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

$𝑓\left(𝑥\right)=\sqrt{\surd \frac{\left|𝑥\right|-2}{\left|𝑥\right|-3}}$

$\Rightarrow \frac{\left|𝑥\right|-2}{\left|𝑥\right|-3}\ge 0$

Case I $\left|𝑥\right|-2\ge 0$ and $\left|𝑥\right|-3>0$

$\begin{array}{cc}\Rightarrow & \left|𝑥\right|\ge 2\text{ and }\left|𝑥\right|>3\\ \Rightarrow & \left|𝑥\right|>3\\ \Rightarrow & 𝑥>3\text{ or }𝑥<-3\end{array}$

Case II $\left|𝑥\right|-2\le 0$ and $\left|𝑥\right|-3<0$

$\begin{array}{cc}& \left|𝑥\right|\le 2\text{ and }\left|𝑥\right|<3\\ & \Rightarrow \left|𝑥\right|\le 2\\ & -2\le 𝑥\le 2\\ & \left|𝑥\right|-3\ne 0\Rightarrow \left|𝑥\right|\ne 3\\ & \Rightarrow 𝑥\ne 3\text{ and }𝑥\ne -3\end{array}$

Combining these, we get

$(-\infty ,-3)\cup [-2,2]\cup (3,\infty )$

Finally we write

$𝑅-[-3,-2)\cup (2,3]$

$𝑓\left(𝑥\right)=𝑎{𝑥}^2+𝑏𝑥+𝑐$

substitute $𝑓\left(𝑥\right)$ into the given equation

$\begin{array}{cc}& 𝑎{𝑥}^2+𝑏𝑥+𝑐+𝑎{\left(\frac{1}{𝑥}\right)}^2+𝑏\left(\frac{1}{𝑥}\right)+𝑐\\ & =(𝑎{𝑥}^2+𝑏𝑥+𝑐)(𝑎{\left(\frac{1}{𝑥}\right)}^2+𝑏\left(\frac{1}{𝑥}\right)+𝑐)\\ & \begin{array}{c}\Rightarrow 𝑎{𝑥}^2+𝑏𝑥+𝑐+\frac{𝑎}{{𝑥}^2}+\frac{𝑏}{𝑥}+𝑐\\ =(𝑎{𝑥}^2+𝑏𝑥+𝑐)(\frac{𝑎}{{𝑥}^2}+\frac{𝑏}{𝑥}+𝑐)\end{array}\\ & \begin{array}{c}\Rightarrow 𝑎{𝑥}^4+𝑏{𝑥}^3+𝑐{𝑥}^2+𝑎+𝑏𝑥+𝑐{𝑥}^2\\ =(𝑎{𝑥}^2+𝑏𝑥+𝑐)(𝑎+𝑏𝑥+𝑐{𝑥}^2)\end{array}\\ & \Rightarrow 𝑎{𝑥}^4+𝑏{𝑥}^3+2𝑐{𝑥}^2+𝑏𝑥+𝑎\\ & ={𝑎}^2{𝑥}^2+𝑎𝑏{𝑥}^3+𝑎𝑐{𝑥}^4+𝑎𝑏𝑥+{𝑏}^2{𝑥}^2+𝑏𝑐{𝑥}^3+𝑎𝑐+𝑏𝑐𝑥+{𝑐}^2{𝑥}^2\end{array}$

Compare coefficients : $𝑎=𝑎𝑐$,

$\begin{array}{cc}𝑏& =𝑎𝑏+𝑏𝑐\\ 2𝑐& ={𝑎}^2+{𝑏}^2+{𝑐}^2,𝑏=𝑎𝑏+𝑏𝑐,𝑎=𝑎𝑐\end{array}$

From $𝑎=𝑎𝑐$, we have $𝑐=1$ or $𝑎=0$, if $𝑎=0$ then $𝑓\left(𝑥\right)$ is not quadratic, so $𝑐=1$ from $𝑏=𝑎𝑏+𝑏𝑐$, we have $𝑏=𝑎𝑏+𝑏$, so $𝑎𝑏=0$, since $𝑎\ne 0,𝑏=0$

From $2𝑐={𝑎}^2+{𝑏}^2+{𝑐}^2$, we have

$2={𝑎}^2+0+1$

So, ${𝑎}^2=1$ and $𝑎=\pm 1$

thus, $𝑓\left(𝑥\right)={𝑥}^2+1$ or $𝑓\left(𝑥\right)=-{𝑥}^2+1$

Since, $𝑓(-1)=0$, we have $𝑓\left(𝑥\right)=-{𝑥}^2+1$

$\begin{array}{cc}& \Rightarrow 𝑓\left(𝑥\right)=-{𝑥}^2+1\\ & \Rightarrow \text{ Range }=(-\infty ,1]\end{array}$

$\begin{array}{cc}& \text{ From the given, we have }\\ & \begin{array}{cc}\sum _{𝑘=1}^{𝑛}𝑘& (𝑘+1)(𝑘+2\dots (𝑘+𝑟-1)\\ & =\sum _{𝐾=1}^{𝑛}\frac{(𝑘+𝑟-1)!}{(𝑘-1)!}\\ & =𝑟!\sum _{𝑘=1}^{𝑛}{}^{𝑘+𝑟-1}{𝐶}_{𝑟}\\ & =𝑟!{}^{𝑛+𝑟}{𝐶}_{𝑟+1}\\ & =\frac{𝑛(𝑛+1)(𝑛+2)\dots (𝑛+𝑟)}{(𝑟+1)}\end{array}\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \left|𝐴\right|=1(18-5)-2(6-10)+3(1-6)\\ & =13+8-15=6\\ & |\mathrm{adj}(\mathrm{adj}𝐴\left)\right|=|𝐴{|}^4=6^4\\ & \Rightarrow (\mathrm{adj}𝐴{)}^{-1}=\frac{1}{6}𝐴\\ & \Rightarrow |\mathrm{adj}(\mathrm{adj}𝐴\left)\right|(\mathrm{adj}𝐴{)}^{-1}=𝑘𝐴\\ & \Rightarrow 6^4\cdot \frac{1}{6}𝐴=𝑘𝐴\\ & \therefore 𝑘=216\end{array}\end{array}$

$𝛼+2𝛽+3𝛾=4...\left(𝑖\right)$

$\begin{array}{cc}& 3𝛼+𝛽+𝛾=3...\left(𝑖𝑖\right)\\ & 𝛼+3𝛽+3𝛾=2...\left(𝑖𝑖𝑖\right)\end{array}$

From Eqs. (ii), $3𝛼+𝛾=3-𝛽...\left(𝑖𝑣\right)$

Substracting Eq. (iii) from Eq. (i)

$\begin{array}{cc}& 𝛼+2𝛽+3𝛾=4\\ & 𝛼+3𝛽+3𝛾=2\\ & ----\\ & -𝛽=2\Rightarrow 𝛽=-2\end{array}$

Bu Eq. (iv)

$\begin{array}{cc}3𝛼+𝛾& =3-(-2)=5\\ \Rightarrow 3𝛼+𝛾& =1-2𝛽\end{array}$

$\begin{array}{cc}& 2𝑥+𝑦-𝑧=7,𝑥-3𝑦+2𝑧=1\text{, }\\ & \begin{array}{cc}& 𝑥+4𝑦-3𝑧=5\\ & 𝐴=⎡⎣⎢\begin{array}{ccc}2& 1& -1\\ 1& -3& 2\\ 1& 4& -3\end{array}⎤⎦⎥\\ & \Rightarrow \left|𝐴\right|=\mid \begin{array}{ccc}2& 1& -1\\ 1& -3& 2\\ 1& 4& -3\end{array}\mid \\ & =2(9-8)-1(-3-2)-1(4+3)\\ & =2+5-7=0\\ & \Rightarrow \text{ Number of solution = 0 }\end{array}\end{array}$

$4\widehat{𝐢}+\widehat{𝐣}+3\widehat{𝐤},6\widehat{𝐢}-2\widehat{𝐣}-3\widehat{𝐤},\widehat{𝐢}-\widehat{𝐣}-3\widehat{𝐤}$

$\Rightarrow$ points are $𝑃(4,1,3),𝑄(6,-2,-3)$ and $𝑅(1,-1,-3)$

$\begin{array}{cc}& \begin{array}{cc}\Rightarrow 𝑃𝑄& =\sqrt{\surd (6-4{)}^2+(-3{)}^2+(-6{)}^2}\\ & =\sqrt{\surd 4+9+36}=\sqrt{\surd 49}=7\\ \Rightarrow 𝑄𝑅& =\sqrt{\surd (1-6{)}^2+(-1+2{)}^2+(-3+3{)}^2}\\ & =\sqrt{\surd (-5{)}^2+(1{)}^2+0}=\sqrt{\surd 26}\\ \Rightarrow 𝑃𝑅& =\sqrt{\surd (1-4{)}^2+(-1-1{)}^2+(-3-3{)}^2}\\ & =\sqrt{\surd 9+4+36}=\sqrt{\surd 49}=7\\ \Rightarrow 𝑃𝑄& =𝑃𝑅=7\text{ units }\end{array}\\ & \Rightarrow △𝑃𝑄𝑅\text{ is an isosceles triangle. }\end{array}$

$\begin{array}{cc}& 𝑧=𝑥+𝑖𝑦\text{ and }{𝑥}^2+{𝑦}^2=1\\ & \begin{array}{cc}\frac{1+𝑥+𝑖𝑦}{1+𝑥-𝑖𝑦}& =\frac{\left(\right(1+𝑥)+𝑖𝑦)\left(\right(1+𝑥)+𝑖𝑦)}{(1+𝑥{)}^2-{𝑖}^2{𝑦}^2}\\ & =\frac{(1+𝑥{)}^2+{𝑖}^2{𝑦}^2+2(1+𝑥)𝑦𝑖}{1+{𝑥}^2+2𝑥+{𝑦}^2}\\ & =\frac{1+{𝑥}^2+2𝑥-{𝑦}^2+2(1+𝑥)𝑦𝑖}{2+2𝑥}\\ & =\frac{2{𝑥}^2+2𝑥+2(1+𝑥)𝑦𝑖}{2(1+𝑥)}\\ & =\frac{2𝑥(1+𝑥)+2(1+𝑥)𝑦𝑖}{2(1+𝑥)}\\ & =𝑥+𝑖𝑦=𝑧\end{array}\end{array}$

$\begin{array}{cc}& \sqrt{\surd 3}-𝑖=2(\cos (-\frac{𝜋}{6})+𝑖\sin (-\frac{𝜋}{6}))\\ & =2{𝑒}^{-𝑖𝜋/6}\\ & \Rightarrow (\sqrt{\surd 3}-𝑖{)}^5={\left(2{𝑒}^{-\frac{𝑖𝜋}{6}}\right)}^5=32{𝑒}^{-\frac{𝑖5𝜋}{6}}\end{array}$

The product of the root of ${𝑥}^6=𝑐$ is

$(-1{)}^{6-1}𝑐=-𝑐$

The product of the root is $-32{𝑒}^{-\frac{𝑖5𝜋}{6}}$

$\begin{array}{cc}& =-32(\cos \left(\frac{-5𝜋}{6}\right)+𝑖\sin (-\frac{5𝜋}{6}))\\ & =-32(-\frac{\sqrt{\surd 3}}{2}-\frac{1}{2}𝑖)\\ & =16\sqrt{\surd 3}+16𝑖\\ & =\frac{2^6}{\sqrt{\surd 3}-𝑖}\end{array}$

If $𝛼\ne 0$ and $0$ are the roots of

${𝑥}^2-5𝑘𝑥+(6{𝑘}^2-2𝑘)=0$, then find $𝛼$.

Step 1: Use the fact that 0 is a root.

If $𝑥=0$ is a root, it should satisfy the equation:

$0^2-5𝑘\left(0\right)+(6{𝑘}^2-2𝑘)=0$

$\Rightarrow 6{𝑘}^2-2𝑘=0$

$\Rightarrow 2𝑘(3𝑘-1)=0$

So $𝑘=0$ or $𝑘=\frac{1}{3}$.

But if $𝑘=0$, the equation becomes ${𝑥}^2=0$, giving a double root 0 — contradicting that one root is nonzero ($𝛼\ne 0$).

Hence,

$𝑘=\frac{1}{3}.$

Step 2: Substitute $𝑘=\frac{1}{3}$ into the equation.

${𝑥}^2-5\left(\frac{1}{3}\right)𝑥+(6{\left(\frac{1}{3}\right)}^2-2\left(\frac{1}{3}\right))=0$

$\Rightarrow {𝑥}^2-\frac{5}{3}𝑥+(\frac{6}{9}-\frac{2}{3})=0$

$\Rightarrow {𝑥}^2-\frac{5}{3}𝑥+(\frac{2}{3}-\frac{2}{3})=0$

Wait — compute the constant term carefully:

$\frac{6}{9}-\frac{2}{3}=\frac{2}{3}-\frac{2}{3}=0.$

So the equation becomes:

${𝑥}^2-\frac{5}{3}𝑥=0$

$𝑥(𝑥-\frac{5}{3})=0$

Step 3: Identify the roots.

Roots are $𝑥=0$ and $𝑥=\frac{5}{3}$.

Given that one root is 0 and another is $𝛼\ne 0$,

$\overline{)𝛼=\frac{5}{3}}.$

✅ Final Answer:

$\overline{)𝛼=\frac{5}{3}}$

Option C.

$\begin{array}{cc}& \frac{8{𝑥}^2-14𝑥-9}{3{𝑥}^2-7𝑥-6}-2>0\\ & \Rightarrow \frac{8{𝑥}^2-14𝑥-9-6{𝑥}^2+14𝑥+12}{3{𝑥}^2-7𝑥-6}>0\\ & \Rightarrow \frac{2{𝑥}^2+3}{3{𝑥}^2-9𝑥+2𝑥-6}>0\\ & \Rightarrow \frac{2{𝑥}^2+3}{3𝑥(𝑥-3)+2(𝑥-3)}>0\\ & \Rightarrow \frac{2{𝑥}^2+3}{(𝑥-3)(3𝑥+2)}>0\end{array}$

$𝑥\in (-\infty ,\frac{-2}{3})\cup (3,\infty )$

$𝑥({𝑥}^3-6{𝑥}^2+11𝑥-6)=0$

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

The least roots is 0 .

Let the roots of ${𝑥}^3+𝛼{𝑥}^2+𝛽𝑥+6=0$ be ${𝑟}_1,{𝑟}_2,{𝑟}_3$

The roots of ${𝑥}^3+𝛼{𝑥}^2+𝛽𝑥+6=0$

increased by 1 are ${𝑟}_1+1,{𝑟}_2+1,{𝑟}_3+1$

One of these roots is 0 , so ${𝑟}_{𝑖}+1=0$. for some ${𝑖}^0$.

$\Rightarrow {𝑟}_{𝑖}=-1$

Since ${𝑟}_{𝑖}=-1$ is root of

$\begin{array}{cc}& {𝑥}^3+𝛼{𝑥}^2+𝛽𝑥+6=0\\ & \Rightarrow (-1{)}^3+𝛼(-1{)}^2+𝛽(-1)+6=0\\ & \Rightarrow -1+𝛼-𝛽+6=0\\ & \Rightarrow 𝛼-𝛽+5=0\end{array}$

$\begin{array}{cc}& 𝛼+𝛽+𝛾=𝑎\\ & 𝛼𝛽+𝛽𝛾+𝛾𝛼=𝑎\\ & \begin{array}{c}𝛼𝛽𝛾=1\\ \Rightarrow {𝛼}^2+{𝛽}^2+{𝛾}^2=(𝛼+𝛽+𝛾{)}^2-2(𝛼𝛽+𝛽𝛾+𝛾𝛼)\end{array}\\ & \begin{array}{cc}& \Rightarrow {𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2=(𝛼𝛽+𝛽𝛾+𝛾𝛼{)}^2-2𝛼𝛽𝛾(𝛼+𝛽+𝛾)\\ & ={𝑎}^2-2𝑎\\ & \therefore {𝛼}^2{𝛽}^2{𝛾}^2=(𝛼𝛽𝛾{)}^2=1^2=1\end{array}\end{array}$

Now, equate the coefficients of the original and new equations.

$\begin{array}{c}{𝑎}^2-2𝑎=𝑎\\ {𝑎}^2-2𝑎=𝑎\\ 1=1\end{array}$

Solve for $𝒂$

${𝑎}^2-2𝑎=𝑎$

$\begin{array}{cc}& \Rightarrow {𝑎}^2-3𝑎=0\\ & \Rightarrow 𝑎(𝑎-3)=0\\ & \because 𝑎\ne 0\therefore 𝑎=3\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& {}^{27}{𝑃}_{𝑟+7}=7722{}^{25}{𝑃}_{𝑟+4}\\ & \Rightarrow \frac{27!}{20-𝑟!}=7722\cdot \frac{25!}{21-𝑟!}\\ & \Rightarrow \frac{27\cdot 26\cdot 25!}{20-𝑟!}=7722\cdot \frac{25!}{(21-𝑟)20-𝑟!}\\ & \Rightarrow 27\cdot 26=7722\times \frac{1}{(21-𝑟)}\\ & \Rightarrow \text{ By option put }𝑟=10\\ & \Rightarrow 702=702\end{array}\end{array}$

Let number of side $=𝑛$

The number of diagonals of a regular

$\begin{array}{cc}\Rightarrow & \frac{𝑛(𝑛-3)}{2}=35\\ \Rightarrow & {𝑛}^2-3𝑛=70\\ \Rightarrow & {𝑛}^2-3𝑛-70=0\\ \Rightarrow & {𝑛}^2-10𝑛+7𝑛-70=0\\ \Rightarrow & 𝑛(𝑛-10)+7(𝑛-10)=0\\ \Rightarrow & (𝑛-10)(𝑛+7)=0\\ \Rightarrow & 𝑛=10,𝑛=-7\left(\text{ neglected }\right)\\ \therefore & 𝑛=10\end{array}$

ASSIGNMENT

Total $=10$

2 S and 2 N

Different $=8$

A, I, G, M, E, T

Case I All are different

$={}^8{𝐶}_4\times 4!=1680$

Case II Two are same and two are different

$\begin{array}{cc}& ={}^2{𝐶}_1\times {}^7{𝐶}_2\times \frac{4!}{2!}=756\\ & =2!\times \frac{7!}{2!\cdot 5!}\times \frac{4!}{2!}\\ & =\frac{7\times 6\times 5!}{2!\times 5!}\times 4\times 3\times 2!=504\end{array}$

Case III Two are same and two are same

$={}^2{𝐶}_2\times \frac{4!}{2!\times 2!}=6$

Total $=1680+504+6=2190$

General term for ${(𝑥+{𝑦}^2)}^{13}$

$={}^{13}{𝐶}_{𝑛}{𝑥}^{13-𝑛}{𝑦}^{2𝑛}$

The general term for ${({𝑥}^2+𝑦)}^{14}$

$={}^{14}{𝐶}_{{𝑟}_2}{𝑥}^{2{𝑟}_2}{𝑦}^{14-{𝑟}_2}$

Equate the power of $𝑥$ and $𝑦$

$\begin{array}{cc}& 13-{𝑟}_1=2{𝑟}_2\\ & 2{𝑟}_1=14-{𝑟}_2\end{array}$

Solving these equation, we get

${𝑟}_1=5,{𝑟}_2=4$

The common term is ${𝑥}^{13-5}{𝑦}^{2\left(9\right)}={𝑥}^8{𝑦}^{10}$ The common term is ${𝑥}^8{𝑦}^{10}$

$\Rightarrow 𝑟=8,𝑠=10$

$𝛼=1$, since there is only one common term

$\begin{array}{cc}& \Rightarrow 𝛼(𝑟+𝑠)=1(8+10)=18\\ & \therefore \text{ Sum }𝛼\sum _{𝑟,𝑠}(𝑟+𝑠)=18\end{array}$

$\frac{1+4𝑥-3{𝑥}^2}{(1+3𝑥{)}^3}=(1+4𝑥-3{𝑥}^2)(1+3𝑥{)}^{-3}$

$\begin{array}{c}\Rightarrow (1+3𝑥{)}^{-3}=1-3(3𝑥)+\frac{(-3)(-4)(3𝑥{)}^2}{2!}\\ +\frac{(-3)(-4)(-5)(3𝑥{)}^3}{3!}+\dots \\ \Rightarrow (1+3𝑥{)}^{-3}=1-9𝑥+54{𝑥}^2-270{𝑥}^3+\dots \\ \begin{array}{cc}(1+4𝑥& -3{𝑥}^2)(1-9𝑥+54{𝑥}^2-270{𝑥}^3+\dots )\\ =& 1(-270{𝑥}^3)+4𝑥\left(5{𝑥}^2\right)-3{𝑥}^2(-9𝑥)\\ =& -270{𝑥}^3+216{𝑥}^3+27{𝑥}^3\\ =& -270+216+27=-27\end{array}\end{array}$

$\Rightarrow$ The coefficient of ${𝑥}^3=-27$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \frac{𝑎𝑥+5}{({𝑥}^2+𝑏)(𝑥+3)}=\frac{𝑥+21}{12({𝑥}^2+𝑏)}+\frac{𝑐}{12(𝑥+3)}\\ & \Rightarrow 12(𝑎𝑥+5)=(𝑥+21)(𝑥+3)+𝑐({𝑥}^2+𝑏)\\ & \Rightarrow 12𝑎𝑥+60={𝑥}^2+3𝑥+21𝑥+63+𝑐{𝑥}^2+𝑏𝑐\\ & \Rightarrow 12𝑎𝑥+60=(1+𝑐){𝑥}^2+24𝑥+63+𝑏𝑐\end{array}\end{array}$

$\begin{array}{cc}& \text{ Comparing coefficient }\\ & \begin{array}{cccccc}\Rightarrow 1+𝑐& =0& 12𝑎& =24& 63+𝑏𝑐& =60\\ 𝑐& =-1& 𝑎& =2& 63-𝑏& =60\end{array}\end{array}$

$\begin{array}{c}\Rightarrow 𝑏=3\\ {𝑏}^2=9={𝑎}^3-𝑐\\ \therefore {𝑏}^2={𝑎}^3-𝑐\end{array}$

$\begin{array}{cc}& \sin 𝛽=\frac{5}{6}\sin 𝛼\\ & \begin{array}{cc}& \Rightarrow \cos 𝛽=\frac{5\sqrt{\surd 5}}{9}\cos 𝛼\\ & \Rightarrow {\sin }^2𝛽+{\cos }^2𝛽=1\\ & \Rightarrow {(\frac{5}{6}\sin 𝛼)}^2+{(\frac{5\sqrt{\surd 5}}{9}\cos 𝛼)}^2=1\\ & \Rightarrow \frac{25}{36}{\sin }^2𝛼+\frac{125}{81}{\cos }^2𝛼=1\\ & \Rightarrow \frac{25}{36}{\sin }^2𝛼+\frac{125}{81}-\frac{125}{81}{\sin }^2𝛼=1\\ & \Rightarrow \left(\frac{225-500}{324}\right){\sin }^2𝛼=\frac{81-125}{81}\\ & \Rightarrow {\sin }^2𝛼=\frac{-44}{81}\times \frac{324}{-275}\\ & \Rightarrow {\sin }^2𝛼=\frac{4}{9}\times \frac{11}{9}\times \frac{4}{25}\times \frac{81}{11}\\ & \Rightarrow {\sin }^2𝛼=\frac{16}{25}\Rightarrow \sin 𝛼=\frac{4}{5}\end{array}\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& 2\sin 𝑥-\cos 2𝑥=1\\ \Rightarrow & 2\sin 𝑥-(1-2{\sin }^2𝑥)=1\\ \Rightarrow & 2\sin 𝑥-1+2{\sin }^2𝑥=1\\ \Rightarrow & 2{\sin }^2𝑥+2\sin 𝑥-2=0\\ \Rightarrow & {\sin }^2𝑥+\sin 𝑥-1=0\\ \Rightarrow & \sin 𝑥=\frac{-1\pm \sqrt{\surd 1+4}}{2\times 1}\\ \Rightarrow & \sin 𝑥=\frac{-1\pm \sqrt{\surd 5}}{2}\\ \Rightarrow & \sin 𝑥=\frac{\sqrt{\surd 5}-1}{2}\\ \Rightarrow & 3-2{\sin }^2𝑥=3-2\frac{(\sqrt{\surd 5}-1{)}^2}{4}\\ & =\frac{12-2(5+1-2\sqrt{\surd 5})}{4}\\ & =\frac{12-12+4\sqrt{\surd 5}}{4}=\sqrt{\surd 5}\end{array}\end{array}$

$\begin{array}{cc}& {\left(\frac{\sin 3𝜃}{\sin 𝜃}\right)}^2-{\left(\frac{\cos 3𝜃}{\cos 𝜃}\right)}^2=𝑎\cos 𝑏𝜃\\ & \Rightarrow \left(\frac{3\sin 𝜃-4{\sin }^3𝜃}{\sin 𝜃}\right)-\left(\frac{4{\cos }^3𝜃-3\cos 𝜃}{\cos 𝜃}\right)=𝑎\cos 𝑏𝜃\\ & \Rightarrow {(3-4{\sin }^2𝜃)}^2-{(4{\cos }^2𝜃-3)}^2=𝑎\cos 𝑏𝜃\\ & \Rightarrow 9+16{\sin }^4𝜃-24{\sin }^2𝜃-16{\cos }^4𝜃-9+24{\cos }^2𝜃=𝑎\cos 𝑏𝜃\\ & \Rightarrow 24({\cos }^2𝜃-{\sin }^2𝜃)-16({\cos }^4𝜃-{\sin }^4𝜃)=𝑎{\cos }^{𝑏}𝜃\\ & \Rightarrow 8({\cos }^2𝜃-{\sin }^2𝜃)(3-2({\cos }^2𝜃+{\sin }^2𝜃))=𝑎\cos 𝑏𝜃,\\ & \Rightarrow 8\cos 2𝜃(3-2)=𝑎\cos 𝑏𝜃\\ & \Rightarrow 8\cos 2𝜃=𝑎\cos 𝑏𝜃\Rightarrow 𝑎=8,𝑏=2\\ & \therefore 𝑎:𝑏=8:2=4:1\end{array}$