AP EAPCET 2024 23rd May Morning Shift Question Paper with Solutions

We have, $𝑓\left(𝑥\right)=\frac{1}{\sqrt{\surd 64-(0.5{)}^{24+𝑥-{𝑥}^2}}}$

$𝑓\left(𝑥\right)$ will be defined, if $64-(0.5{)}^{24+𝑥-{𝑥}^2}$ $>0$

$\begin{array}{cc}& =64>(0.5{)}^{24+𝑥-{𝑥}^2}\\ & =64>2^{{𝑥}^2-𝑥-24}\\ & =2^6>2^{{𝑥}^2-𝑥-24}\\ & ={𝑥}^2-𝑥-24<6\\ & ={𝑥}^2-𝑥-30<0\\ & =(𝑥+5)(𝑥-6)<0\\ & =-5<𝑥<6\end{array}$

$\because 𝑥=-4,-3,-2-1,0,1,2,3,4,5$

Then, $𝐴=\{-4,-3,-2,-1,0,1,2,3,4,5\}$

Sum of all absolute values of $𝐴$ is

$\begin{array}{cc}& =4+3+2+1+0+1+2+3+4+5\\ & =25\end{array}$

We have, $𝑓\left(𝑥\right)=\log (𝑥+\sqrt{\surd {𝑥}^2+1})$

$\begin{array}{cc}\Rightarrow & 𝑓(-𝑥)=\log (-𝑥+\sqrt{\surd {𝑥}^2+1})\\ & =\log [\sqrt{\surd {𝑥}^2+1}-𝑥\times \frac{\sqrt{\surd {𝑥}^2+1}+𝑥}{\sqrt{\surd {𝑥}^2+1}+𝑥}]\\ & =\log \left[\frac{{𝑥}^2+1-{𝑥}^2}{\sqrt{\surd {𝑥}^2+1}+𝑥}\right]\\ & =\log \left(\frac{1}{\sqrt{\surd {𝑥}^2+1}+𝑥}\right)\\ & =-\log (𝑥+\sqrt{\surd {𝑥}^2+1})\\ & =-𝑓\left(𝑥\right)\end{array}$

$\therefore 𝑓\left(𝑥\right)$ is an odd function.

We have,

$1+(3+5+7)+(9+11+13+15+17)+\dots$

Total number of terms

$\begin{array}{cc}& =1+3+5+7+\dots +𝑛\text{ terms }\\ & =\frac{𝑛}{2}[2+(𝑛-1\left)2\right]\\ & ={𝑛}^2\\ {𝑆}_{𝑛}=1& +3+5+7+9+11+13+15+17+\dots \end{array}$

upto ${𝑛}^2$ terms

$\begin{array}{cc}{𝑆}_{𝑛}& =\frac{{𝑛}^2}{2}[2+({𝑛}^2-1)2]\\ & =\frac{{𝑛}^2}{2}\times \left(2{𝑛}^2\right)={𝑛}^4\\ {𝑎}_{𝑛}& ={𝑆}_{𝑛}-{𝑆}_{𝑛-1}\\ & ={𝑛}^4-(𝑛-1{)}^4\\ & =(𝑛-(𝑛-1\left)\right)(𝑛+(𝑛-1\left)\right)({𝑛}^2+(𝑛-1{)}^2)\\ & =(2𝑛-1)[{𝑛}^2+(𝑛-1{)}^2]\end{array}$

$\begin{array}{cc}& \text{ We have, }𝐴=⎡⎣⎢\begin{array}{ccc}83& 74& 41\\ 93& 96& 31\\ 24& 15& 79\end{array}⎤⎦⎥\\ & \begin{array}{cc}{𝐴}^{𝑇}& =⎡⎣⎢\begin{array}{ccc}83& 93& 24\\ 74& 96& 15\\ 41& 31& 79\end{array}⎤⎦⎥.\\ \therefore 𝐴-{𝐴}^{𝑇}& =⎡⎣⎢\begin{array}{ccc}83& 74& 41\\ 93& 96& 31\\ 24& 15& 79\end{array}⎤⎦⎥-⎡⎣⎢\begin{array}{ccc}83& 93& 24\\ 74& 96& 15\\ 41& 31& 79\end{array}⎤⎦⎥\\ & =⎡⎣⎢\begin{array}{ccc}0& -19& 17\\ 19& 0& 16\\ -17& -16& 0\end{array}⎤⎦⎥\end{array}\\ & \begin{array}{cc}& \therefore \det (𝐴-{𝐴}^{𝑇})=\mid \begin{array}{ccc}0& -19& 17\\ 19& 0& 16\\ -17& -16& 0\end{array}\mid \\ & 0(0+16\times 16)+19(9\times 0+16\times 17)\\ & +17(-19\times 16-0)\\ & =19\times 16\times 17-17\times 19\times 16\\ & =0\end{array}\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \text{ We have, }\mid \begin{array}{ccc}𝑎& 1& 1\\ 1& 𝑏& 1\\ 1& 1& 𝑐\end{array}\mid >0\\ & =\left\{\right(𝑎𝑏𝑐-𝑎)-1(𝑐-1)+1(1-𝑏\left)\right\}>0\\ & =(𝑎𝑏𝑐-𝑎-𝑐+1+1-𝑏)>0\\ & =(𝑎𝑏𝑐-(𝑎+𝑏+𝑐)+2)>0\\ & =𝑎𝑏𝑐+2>(𝑎+𝑏+𝑐)\\ & \begin{array}{c}𝑎𝑏𝑐+2>3(𝑎𝑏𝑐{)}^{\frac{1}{3}}\end{array}\\ & [\because 𝐴𝑚>𝐺𝑚\Rightarrow \frac{𝑎+𝑏+𝑐}{3}>(𝑎+𝑏{)}^{1/3}]\\ & \text{ let }𝑥=(𝑎𝑏𝑐{)}^{1/3}\Rightarrow 𝑎𝑏𝑐={𝑥}^3\\ & {𝑥}^3+2>3𝑥\\ & \begin{array}{c}{𝑥}^3-3𝑥+2>0\\ (𝑥-1{)}^2(𝑥+2)>0\\ (𝑥+2)>0=𝑥>-2\\ 1\\ =(𝑎𝑏𝑐{)}^{\frac{1}{3}}>-2=(𝑎𝑏𝑐)>-8\end{array}\end{array}\end{array}$

The given system of equations are homogeneous and has only trival solution, then

$\mid \begin{array}{ccc}{𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\\ {𝑎}_3& {𝑏}_3& {𝑐}_3\end{array}\mid =0$

$\mathrm{Rank}\left[𝐴\right]=𝑛$, where $𝑛$ is the order of square matrix and

$𝐴=⎡⎣⎢\begin{array}{ccc}{𝑎}_1& {𝑏}_1& {𝑐}_1\\ {𝑎}_2& {𝑏}_2& {𝑐}_2\\ {𝑎}_3& {𝑏}_3& {𝑐}_3\end{array}⎤⎦⎥$

$\therefore \mathrm{Rank}\left[𝐴\right]=3$

We have, $|𝑧-1|\le 2...\left(𝑖\right)$

$\begin{array}{cc}& \text{ and }\mid {𝜔}^2𝑧-1-𝜔\mid =𝑎\\ & \begin{array}{c}{𝜔}^2𝑧+{𝜔}^2<text>\mid </text><text>=</text>𝑎[\because {𝜔}^2=-1-𝜔]\\ \mid {𝜔}^2\mid |𝑧+1|=𝑎\\ |𝑧-1+2|=𝑎\\ |𝑧-1|+2\ge 𝑎\end{array}\end{array}$

From Eq. (i), we get

$\begin{array}{cc}& |𝑧-1|+2=4\\ \Rightarrow & 0\le 𝑎\le 4\end{array}$

We have, ${𝑧}^3+𝑖{𝑧}^2+2𝑖=0$

One root of ${𝑧}^3+𝑖{𝑧}^2+2𝑖=0$ is $𝑖$.

$\begin{array}{cc}& [\because {𝑖}^3+𝑖\left({𝑖}^2\right)+2𝑖=-𝑖-𝑖+2𝑖=0]\\ & {𝑧}^3+𝑖{𝑧}^2+2𝑖=0\\ \Rightarrow & (𝑧-𝑖)({𝑧}^2+2𝑖𝑧-2)=0\\ \Rightarrow & (𝑧-𝑖)({𝑧}^2+2𝑖𝑧+{𝑖}^2+1-2)=0\\ \Rightarrow & (𝑧-𝑖)\left(\right(𝑧+𝑖{)}^2-1)=0\\ \Rightarrow & (𝑧-𝑖)(𝑧+(𝑖+1\left)\right)(𝑧+(𝑖-1\left)\right)=0\\ \Rightarrow & 𝑧=𝑖,-𝑖-1,1-𝑖\end{array}$

$\therefore$ The triangle formed by the vertices $𝐴(0,1),𝐵(-1,-1)$ and $𝐶(1,-1)$.

$𝐴𝐵=\sqrt{\surd 1+4}=\sqrt{\surd 5}\text{ units }$

$𝐵𝐶=2$ units

$𝐴𝐶=\sqrt{\surd 1+4}=\sqrt{\surd 5}$ units

$\Rightarrow △𝐴𝐵𝐶$ is an isosceles triangle.

$[\because 𝐴𝐵=𝐴𝐶]$

Since, the non-equal side is less than the equal side, so $𝐴𝐵𝐶$ is not a right-angled triangle.

In the complex plane, any point can be represented in polar form as $(𝑟,𝜃)=𝑟(\cos 𝜃+𝑖\sin 𝜃)$. Here, $𝑥=(1,𝛼)$, $𝑦=(1,𝛽)$, and $𝑧=(1,𝛾)$. Given the condition $𝑥+𝑦+𝑧=0$, we can express

$𝑥=\cos 𝛼+𝑖\sin 𝛼$

Similarly,

$𝑦=\cos 𝛽+𝑖\sin 𝛽$

$𝑧=\cos 𝛾+𝑖\sin 𝛾$

These imply:

$𝑥+𝑦+𝑧=0\Rightarrow \cos 𝛼+\cos 𝛽+\cos 𝛾=0$

$\sin 𝛼+\sin 𝛽+\sin 𝛾=0$

From the above, we isolate:

$\cos 𝛼+\cos 𝛽=-\cos 𝛾\text{...(i)}$

$\sin 𝛼+\sin 𝛽=-\sin 𝛾\text{...(ii)}$

Squaring and adding equations (i) and (ii), we get:

$\begin{array}{cc}(\cos 𝛼+\cos 𝛽{)}^2+(\sin 𝛼+\sin 𝛽{)}^2& ={\cos }^2𝛾+{\sin }^2𝛾\\ {\cos }^2𝛼+{\cos }^2𝛽+2\cos 𝛼\cos 𝛽+{\sin }^2𝛼+{\sin }^2𝛽+2\sin 𝛼\sin 𝛽& =1\end{array}$

This simplifies to:

$2\cos (𝛼-𝛽)=-1\Rightarrow \cos (𝛼-𝛽)=-\frac{1}{2}$

By symmetry,

$\cos (𝛽-𝛾)=-\frac{1}{2}$

and

$\cos (𝛾-𝛼)=-\frac{1}{2}$

The corresponding sine values are:

$\sin (𝛼-𝛽)=\sin (𝛽-𝛾)=\sin (𝛾-𝛼)=\frac{\sqrt{\surd 3}}{2}$

Consider $\cos (2𝛼-𝛽-𝛾)$:

$\begin{array}{cc}\cos (2𝛼-𝛽-𝛾)& =\cos \left(\right(𝛼-𝛽)-(𝛾-𝛼\left)\right)\\ & =\cos (𝛼-𝛽)\cos (𝛾-𝛼)+\sin (𝛼-𝛽)\sin (𝛾-𝛼)\\ & ={(-\frac{1}{2})}^2+{\left(\frac{\sqrt{\surd 3}}{2}\right)}^2\\ & =\frac{1}{4}+\frac{3}{4}\\ & =1\end{array}$

Similarly,

$\cos (2𝛽-𝛾-𝛼)=1$

$\cos (2𝛾-𝛼-𝛽)=1$

Thus, we find:

$\Sigma \cos (2𝛼-𝛽-𝛾)=1+1+1=3$

We have, $\frac{7{𝑥}^2-5𝑥-18}{2{𝑥}^2+𝑥-6}<2$

$\begin{array}{cc}& \frac{7{𝑥}^2-5𝑥-18}{2{𝑥}^2+𝑥-6}-2<0\\ & \frac{7{𝑥}^2-5𝑥-18-4{𝑥}^2-2𝑥+12}{2{𝑥}^2+𝑥-6}<0\\ & \frac{3{𝑥}^2-7𝑥-6}{2{𝑥}^2+𝑥-6}<0\\ & \frac{3{𝑥}^2-9𝑥+2𝑥-6}{2{𝑥}^2+4𝑥-3𝑥-6}<0\\ & \frac{(3𝑥+2)(𝑥-3)}{(2𝑥-3)(𝑥+2)}<0\end{array}$

Case I

$\begin{array}{cc}& (3𝑥+2)(𝑥-3)<0\text{ and }\\ & (2𝑥-3)(𝑥+2)>0\\ & \Rightarrow 𝑥\in (\frac{-2}{3},3)\text{ and }𝑥\in (-\infty ,-2)\cup (\frac{3}{2},0)\\ & \Rightarrow 𝑥\in (\frac{3}{2},3)\end{array}$

$\begin{array}{cc}& \text{ Case II }\\ & \begin{array}{cc}& (3𝑥+2)(𝑥-3)>0\\ \text{ and }& (2𝑥-3)(𝑥+2)<0\end{array}\end{array}$

$\begin{array}{cc}& \Rightarrow 𝑥\in (-\infty ,-\frac{2}{3})\cup (3,\infty )\text{ and }\\ & \Rightarrow 𝑥\in (-2,\frac{3}{2})\\ & \Rightarrow 𝑥\in (-2,-\frac{2}{3})\\ & \text{ Hence, }𝑥\in (-2,-\frac{2}{3})\cup (\frac{3}{2},3)\end{array}$

To determine the set of all values of $𝑘$ for which the inequality ${𝑥}^2-(3𝑘+1)𝑥+4{𝑘}^2+3𝑘-3>0$ holds for all real values of $𝑥$, we proceed as follows.

First, recall that the inequality $𝑓\left(𝑥\right)>0$ for all $𝑥$ requires that the quadratic equation has no real roots, which corresponds to the discriminant $𝐷$ being less than zero. For a quadratic equation $𝑎{𝑥}^2+𝑏𝑥+𝑐$, the discriminant is given by ${𝑏}^2-4𝑎𝑐$.

In this scenario, the quadratic in $𝑥$ is given by:

$𝑎=1,𝑏=-(3𝑘+1),𝑐=4{𝑘}^2+3𝑘-3.$

The discriminant $𝐷$ is:

$𝐷=(3𝑘+1{)}^2-4(1\left)\right(4{𝑘}^2+3𝑘-3).$

Simplifying, we get:

$𝐷=(3𝑘+1{)}^2-16{𝑘}^2-12𝑘+12.$

Expanding:

$𝐷=9{𝑘}^2+6𝑘+1-16{𝑘}^2-12𝑘+12.$

Simplifying further:

$𝐷=-7{𝑘}^2-6𝑘+13.$

To ensure $𝐷<0$, we need:

$7{𝑘}^2+6𝑘-13>0.$

To solve this inequality, factor the quadratic expression:

$7{𝑘}^2+13𝑘-7𝑘-13=(7𝑘+13)(𝑘-1).$

The solution to $(7𝑘+13)(𝑘-1)>0$ can be found using sign analysis. The critical points are $𝑘=-\frac{13}{7}$ and $𝑘=1$.

The intervals to consider are:

$𝑘<-\frac{13}{7}$

$-\frac{13}{7}<𝑘<1$

$𝑘>1$

Test each interval:

For $𝑘<-\frac{13}{7}$ and $𝑘>1$, both factors $(7𝑘+13)$ and $(𝑘-1)$ have the same sign (either both positive or both negative), making the product positive.

For $-\frac{13}{7}<𝑘<1$, the factors have opposite signs, making the product negative.

Thus, the solution set for $𝑘$ is:

$𝑘\in (-\infty ,-\frac{13}{7})\cup (1,\infty ).$

To find the cubic equation whose roots are the squares of the roots of the given equation, we start by analyzing the equation:

$12{𝑥}^3-20{𝑥}^2+𝑥+3=0$

Let's consider $𝛼$, $𝛽$, and $𝛾$ as the roots of this equation. From the properties of polynomial roots, we have:

$𝛼+𝛽+𝛾=\frac{20}{12}=\frac{5}{3}$

$𝛼𝛽+𝛽𝛾+𝛾𝛼=\frac{1}{12}$

$𝛼𝛽𝛾=-\frac{3}{12}=-\frac{1}{4}$

Next, we need to find relationships involving the squares of these roots ${𝛼}^2$, ${𝛽}^2$, and ${𝛾}^2$.

Sum of squares of the roots:

${𝛼}^2+{𝛽}^2+{𝛾}^2=(𝛼+𝛽+𝛾{)}^2-2(𝛼𝛽+𝛽𝛾+𝛾𝛼)$

$={\left(\frac{5}{3}\right)}^2-2\times \frac{1}{12}=\frac{25}{9}-\frac{1}{6}=\frac{50-3}{18}=\frac{47}{18}$

Product of squares of the roots:

${𝛼}^2{𝛽}^2{𝛾}^2=(𝛼𝛽𝛾{)}^2={(-\frac{1}{4})}^2=\frac{1}{16}$

Sum of products of squares of roots taken two at a time:

${𝛼}^2{𝛽}^2+{𝛽}^2{𝛾}^2+{𝛾}^2{𝛼}^2=(𝛼𝛽+𝛽𝛾+𝛾𝛼{)}^2-2𝛼𝛽𝛾(𝛼+𝛽+𝛾)$

$={\left(\frac{1}{12}\right)}^2-2\times (-\frac{1}{4})\times \frac{5}{3}$

$=\frac{1}{144}+\frac{5}{6}=\frac{1+120}{144}=\frac{121}{144}$

Using these calculated values, the cubic equation whose roots are the squares of the original roots is given by:

${𝑥}^3-\frac{47}{18}{𝑥}^2+\frac{121}{144}𝑥-\frac{1}{16}=0$

Multiplying through by 144 to clear fractions, we obtain:

$144{𝑥}^3-376{𝑥}^2+121𝑥-9=0$

Let $𝛼,𝛽$ and $𝛾$ be the roots of equation ${𝑥}^3+3{𝑥}^2-10𝑥-24=0$

$𝛼+𝛽+𝛾=-3,𝛼𝛽+𝛽𝛾+𝛾𝛼=-10$

and $𝛼𝛽𝛾=24$

$\Rightarrow 𝛼=-2,𝛽=-4,𝛾=3$

As, $𝛼(𝛽+𝛾),𝛽(𝛾+\infty )$ and $𝛾(𝛼+𝛽)$ are the roots of equation

$\begin{array}{cc}{𝑥}^3& +𝑝{𝑥}^2+𝑞𝑥+𝑟=0\\ -𝑝& =𝛼(𝛽+𝛾)+𝛽(𝛾+𝛼)+𝛾(𝛼+𝛽)\\ & =-2(-4+3)-4(-2+3)+3(-2-4)\\ & =2-4-18\Rightarrow -20\\ \therefore 𝑝& =20\\ 𝑞& =2(-4)+(-4)(-18)+(-18)2\\ & =-8+72-36=28\end{array}$

Case I When the unit digit is either 0 or 4 . Then, this digit should be 0,2 or 4.

$\therefore 4$-digit number in case 1 is

$=4\times 5\times 3\times 2=120$

Case II When, the unit digit is 2 . Then, the ten's digit is either 1 or 3 .

$\therefore 4$-digit number in case $2=4\times 5\times 2\times 1$ $=40$

Hence, required number

$=120+40=160.$

To determine the number of ways to arrange 2 red roses, 3 white roses, and 5 yellow roses of different sizes into a garland, ensuring that no two yellow roses are adjacent, we follow these steps:

Arrange Red and White Roses:

Since a garland has no fixed starting point, arranging the 2 red and 3 white roses in a circular fashion is equivalent to arranging them linearly with a rotational symmetry. This can be calculated using the formula for circular permutations, adjusted for identical elements:

$\frac{(5-1)!}{2}=\frac{4!}{2}$

Calculate Available Spaces:

After arranging the 2 red and 3 white roses, we create "gaps" between them and at the ends. This creates 5 gaps.

Arrange Yellow Roses:

We now need to place the 5 yellow roses in these 5 gaps without any two yellow roses being adjacent, which can be done in $5!$ ways (as we are using all available spaces).

The calculation for this arrangement is done using permutations:

${}^5{𝑃}_5=5!$

Total Arrangements:

The total number of ways to arrange the garland is the product of ways to arrange the red and white roses and the distinct ways to fill the gaps with yellow roses:

$\frac{4!}{2}{\times }^5{𝑃}_5=\frac{4!\times 5!}{2}$

Calculation:

Calculate the factorials:

$4!=24,5!=120$

Calculate the total number of arrangements:

$\frac{24\times 120}{2}=1440$

Thus, the required number of ways to arrange the roses into a garland such that no two yellow roses are adjacent is 1440.

Here, a i.e. first term $\ge 1$

$𝑎{𝑟}^2($ third term $)\le 100$

$\begin{array}{cc}& {𝑟}^2\le 100\\ \Rightarrow & 𝑟\le 10\end{array}$

When $𝑟=2$, then $𝑎\le \frac{100}{4}=25$

$\therefore 𝑎$ can have 25 values

When $𝑟=3$, then $𝑎\le \frac{100}{9}\Rightarrow 𝑎\le 11$

$\therefore 𝑎$ can have 11 values

When $𝑟=4$, then $𝑎\le \frac{100}{16}\Rightarrow 𝑎\le 6$

$\therefore 𝑎$ can have 6 values.

When $𝑟=5$, then $𝑎\le \frac{100}{25}=4$

$\therefore 𝑎$ can have 4 values.

When $𝑟=6$, then $𝑎\le \frac{100}{36}\Rightarrow 𝑎\le 2$

$\therefore 𝑎$ can have 2 values.

When $𝑟=7$, then $𝑎\le \frac{100}{49}\Rightarrow 𝑎\le 2$

$\therefore 𝑎$ can have 2 values.

When $𝑟=8,9$ and 10 , then

$𝑎\le \frac{100}{64},𝑎\le \frac{100}{81}$ and $𝑎\le \frac{100}{100}$,

respectively.

$\therefore 𝑎$ can have 1 value in each case. Hence, total number of ways.

$\begin{array}{cc}& =25+11+6+4+2+2+1+1+1\\ & =53\end{array}$

$\begin{array}{cc}& \\ & \begin{array}{cc}& \text{ We have, }(1+𝑥+2{𝑥}^2){(\frac{3{𝑥}^2}{2}-\frac{1}{3𝑥})}^9\\ & =(1+𝑥+2{𝑥}^2)\\ & \left\{\sum _{𝑟=0}^9{}^9{𝐶}_{𝑟}{\left(\frac{3}{2}{𝑥}^2\right)}^{9-𝑟}{(-\frac{1}{3𝑥})}^{𝑟}\right\}\\ & =(1+𝑥+2{𝑥}^2)\\ & \left\{\sum _{𝑟=0}^9{}^9{𝐶}_{𝑟}{\left(\frac{3}{2}\right)}^{9-𝑟}{𝑥}^{18-2𝑟}{(-\frac{1}{3})}^1{𝑥}^{-𝑟}\right\}\\ & =(1+𝑥+2{𝑥}^2)\\ & \left\{\sum _{𝑟=0}^9{}^9{𝐶}_{𝑟}{\left(\frac{3}{2}\right)}^{9-𝑟}{(-\frac{1}{3})}^{𝑟}{𝑥}^{18-3𝑟}\right\}\end{array}\end{array}$

$\begin{array}{cc}=\sum _{𝑟=0}^9{}^9{𝐶}_{𝑟}& {\left(\frac{3}{2}\right)}^{9-𝑟}{(-\frac{1}{3})}^{𝑟}{𝑥}^{18-3𝑟}\\ & +\sum _{𝑟=0}^9{}^9{𝐶}_{𝑟}{\left(\frac{3}{2}\right)}^{9-𝑟}{(-\frac{1}{3})}^{𝑟}{𝑥}^{19-3𝑟}\\ & +2\left\{\sum _{𝑟=0}^9{}^9{𝐶}_{𝑟}{\left(\frac{3}{2}\right)}^{9-𝑟}{(-\frac{1}{3})}^{𝑟}{𝑥}^{20-3𝑟}\right\}\end{array}$

Clearly, first term independent of $𝑥$ by

$18-3𝑟=0\Rightarrow 𝑟=6$

$\therefore$ Coefficient of independent of $𝑥$ is

$\begin{array}{cc}& {}^9{𝐶}_6{\left(\frac{3}{2}\right)}^3{(-\frac{1}{3})}^6={}^9{𝐶}_3\times \frac{3^3}{8}\times \frac{1}{3^6}\\ & =\frac{9\times 8\times 7}{3\times 2}\times \frac{1}{8}\times \frac{1}{3^3}=\frac{7}{18}.\end{array}$

To find the coefficient of $𝑥$ in the expansion of

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

we proceed by approximating each component in the product.

Expansion of $(1-4𝑥{)}^2$:

$(1-4𝑥{)}^2=1-8𝑥+16{𝑥}^2$

Approximation of ${(1-2{𝑥}^2)}^{\frac{1}{2}}$:

Using the binomial approximation for small ${𝑥}^2$, we have:

${(1-2{𝑥}^2)}^{\frac{1}{2}}\approx 1-{𝑥}^2$

Approximation of ${(1-\frac{𝑥}{4})}^{-\frac{3}{2}}$:

Again, using binomial expansion, we write:

${(1-\frac{𝑥}{4})}^{-\frac{3}{2}}\approx 1+\frac{3}{2}\times \frac{𝑥}{4}=1+\frac{3𝑥}{8}$

Putting it all together, we compute the product:

$\frac{1}{8}\left[\right(1-8𝑥+16{𝑥}^2\left)\right(1-{𝑥}^2\left)\right(1+\frac{3𝑥}{8}\left)\right]$

Next, we multiply these expressions and collect the terms involving $𝑥$:

$\begin{array}{cc}& =\frac{1}{8}\left[\right(1-8𝑥+16{𝑥}^2\left)\right(1-{𝑥}^2\left)\right(1+\frac{3𝑥}{8}\left)\right]\\ & =\frac{1}{8}\left[\right(1-8𝑥+16{𝑥}^2\left)\right(1-{𝑥}^2+\frac{3𝑥}{8}-\frac{3{𝑥}^3}{8}+\cdots \left)\right]\end{array}$

We are interested only in terms involving $𝑥$. Consider:

The coefficient of $𝑥$ from $(1-8𝑥+16{𝑥}^2)\left(1\right)$ is $-8$.

The contribution due to combining $1$ from $(1-8𝑥+16{𝑥}^2)$ with the $\frac{3𝑥}{8}$ term from $(1+\frac{3𝑥}{8})$ is $\frac{3}{8}$.

Thus, the total coefficient of $𝑥$ from this product is:

$\frac{1}{8}(\frac{3}{8}-8)$

Simplifying, we calculate:

$\frac{1}{8}\cdot (-\frac{61}{8})=-\frac{61}{64}$

Therefore, the coefficient of $𝑥$ is $-\frac{61}{64}$.

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \frac{4{𝑥}^2+5}{(𝑥-2{)}^4}=\frac{𝐴}{𝑥-2}+\frac{𝐵}{(𝑥-2{)}^2}\\ & +\frac{𝐶}{(𝑥-2{)}^3}+\frac{𝐷}{(𝑥-2{)}^4}\\ & \Rightarrow 4{𝑥}^2+5=𝐴(𝑥-2{)}^3+𝐵(𝑥-2{)}^2\\ & +𝐶(𝑥-2)+𝐷\\ & \Rightarrow 4{𝑥}^2+5=𝐴({𝑥}^3-8-3\left({𝑥}^2\right)(2)\\ & +3\left(𝑥\right)\left(4\right)+𝐵({𝑥}^2+4-4𝑥)\\ & +𝐶(𝑥-2)+𝐷\\ & \Rightarrow 4{𝑥}^2+5=𝐴{𝑥}^3+{𝑥}^2(-6𝐴+𝐵)\\ & +𝑥(12𝐴-4𝐵+𝐶)\\ & +(-8𝐴+4𝐵-2𝐶+𝐷)\\ & \Rightarrow 𝐴=-6𝐴+𝐵=4\\ & \Rightarrow 𝐵=4\\ & 12𝐴-4𝐵+𝐶=0\\ & \Rightarrow 0-16+𝐶=0\Rightarrow 𝐶=16\end{array}\end{array}$

$\begin{array}{cc}& \text{ and, }-8𝐴+4𝐵-2𝐶+𝐷=5\\ & \Rightarrow 0+16-32+𝐷=5\\ & \Rightarrow 𝐷=21\\ & \therefore \sqrt{\surd \frac{𝐴+𝐵+𝐷}{𝐶}}=\sqrt{\surd \frac{0+4+21}{16}}=\sqrt{\surd \frac{25}{16}}=\frac{5}{4}\end{array}$