TG EAPCET 2024 10th May Evening Shift Question Paper with Solutions

To find the domain of the function $𝑓\left(𝑥\right)={\sin }^{-1}({\log }_2\left(\frac{{𝑥}^2}{2}\right))$, we need to determine where the expression ${\log }_2\left(\frac{{𝑥}^2}{2}\right)$ falls within the domain of the inverse sine function, which is $[-1,1]$.

First, ensure that:

$-1\le {\log }_2\left(\frac{{𝑥}^2}{2}\right)\le 1$

This inequality translates to:

$2^{-1}\le \frac{{𝑥}^2}{2}\le 2$

Solving the inequality:

$\frac{1}{2}\le \frac{{𝑥}^2}{2}\le 2$

Multiply through by 2 to clear the fraction:

$1\le {𝑥}^2\le 4$

This implies that $𝑥$ must satisfy:

$𝑥\in [-2,-1]\cup [1,2]$

We also need to ensure that $\frac{{𝑥}^2}{2}\ge 0$, which is always true since ${𝑥}^2\ge 0$.

Therefore, considering the intersection of the feasible solutions:

$𝑥\in [-2,-1]\cup [1,2]$

Thus, the domain of the function is:

$𝑥\in [-2,-1]\cup [1,2]$

To determine the range of the function $𝑓\left(𝑥\right)={\log }_3(5+4𝑥-{𝑥}^2)$, we need to analyze the behavior of the quadratic expression $5+4𝑥-{𝑥}^2$.

Analysis of $5+4𝑥-{𝑥}^2$

The expression $5+4𝑥-{𝑥}^2$ is a quadratic function of the form $𝑎{𝑥}^2+𝑏𝑥+𝑐$, where $𝑎=-1$, $𝑏=4$, and $𝑐=5$.

Determining the Range of the Quadratic

Since $𝑎=-1<0$, the parabola opens downwards, and the quadratic function achieves its maximum value at its vertex. The maximum (or minimum for upward-opening parabolas) value of a quadratic function $𝑎{𝑥}^2+𝑏𝑥+𝑐$ is given by $-\frac{𝐷}{4𝑎}$, where $𝐷={𝑏}^2-4𝑎𝑐$ is the discriminant.

Calculate the discriminant:

$𝐷=4^2-4\times (-1)\times 5=16+20=36$

The maximum value of the quadratic $5+4𝑥-{𝑥}^2$ is:

$-\frac{𝐷}{4𝑎}=-\frac{36}{4\times (-1)}=9$

Hence, the range of $5+4𝑥-{𝑥}^2$ is $(-\infty ,9]$.

Translating to the Logarithmic Function

Considering the function $𝑓\left(𝑥\right)={\log }_3(5+4𝑥-{𝑥}^2)$, the possible values of $5+4𝑥-{𝑥}^2$ determine the input values for the logarithm.

Since $5+4𝑥-{𝑥}^2\le 9$, we take the logarithm base 3 of each side:

${\log }_3(5+4𝑥-{𝑥}^2)\le {\log }_3\left(9\right)$

Given that ${\log }_3\left(9\right)=2$, we conclude:

$𝑓\left(𝑥\right)={\log }_3(5+4𝑥-{𝑥}^2)\le 2$

Thus, the range of $𝑓\left(𝑥\right)$ is $(-\infty ,2]$.

Given the expression $3^{2𝑛+2}-8𝑛-9$, we want to determine the maximum power of $2$ that divides this expression for all $𝑛\in \mathbb{N}$.

First, we can rewrite $3^{2𝑛+2}$ as $9^{𝑛+1}$:

$9^{𝑛+1}-8𝑛-9$

Considering $9=1+8$, we can apply the binomial expansion to $(1+8{)}^{𝑛+1}$:

$(1+8{)}^{𝑛+1}={}^{𝑛+1}{𝐶}_0+{}^{𝑛+1}{𝐶}_1\cdot 8+{}^{𝑛+1}{𝐶}_2\cdot 8^2+{}^{𝑛+1}{𝐶}_3\cdot 8^3+\dots$

Simplifying, we get:

$1+8(𝑛+1)+8^2{}^{𝑛+1}{𝐶}_2+{}^{𝑛+1}{𝐶}_3\cdot 8^3+\dots$

Subtracting $8𝑛+9$ from this sequence, we have:

$1+8𝑛+8+8^2{}^{𝑛+1}{𝐶}_2+\dots -8𝑛-9$

Which simplifies to:

$8^2({}^{𝑛+1}{𝐶}_2+8{}^{𝑛+1}{𝐶}_3+\dots )$

The expression $8^2$ can be further evaluated as $2^6$. Thus, the binomial expansion shows that after simplification, the given expression is a multiple of $2^6$.

Therefore, the maximum value of $𝑝$ such that the expression is divisible by $2^{𝑝}$ is:

$\overline{)6}$

Let's consider the $3\times 3$ matrix $𝐴=⎡⎣⎢\begin{array}{ccc}{𝑎}_{11}& {𝑎}_{12}& {𝑎}_{13}\\ {𝑎}_{21}& {𝑎}_{22}& {𝑎}_{23}\\ {𝑎}_{31}& {𝑎}_{32}& {𝑎}_{33}\end{array}⎤⎦⎥$ where each element is a positive integer and each row sums to 6, with ${𝑎}_{22}=2$.

The matrix properties provide the following conditions:

${𝑎}_{22}={𝑎}_{23}+{𝑎}_{32}$ implies ${𝑎}_{23}+{𝑎}_{32}=2$.

${𝑎}_{11}={𝑎}_{12}+{𝑎}_{21}$.

${𝑎}_{33}={𝑎}_{31}+{𝑎}_{13}$.

Substituting the conditions about row sums, we get:

${𝑎}_{11}+{𝑎}_{12}+{𝑎}_{13}=6$.

${𝑎}_{21}+{𝑎}_{22}+{𝑎}_{23}=6$.

${𝑎}_{31}+{𝑎}_{32}+{𝑎}_{33}=6$.

From ${𝑎}_{23}+{𝑎}_{32}=2$, if ${𝑎}_{23}=1$, then ${𝑎}_{32}=1$.

From the second row sum, we have: ${𝑎}_{21}+2+1=6$ which gives ${𝑎}_{21}=3$.

Given ${𝑎}_{11}={𝑎}_{12}+{𝑎}_{21}$ and the first row sum,

with $2{𝑎}_{12}+{𝑎}_{13}=3$,

choosing ${𝑎}_{12}=1$ and ${𝑎}_{13}=1$, results in ${𝑎}_{11}=4$.

Finally, using the third row equation $2{𝑎}_{31}+{𝑎}_{32}+{𝑎}_{13}=6$,

substituting ${𝑎}_{32}=1$ and ${𝑎}_{13}=1$ results in ${𝑎}_{31}=2$.

Now, from ${𝑎}_{31}+{𝑎}_{32}+{𝑎}_{33}=6$, with ${𝑎}_{31}=2$ and ${𝑎}_{32}=1$,

results in ${𝑎}_{33}=3$.

Matrix $𝐴$ is:

$𝐴=⎡⎣⎢\begin{array}{ccc}4& 1& 1\\ 3& 2& 1\\ 2& 1& 3\end{array}⎤⎦⎥$

To find the determinant $\left|𝐴\right|$, we calculate:

$\left|𝐴\right|=4(2\cdot 3-1\cdot 1)-1(3\cdot 3-1\cdot 2)+1(3\cdot 1-2\cdot 2)$

$=4(6-1)-1(9-2)+1(3-4)$

$=4\cdot 5-1\cdot 7+1\cdot (-1)$

$=20-7-1=12$

Thus, the determinant $\left|𝐴\right|=12$.

Given that $|\mathrm{adj}𝐴|=𝑥$ and $|\mathrm{adj}𝐵|=𝑦$, we are tasked with finding $\mid (\mathrm{adj}(𝐴𝐵){)}^{-1}\mid$.

First, recognize that the determinant of the adjugate of a matrix can be expressed with respect to the determinants of the matrix itself. Using this relation, we have:

$|\mathrm{adj}(𝐴𝐵\left)\right|=|\mathrm{adj}(\mathrm{adj}\left(𝐴\right)\left)\right|\cdot |\mathrm{adj}(\mathrm{adj}\left(𝐵\right)\left)\right|$

The inverse of the adjugate adjoint product can be decomposed as:

$\mid (\mathrm{adj}(𝐴𝐵){)}^{-1}\mid =\mid (\mathrm{adj}𝐵{)}^{-1}(\mathrm{adj}𝐴{)}^{-1})\mid$

Because $|\mathrm{adj}𝐴|=𝑥$ and $|\mathrm{adj}𝐵|=𝑦$, the properties of determinants give us:

$\mid (\mathrm{adj}𝐴{)}^{-1}\mid =\frac{1}{|\mathrm{adj}𝐴|}=\frac{1}{𝑥}$

$\mid (\mathrm{adj}𝐵{)}^{-1}\mid =\frac{1}{|\mathrm{adj}𝐵|}=\frac{1}{𝑦}$

Thus, the determinant of the inverse of the product is:

$\mid (\mathrm{adj}(𝐴𝐵){)}^{-1}\mid =\frac{1}{𝑥}\times \frac{1}{𝑦}=\frac{1}{𝑥𝑦}$

Therefore, the result is $\frac{1}{𝑥𝑦}$.

Given, $𝑥+3𝑏𝑦+𝑏𝑧=0$

$𝑥+2𝑎𝑦+𝑎𝑧=0$

$𝑥+4𝑐𝑦+𝑐𝑧=0$

$\Rightarrow \Delta =\mid \begin{array}{ccc}1& 3𝑏& 𝑏\\ 1& 2𝑎& 𝑎\\ 1& 4𝑐& 𝑐\end{array}\mid$

$\Delta =2𝑎𝑐+3𝑎𝑏-2𝑎𝑏+4𝑏𝑐-4𝑎𝑐-3𝑏𝑐$

$\Delta =-2𝑎𝑐+𝑎𝑏+𝑏𝑐$

For non-zero solution $\Delta =0$

$\Rightarrow 2𝑎𝑐=𝑎𝑏+𝑏𝑐$

To evaluate the determinant of the matrix:

$\mid \begin{array}{ccc}\frac{-𝑏𝑐}{{𝑎}^2}& \frac{𝑐}{𝑎}& \frac{𝑏}{𝑎}\\ \frac{𝑐}{𝑏}& -\frac{𝑎𝑐}{{𝑏}^2}& \frac{𝑎}{𝑏}\\ \frac{𝑏}{𝑐}& \frac{𝑎}{𝑐}& -\frac{𝑎𝑏}{{𝑐}^2}\end{array}\mid$

we can simplify the process by factoring out $\frac{1}{{𝑎}^2},\frac{1}{{𝑏}^2},$ and $\frac{1}{{𝑐}^2}$ from rows ${𝑅}_1,{𝑅}_2,$ and ${𝑅}_3$ respectively. This allows us to focus on a simpler determinant:

$\Rightarrow \frac{1}{{𝑎}^2{𝑏}^2{𝑐}^2}\mid \begin{array}{ccc}-𝑏𝑐& 𝑎𝑐& 𝑎𝑏\\ 𝑏𝑐& -𝑎𝑐& 𝑎𝑏\\ 𝑏𝑐& 𝑎𝑐& -𝑎𝑏\end{array}\mid$

Next, apply row operations to simplify further: replace ${𝑅}_2$ with ${𝑅}_2+{𝑅}_1$ and ${𝑅}_3$ with ${𝑅}_3+{𝑅}_1$:

$\Rightarrow \frac{1}{{𝑎}^2{𝑏}^2{𝑐}^2}\mid \begin{array}{ccc}-𝑏𝑐& 𝑎𝑐& 𝑎𝑏\\ 0& 0& 2𝑎𝑏\\ 0& 2𝑎𝑐& 0\end{array}\mid$

This matrix is now triangular, allowing us to compute the determinant by multiplying the diagonal elements:

$\Rightarrow \frac{1}{{𝑎}^2{𝑏}^2{𝑐}^2}\times \left(4{𝑎}^2{𝑏}^2{𝑐}^2\right)=4$

Thus, the value of the determinant is 4.

To solve the given equation ${𝑧}^2+𝑎𝑧+{𝑎}^2=0$, where $𝑧=𝑥+𝑖𝑦$ and $𝑎\in ℝ$, follow these steps:

First, multiply both sides of the equation by $(𝑧-𝑎)$:

$(𝑧-𝑎)({𝑧}^2+𝑎𝑧+{𝑎}^2)=0$

Apply the identity for the difference of cubes:

${𝑧}^3-{𝑎}^3=0$

This implies:

${𝑧}^3={𝑎}^3$

Taking the modulus (or absolute value) of both sides:

$|𝑧{|}^3=|𝑎{|}^3$

Thus, we conclude:

$\left|𝑧\right|=\left|𝑎\right|$

Given the complex numbers ${𝑧}_1,{𝑧}_2,{𝑧}_3$ with unit modulus, we know:

${𝑧}_1{\overline{𝑧}}_1={𝑧}_2{\overline{𝑧}}_2={𝑧}_3{\overline{𝑧}}_3=1$

This implies that each complex number has a magnitude of 1. We are also given the condition:

$|{𝑧}_1-{𝑧}_2{|}^2+|{𝑧}_1-{𝑧}_3{|}^2=4$

We can expand the squared magnitudes as follows:

$\begin{array}{cc}|{𝑧}_1-{𝑧}_2{|}^2& =({𝑧}_1-{𝑧}_2)({\overline{𝑧}}_1-{\overline{𝑧}}_2)={𝑧}_1{\overline{𝑧}}_1-{𝑧}_1{\overline{𝑧}}_2-{\overline{𝑧}}_1{𝑧}_2+{𝑧}_2{\overline{𝑧}}_2,\\ |{𝑧}_1-{𝑧}_3{|}^2& =({𝑧}_1-{𝑧}_3)({\overline{𝑧}}_1-{\overline{𝑧}}_3)={𝑧}_1{\overline{𝑧}}_1-{𝑧}_1{\overline{𝑧}}_3-{\overline{𝑧}}_1{𝑧}_3+{𝑧}_3{\overline{𝑧}}_3.\end{array}$

Substitute the expansions back into the equation:

$\begin{array}{cc}& ({𝑧}_1{\overline{𝑧}}_1-{𝑧}_1{\overline{𝑧}}_2-{\overline{𝑧}}_1{𝑧}_2+{𝑧}_2{\overline{𝑧}}_2)\\ & +({𝑧}_1{\overline{𝑧}}_1-{𝑧}_1{\overline{𝑧}}_3-{\overline{𝑧}}_1{𝑧}_3+{𝑧}_3{\overline{𝑧}}_3)=4.\end{array}$

Since ${𝑧}_1{\overline{𝑧}}_1={𝑧}_2{\overline{𝑧}}_2={𝑧}_3{\overline{𝑧}}_3=1$, we simplify:

$(1-{𝑧}_1{\overline{𝑧}}_2-{\overline{𝑧}}_1{𝑧}_2+1)+(1-{𝑧}_1{\overline{𝑧}}_3-{\overline{𝑧}}_1{𝑧}_3+1)=4$

Simplifying further gives:

$4-({𝑧}_1{\overline{𝑧}}_2+{\overline{𝑧}}_1{𝑧}_2+{𝑧}_1{\overline{𝑧}}_3+{\overline{𝑧}}_1{𝑧}_3)=4$

Thus:

${𝑧}_1{\overline{𝑧}}_2+{\overline{𝑧}}_1{𝑧}_2+{𝑧}_1{\overline{𝑧}}_3+{\overline{𝑧}}_1{𝑧}_3=0$

This is the required result.

Given the equation:

${\left(\frac{𝑎+𝑏𝜔+𝑐{𝜔}^2}{𝑐+𝑎𝜔+𝑏{𝜔}^2}\right)}^{𝑘}+{\left(\frac{𝑎+𝑏𝜔+𝑐{𝜔}^2}{𝑏+𝑎{𝜔}^2+𝑐𝜔}\right)}^{𝑙}=2$

where $𝜔$ is the complex cube root of unity, meaning ${𝜔}^3=1$ and $1+𝜔+{𝜔}^2=0$.

To solve this, we first manipulate the terms by using properties of complex numbers. Consider multiplying and dividing:

Multiply and divide the first term by ${𝜔}^2$:

${\left(\frac{{𝜔}^2(𝑎+𝑏𝜔+𝑐{𝜔}^2)}{{𝜔}^2(𝑐+𝑎𝜔+𝑏{𝜔}^2)}\right)}^{𝑘}$

Multiply and divide the second term by $𝜔$:

${\left(\frac{𝜔(𝑎+𝑏𝜔+𝑐{𝜔}^2)}{𝜔(𝑏+𝑎{𝜔}^2+𝑐𝜔)}\right)}^{𝑙}$

This simplifies to:

$\left({𝜔}^{2𝑘}\right)+\left({𝜔}^{𝑙}\right)=2$

For these powers to sum to 2, and knowing the properties of cube roots of unity, we have specific implications for the exponents:

Since ${𝜔}^{𝑛}$ can only take values 1, $𝜔$, or ${𝜔}^2$, for the sum ${𝜔}^{2𝑘}+{𝜔}^{𝑙}=2$ to be a real number (i.e., 2), both ${𝜔}^{2𝑘}$ and ${𝜔}^{𝑙}$ must equal 1.

Thus, this requires both 2k and l to be multiples of 3 (since ${𝜔}^3=1$).

Hence, the expression $2𝑘+𝑙$ is also a multiple of 3, indicating that $2𝑘+𝑙$ is always divisible by 3.

Given, ${𝑧}_1=\sqrt{\surd 3}+𝑖\sqrt{\surd 3}$

$\begin{array}{cc}{𝑧}_2& =\sqrt{\surd 3}+𝑖\\ \frac{{𝑧}_1}{{𝑧}_2}& =\frac{\sqrt{\surd 3}(1+𝑖)}{\sqrt{\surd 3}+𝑖}=\frac{\sqrt{\surd 3}(1+𝑖)}{\sqrt{\surd 3}+𝑖}\frac{(\sqrt{\surd 3}-𝑖)}{(\sqrt{\surd 3}-𝑖)}\\ & =\frac{\sqrt{\surd 3}(\sqrt{\surd 3}-𝑖+\sqrt{\surd 3}𝑖-{𝑖}^2)}{3-{𝑖}^2}\\ & =\frac{\sqrt{\surd 3}}{4}[\sqrt{\surd 3}+1+𝑖(\sqrt{\surd 3}-1\left)\right]\end{array}$

multiply and divide by $2\sqrt{\surd 2}$.

$\begin{array}{cc}\Rightarrow \frac{{𝑧}_1}{{𝑧}_2}& =\frac{2\sqrt{\surd 2}\sqrt{\surd 3}}{4}[\cos {15}^{\circ }+𝑖\sin {15}^{\circ }]\\ \frac{{𝑧}_1}{{𝑧}_2}& =\frac{\sqrt{\surd 6}}{2}\left[{𝑒}^{𝑖{15}^{\circ }}\right]\\ \Rightarrow {\left(\frac{{𝑧}_1}{{𝑧}_2}\right)}^{50}& ={\left(\frac{\sqrt{\surd 6}}{2}\right)}^{50}{𝑒}^{𝑖{750}^{\circ }}\\ 𝜃& ={750}^{\circ }-{720}^{\circ }={30}^{\circ }\end{array}$

$\because$ First quadrant

To find the solution set of the inequality $3^{𝑥}+3^{1-𝑥}-4<0$, we begin by rewriting it as:

$3^{𝑥}+\frac{3}{3^{𝑥}}-4<0$

Let $3^{𝑥}=𝑡$. Since $3^{𝑥}$ is always positive, we have $𝑡>0$.

Substitute into the equation:

$𝑡+\frac{3}{𝑡}-4<0$

Multiply the entire inequality by $𝑡$ (noting that $𝑡>0$ so the inequality sign does not change):

${𝑡}^2-4𝑡+3<0$

Factoring the quadratic, we get:

$(𝑡-3)(𝑡-1)<0$

The solution to this inequality is:

$𝑡\in (1,3)$

Since $𝑡=3^{𝑥}$, substitute back:

$3^{𝑥}\in (1,3)$

Taking the logarithm with base 3 of both sides gives:

$𝑥\in ({\log }_{3}1,{\log }_{3}3)$

Since ${\log }_{3}1=0$ and ${\log }_{3}3=1$, we find:

$𝑥\in (0,1)$

To find the common solution set for the inequalities ${𝑥}^2-4𝑥\le 12$ and ${𝑥}^2-2𝑥\ge 15$, we'll begin by rewriting them in a standard form:

${𝑥}^2-4𝑥-12\le 0$

${𝑥}^2-2𝑥-15\ge 0$

Next, we factor each equation:

$(𝑥-6)(𝑥+2)\le 0$

$(𝑥-5)(𝑥+3)\ge 0$

Now, we determine the solution sets for each inequality:

For $(𝑥-6)(𝑥+2)\le 0$, the interval is $𝑥\in [-2,6]$.

For $(𝑥-5)(𝑥+3)\ge 0$, the intervals are $𝑥\in (-\infty ,-3]\cup [5,\infty )$.

To find the common solution set, we identify where these intervals overlap:

The intersection of $[-2,6]$ and $(-\infty ,-3]\cup [5,\infty )$ is $𝑥\in [5,6]$.

Thus, the common solution set for the given inequalities is $𝑥\in [5,6]$.

${𝑥}^3-3{𝑥}^2+3𝑥+7=0$

$\begin{array}{c}\Rightarrow (𝑥+1)({𝑥}^2-4𝑥+7)=0\\ \Rightarrow 𝑥=-1\text{ or }\frac{4\pm \sqrt{\surd 16-28}}{2\left(1\right)}\\ \Rightarrow 𝑥=-1\text{ or }2\pm \sqrt{\surd 3}𝑖\\ \therefore 𝛼=-1,𝛽=2+\sqrt{\surd 3}𝑖\text{ and }\\ 𝛾=2-\sqrt{\surd 3}𝑖\end{array}$

New root are $-1-{ℎ}_{,}(2-ℎ)+\sqrt{\surd 3}𝑖$ and $(2-ℎ)-\sqrt{\surd 3}𝑖$

Equation which have these roots is $(𝑥+1+ℎ)(𝑥+ℎ-2-\sqrt{\surd 3}𝑖)$

$\begin{array}{c}(𝑥+ℎ-2+\sqrt{\surd 3}𝑖)=0\\ \Rightarrow (𝑥+1+ℎ)\left[\right(𝑥+ℎ-2{)}^2-\left(\sqrt{\surd 3}𝑖{)}^2\right]=0\\ \Rightarrow (𝑥+1+ℎ)[{𝑥}^2+(ℎ-2{)}^2\\ +2𝑥(ℎ-2)+3]=0\end{array}$

In this equation terms containing ${𝑥}^2$ and $𝑥$ should be missing.

$\therefore (1+ℎ)+2(ℎ-2)=0$ and

$2(1+ℎ)(ℎ-2)+3+(ℎ-2{)}^2=0$

Now, $\frac{𝛼-ℎ}{𝛽-ℎ}+\frac{𝛽-ℎ}{𝛾-ℎ}+\frac{𝛾-ℎ}{𝛼-ℎ}$

$=\frac{-1-1}{2+\sqrt{\surd 3}𝑖-1}+\frac{2+\sqrt{\surd 3}𝑖-1}{2-\sqrt{\surd 3}𝑖-1}+\frac{2-\sqrt{\surd 3}𝑖-1}{-1-1}$

$=\frac{-2}{1+\sqrt{\surd 3}𝑖}+\frac{1+\sqrt{\surd 3}𝑖}{1-\sqrt{\surd 3}𝑖}+\frac{1-\sqrt{\surd 3}𝑖}{-2}$

$\begin{array}{c}=\frac{-2(1-\sqrt{\surd 3}𝑖)}{4}+\frac{(1+\sqrt{\surd 3}𝑖{)}^2}{4}+\frac{(-2)(1-\sqrt{\surd 3}𝑖)}{4}\\ =\frac{1}{4}[-2+2\sqrt{\surd 3}𝑖+1-3+2\sqrt{\surd 3}𝑖-2+2\sqrt{\surd 3}𝑖]\\ =\frac{1}{4}[-6+6\sqrt{\surd 3}𝑖]\\ =3[-\frac{1}{2}+\frac{\sqrt{\surd 3}}{2}𝑖]\end{array}$

Let $𝑓\left(𝑥\right)=3{𝑥}^3+𝑏{𝑥}^2+𝑏𝑥+3$

(A) All the roots are negative. So, the number of sign change in $𝑓(-𝑥)$ should be 3 , which is possible only when $𝑏$ is positive. According to options (A) $\to$ IV.

(B) Clearly for any value of $𝑏,𝑓\left(𝑥\right)=0$ has root $𝑥=-1$.

$\begin{array}{c}\therefore 𝑓\left(𝑥\right)=0\Rightarrow (𝑥+1)(3{𝑥}^2+(𝑏-3)𝑥+3)=0\\ \Rightarrow 3{𝑥}^2+(𝑏-3)𝑥+3\ne 0\end{array}$

[ $\because$ two roots are complex]

$\Rightarrow (𝑏-3{)}^2-4(3\left)\right(3)<0$

$\Rightarrow {𝑏}^2-6𝑏-27<0$

$\Rightarrow (𝑏+3)(𝑏-9)<0$

$\Rightarrow -3<𝑏<9$

Hence, $B\to$ II.

(C) We know that one root is $𝑥=-1$

(Negative).

Two roots are complex if $-3<𝑏<9$.

If $𝑏=-3$, then $𝑓\left(𝑥\right)=0$

$\Rightarrow (𝑥+1)(𝑥-1{)}^2=0$

In this case two roots are positive and equal.

(D) All roots will be real and distinct if and only if $3{𝑥}^2+(𝑏-3)𝑥+3=0$ have two distinct real solutions other, than -1 .

$\begin{array}{cc}& (𝑏-3{)}^2-4(3\left)\right(3)>0\\ \Rightarrow & {𝑏}^2-6𝑏-27>0\\ \Rightarrow & (𝑏+3)(𝑏-9)>0\\ \Rightarrow & 𝑏\in (-\infty ,-3)\cup (9,\infty )\end{array}$

Thus, D $\to$ III

Therefore, $A\to \mathrm{IV},B\to \mathrm{II},C\to V,D\to \mathrm{III}$

To find the number of ways to arrange all the letters of the word 'COMBINATIONS' around a circle such that no two vowels are together, let's break it down step-by-step.

Identify Vowels and Consonants:

Vowels: O, O, I, I, A (total of 5 vowels)

Consonants: C, M, B, N, T, N, S (total of 7 consonants)

Arrangement of Consonants:

Since the arrangement is in a circle, we arrange the consonants in $(7-1)!=6!$ ways because in circular permutations, we fix one position to break the circle and linearly order the rest.

Interleaving Vowels:

Arranging the consonants creates 7 gaps (before the first consonant, between each pair, and after the last consonant) to place the vowels. We need to select 5 gaps for the vowels, ensuring no two vowels are adjacent, which is crucial since we want them not together.

Arrangement of Vowels:

Arrange the vowels O, O, I, I, A in these gaps.

The arrangement of the vowels is affected by repetition. The formula for arranging 'n' items where there are repetitions is given by $\frac{𝑛!}{𝑘1!\cdot 𝑘2!\cdot \dots \cdot 𝑘𝑟!}$, where ${𝑘}_{𝑖}$ represents the number of times each repeated letter occurs.

The necessary calculation for the vowels: $\frac{5!}{2!\cdot 2!}$ because O and I each repeat twice.

Total Arrangements:

Combine the arrangements of consonants and vowels as follows:

$\text{Total ways}=6!\times \frac{5!}{2!\times 2!}$

By further expanding this calculation based on the repetitions:

$\frac{7!\times 6!}{(2!{)}^4}$

This gives the total number of arrangements of the letters in 'COMBINATIONS' around a circle, ensuring that no two vowels are adjacent.

Explanation

To find the difference between the number of odd and even numbers formed by the digits 3, 5, 6, 7, and 8, where the numbers are greater than 6000 and less than 10000, we consider the following process:

Odd Numbers:

Set I: When the first digit (thousands place) is 6, 7, or 8, and the last digit (units place) is 3.

Choices:

3 choices for the first digit (6, 7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (3).

Combinations:

$3\times 3\times 2\times 1=18$

Set II: Similar to Set I, but the last digit (units place) is 5.

Choices:

3 choices for the first digit (6, 7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (5).

Combinations:

$3\times 3\times 2\times 1=18$

Set III: The first digit (thousands place) is 6 or 8, and the last digit is 7.

Choices:

2 choices for the first digit (6, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (7).

Combinations:

$2\times 3\times 2\times 1=12$

Total Odd Numbers:

$18+18+12=48$

Even Numbers:

Set I: The first digit (thousands place) is 7 or 8, and the last digit (units place) is 6.

Choices:

2 choices for the first digit (7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (6).

Combinations:

$2\times 3\times 2\times 1=12$

Set II: The first digit (thousands place) is 6 or 7, and the last digit (units place) is 8.

Choices:

2 choices for the first digit (6, 7).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (8).

Combinations:

$2\times 3\times 2\times 1=12$

Total Even Numbers:

$12+12=24$

Calculating the Difference:

The required difference between the number of odd and even numbers is:

$48-24=24$

This result can be expressed as ${}^4{𝑃}_3$.

There are 4 possibilities for inviting the relatives to the party:

3 ladies from the husband's side and 3 gentlemen from the wife's side:

Total number of ways:

$={}^4{𝐶}_3\times {}^4{𝐶}_3=16$

3 gentlemen from the husband's side and 3 ladies from the wife's side:

Total number of ways:

$={}^3{𝐶}_3\times {}^3{𝐶}_3=1$

2 ladies and 1 gentleman from the husband's side and 2 gentlemen and 1 lady from the wife's side:

Total number of ways:

$=({}^4{𝐶}_2\times {}^3{𝐶}_1)\times ({}^3{𝐶}_1\times {}^4{𝐶}_2)=324$

1 lady and 2 gentlemen from the husband's side, and 2 ladies and 1 gentleman from the wife's side:

Total number of ways:

$=({}^4{𝐶}_1\times {}^3{𝐶}_2)\times ({}^3{𝐶}_2\times {}^4{𝐶}_1)=144$

The total number of ways to invite 3 of the man's relatives and 3 of the wife's relatives is:

$=16+1+324+144=485$

Given the polynomial expression $(1+𝑥+{𝑥}^2+{𝑥}^3{)}^{100}$, we are interested in the coefficient of ${𝑥}^{𝑟}$, denoted by ${𝑎}_{𝑟}$. To find the sum of the products $𝑟\cdot {𝑎}_{𝑟}$, we perform the following steps:

First, evaluate the polynomial at $𝑥=1$:

$(1+1+1^2+1^3{)}^{100}=(4{)}^{100}$

This shows that the sum of all coefficients, $\sum _{300}^{𝑟=0}{𝑎}_{𝑟}$, equals $4^{100}$.

Next, differentiate $(1+𝑥+{𝑥}^2+{𝑥}^3{)}^{100}$ with respect to $𝑥$:

$\frac{𝑑}{𝑑𝑥}\left(\right(1+𝑥+{𝑥}^2+{𝑥}^3{)}^{100})=100(1+𝑥+{𝑥}^2+{𝑥}^3{)}^{99}(1+2𝑥+3{𝑥}^2)$

Substitute $𝑥=1$ into the derivative:

$\begin{array}{cc}\sum _{𝑟=0}^{300}𝑟\cdot {𝑎}_{𝑟}& =100\left(4{)}^{99}\right(1+2\cdot 1+3\cdot 1^2)\\ & =100\left(4^{99}\right)\cdot 6\end{array}$

Simplify the expression to find:

$\sum _{300}^{𝑟=0}𝑟\cdot {𝑎}_{𝑟}=600\left(4^{99}\right)=150\times 4^{100}$

Since $4^{100}=\sum _{300}^{𝑟=0}{𝑎}_{𝑟}=𝑆$, we have:

$\sum _{300}^{𝑟=0}𝑟\cdot {𝑎}_{𝑟}=150𝑆$

Thus, $\sum _{300}^{𝑟=0}𝑟\cdot {𝑎}_{𝑟}$ is equal to $150𝑆$.