AP EAPCET 2024 22nd May Morning Shift Question Paper with Solutions

For real values of $𝑓\left(𝑥\right)$

$\sqrt{\surd 9-\sqrt{\surd {𝑥}^2-144}}\ge 0$

On squaring both sides, we get

$\begin{array}{cc}9-\sqrt{\surd {𝑥}^2-144}& \ge 0\\ 9& \ge \sqrt{\surd {𝑥}^2-144}\end{array}$

Again, squaring on both sides, we get

$\begin{array}{cc}{9}^2& \ge {𝑥}^2-144\\ 81+144& \ge {𝑥}^2\\ 225& \ge {𝑥}^2\\ {𝑥}^2& \le 225\end{array}$

On taking principal square root

$\left|𝑥\right|\le 15$

i.e. $𝑥\le 15$ or $𝑥\ge -15$

$\begin{array}{cc}𝑁𝑜𝑤,{𝑥}^2-144& \ge 0\\ {𝑥}^2& \ge 144\end{array}$

On taking principal square root

$\begin{array}{c}\sqrt{\surd {𝑥}^2}\ge \sqrt{\surd {12}^2}\\ \left|𝑥\right|\ge 12\end{array}$

i.e. $𝑥\le -12$ or $𝑥\ge 12$

We can see on number line the domain of the given function will be cross shadow area.

i.e. $[-15,-12]\cup [12,15]$

Set $𝐴$ has 5 elements.

Set $𝐵$ has 7 elements.

In a function from $𝐴$ to $𝐵$, every element of $𝐴$ must be mapped to an element of $𝐵$. Therefore, total number of functions will be

$=7\times 7\times 7\times 7\times 7=7^5$

Now, number of many one functions $=$ Total functions-number of one-one functions

$\begin{array}{cc}& =7^5-{}^7{𝐶}_5\times 5!\\ & =7^5-\frac{7!}{(7-5)!5!}\times 5!\\ & =7^5-\frac{7!}{(7-5)!}=7^5-{}^7{𝑃}_5\\ & (\because {}^{𝑛}{𝑃}_{𝑟}=\frac{𝑛!}{(𝑛-𝑟)!})\\ & =7^5-{}^7{𝑃}_5\end{array}$

$2+3+5+6+8+9\dots 2𝑛$ terms

Observe the sequence and note that every third term is missing.

So, we can split this summation into two Arithmatic Progiression (AP's).

First $𝐴{𝑃}_1=2+5+8+\dots 𝑛$ terms

Then, ${𝑆}_1=\frac{𝑛}{2}[2{𝑎}_1+(𝑛-1\left){𝑑}_1\right]$

Similarly, $𝐴{𝑃}_2=3+6+9+\dots 𝑛$ terms

Then, ${𝑆}_2=\frac{𝑛}{2}[2{𝑎}_2+(𝑛-1\left){𝑑}_2\right]$

Therefore, $𝑆={𝑆}_1+{𝑆}_2$

$\begin{array}{cc}𝑆=& \frac{𝑛}{2}[2\times 2+(𝑛-1\left)3\right]+\frac{𝑛}{2}[2\times 3+(𝑛-1)\times 3]\\ =& \frac{𝑛}{2}[4+3(𝑛-1)+6+3(𝑛-1\left)\right]\\ =& \frac{𝑛}{2}[10+3𝑛-3+3𝑛-3]\\ =& \frac{𝑛}{2}[4+6𝑛]=\frac{𝑛}{2}[2+3𝑛]\times 2\\ =& 2𝑛+3{𝑛}^2\\ 𝑆& =3{𝑛}^2+2𝑛\end{array}$

To determine the condition for $𝑎$ and $𝑏$ so, that the set of equations

$\begin{array}{cc}& 𝑥+2𝑦+3𝑧=6\\ & 𝑥+3𝑦+5𝑧=9\end{array}$

$2𝑥+5𝑦+𝑎𝑧=𝑏$ has a unique solution. We need to ensure that the coefficient matrix has a non zero determinant.

Let $𝐴=⎛⎝⎜\begin{array}{ccc}1& 2& 3\\ 1& 3& 5\\ 2& 5& 𝑎\end{array}⎞⎠⎟$

Then, $\det \left(𝐴\right)$

$\begin{array}{cc}& =1(3𝑎-25)-2(\cdot 𝑎-10)+3(5-6)\\ & =1(3𝑎-25)-2(𝑎-10)+3(-1)\\ & =3𝑎-25-2𝑎+20-3\\ & =𝑎-8\end{array}$

For system to have a unique solution. $\det \left(𝐴\right)$ must be non-zero.

i.e. $𝑎-8\ne 0$

Thus, $𝑎\ne 8$

So, $𝑏$ can be other number which should be a non-zero real number and therefore $𝑏$ must be a real number.

Thus, $𝑎\ne 8,𝑏\in 𝑅$.

Given, $\left|𝑃𝑄\right|=1$ and $\left|𝑃\right|=9$.

For any $𝑛\times 𝑛$ matrix $𝐴$, the determinant of $\mathrm{adj}\left(𝐴\right)$ is $|𝐴{|}^{𝑛-1}$, where $𝑛$ is the dimension of the square matrix.

If $𝐵=𝑘𝐴$, where $𝑘$ is a scalar, then $\left|𝐵\right|={𝑘}^{𝑛}\left|𝐴\right|$ for an $𝑛\times 𝑛$ matrix $𝐴$.

Also, $\left|𝑃𝑄\right|=\left|𝑃\right|\left|𝑄\right|$.

Utilizing these properties, for a $3\times 3$ matrix we have:

$\left|3𝑄\right|=3^3\left|𝑄\right|=27\left|𝑄\right|.$

Now, considering the matrix $\mathrm{adj}\left(3𝑄\right)$:

$\mathrm{Det}(\mathrm{adj}(3𝑄\left)\right)=\left(27\right|𝑄|{)}^2.$

Since $\left|𝑃𝑄\right|=1$ and $\left|𝑃\right|=9$, we have:

$\begin{array}{cc}1& =9\cdot \left|𝑄\right|,\\ \left|𝑄\right|& =\frac{1}{9}.\end{array}$

Finally, compute $\det (\mathrm{adj}(3𝑄\left)\right)$:

$\begin{array}{cc}\det (\mathrm{adj}(3𝑄\left)\right)& =\left(27\right|𝑄|{)}^2\\ & =(27\times \frac{1}{9}{)}^2\\ & =3^2\\ & =9.\end{array}$

For $\mathrm{adj}(𝑃\cdot \mathrm{adj}(3𝑄\left)\right)$:

$\begin{array}{cc}\mathrm{adj}(𝑃\cdot \mathrm{adj}(3𝑄\left)\right)& =(9\times 9{)}^{3-1}\\ & =(9^2{)}^2\\ & =9^4.\end{array}$

We know that for a square matrix the product $𝐴\cdot \mathrm{adj}\left(𝐴\right)=\det \left(𝐴\right)$. I where I is the identity matrix of the same order as $𝐴$.

If $𝐴$ is a $3\times 3$ matrix, then the product should be : $𝐴\cdot \mathrm{adj}\left(𝐴\right)=\det \left(𝐴\right)\cdot {𝐼}_3$

Given matrix

$\begin{array}{cc}𝐴& =⎡⎣⎢\begin{array}{ccc}𝑎& 1& 2\\ 1& 2& 𝑏\\ 𝑐& 1& 3\end{array}⎤⎦⎥\\ \mathrm{adj}\left(𝐴\right)& =⎡⎣⎢\begin{array}{ccc}7& -1& -5\\ -3& 9& 5\\ 1& -3& 5\end{array}⎤⎦⎥\end{array}$

Let's multiply $𝐴$ with adj $\left(𝐴\right)$

$𝐴\cdot \mathrm{adj}\left(𝐴\right)=⎡⎣⎢\begin{array}{ccc}𝑎& 1& 2\\ 1& 2& 𝑏\\ 𝑐& 1& 3\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}7& -1& -5\\ -3& 9& 5\\ 1& -3& 5\end{array}⎤⎦⎥$

$=⎡⎣⎢\begin{array}{ccc}7𝑎-1& 𝑏+1& 7𝑐\\ -𝑎+3& -3𝑏+17& -𝑐\\ -5𝑎+15& 5𝑏+5& -5+20\end{array}⎤⎦⎥$

Setting these elements equal to identity matrix adjusted by determinant

We get, $-𝑎+3=0$

$\begin{array}{cc}\Rightarrow 𝑎& =3,𝑏+1=0\Rightarrow 𝑏=-1\\ 7𝑐& =0\Rightarrow 𝑐=0\end{array}$

Thus, $𝑎=3,𝑏=-1$ and $𝑐=0$

Hence, ${𝑎}^2+{𝑏}^2+{𝑐}^2$

$=3^2+(-1{)}^2+0=9+1=10$

$\left|𝑍\right|\le 3\Rightarrow$ Complex number Z lies within a circle centered at the origin with radius 3 .

$\frac{-𝜋}{2}\le \arg \left(𝑧\right)\le \frac{𝜋}{2}$. This restricts z to the right half of the complex plane.

The intersection of these region is a semicircle with radius 3 . The area of semicircle is given by half the area of the full circle

The area of circle with radius $𝑟$ is $=𝜋{𝑟}^2$

$=𝜋(3{)}^2=9𝜋$

Half of this area $=\frac{9𝜋}{2}$

So, area of region $=\frac{9𝜋}{2}$ sq units.

To determine the locus of the complex number $𝑍$, where $𝑍=𝑥+𝑖𝑦$ (with $𝑥$ and $𝑦$ as real numbers), we analyze the given condition:

$\arg \left(\frac{𝑍-1}{𝑍+1}\right)=\frac{𝜋}{4}$

This equation implies that the complex number $𝐴=\frac{𝑍-1}{𝑍+1}$ forms an angle of $\frac{𝜋}{4}$ with the positive real axis. Therefore, $𝐴$ lies on a line making an angle of $\frac{𝜋}{4}$.

To simplify, we express:

$\frac{𝑍-1}{𝑍+1}={𝑒}^{𝑖\frac{𝜋}{4}}=\cos \frac{𝜋}{4}+𝑖\sin \frac{𝜋}{4}=\frac{1}{\sqrt{\surd 2}}+\frac{𝑖}{\sqrt{\surd 2}}$

Thus, we have:

$\mid \frac{𝑍-1}{𝑍+1}\mid =\mid \frac{1}{\sqrt{\surd 2}}+𝑖\cdot \frac{1}{\sqrt{\surd 2}}\mid =\sqrt{\surd {\left(\frac{1}{\sqrt{\surd 2}}\right)}^2+{\left(\frac{1}{\sqrt{\surd 2}}\right)}^2}=1$

This calculation indicates that the complex number $\frac{𝑍-1}{𝑍+1}$ is on the unit circle. Therefore, we equate:

$\mid \frac{𝑍-1}{𝑍+1}\mid =1\Rightarrow |𝑍-1|=|𝑍+1|$

This describes the locus as the perpendicular bisector of the line segment joining 1 and -1, which corresponds to the imaginary axis $𝑥=0$. Thus, the locus is a straight line on the imaginary axis.

To find the values of $(8𝑖{)}^{\frac{1}{3}}$, we utilize De Moivre's theorem. Express $8𝑖$ in polar form as $8{𝑒}^{𝑖(\frac{𝜋}{2}+2𝑘𝜋)}$, then take the cube root:

$(8𝑖{)}^{\frac{1}{3}}=8^{\frac{1}{3}}{𝑒}^{𝑖\left(\frac{𝜃+2𝑘𝜋}{3}\right)}=2{𝑒}^{𝑖(\frac{𝜋}{6}+\frac{2𝑘𝜋}{3})}$

Next, calculate this for different values of $𝑘$:

For $𝑘=0$:

$2{𝑒}^{𝑖\left(\frac{𝜋}{6}\right)}=2(\cos \frac{𝜋}{6}+𝑖\sin \frac{𝜋}{6})=2(\frac{\sqrt{\surd 3}}{2}+𝑖\frac{1}{2})=\sqrt{\surd 3}+𝑖$

For $𝑘=1$:

$2{𝑒}^{𝑖(\frac{𝜋}{6}+\frac{2𝜋}{3})}=2{𝑒}^{𝑖\left(\frac{5𝜋}{6}\right)}=2(\cos \frac{5𝜋}{6}+𝑖\sin \frac{5𝜋}{6})=2(-\frac{\sqrt{\surd 3}}{2}+𝑖\frac{1}{2})=-\sqrt{\surd 3}+𝑖$

For $𝑘=2$:

$2{𝑒}^{𝑖(\frac{𝜋}{6}+\frac{4𝜋}{3})}=2{𝑒}^{𝑖\left(\frac{3𝜋}{2}\right)}=2(0-𝑖)=-2𝑖$

Therefore, all the values of $(8𝑖{)}^{\frac{1}{3}}$ are $\sqrt{\surd 3}+𝑖$, $-\sqrt{\surd 3}+𝑖$, and $-2𝑖$, or in other words, $\pm \sqrt{\surd 3}+𝑖$, $-2𝑖$.

When solving the quadratic equation ${𝑥}^2-6𝑥-2=0$, the roots $𝛼$ and $𝛽$ can be found using the quadratic formula:

$𝛼,𝛽=\frac{6\pm \sqrt{\surd 6^2-4\cdot 1\cdot (-2)}}{2}=\frac{6\pm \sqrt{\surd 36+8}}{2}=\frac{6\pm \sqrt{\surd 44}}{2}=3\pm \sqrt{\surd 11}$

Given that ${𝑎}_{𝑛}={𝛼}^{𝑛}-{𝛽}^{𝑛}$ for $𝑛\ge 1$, and knowing $𝛼$ and $𝛽$ are roots, they satisfy:

${𝑡}^2-6𝑡-2=0⟹{𝑡}^2=6𝑡+2$

Multiplying by ${𝑡}^{𝑛-2}$, we have:

${𝑡}^{𝑛}=6{𝑡}^{𝑛-1}+2{𝑡}^{𝑛-2}$

Therefore, for $𝛼$:

${𝛼}^{𝑛}=6{𝛼}^{𝑛-1}+2{𝛼}^{𝑛-2}$

And similarly for $𝛽$:

${𝛽}^{𝑛}=6{𝛽}^{𝑛-1}+2{𝛽}^{𝑛-2}$

This gives us for ${𝑎}_{𝑛}$:

$\begin{array}{cc}{𝑎}_{𝑛}& ={𝛼}^{𝑛}-{𝛽}^{𝑛}\\ & =6({𝛼}^{𝑛-1}-{𝛽}^{𝑛-1})+2({𝛼}^{𝑛-2}-{𝛽}^{𝑛-2})\\ {𝑎}_{𝑛}& =6{𝑎}_{𝑛-1}+2{𝑎}_{𝑛-2}\end{array}$

To find $\frac{{𝑎}_{10}-2{𝑎}_8}{2{𝑎}_9}$, we apply the recurrence relation:

Given ${𝑎}_{10}=6{𝑎}_9+2{𝑎}_8$, we have:

${𝑎}_{10}-2{𝑎}_8=6{𝑎}_9+2{𝑎}_8-2{𝑎}_8=6{𝑎}_9$

Thus,

$\frac{{𝑎}_{10}-2{𝑎}_8}{2{𝑎}_9}=\frac{6{𝑎}_9}{2{𝑎}_9}=3$

Therefore, the value is $3$.

Given, ${𝑥}^2-6𝑎𝑥+2-2𝑎+9{𝑎}^2=0$

$\Rightarrow {𝑥}^2-6𝑎𝑥+(2-2𝑎+9{𝑎}^2)=0$

For both roots to exceed 3, $𝑥=3$ should be a root or both roots should be greater than 3.

Sum of roots $=-\frac{𝑏}{𝑎}=-\left(\frac{-6𝑎}{1}\right)=6𝑎$

$6𝑎>18\Rightarrow 𝑎>3$

Product of roots $=\frac{𝑐}{𝑎}=\frac{9{𝑎}^2-2𝑎+2}{1}$

Since, $𝑎>3$ so product, also should be greater than 3 .

That leads us to the correct condition

$𝑎>\frac{11}{9}$

We begin with the polynomial:

${𝑥}^5-5{𝑥}^3+5{𝑥}^2-1=(𝑥-1{)}^3({𝑥}^2+3𝑥+1)$

We are given that $𝛼$ and $𝛽$ are distinct negative roots of the equation. Therefore, $𝛼$ and $𝛽$ must be the roots of the quadratic equation:

${𝑥}^2+3𝑥+1=0$

Given that $𝛼$ and $𝛽$ are negative, we establish that:

$𝛼𝛽=1\text{(as derived from the product of the roots of }{𝑥}^2+3𝑥+1=0\text{)}$

We need to find the polynomial equation for which $\sqrt{\surd -𝛼}$ and $\sqrt{\surd -𝛽}$ are roots. Note:

$\sqrt{\surd -𝛼}\cdot \sqrt{\surd -𝛽}=\sqrt{\surd 𝛼\cdot 𝛽}=\sqrt{\surd 1}=1$

Thus, the product of $\sqrt{\surd -𝛼}$ and $\sqrt{\surd -𝛽}$ is 1. The equation with these roots and integer coefficients is:

${𝑥}^4-3{𝑥}^2+1=0$

The product of the roots for this equation equals 1, satisfying the condition $\sqrt{\surd -𝛼}\cdot \sqrt{\surd -𝛽}=1$, confirming that $\sqrt{\surd -𝛼}$ and $\sqrt{\surd -𝛽}$ are roots of the equation.

${𝑥}^9-{𝑥}^5+{𝑥}^4-1=0$

$\begin{array}{ccc}\Rightarrow & {𝑥}^5({𝑥}^4-1)+1({𝑥}^4-1)& =0\\ \Rightarrow & ({𝑥}^5+1)({𝑥}^4-1)& =0\\ \Rightarrow & {𝑥}^5+1=0\text{ or }{𝑥}^4-1=0\\ \Rightarrow & {𝑥}^5=-1\text{ or }{𝑥}^4=1\\ \Rightarrow & {𝑥}^5=-1\text{ or }𝑥=1,-1,𝑖,-𝑖\end{array}$

$\therefore$ Number of real roots $\left(𝑛\right)=3$

Number of complex roots $\left(𝑚\right)=2$

Number of complex roots having argument in 2nd quadrant $\left(𝑘\right)=1$

$\therefore 𝑚\cdot 𝑛\cdot 𝑘=3\left(2\right)\left(1\right)=6$

Write the word TABLE

alphabetically A, B, E, L, T

To finding the rank of BLATE

Fixing B, arrangements with $𝐴$ in the

first position $4!=24(𝐴+$ permutations

of $𝐵,𝐸,𝐿,𝑇$ )

If permutation starting with $𝐵$, then

fixing $𝐵$ and then $𝐿$, arrangements of,

$𝐴,𝐵,𝐸,𝑇:3!=6(𝐵𝐿+$ permutations of

$𝐴,𝐵,𝐸,𝑇$ )

if permuatations starting with $𝐵𝐿$

Fixing BL, arrangements of A, B, E, T

$=3!=6$

(BLA + permutations)

arrangements of $𝐴,𝐸,𝐵,𝑇:2!=2$

(BLAE + permutations)

arrangement of $𝐴,𝐸,𝑇:1!=1$

(BLATE + permutations)

Thus, fixing $𝐴=24$

fixing $𝐵=𝐴+6=30$

Fixing L: BLA $=6$

Final E: BLAE $+1=31$

Therefore, total permutations of BLATE excluding irredevent permutations.

Thus, BLATE +31 : The possible rank of Table are permutations $=30+31=61$

Linear arrangements where no two boys sit together (X):

First, arrange the 6 girls. This can be done in $5!$ (120) ways since the arrangement of the boys around these gaps is what's crucial. Once the girls are arranged, they create 7 possible gaps (1 before each girl and 1 after the last girl) to place the boys.

Next, select 5 of these 7 gaps to place a boy. This can be done in ${}^7{𝐶}_5$ ways. Each selection of gaps can be filled by arranging the boys in $5!$ different ways.

Therefore, the total number of linear arrangements where no two boys sit together (X) is:

$𝑋=5!\times {}^7{𝐶}_5\times 5!=5!\times \frac{7!}{5!2!}\times 5!$

Simplifying,

$𝑋=\frac{7\times 6}{2}\times 5!\times 5!=21\times 5!\times 5!$

Linear arrangements where no two girls sit together (Y):

This problem can be solved similarly to the arrangement for X. Here, arrange the boys first, which can also be done in $5!$ ways. This arrangement creates 6 gaps.

All 6 girls can be placed into these exact 6 gaps in ${}^6{𝐶}_6$ ways, and then arrange those girls among themselves in $6!$ (720) ways.

Thus, the total number of linear arrangements where no two girls sit together (Y) is:

$𝑌=5!\times {}^6{𝐶}_6\times 6!=5!\times 1\times 6!$

Since $𝑌=5!\times 6!$, and recognizing $6!=6$, it remains:

$𝑌=5!\times 5!$

Circular arrangements where no two boys sit together (Z):

For arranging 11 people in a circle, consider fixing one girl as a reference point, leaving 10 people to arrange, which alters our calculation by one less arrangement over linear. Arrange the remaining 5 boys and 5 girls around this fixed position.

The number of circular arrangements, where no two boys sit together for them is:

$𝑍=5!\times 5!$

For the final result, calculating the ratio $𝑋:𝑌:𝑍$ gives us:

$𝑋:𝑌:𝑍=21\times 5!\times 5!:1\times 5!\times 5!:1\times 5!\times 5!=21:1:1$

Distribute the Minimum Apples:

$𝐴$ receives 2 apples, and $𝐶$ receives 2 apples.

This leaves $15-2-2=11$ apples to be distributed among $𝐴$, $𝐵$, and $𝐶$.

Define Variables for Remaining Distribution:

Let $𝑥$ be the number of apples $𝐴$ gets over the initial 2.

Let $𝑦$ be the number of apples $𝐵$ receives.

Let $𝑧$ be the number of apples $𝐶$ gets over the initial 2.

The equation to solve is $𝑥+𝑦+𝑧=11$.

Apply Constraints:

Since $𝐵$ can receive at most 5 apples, $𝑦\le 5$.

Use the Stars and Bars Method:

Calculate the total number of solutions to $𝑥+𝑦+𝑧=11$ without any constraints:

$(\frac{11+2}{2})=(\frac{13}{2})=78$

Account for the Constraint on $𝐵$:

For cases where $𝑦>5$, set ${𝑦}^{\prime }=𝑦-6$, where ${𝑦}^{\prime }\ge 0$.

Substitute $𝑦$ with ${𝑦}^{\prime }$ to get the equation $𝑥+{𝑦}^{\prime }+𝑧=5$.

Count the number of solutions to $𝑥+{𝑦}^{\prime }+𝑧=5$:

$(\frac{5+2}{2})=(\frac{7}{2})=21$

Calculate the Final Result:

Subtract the invalid cases from the total solutions:

$78-21=57$

Thus, there are 57 ways to distribute the apples based on the given conditions.

We know that $𝑟$ th term in the expansion $(𝑥+𝑎{)}^{𝑛}$ is given by

${𝑇}_{𝑟+1}={}^{𝑛}{𝐶}_{𝑟}{𝑥}^{𝑛-𝑟}{𝑎}^{𝑟}$

$\begin{array}{cc}𝑔𝑖𝑣𝑒𝑛,& {𝑇}_2={}^{𝑛}{𝐶}_1{𝑥}^{𝑛-1}𝑎=96...\left(𝑖\right)\\ & {𝑇}_3={}^{𝑛}{𝐶}_2{𝑥}^{𝑛-2}{𝑎}^2=216...\left(𝑖𝑖\right)\\ & {𝑇}_4={}^{𝑛}{𝐶}_3{𝑥}^{𝑛-3}{𝑎}^3=216...\left(𝑖𝑖𝑖\right)\end{array}$

On dividing Eq. (ii) from Eq. (i), we get

$\frac{{}^{𝑛}{𝐶}_2{𝑥}^{𝑛-2}{𝑎}^2}{{}^{𝑛}{𝐶}_1{𝑥}^{𝑛-1}𝑎}=\frac{216}{96}$

$\begin{array}{cc}& \frac{𝑛!}{(𝑛-2)!2!}{𝑥}^{𝑛-1-1}{𝑎}^2\\ & \frac{𝑛!}{(𝑛-1)!1!}{𝑥}^{𝑛-1}𝑎=\frac{9}{4}\\ & \frac{1}{2}\cdot \frac{(𝑛-1)(𝑛-2)!}{(𝑛-2)!}\frac{𝑎}{𝑥}=\frac{9}{4}\\ & (𝑛-1)\frac{𝑎}{𝑥}=\frac{9}{2}\\ & (𝑛-1)𝑎=\frac{9}{2}𝑥...\left(𝑖𝑣\right)\end{array}$

Similarly, on dividing Eq. (iii) from Eq. (ii), we get

$\begin{array}{cc}\frac{{}^{𝑛}{𝐶}_3{𝑥}^{𝑛-3}{𝑎}^3}{{}^{𝑛}{𝐶}_2{𝑥}^{𝑛-2}{𝑎}^2}& =\frac{216}{216}\\ \Rightarrow \frac{\frac{𝑛!}{(𝑛-3)!3!}{𝑥}^{𝑛-2-1}{𝑎}^3}{\frac{𝑛!}{(𝑛-2)!2!}{𝑥}^{𝑛-2}{𝑎}^2}& =1\\ \Rightarrow \frac{2\cdot 1(𝑛-2)(𝑛-3)!𝑎}{3\cdot 2\cdot 𝑙(𝑛-3)!}\frac{𝑎}{𝑥}& =1\\ \Rightarrow (𝑛-2)𝑎& =3𝑥...\left(𝑣\right)\end{array}$

Now, on dividing Eq. (v) from Eq. (iv) we get,

$\begin{array}{cc}\frac{(𝑛-2)𝑎}{(𝑛-1)𝑎}=\frac{3𝑥}{\frac{9}{2}𝑥}& =\frac{2}{3}\\ \Rightarrow 3𝑛-6& =2𝑛-2\\ 𝑛& =4\end{array}$

From Eq. (v), we get

$\begin{array}{cc}(4-2)𝑎& =3𝑥\\ 𝑎& =\frac{3}{2}𝑥...\left(𝑣𝑖\right)\end{array}$

From Eq (i), we get

$\begin{array}{cc}{}^4{𝐶}_1{𝑥}^{4-1}{𝑎}^1& =96\\ \frac{4!}{3!1!}{𝑥}^3𝑎& =96\\ 4{𝑥}^3\cdot \frac{3}{2}𝑥& =96[𝐹𝑟𝑜𝑚𝐸𝑞.(𝑣𝑖\left)\right]\\ {𝑥}^4=\frac{96}{6}& =16\\ 𝑥& =2\end{array}$

Using Eq. (vi), we get

$𝑎=3$

$\begin{array}{cc}𝐶𝑙𝑒𝑎𝑟𝑙𝑦,& 𝑥+𝑎=2+3=4+1\\ & 𝑥+𝑎=𝑛+1\end{array}$

Given the expression:

$\frac{1}{2}{(1\cdot 2+2\cdot 3𝑥+3\cdot 4{𝑥}^2+\dots +\infty )}^{-25}$

First, consider the series inside the parentheses:

$1\cdot 2+2\cdot 3𝑥+3\cdot 4{𝑥}^2+\dots +\infty$

This can be rewritten as:

$2(1+3𝑥+6{𝑥}^2+\dots +\infty )$

Recognizing the series as a result of differentiating and manipulating a geometric series, it sums to:

$2\frac{1}{(1-𝑥{)}^3}$

So,

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

Thus, the expression evaluates to:

$\frac{1}{2}{(1\cdot 2+2\cdot 3𝑥+3\cdot 4{𝑥}^2+\dots +\infty )}^{-25}={\left(\frac{1}{(1-𝑥{)}^3}\right)}^{-25}=(1-𝑥{)}^{75}$

Therefore, the number of terms in the expansion of $(1-𝑥{)}^{75}$ is:

$75+1=76$

$\begin{array}{cc}& \frac{{𝑥}^4}{(𝑥-2)(𝑥+1)}={𝑥}^2+𝑥+3+\frac{-\frac{1}{3}}{𝑥+1}+\frac{\frac{16}{3}}{𝑥-2}\\ & ={𝑥}^2+𝑥+3+(-\frac{1}{3})(1+𝑥{)}^{-1}+\frac{16}{3}(𝑥-2{)}^{-1}\end{array}$

$\begin{array}{c}={𝑥}^2+𝑥+3-\frac{1}{3}[1-𝑥+{𝑥}^2-{𝑥}^3+\dots ]\\ -\frac{8}{3}{(1-\frac{𝑥}{2})}^1\end{array}$

$\begin{array}{cc}& \begin{array}{cc}={𝑥}^2+𝑥+3-& \frac{1}{3}[1-𝑥+{𝑥}^2-{𝑥}^3+\dots ]\\ & -\frac{8}{3}[1+\frac{𝑥}{2}+\frac{{𝑥}^2}{4}+\dots ]\end{array}\\ & \text{ Hence, coefficient of }{𝑥}^2\text{ is }(1-\frac{1}{3}-\frac{2}{3})\\ & =(1-1)=0\end{array}$

$\frac{1}{11\cos 2𝑥+60\sin 2𝑥+69}$

denominator $11\cos 2𝑥+60\sin 2𝑥+69$

To find extrema of denominator

$11\cos 2𝑥+60\sin 2𝑥=\sqrt{\surd {11}^2+{60}^2}\sin (2𝑥+𝜃)$

where, $𝜃={\tan }^{-1}\frac{11}{60}$

$=61\sin (2𝑥+𝜃)$

This minimum value is -61 and the maximum value is 61 .

Therefore, $11\cos 2𝑥+60\sin 2𝑥+69$

$=69-61=8\text{ to }69+61=130$

The maximum value is 130 and minimum is 8 of denominator

Hence, ${𝑀}_1=\frac{1}{8}$

For, $3{\cos }^{2}5𝑥+4{\sin }^{2}5𝑥$

$\begin{array}{cc}& =3(1-{\sin }^{2}5𝑥)+4{\sin }^{2}5𝑥\\ & =3+{\sin }^{2}5𝑥\end{array}$

The maximum value of ${\sin }^{2}5𝑥$ is 1 , hence

$3+{\sin }^{2}5𝑥=3+1=4$

because the range of (sine) ${}^2$ is between 0 and 1

The minimum value of ${\sin }^{2}5𝑥$ is 0 .

Hence, $3+{\sin }^{2}5𝑥\ge 3+0=3$

Thus, the maximum value ${𝑀}_2=4$

Finally the ratio is

$\frac{{𝑀}_1}{{𝑀}_2}=\frac{\frac{1}{8}}{4}=\frac{1}{8\times 4}\Rightarrow \frac{{𝑀}_1}{{𝑀}_2}=\frac{1}{32}$

To solve the expression:

$4\cos \frac{𝜋}{7}\cos \frac{𝜋}{5}\cos \frac{2𝜋}{7}\cos \frac{2𝜋}{5}\cos \frac{4𝜋}{7}$

we can use known trigonometric identities and values. Direct computation might be cumbersome, so we leverage a known result for products of cosines.

Here is the important formula:

$\cos \left(\frac{𝜋}{7}\right)\cos \left(\frac{2𝜋}{7}\right)\cos \left(\frac{3𝜋}{7}\right)=\frac{1}{8}$

Given the symmetry properties of cosine, specifically that:

$\cos \left(\frac{4𝜋}{7}\right)=-\cos \left(\frac{3𝜋}{7}\right),$

we can deduce:

$\cos \frac{𝜋}{7}\cos \frac{2𝜋}{7}\cos \frac{4𝜋}{7}=\frac{-1}{8}$

Additionally, using the identity for the product of cosines:

$\cos \frac{𝜋}{5}\cos \frac{2𝜋}{5}=\frac{1}{4}$

Combining these results from equations (i) and (ii):

$\begin{array}{cc}4\cos \frac{𝜋}{7}\cos \frac{𝜋}{5}\cos \frac{2𝜋}{7}\cos \frac{2𝜋}{5}\cos \frac{4𝜋}{7}& =4\times (-\frac{1}{8})\times \frac{1}{4}\\ & =\frac{-1}{8}\end{array}$

Hence, the value of the expression is $-\frac{1}{8}$.

Given $△𝐴𝐵𝐶,𝐴,𝐵,𝐶$ in $𝐴𝑃$ which means $𝐵-𝐶=𝐴-𝐵$ where $𝐶>𝐵>𝐴$