AP EAPCET 2024 21st May Evening Shift Question Paper with Solutions

Given that

$𝑓\left(𝑥\right)={\sin }^{-1}\left(\frac{1+{𝑥}^2}{2𝑥}\right)+{\cos }^{-1}\left(\frac{2𝑥}{1+{𝑥}^2}\right)$

$\text{ Let }𝑡=\frac{1+{𝑥}^2}{2𝑥}\text{ and }𝑢=\frac{2𝑥}{1+{𝑥}^2}$

$\because 𝑡=\frac{1}{𝑢}$

$\therefore 𝑡\cdot 𝑢\equiv 1$

We know that

${\sin }^{-1}\left(𝑎\right)+{\cos }^{-1}\left(𝑎\right)=\frac{𝜋}{2},𝑎\in [-1,1]$

Thus, ${\sin }^{-1}\left(𝑡\right)+{\cos }^{-1}\left(𝑢\right)=\frac{𝜋}{2}$

So, the range of the function

$𝑓\left(𝑥\right)={\sin }^{-1}\left(\frac{1+{𝑥}^2}{2𝑥}\right)+{\cos }^{-1}\left(\frac{2𝑥}{1+{𝑥}^2}\right)\text{ is }\left\{\frac{𝜋}{2}\right\}$

Given function $𝑓:𝑅\to [\frac{5}{2},\infty ]$ is defined by

$𝑓\left(𝑥\right)=|2𝑥+1|+|𝑥-2|$

Consider the graph

There are two points, where the expressions inside the absolute values change signs.

$𝑥=-\frac{1}{2}\text{ or }-0.5\text{ and }𝑥=2$

If $𝑥<-0.5$

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

If -0.5 < x < 2

$𝑓\left(𝑥\right)=2𝑥+1-𝑥+2=𝑥+3$

If $𝑥>2$

$𝑓\left(𝑥\right)=2𝑥+1+𝑥-2=3𝑥-1$

Each vertical line should intersect the graph at exactly one point. If not then function is not onto.

If a horizontal line intersects the graph more than once, the function is not one-one.

Clearly, from graph, through horizontal line two cut points implies that not one-one and through vertical line one cut point implies that onto.

Hence, function is onto but not one-one.

To solve the given problem, we need to analyze the expression for the sum of terms and match it with the provided equation.

The sequence is structured as:

$1\cdot 3\cdot 5+3\cdot 5\cdot 7+5\cdot 7\cdot 9+\dots$

We define the $𝑟$-th term as:

${𝑎}_{𝑟}=(2𝑟-1)(2𝑟+1)(2𝑟+3)$

Calculating ${𝑎}_{𝑟}$, we expand:

${𝑎}_{𝑟}=(2𝑟-1)(2𝑟+1)(2𝑟+3)=8{𝑟}^3+12{𝑟}^2-2𝑟-3$

The sum ${𝑆}_{𝑛}$ of the first $𝑛$ terms is:

${𝑆}_{𝑛}=\sum _{𝑛}^{𝑟=1}{𝑎}_{𝑟}=\sum _{𝑛}^{𝑟=1}(8{𝑟}^3+12{𝑟}^2-2𝑟-3)$

Breaking it down, we compute:

${𝑆}_{𝑛}=8\sum _{𝑛}^{𝑟=1}{𝑟}^3+12\sum _{𝑛}^{𝑟=1}{𝑟}^2-2\sum _{𝑛}^{𝑟=1}𝑟-3\sum _{𝑛}^{𝑟=1}1$

Substituting the formula for each sum, we have:

$8{\left[\frac{𝑛(𝑛+1)}{2}\right]}^2+12\left[\frac{𝑛(𝑛+1)(2𝑛+1)}{6}\right]-2\left[\frac{𝑛(𝑛+1)}{2}\right]-3𝑛$

Simplifying, this becomes:

${𝑆}_{𝑛}=𝑛(𝑛+1)(2{𝑛}^2+6𝑛+1)-3𝑛$

Here, the function $𝑓\left(𝑛\right)$ is:

$𝑓\left(𝑛\right)=2{𝑛}^2+6𝑛+1$

To find $𝑓\left(1\right)$, substitute $𝑛=1$:

$𝑓\left(1\right)=2(1{)}^2+6(1)+1=9$

Thus, the function $𝑓\left(𝑙\right)$ evaluates to $9$.

$\begin{array}{cc}& \text{ Given that }3𝐴=⎡⎣⎢\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ 𝑎& 2& 𝑏\end{array}⎤⎦⎥\\ & \begin{array}{cc}\Rightarrow 𝐴& =⎡⎣⎢\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{𝑎}{3}& \frac{2}{3}& \frac{𝑏}{3}\end{array}⎤⎦⎥\\ \therefore {𝐴}^{𝑇}& =⎡⎣⎢\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{𝑎}{3}\\ \frac{2}{3}& \frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{𝑏}{3}\end{array}⎤⎦⎥\end{array}\end{array}$

$\begin{array}{cc}& \text{ Now, }𝐴{𝐴}^{𝑇}=𝐼\Rightarrow ⎡⎣⎢\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{2}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{𝑎}{3}& \frac{2}{3}& \frac{𝑏}{3}\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}\frac{1}{3}& \frac{2}{3}& \frac{𝑎}{3}\\ \frac{2}{3}& \frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{2}{3}& \frac{𝑏}{3}\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥\\ & \Rightarrow ⎡⎣⎢\begin{array}{ccc}\frac{1}{9}+\frac{4}{9}+\frac{4}{9}& \frac{2}{9}+\frac{2}{9}-\frac{4}{9}& \frac{𝑎}{9}+\frac{4}{9}+\frac{2𝑏}{9}\\ \frac{2}{9}+\frac{2}{9}-\frac{4}{9}& \frac{4}{9}+\frac{1}{9}+\frac{4}{9}& \frac{2𝑎}{9}+\frac{2}{9}-\frac{2𝑏}{9}\\ \frac{𝑎}{9}+\frac{4}{9}+\frac{2𝑏}{9}& \frac{2𝑎}{9}+\frac{2}{9}-\frac{2𝑏}{9}& \frac{{𝑎}^2}{9}+\frac{4}{9}+\frac{{𝑏}^2}{9}\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& \Rightarrow ⎡⎣⎢\begin{array}{ccc}1& 0& \frac{𝑎+4+2𝑏}{9}\\ \frac{𝑎+4+2𝑏}{9}& \frac{2𝑎+2-2𝑏}{9}& \frac{\frac{{𝑎}^2+4+{𝑏}^2}{9}}{0}\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \therefore {𝑎}^2+{𝑏}^2+4=9...\left(𝑖\right)\\ & \Rightarrow {𝑎}^2+{𝑏}^2=5\\ & \text{ and }2𝑎+2-2𝑏=0\\ & \Rightarrow 𝑏=𝑎+1...\left(𝑖𝑖\right)\\ & \text{ and }𝑎+4+2𝑏=0\\ & \Rightarrow 𝑎+2𝑏=-4...\left(𝑖𝑖𝑖\right)\end{array}\\ & \text{ On solving Eqs. (ii) and (iii), we get }\\ & 𝑎+2𝑏=-4\end{array}$

$\begin{array}{cccc}\Rightarrow & & 𝑎+2(𝑎+1)& =-4\\ \Rightarrow & & 𝑎+2𝑎+2& =-4\\ \Rightarrow & & 3𝑎& =-6\\ \Rightarrow & & 𝑎& =-2\text{ and }𝑏=-2+1\\ \Rightarrow & & 𝑏& =-1\end{array}$

$\begin{array}{cc}& \\ & 𝑁𝑜𝑤,\frac{𝑎}{𝑏}+\frac{𝑏}{𝑎}=\frac{-2}{-1}+\frac{-1}{-2}=2+\frac{1}{2}=\frac{5}{2}\end{array}$

$\begin{array}{cc}& \text{ We have, }\\ & \begin{array}{cc}& \mid \begin{array}{ccc}𝑎+𝑏+2𝑐& 𝑎& 𝑏\\ 𝑐& 𝑏+𝑐+2𝑎& 𝑏\\ 𝑐& 𝑎& 𝑐+𝑎+2𝑏\end{array}\mid \\ & \text{ Applying }{𝐶}_1\to {𝐶}_1+{𝐶}_2+{𝐶}_3\\ & =\mid \begin{array}{ccc}2(𝑎+𝑏+𝑐)& 𝑎& 𝑏\\ 2(𝑏+𝑐+𝑎)& 𝑏+𝑐+2𝑎& 𝑏\\ 2(𝑐+𝑎+𝑏)& 𝑎& 𝑐+𝑎+2𝑏\end{array}\mid \end{array}\end{array}$

$\begin{array}{cc}& =2(𝑎+𝑏+𝑐)\mid \begin{array}{ccc}1& 𝑎& 𝑏\\ 1& 𝑏+𝑐+2𝑎& 𝑏\\ 1& 𝑎& 𝑐+𝑎+2𝑏\end{array}\mid \\ & \text{ Applying }{𝑅}_1\to {𝑅}_1-{𝑅}_2\text{ and }{𝑅}_2\to {𝑅}_2-{𝑅}_3\\ & \begin{array}{cc}& =2(𝑎+𝑏+𝑐)\mid \begin{array}{ccc}0& -(𝑎+𝑏+𝑐)& 0\\ 0& 𝑎+𝑏+𝑐& -(𝑎+𝑏+𝑐)\\ 1& 𝑎& 𝑐+𝑎+2𝑏\end{array}\mid \\ & =2(𝑎+𝑏+𝑐{)}^3\mid \begin{array}{ccc}0& -1& 0\\ 0& 1& -1\\ 1& 𝑎& 𝑐+𝑎+2𝑏\end{array}\mid \\ & =2(𝑎+𝑏+𝑐{)}^3[1(1-0)]=2(𝑎+𝑏+𝑐{)}^3\end{array}\end{array}$

To clarify:

When we have a matrix $𝐵$ of order 3 with a determinant $\left|𝐵\right|=6$, the formula to find the determinant of its adjugate matrix, $|\mathrm{adj}(𝐵\left)\right|$, is given by:

$|\mathrm{adj}(𝐵\left)\right|=|𝐵{|}^{𝑛-1}$

where $𝑛$ is the order of the matrix $𝐵$.

In this case, since $𝑛=3$, we have:

$|\mathrm{adj}(𝐵\left)\right|=6^{3-1}=6^2=36$

This confirms that the assertion is true.

However, the reason provided, $|\mathrm{adj}(𝐵\left)\right|=|𝐵{|}^{𝑛}$, is incorrect because it states that the determinant of the adjugate matrix is raised to the power of $𝑛$ rather than $𝑛-1$. The correct relationship involves raising the determinant of $𝐵$ to the power of $𝑛-1$.

Thus, the assertion is true, but the given reason is false.

We have,

$\begin{array}{cc}& \frac{(1-𝑖{)}^3}{(2-𝑖)(3-2𝑖)}=\frac{1^3-{𝑖}^3-3.1.𝑖(1-𝑖)}{(6-4𝑖-3𝑖+2{𝑖}^2)}\\ & [\because (𝑎-𝑏{)}^3={𝑎}^3-{𝑏}^3-3𝑎𝑏(𝑎-𝑏)]\\ & =\frac{1-(-𝑖)-3𝑖+3{𝑖}^2}{(6-7𝑖-2)}=\frac{1+𝑖-3𝑖-3}{4-7𝑖}\\ & =\frac{-2-2𝑖}{4-7𝑖}\\ & =\frac{-2-2𝑖}{4-7𝑖}\times \frac{4+7𝑖}{4+7𝑖}\left[𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙𝑖𝑠𝑖𝑛𝑔\right]\\ & =\frac{-(2+2𝑖)(4+7𝑖)}{4^2-(7𝑖{)}^2}\\ & =\frac{-(8+14𝑖+8𝑖-14)}{16+49}=\frac{6-22𝑖}{65}\end{array}$

On comparing with $𝑍=𝑥+𝑖𝑦$, we get

Imaginary part is $\frac{-22}{65}$.

To find the square root of $7+24𝑖$, set it equal to $𝑥+𝑦𝑖$, where $𝑥,𝑦\in ℝ$.

Squaring both sides gives:

$(𝑥+𝑦𝑖{)}^2={𝑥}^2-{𝑦}^2+2𝑥𝑦𝑖$

This leads to:

${𝑥}^2-{𝑦}^2+2𝑥𝑦𝑖=7+24𝑖$

By equating the real and imaginary parts, we obtain:

${𝑥}^2-{𝑦}^2=7\text{(equation 1)}$

$2𝑥𝑦=24\Rightarrow 𝑥𝑦=12\text{(equation 2)}$

From equation 2, solve for $𝑦$:

$𝑦=\frac{12}{𝑥}$

Substitute $𝑦=\frac{12}{𝑥}$ into equation 1:

${𝑥}^2-{\left(\frac{12}{𝑥}\right)}^2=7$

Simplifying the equation, we get:

${𝑥}^2-\frac{144}{{𝑥}^2}=7$

${𝑥}^4-144=7{𝑥}^2$

${𝑥}^4-7{𝑥}^2-144=0$

This can be factored as:

$({𝑥}^2-16)({𝑥}^2+9)=0$

Since $𝑥$ is real, we discard ${𝑥}^2+9=0$ as it does not lead to real solutions for $𝑥$. Therefore:

${𝑥}^2=16\Rightarrow 𝑥=\pm 4$

For $𝑥=4$, $𝑦=3$.

For $𝑥=-4$, $𝑦=-3$.

Thus, the square roots of $7+24𝑖$ are $\pm (4+3𝑖)$.

Therefore, the correct solution is $4+3𝑖$.

We have,

$𝑍=\cos 𝜃+𝑖\sin 𝜃,𝜃\ne (2𝑛+1)\frac{𝜋}{2}$

Now, $\frac{1+{𝑍}^{2𝑛}}{1-{𝑍}^{2𝑛}}=\frac{1+(\cos 𝜃+𝑖\sin 𝜃{)}^{2𝑛}}{1-(\cos 𝜃+𝑖\sin 𝜃{)}^{2𝑛}}$

$=\frac{1+(\cos 2𝑛𝜃+𝑖\sin 2𝑛𝜃)}{1-(\cos 2𝑛𝜃+𝑖\sin 2𝑛𝜃)}$

[By De Moivre's theorem]

$\begin{array}{cc}& =\frac{1+(2{\cos }^2𝑛𝜃-1)+2𝑖\sin 𝑛𝜃\cos 𝑛𝜃}{1-(1-2{\sin }^2𝑛𝜃)-2𝑖\sin 𝑛𝜃\cos 𝑛𝜃}\\ & [\cos 2𝐴=2{\cos }^2𝐴-1=1-2{\sin }^2𝐴]\\ & =\frac{2\cos 𝑛𝜃[\cos 𝑛𝜃+𝑖\sin 𝑛𝜃]}{2\sin 𝑛𝜃[\sin 𝑛𝜃-𝑖\cos 𝑛𝜃]}\\ & =-𝑖\cot 𝑛𝜃\left[\frac{\cos 𝑛𝜃+𝑖\sin 𝑛𝜃}{\cos 𝑛𝜃+𝑖\sin 𝑛𝜃}\right]\\ & =-𝑖\cot 𝑛𝜃.\end{array}$

Given that $𝑥$ is real and $𝛼,𝛽$ are the maximum and minimum values of $\frac{{𝑥}^2-𝑥+1}{{𝑥}^2+𝑥+1}$, respectively.

$𝐿𝑒𝑡𝑦=\frac{{𝑥}^2-𝑥+1}{{𝑥}^2+𝑥+1}$

$\Rightarrow 𝑦{𝑥}^2+𝑦𝑥+𝑦={𝑥}^2-𝑥+1$

$\Rightarrow (𝑦-1){𝑥}^2+(𝑦+1)𝑥+𝑦-1=0$

$(𝑥\in 𝑅)$

$\begin{array}{cc}& \text{ Consider }{𝑏}^2-4𝑎𝑐\ge 0\\ & \Rightarrow (𝑦+1{)}^2-4(𝑦-1{)}^2\ge 0\\ & \Rightarrow (𝑦+1+2𝑦-2)(𝑦+1-2𝑦+2)\ge 0\\ & \Rightarrow (3𝑦-1)(𝑦-3)\le 0\end{array}$

$\begin{array}{cc}& \text{ Range }=[\frac{1}{3},3]\\ & \begin{array}{cc}& \because \text{ Minimum value }=\frac{1}{3}=𝛽\\ & \text{ Maximum value }=3=𝛼\\ & 𝛼+𝛽=3+\frac{1}{3}=\frac{10}{3}\end{array}\end{array}$

Given,

$\begin{array}{cc}& {𝑥}^2-5𝑥+𝜆=0...\left(𝑖\right)\\ & {𝑥}^2-8𝑥-2𝜆=0...\left(𝑖𝑖\right)\end{array}$

$\because 𝛼$ is a common factor of Eqs (i) and (ii), we get

And $𝛽,𝛾$ are the other roots of them,

So

$\begin{array}{cc}& {𝛼}^2-5𝛼+𝜆=0\\ & {𝛼}^2-8𝛼-2𝜆=0\\ & -(+)(+)\\ & 3𝛼+3𝜆=0\Rightarrow 𝜆=-𝛼\end{array}$

$\text{ Then, }{𝛼}^2-5𝛼-𝛼=0\Rightarrow {𝛼}^2-6𝛼=0$

$\begin{array}{cc}& \begin{array}{cc}& \Rightarrow 𝛼(𝛼-6)=0\Rightarrow 𝛼=0,6\\ & 𝛼\ne 0\\ & \Rightarrow 𝛼=6\text{ and }𝜆=-6\end{array}\\ & \text{ Now, from Eq. (i), we get }\end{array}$

$\begin{array}{cccc}& & {𝑥}^2-5𝑥-6& =0\\ \Rightarrow & & {𝑥}^2-6𝑥+𝑥-6& =0\\ \Rightarrow & & (𝑥-6)(𝑥+1)& =0\\ \Rightarrow & & 𝑥& =6,𝑥=-1\end{array}$

$\begin{array}{cc}& \therefore 𝛽=-1\\ & \text{ From Eq. (ii), we get }\end{array}$

$\text{begin{array} \& x^2-8 x-2 times(-6) \& =0 Rightarrow quad x^2-8 x+12 \& =0 Rightarrow quad(x-6)(x-2) \& =0 Rightarrow x=6,2 end{array}}$

$\therefore 𝛾=2$

$\begin{array}{cc}& \\ & \begin{array}{cc}𝑁𝑜𝑤,𝛼+𝛽+𝛾+𝜆& =6+(-1)+2+(-𝜃)\\ & =1\end{array}\end{array}$

To solve the given polynomial equation ${𝑥}^4-{𝑥}^3-6{𝑥}^2+4𝑥+8=0$ with two equal roots, we proceed as follows:

The equation can be expressed as:

${𝑥}^3(𝑥+1)-2{𝑥}^2(𝑥+1)-4𝑥(𝑥+1)+8(𝑥+1)=0$

Factoring out $(𝑥+1)$ gives:

$(𝑥+1)({𝑥}^3-2{𝑥}^2-4𝑥+8)=0$

Further factoring results in:

$(𝑥+1)(𝑥-2)({𝑥}^2-4)=0$

This simplifies to:

$𝑥+1=0,𝑥-2=0,{𝑥}^2-4=0$

Hence, the roots are:

$𝑥=-1,𝑥=2,𝑥=\pm 2$

The distinct roots of the equation are $-2$, $-1$, and $2$. Since one root, $2$, repeats, we determine the remaining roots as $𝛼=-2$ and $𝛽=-1$.

To find ${𝛼}^2+{𝛽}^2$, compute:

${𝛼}^2+{𝛽}^2=(-2{)}^2+(-1{)}^2=4+1=5$

Thus, ${𝛼}^2+{𝛽}^2=5$.

Given the polynomial equation ${𝑥}^3-𝑏{𝑥}^2+𝑐𝑥-𝑑=0$, we need to find the condition for its roots to be in arithmetic progression.

Let the roots of this polynomial be $𝛼$, $𝛽$, and $𝛾$. Since these roots are in arithmetic progression, they satisfy the condition:

$2𝛽=𝛼+𝛾$

We also know from the properties of polynomials that the sum of the roots is equal to the coefficient of the ${𝑥}^2$ term, which is $𝑏$. Thus,

$𝛼+𝛽+𝛾=𝑏$

Substituting the condition of arithmetic progression:

$3𝛽=𝑏\Rightarrow 𝛽=\frac{𝑏}{3}$

Next, we use the condition for the sum of the products of the roots taken two at a time:

$𝛼𝛽+𝛽𝛾+𝛾𝛼=𝑐$

And for the product of the roots:

$𝛼𝛽𝛾=𝑑$

Substituting $𝛽=\frac{𝑏}{3}$, the product becomes:

$𝛼𝛾\cdot \frac{𝑏}{3}=𝑑\Rightarrow 𝛼𝛾=\frac{3𝑑}{𝑏}$

Now substituting the values of $𝛽$ and $𝛼𝛾$ into the equation for the sum of the products of the roots taken two at a time:

$𝛼\cdot \frac{𝑏}{3}+\frac{𝑏}{3}𝛾+\frac{3𝑑}{𝑏}=𝑐$

Simplifying further:

$\frac{𝑏}{3}(𝛼+𝛾)+\frac{3𝑑}{𝑏}=𝑐$

Using the condition $𝛼+𝛾=2𝛽$:

$\frac{𝑏}{3}\cdot 2𝛽+\frac{3𝑑}{𝑏}=𝑐$

Substituting $𝛽=\frac{𝑏}{3}$:

$\frac{𝑏}{3}\cdot \frac{2𝑏}{3}+\frac{3𝑑}{𝑏}=𝑐$

Simplifying this expression leads to:

$\frac{2{𝑏}^2}{9}+\frac{3𝑑}{𝑏}=𝑐$

By multiplying through by $9𝑏$ to clear denominators, we become:

$2{𝑏}^3+27𝑑=9𝑏𝑐$

Thus, the condition for the roots of the polynomial to be in arithmetic progression is:

$9𝑏𝑐=2{𝑏}^3+27𝑑$

and number of poetry $=3$

$\therefore$ Number of ways of selecting 4 novels from 6

$={}^6{𝐶}_4\text{ way }$

Number of ways of selecting 1 poetry from 3.

$={}^3{𝐶}_1\text{ way }$

When poetry is always in middle,

So, 4 nowels can be arrange in 4 ! ways.

$\therefore$ Required number of ways

$\begin{array}{cc}& ={}^6{𝐶}_4\times {}^3{𝐶}_1\times 4!\\ & =\frac{6\times 5}{2\times 1}\times 3\times 24\\ & =15\times 3\times 24\\ & =1080\end{array}$

To form a five-digit number using the digits 0, 1, 2, 3, 4, and 5, which must be divisible by 3, we need to consider the divisibility rule for 3. A number is divisible by 3 if the sum of its digits is a multiple of 3. This gives us two possible scenarios to examine:

Case I: Using the digits 0, 1, 2, 4, and 5.

The sum of these digits is $0+1+2+4+5=12$, which is divisible by 3.

To avoid starting with 0 (as it wouldn't be a five-digit number), the first digit has 4 options (1, 2, 4, or 5).

The remaining four digits can be arranged in $4!$ (4 factorial) ways.

Thus, the number of ways to arrange these digits is:

$4\times 4\times 3\times 2\times 1=96$

Case II: Using the digits 1, 2, 3, 4, and 5.

The sum of these digits is $1+2+3+4+5=15$, which is also divisible by 3.

Any of these five digits can be at the start, providing 5 choices for the first digit.

The remaining four digits are arranged in $4!$ ways.

Thus, the number of ways to arrange these digits is:

$5\times 4\times 3\times 2\times 1=120$

Conclusion:

Adding the numbers from both scenarios gives the total number of five-digit numbers divisible by 3:

$96+120=216$

Four-digit number formed using the digits $\{1,2,3,4,5,6,7\}$ with all distinct digits.

$𝑝={}^7{𝑃}_4=7\times 6\times 5\times 4=840$

Case I Leading digit is 3 .

For the number to be greater than 3400 , the second digit must be $4,5,6$ or 7 .

(i) Fix 3 as first digit.

(ii) Choose one of 4, 5, 6 or 7 for the second digits ( 4 choices).

(iii) For the third and fourth digits, Choose from the remaining 5 digits So, for each choice of the second digit $5\times 4=20$

There are 4 choices for the second digits so, $4\times 20=80$

Case II Leading digit is $4,5,6$ or 7 .

Any number starting with $4,5,6$ or 7 is automatically greater than 3400 .

There are 4 choices for the first digit (4,5,6 or 7 )

Choose any 3 digits from the remaining 6 digits.

So, for each choice of the first digit

$6\times 5\times 4=120$

There are 4 choices for the first digit, so

$4\times 120=480$

Total numbers greater than 3400

$𝑞=80+480=560$

Now, $\frac{𝑝}{𝑞}=\frac{840}{560}=\frac{3}{2}$

So, the ratio $𝑝:𝑞=3:2$.

In the expansion of $(1+𝑥{)}^{12}$, we need to determine the sum of all its terms given the ratio of terms equidistant from the middle is $\frac{1}{256}$.

Let's find the middle term and the equidistant terms:

The $𝑘$-th term is denoted by:

${𝑇}_{𝑘}=(\frac{12}{𝑘}){𝑥}^{𝑘}$

The equidistant terms from the middle are:

${𝑇}_{12-𝑘}=(\frac{12}{12-𝑘}){𝑥}^{12-𝑘}$

Since, the ratio of equidistant terms is given by:

$\frac{{𝑇}_{𝑘}}{{𝑇}_{12-𝑘}}=\frac{(\frac{12}{𝑘}){𝑥}^{𝑘}}{(\frac{12}{12-𝑘}){𝑥}^{12-𝑘}}=\frac{1}{256}$

Since $(\frac{12}{12-𝑘})=(\frac{12}{𝑘})$, the equation simplifies to:

$\frac{{𝑥}^{𝑘}}{{𝑥}^{12-𝑘}}={𝑥}^{-12+2𝑘}=\frac{1}{256}=2^{-8}$

This gives us:

$-12+2𝑘=-8$

Solving for $𝑘$, we find:

$2𝑘=4⟹𝑘=2$

Thus, the equidistant terms are ${𝑇}_2$ and ${𝑇}_{10}$.

Now, calculate the sum of all terms in the expansion:

When $𝑥=2$:

$𝑆=(1+2{)}^{12}=3^{12}$

When $𝑥=4$:

$𝑆=(1+4{)}^{12}=5^{12}$

Therefore, the sum of all terms could either be $3^{12}$ or $5^{12}$ based on the value of $𝑥$.

Given expression $\frac{2𝑥+1}{(1+𝑥)(1-2𝑥)}$

By partial fraction

$\begin{array}{cc}\frac{(2𝑥+1)}{(1+𝑥)(1-2𝑥)}& =\frac{𝐴}{(1+𝑥)}+\frac{𝐵}{(1-2𝑥)}...\left(𝑖\right)\\ 2𝑥+1& =𝐴(1-2𝑥)+𝐵(1+𝑥)\\ 2𝑥+1& =𝐴+𝐵+(-2𝐴+𝐵)𝑥\end{array}$

On comparing coefficients both sides, we get

$\begin{array}{c}𝐴+𝐵=1\Rightarrow 𝐴=1-𝐵...\left(𝑖𝑖\right)\\ -2𝐴+𝐵=2...\left(𝑖𝑖𝑖\right)\end{array}$

From Eqs. (ii) and (iii), we get

$\begin{array}{cccc}& & -2+2𝐵+𝐵& =2\\ \Rightarrow & & 𝐵& =\frac{4}{3}\text{ and }𝐴=-\frac{1}{3}\end{array}$

On putting the value of $𝐴$ and $𝐵$ in Eq.

(i), we get

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

$\begin{array}{cc}& =-\frac{1}{3}[1-𝑥+{𝑥}^2-{𝑥}^3+{𝑥}^4-{𝑥}^5+{𝑥}^6-{𝑥}^7+{𝑥}^8-{𝑥}^9]+\frac{4}{3}[1+2𝑥+4{𝑥}^2+8{𝑥}^3+16{𝑥}^4+32{𝑥}^5]\\ & 64{𝑥}^6+128{𝑥}^7+256{𝑥}^8+512{𝑥}^9+\dots ]\end{array}$

$\begin{array}{cc}& =(-\frac{1}{3}+\frac{4}{3})+(\frac{1}{3}+\frac{1}{8})𝑥+(-\frac{1}{3}+\frac{16}{3}){𝑥}^2+(\frac{1}{3}+\frac{32}{3}){𝑥}^3+(-\frac{1}{3}+\frac{64}{3}){𝑥}^4+(\frac{1}{3}+\frac{128}{3}){𝑥}^5\\ & +(-\frac{1}{3}+\frac{256}{3}){𝑥}^6+(\frac{1}{3}+\frac{512}{3}){𝑥}^7+(-\frac{1}{3}+\frac{1024}{3}){𝑥}^8+(\frac{1}{3}+\frac{2024}{3}){𝑥}^5\end{array}$

Now, sum of coefficients of first 5 odd power of $𝑥$, we get

$\begin{array}{cc}& =(\frac{1}{3}+\frac{8}{3})+(\frac{1}{3}+\frac{32}{3})+(\frac{1}{3}+\frac{128}{3})+(\frac{1}{3}+\frac{512}{3})+(\frac{1}{3}+\frac{2048}{3})\\ & =[\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}]+[\frac{8}{3}+\frac{32}{3}+\frac{128}{3}+\frac{512}{3}+\frac{2048}{3}]\end{array}$