3+\dots +12\text{ terms }\\ =\left[\right(1{)}^3+(1{)}^3+(-2{)}^3+(-1{)}^3+(-1{)}^3+\left(2{)}^3\right]2=0\end{array}$

$\begin{array}{cc}& {𝑥}^2+𝑝𝑥+2=0\\ & {𝑥}^2+𝑥+2𝑝=0\end{array}$

Let ' $𝛼$ ' be the common root

$\begin{array}{cc}\therefore & 𝛼=2\\ & 4+2𝑝+2=0\\ & 2𝑝=-6\\ & 𝑝=-3\end{array}$

∴ Sum of roots of ${𝑥}^2+2𝑝𝑥+8=0$ (i.e., ${𝑥}^2-6𝑥+8=0$ ) is 6 .

${𝑥}^2-5𝑎𝑥+6𝑎=0$

∵ Both roots are greater than 2 .

$\begin{array}{cc}& 𝐷\ge 0,\frac{-𝑏}{2𝑎}>1,𝑓\left(0\right)>0\\ & (5𝑎{)}^2-4(1\left)\right(6𝑎)\ge 0,\frac{5𝑎}{2}>1,6𝑎>0\\ & \Rightarrow 𝑎(25𝑎-24)\ge 0,𝑎>\frac{2}{5},𝑎>0\\ & \therefore 𝑎\in [\frac{24}{25},\infty )\end{array}$

$\begin{array}{cc}& \Rightarrow {𝑥}^3(𝑥-1)-3{𝑥}^2(𝑥-1)+2(𝑥-1)=0\\ & \Rightarrow (𝑥-1)({𝑥}^3-3{𝑥}^2+2)=0\end{array}$

$\begin{array}{cc}& \Rightarrow (𝑥-1)\left[{𝑥}^2\right(𝑥-1)-2𝑥(𝑥-1)-2(𝑥-1\left)\right]=0\\ & \Rightarrow (𝑥-1)(𝑥-1)({𝑥}^2-2𝑥-2)=0\\ & \therefore 𝛼=1,𝛾+𝛿=2\\ & 𝛽=1,𝛾𝛿=-2\\ & \therefore 𝛼+2𝛽+{𝛾}^2+{𝛿}^2\end{array}$

$\begin{array}{cc}& =1+2+(𝛾+𝛿{)}^2-2𝛾𝛿\\ & =3+4+4=11\end{array}$

We have, ${𝑥}^4+4{𝑥}^3-16𝑥-16=0$

$\begin{array}{cc}& \Rightarrow {𝑥}^3(𝑥-2)+6{𝑥}^2(𝑥-2)+12𝑥(𝑥-2)+8(𝑥-2)=0\\ & \Rightarrow (𝑥-2)({𝑥}^3+6{𝑥}^2+12𝑥+8)=0\\ & \Rightarrow (𝑥-2)\left[{𝑥}^2\right(𝑥+2)+4𝑥(𝑥+2)+4(𝑥+2\left)\right]=0\\ & \Rightarrow (𝑥-2)(𝑥+2)({𝑥}^2+4𝑥+4)=0\\ & \Rightarrow (𝑥-2)(𝑥+2{)}^3=0\end{array}$

∴ Multiply roots is -2 .

(a) ${𝑥}^2+2𝑥-3=0\Rightarrow (𝑥+3)(𝑥-1)=0$

(b) ${𝑥}^2-3𝑥+2=0\Rightarrow (𝑥-2)(𝑥-1)=0$

(c) ${𝑥}^2+𝑥-2=0\Rightarrow (𝑥+2)(𝑥-1)=0$

(d) ${𝑥}^2-4𝑥+3=0\Rightarrow (𝑥-3)(𝑥-1)=0$

∵ Total number of ways in which train has stopped at exactly 5 stations $={}^{15}{𝐶}_5$

And total number of ways in which train has not stopped in consecutive stations

${𝑥}_1+{𝑥}_2+{𝑥}_3+{𝑥}_4+{𝑥}_5+𝐿=15-4=11$

${𝑥}_{𝑖}=$ number of station left before ${𝑖}^{\text{th }}$ station

${𝑥}_1\ge 0{𝑥}_2,{𝑥}_3,{𝑥}_4,{𝑥}_5\ge 1\text{, and }𝐿=\text{ station }$

left after 4 stations

$\begin{array}{cc}{𝑥}_1+({𝑥}_2-1)+& ({𝑥}_3-1)+({𝑥}_4-1)+({𝑥}_5-1)+𝐿\\ =& 11-4=7\end{array}$

$\begin{array}{cc}\therefore \text{ Total number of solution }& ={}^{7+5-1}{𝐶}_{6-1}\\ & ={}^{11}{𝐶}_5\end{array}$

$\text{ ∴ Required ways }={}^{15}{𝐶}_5-{}^{11}{𝐶}_5$

Number of all possible ways of distributing eight identical apples among 3 persons is number of non-negative integral solution of equation is

$\begin{array}{cc}& 𝑥+𝑦+𝑧=8\\ & \therefore {}^{8+3-1}{𝐶}_{3-1}\\ & ={}^{10}{𝐶}_2\\ & =\frac{10\times 9}{2\times 1}=45\end{array}$

The number of words in CAB and NET both appears $=$ CAB NET $𝐼=3$ !

∴ Required number of ways

$\begin{array}{cc}& =7!-(5!+5!-3!)\\ & =5040-(120+120-6)\\ & =5040-234\\ & =4806\end{array}$

For numerically greatest term in the expansion of $(2𝑥-3𝑦{)}^{𝑛}$

$\begin{array}{cc}𝑟& =\frac{𝑛+1}{1+\mid \frac{2𝑥}{3𝑦}\mid }=\frac{13+1}{1+\mid \frac{2\times \frac{7}{2}}{3\times \frac{3}{7}}\mid }=\frac{14}{1+\frac{49}{9}}\\ & =\frac{14\times 9}{9+49}=\frac{126}{58}\\ \therefore \left[𝑟\right]& =2\end{array}$

$\therefore {𝑇}_3$ will be numerically greatest term

$\begin{array}{cc}\therefore {𝑇}_3& ={}^{13}{𝐶}_2\left(2𝑥{)}^{11}\right(-3𝑦{)}^2\\ & =\frac{13\times 12}{2}\times {(2\times \frac{7}{2})}^{11}{(-3\times \frac{3}{7})}^2\\ & =13\times 6\times 7^{11}\times \frac{3^4}{7^2}\\ & ={263}^5{7}^9\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \sum _{𝑟=1}^8{𝑟}^3\cdot \frac{{𝐶}_{𝑟}}{{𝐶}_{𝑟-1}}=\sum _{𝑟=1}^8{𝑟}^3\cdot \frac{8-𝑟+1}{𝑟}\\ =& \sum _{𝑟=1}^8(9{𝑟}^2-{𝑟}^3)[\because \frac{{}^{𝑛}{𝐶}_{𝑟}}{{}^{𝑛}{𝐶}_{𝑟-1}}=\frac{𝑛-𝑟+1}{𝑟}]\\ =& 9\left(\frac{8}{6}\right(8+1\left)\right(17\left)\right)-{(\frac{8}{2}\times 9)}^2\\ =& 3\times 4\times 9\times 17-4\times 4\times 9\times 9\\ =& 4\times 9(51-36)=36\times 15=540\end{array}\end{array}$

$\begin{array}{cc}& \frac{𝑥+3}{(𝑥+1)({𝑥}^2+2)}=\frac{𝑎}{𝑥+1}+\frac{𝑏𝑥+𝑐}{{𝑥}^2+2}\\ & \Rightarrow 𝑥+3=𝑎({𝑥}^2+2)+(𝑏𝑥+𝑐)(𝑥+1)\\ & \text{ put }𝑥=-1\\ & 2=𝑎\left(3\right)\Rightarrow 𝑎=\frac{2}{3}\end{array}$

put $𝑥=0$

$\begin{array}{cc}& 3=2𝑎+𝑐\\ & 𝑐=3-\frac{4}{3}=\frac{5}{3}\end{array}$

put $𝑥=1$

$\begin{array}{c}4=3𝑎+(𝑏+𝑐)2\\ 4=2+(𝑏+\frac{5}{3})2\\ 1=𝑏+\frac{5}{3}\Rightarrow 𝑏=\frac{-2}{3}\\ \therefore 𝑎-𝑏+𝑐=\frac{2}{3}+\frac{2}{3}+\frac{5}{3}=\frac{9}{3}=3\end{array}$

$3\sin 𝜃+4\cos 𝜃=3$

On squaring both sides, we get $79{\sin }^2𝜃+16{\cos }^2𝜃+24\sin 𝜃\cos 𝜃=9$

$\begin{array}{cc}& \Rightarrow 9+7{\cos }^2𝜃+24\sin 𝜃\cos 𝜃=9\\ & \Rightarrow \cos 𝜃(7\cos 𝜃+24\sin 𝜃)=0\\ & \Rightarrow \cos 𝜃\ne 0\\ & 𝜃\ne (2𝑛+1)\frac{𝜋}{2}\\ & \text{ and }7\cos 𝜃+24\sin 𝜃=0\\ & \begin{array}{cc}& \tan 𝜃=\frac{-7}{24}\\ & \therefore \sin 2𝜃=\frac{2\tan 𝜃}{1+{\tan }^2𝜃}\\ & =\frac{2\times \left(\frac{-7}{24}\right)}{1+\frac{49}{576}}=\frac{-14\times 24}{625}\\ & =\frac{-336}{625}\end{array}\end{array}$

$\frac{\cos {15}^{\circ }{\cos }^{2}22{\frac{1}{2}}^{\circ }-\sin {75}^{\circ }{\sin }^{2}52{\frac{1}{2}}^{\circ }}{{\cos }^2{15}^{\circ }-{\cos }^2{75}^{\circ }}$

$=\frac{\cos {15}^{\circ }({\cos }^{2}22{\frac{1}{2}}^{\circ }-{\sin }^{2}52{\frac{1}{2}}^{\circ })}{{\cos }^2{15}^{\circ }-{\sin }^2{15}^{\circ }}(\frac{\because \sin {75}^{\circ }=\sin ({90}^{\circ }-{15}^{\circ })}{=\cos {15}^{\circ }})$

$\begin{array}{cc}& =\frac{\cos {15}^{\circ }(\frac{1+\cos {45}^{\circ }}{2}-\frac{1-\cos {105}^{\circ }}{2})}{\cos {30}^{\circ }}\\ & =\frac{\cos {15}^{\circ }}{2}\frac{[\cos {45}^{\circ }+\cos (90+{15}^{\circ })]}{\frac{\sqrt{\surd 3}}{2}}\\ & =\frac{1}{\sqrt{\surd 3}}(\cos {15}^{\circ }\cos {45}^{\circ }-\sin {15}^{\circ }\cos {15}^{\circ })\\ & =\frac{1}{2\sqrt{\surd 3}}(2\cos {15}^{\circ }\cos {45}^{\circ }-2\cos {15}^{\circ }\sin {15}^{\circ })\\ & =\frac{1}{2\sqrt{\surd 3}}[\cos \left({60}^{\circ }\right)+\cos {30}^{\circ }-\sin {30}^{\circ }]\\ & =\frac{1}{2\sqrt{\surd 3}}\left(\frac{\sqrt{\surd 3}}{2}\right)=\frac{1}{4}\end{array}$