Quadratic Equations

The other two roots are $(1-\sqrt{2})$ and $(2+i)$.

[ ∵ these roots occur in conjugate pair]

$ \begin{aligned} & \therefore x^4+b x^3+c x^2+d x+e \\ & =\left[x^2-\{(1-\sqrt{2})+(1+\sqrt{2})\}\right. \\ & \quad x+(1-\sqrt{2})(1+\sqrt{2})] \times\left[x^2-\{(2-i)\right. \\ & \quad+(2+i)\} x+(2-i)(2+i)] \end{aligned} $

$ \begin{aligned} & =\left(x^2-2 x-1\right)\left(x^2-4 x+5\right) \\ & =x^4-6 x^3+12 x^2-6 x-5 \\ & \Rightarrow b=-6, c=12, d=-6 \end{aligned} $

Now, the given quadratic equation becomes $-6 x^2+12 x-6=0$

$ \begin{array}{r}i.e\,\,\,\, x^2-2 x+1=0 \\ (x-1)^2=0 \end{array} $

Hence, the roots are real and equal.

Replace $x$ by $x+k$

$ \begin{array}{r} 2(x+k)^4-8(x+k)^3+3(x+k)^2-1=0 \\ \Rightarrow 2 x^4+(8 k-8) x^3+\left(12 k^2-24 k+3\right) x^2 \\ +\left(8 k^3-24 k^2+6 k\right) x \\ +\left(2 k^4-8 k^3+3 k^2-1\right)=0 \end{array} $

∵ Coefficient of $x^3$ is 0 .

$ \begin{aligned} & \therefore 8 k-8=0 \Rightarrow k=1 \\ & \Rightarrow 2 x^4-9 x^2-10 x-4=0 \\ & \therefore b=-9 \end{aligned} $

According to the question,

$ \begin{aligned} &\begin{aligned} -p & =\tan 15^{\circ}+\tan 30^{\circ} \\ \text { and } \quad q & =\tan 15^{\circ} \tan 30^{\circ} \\ \Rightarrow \quad(-p)(q) & =\left(\tan 15^{\circ}+\tan 30^{\circ}\right. \end{aligned}\\ &\left(\tan 15^{\circ} \tan 30^{\circ}\right) \end{aligned} $

$ \begin{aligned} \Rightarrow-p q & =\left[(2-\sqrt{3})+\frac{1}{\sqrt{3}}\right]\left[(2-\sqrt{3}) \frac{1}{\sqrt{3}}\right] \\ & =\frac{(2-\sqrt{3})^2}{\sqrt{3}}+\frac{2-\sqrt{3}}{3} \\ & =\frac{(7-4 \sqrt{3})}{\sqrt{3}}+\frac{2-\sqrt{3}}{3} \\ & =\frac{21-12 \sqrt{3}+2 \sqrt{3}-3}{3 \sqrt{3}} \\ & =\frac{18-10 \sqrt{3}}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ & =\frac{18 \sqrt{3}-30}{9}=\frac{-3(10-6 \sqrt{3})}{9} \\ \Rightarrow \quad p q & =\frac{10-6 \sqrt{3}}{3} \end{aligned} $

Let $f(x)=3 x-2 x^2+1$

Given, extreme value of $f(x)$ is $k$.

$ \begin{aligned} & \Rightarrow \quad f^{\prime}(k)=0 \quad 3-2(2 k)=0 \quad k=3 / 4 \\ & \because \quad k x^2+2 x+1>0 \\ & \Rightarrow \frac{3}{4} x^2+2 x+1>0 \Rightarrow 3 x^2+8 x+4>0 \\ & \Rightarrow\left(x+\frac{2}{3}\right)(x+2)>0 \\ & \Rightarrow \quad x \in\left(-\infty, \frac{-2}{3}\right) \cup(-2, \infty) \end{aligned} $

Let $\alpha, \beta, \gamma$ be the roots of the equation

$ \begin{array}{r} x^3-5 x^2-2 x+24=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \alpha+\beta+\gamma=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

$ \begin{array}{r} \alpha \beta+\beta \gamma+\gamma \alpha=-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \alpha \beta \gamma=-24 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

$ \begin{aligned} & \therefore \quad \frac{\beta \gamma}{\alpha}+\frac{\gamma \alpha}{\beta}+\frac{\alpha \beta}{\gamma} \\ & =\frac{\beta \gamma}{-24} \times \beta \gamma+\frac{\gamma \alpha}{-24} \times \gamma \alpha+\frac{\alpha \beta}{-24} \times \alpha \beta \\ & \quad \quad \text { using Eq. (iii) }] \\ & =\frac{1}{-24}\left[(\beta \gamma)^2+(\gamma \alpha)^2+(\alpha \beta)^2\right] \\ & =\frac{-1}{24}\left[(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2\left(\alpha \beta^2 \gamma\right.\right. \left.\left.+\alpha \beta \gamma^2+\alpha^2 \beta \gamma\right)\right] \\ & =\frac{-1}{24}\left[(-2)^2-2(-24(\alpha+\beta+\gamma))\right] \\ & =\frac{-1}{24}[4+2 \times 24 \times 5]=-\frac{61}{6} \end{aligned} $

$p(x)$ is a quadratic polynomial having only purely imaginary roots.

Let $p(x)=x^2+a^2=0, a \in R$

whose roots are $x= \pm i a$

$ \begin{aligned} \text { Then, } p(p(x))= & \left(x^2+a^2\right)^2+a^2 \\ = & x^4+2 a^2 x^2+a^4+a^2 \\ \text { Discriminant }= & 4 a^4-4\left(a^4+a^2\right) \\ = & -4 a^2<0 \end{aligned} $

∴ Roots of $p(p(x)$ are non-real.

Given equation, $4 x^3+12 x^2-7 x+165=0$ has roots $\alpha, \beta$ and $\gamma$, respectively.

$ \alpha+\beta+\gamma=-\frac{12}{4} \quad \alpha \beta \gamma=-\frac{165}{4} $

$ \alpha \beta+\beta \gamma+\gamma \alpha=-\frac{7}{4} $

Now, $a x^3+b x^2+c x+d=0$ has roots $\alpha+5, \beta+5$ and $\gamma+5$, respectively. Product of roots $=(\alpha+5)(\beta+5)(\gamma+5)$

$ \begin{aligned} & =\alpha \beta \gamma+5(\alpha \beta+\beta \gamma+\gamma \alpha)+25(\alpha+\beta+\gamma)+125 \\ & =\frac{-165}{4}-\frac{35}{4}-\frac{300}{4}+125=0 \end{aligned} $

Given, equation $x^2-2(1+3 m) x+7(3+2 m)=0$

Here, $a>0$ and $D=b^2-4 a c>0$

Expression is always positive, if roots are distinct then $D>0$,

$\begin{aligned} & {[-2(1+3 m)]^2-4 \times 7(3+2 m)>0 } \\ \Rightarrow & 4\left[1+9 m^2+6 m\right]-84-56 m>0 \\ \Rightarrow & 36 m^2-32 m-80>0 \\ \Rightarrow & 9 m^2-8 m-20>0 \\ \Rightarrow & (m-2)\left(m+\frac{10}{9}\right)>0 \\ \Rightarrow & m \in\left(-\infty, \frac{-10}{9}\right) \cup[2, \infty) \end{aligned}$

$\therefore$ Integral value of $m$ are .......... $3 i, 1,2,3,4,5,6$. So, the number of element in $S$ is infinite.

Considering the question to be, to find the roots of the equation $x^4-2 x^3+x-380=0$

Now, by trial error method we find that $x=5$ is a solution to the equation

$\begin{aligned} & \Rightarrow \quad 5^4-2 \times 5^3+5-380=0 \\ & \quad=625-250+5-380=0 \\ & \quad=380-380=0 \end{aligned}$

Thus, $x=5$ is a solution.

We also find that $x=-4$ is a solution

$\begin{aligned} & (-4)^4-2 \times(-4)^3-4-380=0 \\ & =256+128-4-380 \\ & =380-380=0 \end{aligned}$

The real roots are 5 and $-$4.

Sum of real roots are $5+(-4)=1$

Let $a, 2 a$ and $b$ be roots of the cubic equation $x^3+36-7 x^2=0$

So, sum of roots $=-\frac{(-7)}{1}$

$\begin{aligned} & \therefore & a+2 a+b & =7 \\ &\Rightarrow & 3 a+b & =7 \quad \text{... (i)} \end{aligned}$

$\text { Now, } a(2 a) b=36 \quad\left[\text { product of roots }=\frac{d}{a}\right]$

$\begin{aligned} & 2 a^2 b=36 \text { and } a b+a(2 a)+2 a(b)=0 \\ & \Rightarrow \quad 2 a^2+3 a b=0 \Rightarrow a(2 a+3 b)=0 \\ & \Rightarrow \quad b=\frac{-2 a}{3} \end{aligned}$

$\begin{array}{lll} \Rightarrow & 3 a-\frac{2 a}{3}=7 & \text { [from Eq. (i)] } \end{array}$

$\begin{aligned} & \Rightarrow \quad \frac{9 a-2 a}{3}=7 \Rightarrow 9 a-2 a=21 \\ & \Rightarrow \quad 7 a=21 \Rightarrow a=3 \\ & \text { and } b=7-3 a \\ & =7-3(3)=-2 \quad \text{[from Eq. (i)]}\\ \end{aligned}$

Roots are $3,6,-2$

$\therefore$ Number of negative root is 1.

Given, $f(x)=x^2+a x+b(b \neq 0)$

$\Rightarrow f(f(x))=\left(x^2+a x+b\right)^2+a\left(x^2+a x+b\right)+b$

Since, $f(f(0))=0$

Now, $b^2+a b+b=0$ ..... (i)

Take $a=0$ and $b=-1$, then

Eq. (i) is satisfied.

Hence, $a+b=0+(-1)=-1$

$\begin{aligned} & \text {Let } f(x)=|x-2| \\ & \left\{\begin{array}{cl} x-2, & x>2 \\ -(x-2), & x<2 \end{array}\right. \end{aligned}$

Then, for $x>2$ the equation becomes

$\begin{array}{rlrl} \Rightarrow & & (x-2)^2+(x-2)-2 & =0 \\ \Rightarrow && x^2-3 x & =0 \\ \Rightarrow & & x & =0,3 \end{array}$

Thus, the root of the equation for $x>2$ is 3.

For $x<0$, the equation becomes

$ \begin{array}{rrr} & (x-2)^2+(2-x)-2=0 \\ \Rightarrow & x^2-5 x+4=0 \\ \Rightarrow & (x-4)(x-1)=0 \\ \Rightarrow & x=4,1 \end{array}$

The root which is less than 2 is 1.

Thus, the roots of given equations are 3, 1. Sum will be $3+1=4$

Let $\alpha_1$ and $\alpha_2$ be roots of the equation $x^2+a x+b=0$.

Therefore, $\alpha_1+\alpha_2=-a$ and $\alpha_1 \alpha_2=b$ and $\beta_1$ and $\beta_2$ be the roots of the equation $x^2+b x+a=0$

Therefore, $\beta_1+\beta_2=-b$ and $\beta_1, \beta_2=a$

$\begin{aligned} & \text { Given that } \alpha_1-\alpha_2=\beta_1-\beta_2 \\ & \Rightarrow \quad\left(\alpha_1+\alpha_2\right)^2-4 \alpha_1 \alpha_2=\left(\beta_1+\beta_2\right)^2-4 \beta_1 \beta_2 \\ & \Rightarrow \quad(-a)^2-4 b=t-4 a \\ & \Rightarrow \quad a^2-b^2+4 a-4 b=0 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad(a-b)(a+b)+4(a-b)=0 \\ & \Rightarrow \quad(a-b)(a+b+4)=0 \end{aligned}$

Therefore, $a-b=0$ and $a+b+4=0$

$\begin{aligned} & \text { Given, }(x+a)(x+1991)+1 \\ & \therefore(x+a)(x+1991)-1=0 \\ & \Rightarrow(x+a)(x+1991)=-1 \end{aligned}$

Multiplication of two integers is $-1$, this implies that either $(x+a)$ is +1 and $(x+1991)$ is $-1$ or $(x+a)$ is $-1$ and $(x+1991)$ is +1.

Let $(x+a)=1$, then $x+1991=-1$

$\begin{aligned} & \Rightarrow \quad x=-1-1991 \\ & x=-1992 \\ & \therefore \quad a=1-x=1-(-1992)=1+1992=1993 \\ \end{aligned}$

Let $(x+a)=-1$, then $x+1991=+1$

$\begin{aligned} \Rightarrow \quad x & =1-1991 \\ & =-1990 \end{aligned}$

$\begin{aligned} \therefore \quad a & =-1-x=-1-(-1990) \\ & =-1+1990=1989 \end{aligned}$

So, $a$ can be 1993 or 1989.

Given, $\frac{13 x+43}{2 x^2+17 x+30}=\frac{A}{2 x+5}+\frac{B}{x+6}$

$\Rightarrow \frac{13 x+43}{2 x^2+17 x+30}=\frac{A(x+6)+B(2 x+5)}{(2 x+5)(x+6)}$

Now, $13 x+43=(A+2 B) x+(6 A+5 B)$

$\therefore A+2 B=13$ and $6 A+5 B=43$

Solving these equations, we get

$B=\frac{78-43}{7}=\frac{35}{7}=5$ and $A=3$

$\therefore \quad A^2+B^2=(3)^2+(5)^2=9+25=34$

Given: $ f(x) = ax^2 + bx + c $ for some $ a, b, c \in \mathbb{R} $ with $ a + b + c = 3 $, and $ f(x + y) = f(x) + f(y) + xy $ for all $ x, y \in \mathbb{R} $.

Let's analyze the function step by step:

1. Start with the given quadratic function:

$ f(x) = ax^2 + bx + c $

1. Since $ a + b + c = 3 $, and given the functional equation $ f(x + y) = f(x) + f(y) + xy $, we need to verify and utilize this relationship:

$ f(1) = a + b + c = 3 $

1. Next, consider $ f(2) $:

$ f(2) = f(1 + 1) = f(1) + f(1) + 1 = 2f(1) + 1 = 2 \times 3 + 1 = 7 $

1. Now, calculate $ f(3) $:

$ f(3) = f(2 + 1) = f(2) + f(1) + 2 = 7 + 3 + 2 = 12 $

1. Then compute $ f(4) $:

$ f(4) = f(3 + 1) = f(3) + f(1) + 3 = 12 + 3 + 3 = 18 $

1. For the sum $ \sum_{n=1}^{10} f(n) $:

$ \sum_{n=1}^{10} f(n) = f(1) + f(2) + f(3) + \ldots + f(10) $

Let's express each term:

$ f(1) = 3 $

$ f(2) = 2f(1) + 1 = 7 $

$ f(3) = 3f(1) + 3 = 12 $

$ f(4) = 4f(1) + 6 = 18 $

$ \ldots $

$ f(10) = 10f(1) + 45 = 75 $

1. Summarize the formula:

$ \sum_{n=1}^{10} f(n) = f(1) + f(2) + f(3) + \ldots + f(10) $

$ = f(1)[1 + 2 + 3 + \ldots + 10] + [1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45] $

$ = 3 \left(\frac{10(10 + 1)}{2}\right) + 165 $

$ = 3 \times 55 + 165 $

$ = 165 + 165 $

$ = 330 $

Therefore, the sum $\sum_{n=1}^{10} f(n) = 330$.

$\begin{aligned} & \text { } 3^{x+1}+3^{-x+1}=10 \\ & \Rightarrow \quad 3^x \cdot 3^1+3^{-x} \cdot 3^1=10 \\ & \Rightarrow \quad-3 y+\frac{3}{y}=10 \quad [\text {Let}~ 3^x=y] \\ & \Rightarrow \quad 3 y^2-10 y+3=0 \\ & \Rightarrow \quad 3 y^2-9 y-y+3=0 \\ & \Rightarrow 3 y(y-3-1(y-3)=0 \\ & \Rightarrow \quad(y-3)(3 y-1)=0 \\ & \Rightarrow \quad y=3 \text { and } y=\frac{1}{3} \\ & \Rightarrow \quad 3^x=3^1 \text { and } 3^x=3^{-1} \\ & \Rightarrow \quad x=1 \text { and } x=-1 \end{aligned}$

Number of positive real roots is 1.

$\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$

$\begin{aligned} & \Rightarrow \frac{x+(1-x)}{(\sqrt{1-x})(\sqrt{x})}=\frac{13}{6} \Rightarrow \frac{1}{\sqrt{x-x^2}}=\frac{13}{6} \\ & \Rightarrow \quad 6=13 \sqrt{x-x^2} \\ & \Rightarrow \quad 36=169\left(x-x^2\right) \\ & \Rightarrow \quad 169 x^2-169 x+36=0 \\ & D=(-169)^2-4(169)(36)=4225>0 \\ & \because \quad D>0 \\ \end{aligned}$

$\therefore$ Number of roots will be 2.

Given, equations are

$\begin{aligned} & x^2+a x+b=0 \quad .....(\text{i})\\ & x^2+b x+a=0\quad .....(\text{ii}) \end{aligned}$

Let $\alpha$ be common root of Eqs. (i) and (ii), then

$\begin{aligned} & \alpha^2+a \alpha+b=0 \quad .....(\text{iii})\\ & \alpha^2+b \alpha+a=0\quad .....(\text{iv}) \end{aligned}$

Compare Eqs. (iii) and (iv), we get

$\begin{aligned} & \alpha^2+a \alpha+b=\alpha^2+b \alpha+a \\ & \Rightarrow \quad \alpha(a-b)=(a-b) \\ & \Rightarrow \quad \alpha=1 \\ & \text { Common root }=1 \\ & \text { From Eq. (iii), we get } \\ & 1+a+b=0 \\ & \Rightarrow \quad a+b=-1 \\ \end{aligned}$

Given equation is

$9 x^3+112 x^2-120 x+a=0$

Let $\alpha, \beta, \gamma$ be roots of Eq.(i), then

Product of roots $=\alpha \beta \gamma=-\frac{a}{9}$

Given, $-\frac{a}{9}=12$

$\begin{aligned} & \Rightarrow \quad a=-12 \times 9=-108 \\ & \therefore \quad a=-108 \\ \end{aligned}$

Roots of cubic equations are 1 and $2+\sqrt{5}$.

Since, $2+\sqrt{5}$ is root, then its conjugate also become roots, i.e. $2-\sqrt{5}$ is root of cubic equation.

$\therefore \text { Roots }=1,2+\sqrt{5}, 2-\sqrt{5} \text { i.e. } \alpha, \beta, \gamma$

Let cubic equation be

$x^3+b x^2+c x+d=0$

Then,

$\begin{aligned} & \text { Sum of Roots }=-b=1+2+\sqrt{5}+2-\sqrt{5}=5 \\ & \Rightarrow \quad b=-5 \\ & \text { Product of Roots }=-d=(1)(2+\sqrt{5})(2-\sqrt{5}) \\ & =4-5=-1 \\ & \Rightarrow \quad d=1 \\ & \therefore \quad \alpha \beta+\beta \gamma+\gamma \alpha=c \\ & \text { (l) }(2+\sqrt{5})+(2+\sqrt{5})(2-\sqrt{5})+(2-\sqrt{5})(\mathrm{l})=c \\ & 2+\sqrt{5}+2-\sqrt{5}+4-5=c \\ & \Rightarrow \\ & 3=c \\ \end{aligned}$

$\therefore$ Cubic equation is

$x^3-5 x^2+3 x+1=0$

$x^2+x+1=0$

Roots : $\alpha=\omega$ and $\beta=\omega^2$

$\begin{aligned} \text { where, } \quad \omega & =\frac{-1+\sqrt{3} i}{2} \text { and } \omega^3=1 \\ \alpha^{2021} & =\omega^{2021}=\omega^2 \\ \text { and } \quad \beta^{2021} & =\omega^{4042}=\omega \end{aligned}$

$\therefore$ Equation whose roots are $\alpha^{2021}$ and $\beta^{2021}$ is

$(x-\omega)\left(x-\omega^2\right)=x^2+x+1 $

This is the graph of $f(x)=x^3-4 x^2+5 x-2$

And the graph of $f\left(x+\frac{1}{3}\right)$ will shift towards left by $1 / 3$ units, so the new roots will also shift by $1 / 3$ units towards left.

$\therefore \text { Roots }=\left(1-\frac{1}{3}, 1-\frac{1}{3}, 2-\frac{1}{3}\right)=\left(\frac{2}{3}, \frac{2}{3}, \frac{5}{3}\right)$

$f(x)=2 x^3+m x^2-13 x+n$

Let $\alpha, \beta, \gamma$ are the roots of $f(x)=0$ where $\alpha=2, \beta=3$ and let $\gamma=k$

$\begin{array}{cc} \Rightarrow & \alpha+\beta+\gamma=-\frac{m}{2}=2+3+k \quad \text{......(i)}\\ \Rightarrow & \alpha \beta+\beta \gamma+\gamma \alpha=-\frac{13}{2}=6+5 k \quad \text{......(ii)}\\ \Rightarrow & \alpha \beta \gamma=-\frac{n}{2}=6 k\quad \text{......(iii)} \end{array}$

From Eq. (ii), $-13=12+10 k$

$\Rightarrow \quad \begin{aligned} k & =\frac{-5}{2} \\ n & =-12 k=30 \\ m & =-10-2 k=-5 \end{aligned}$

Given equation, $11 x^2+12 x-13=0$

If $\alpha, \beta$ are the roots, then

$\alpha+\beta=\frac{-12}{11} \text { and } \alpha \beta=-\frac{13}{11}$

Now, $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{(\alpha \beta)^2}$

$\begin{aligned} & =\frac{\frac{144}{121}+\frac{26}{11}}{\frac{169}{121}} \\ & =\frac{144+286}{169}=2.54 \\ \end{aligned}$

Let $\alpha$ be the common root of

$\begin{aligned} & x^3+a x+1=0 \\ & \text { and } \quad x^4+a x^2+1=0 \\ & \text { Then, } \quad \alpha^3+a \alpha+1=0 \\ & \text { and } \quad \alpha^4+a \alpha^2+1=0 \\ \end{aligned}$

subtracting both equations

$\begin{array}{rlrl} \alpha^4-\alpha^3+a \alpha^2-a \alpha & =0 \\ \alpha\left(\alpha^3-\alpha^2+a \alpha-a\right) & =0 \\ \Rightarrow \quad \alpha^2(\alpha-1)+a(\alpha-1) =0 \\ {[\alpha=0 \text { doesn't give a root }]} \\ \Rightarrow \quad (\alpha-1)\left(\alpha^2+a\right) =0 \\ \alpha=1 \text { or } a =-\alpha^2 \end{array}$

Clearly, $\alpha=1$ gives $a=-2$ in both equations.

Hence, for $a=-2$ both equation have a common root.

Given, equation, $7 x^2-13 x+a=0$

$\because$ Roots are rational $\Rightarrow$ Discriminate is a perfect square

$D=169-28 a \Rightarrow D=b^2-4 a c$

$\Rightarrow D$ is perfect square for $a=6$

$e^{4 t}-10 e^{3 t}+29 e^{2 t}-22 e^t+4=0$ ...... (i)

Let $e^t=x$ and roots of (1) be $t_1, t_2, t_3, t_4$

$\therefore \quad x^4-10 x^3+29 x^2-22 x+4=0$

Then, roots are $x_1, x_2, x_3, x_4$

product of roots $x_1 \cdot x_2 \cdot x_3 \cdot x_4=4$

$\begin{aligned} & \Rightarrow e^{t_1} \cdot e^{t_2} \cdot e^{t_3} \cdot e^{t_4}=4 \Rightarrow e^{t_1+t_2+t_3+t_4}=4 \\ & \Rightarrow t_1+t_2+t_3+t_4=\log _e 4=\log _e 2^2=2 \log _e 2\end{aligned}$

Sum of roots $=2 \log _e 2$

Given that,

$ \begin{gathered} \begin{aligned} &(x-n)\left(\left(x^2-2 n x\right)^2+\left(2 n^2-5\right)\left(x^2-2 n x\right)\right. \\ &\left.+\left(n^4-5 n^2+4\right)\right)=0 \\ & \Rightarrow(x-n)\left(x^4-4 n x^3+4 n^2 x^2+2 n^2 x^2\right. \\ &\left.-4 n^3 x-5 x^2+10 n x+n^4-5 n^2+4\right)=0 \\ & \Rightarrow(x-n)\left(x^4-4 n x^3+6 n^2 x^2-4 n^3 x\right. \\ &-\left.5 x^2+10 n x+n^4-5 n^2+4\right)=0 \\ & \Rightarrow x^5-4 n x^4+6 n^2 x^3-4 n^3 x^2 \\ &-5 x^3+10 n x^2+n^4 x-5 n^2 x+4 x-x^4 n \\ &+ 4 n^2 x^3-6 n^3 x^2+4 n^4 x \\ &+5 x^2 n-10 n^2 x-n^5+5 n^3-4 n=0 \\ & \Rightarrow x^5-5 n x^4+5\left(2 n^2-1\right) x^3 \\ &+5 n\left(3-2 n^2\right) x^2 \end{aligned} \\ +\left(5 n^4-15 n^2+4\right) x-n\left(n^4-5 n^2+4\right)=0 \end{gathered} $

Now, product of roots,

$ \begin{aligned} P & =-\frac{\text { Constant term }}{\text { Coefficient of } x^5} \\ & =-\frac{-n\left(n^4-5 n^2+4\right)}{1} \end{aligned} $

$ \begin{aligned} & P=n\left(n^2-1\right)\left(n^2-4\right) \\ & P=n(n-1)(n+1)(n-2)(n+2) \end{aligned} $

Given that, $n>2$

So, putting $n=3$,

$ \begin{aligned} P & =3(3-1)(3+1)(3-1)(3+2) \\ & =3(2)(4)(2)(5) \\ P & =120 \end{aligned} $

Putting $n=4$,

$ \begin{aligned} P & =4(4-1)(4+1)(4-1)(4+2) \\ & =4(3)(5)(3)(6)=1080 \end{aligned} $

Therefore, products of roots is always divisible by 120 .

$ \begin{aligned} &\text { Given, } \alpha, \beta \text { are roots of }\\ &\begin{aligned} & \quad a x^2+b x+c=0 \\ & \Rightarrow \quad a \alpha^2+b \alpha+c=0 \\ & \text { and } \quad a \beta^2+b \beta+c=0 \\ & \quad \alpha(a \alpha+b)=-c \text { and } \\ & \beta(a \beta+b)=-c \\ & \quad a \alpha+b=-\frac{c}{\alpha} \Rightarrow a \beta+b=-\frac{c}{\beta} \end{aligned}\\ &\begin{aligned} & \operatorname{Now}\left(\frac{\alpha}{a \beta+b}\right)^3-\left(\frac{\beta}{a \alpha+b}\right)^3 \\ &=\left(\frac{\alpha}{-c / \beta}\right)^3-\left(\frac{\beta}{-c / \alpha}\right)^3 \\ &=-\left(\frac{\alpha \beta}{c}\right)^3+\left(\frac{\alpha \beta}{c}\right)^3=0 \end{aligned} \end{aligned} $

Given $\left\{x \in \mathbf{R}: \sqrt{x+2}>\sqrt{8-x^2}\right\}$

∴ Defining the function

$ \begin{align*} x+2 \geq 0 & \Rightarrow x \geq-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } 8-x^2 \geq 0 & \Rightarrow x \in[-2 \sqrt{2}, 2 \sqrt{2}] . \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{align*} $

Now, $\sqrt{x+2}>\sqrt{8-x^2}$

∴ Squaring on both sides,

$ \begin{align*} & x+2>8-x^2 \\ \Rightarrow \quad & x^2+x-6>0 \\ \Rightarrow \quad & (x+3)(x-2)>0 \\ \Rightarrow \quad & x \in(-\infty,-3) \cup(2, \infty) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{align*} $

∴ According to Eqs. (i), (ii) and (iii)

$ \Rightarrow x \in[2,2 \sqrt{2}] $

∴ Maximum value of $x=2 \sqrt{2}$.

$ \begin{aligned} &\begin{aligned} & \text { Let } y=\frac{x^2+14 x+9}{x^2+2 x+3} \\ & \Rightarrow x^2 y+2 x y+3 y=x^2+14 x+9 \\ & \Rightarrow x^2(y-1)+2 x(y-7)+3 y-9=0 \end{aligned}\\ &\text { Here, } x \in \mathbf{R} \end{aligned} $

$ \begin{aligned} \therefore \quad 4(y-7)^2-4(y-1)(3 y-9) & \geq 0 \\ (y-7)^2-\left(3 y^2-12 y+9\right) & \geq 0 \\ y^2-14 y+49-3 y^2+12 y-9 & \geq 0 \end{aligned} $

$ \begin{array}{lr} \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 2 y^2+2 y-40 \leq 0 \\ \Rightarrow & y^2+y-20 \leq 0 \\ & (y+5)(y-4) \leq 0 \end{array} $

$ \therefore \quad y \in[-5,4] $

∴ Maximum and minimum value of $\frac{x^2+14 x+9}{x^2+2 x+3}$ are 4 and -5 respectively.

$ \begin{aligned} &\text { We have, }\\ &\begin{array}{lc} & \frac{\sqrt{12-x-x^2}}{x+10} \leq \frac{\sqrt{12-x-x^2}}{2 x+9} \\ \Rightarrow & \sqrt{12-x-x^2}(2 x+9-x-10) \leq 0 \\ \Rightarrow & \sqrt{12-x-x^2}(x-1) \leq 0 \\ \therefore & 12-x-x^2 \geq 0 \text { and } x \leq 1 \\ \Rightarrow & x^2+x-12 \leq 0 \text { and } x \leq 1 \\ \Rightarrow & (x+4)(x-3) \leq 0 \text { and } x \leq 1 \\ \Rightarrow & x \in[-4,3] \text { and } x \in(-\infty, 1] \\ \therefore & x \in[-4,1] \cup\{3\} \end{array} \end{aligned} $

$ \begin{aligned} &\text { Given equation, }\\ &\begin{array}{r} 6 x^6-25 x^5+31 x^4-31 x^2+25 x-6=0 \\ 6\left(x^6-1\right)-25 x\left(x^4-1\right)+31\left(x^4-x^2\right)=0 \\ \left(x^2-1\right)\left[6 x^4+6 x^2+6-25 x^3\right. \left.-25 x+31 x^2\right]=0 \\ 6 x^4+6-25\left(x^3+x\right)+37 x^2=0 \\ 6\left(x^2+\frac{1}{x^2}\right)-25\left(x+\frac{1}{x}\right)+37=0 \\ 6\left(x+\frac{1}{x}\right)^2-12-25\left(x+\frac{1}{x}\right)+37=0 \\ 6\left(x+\frac{1}{x}\right)^2-25\left(x+\frac{1}{x}\right)+25=0 \end{array} \end{aligned} $

$ \begin{aligned} \therefore \quad 6 y^2-25 y+25 & =0 \quad\left[\because x+\frac{1}{x}=y\right] x \\ (6 y-15)(6 y-10) & =0 \\ y & =\frac{5}{2}, \frac{5}{3} \end{aligned} $

$ \begin{aligned} & \Rightarrow & x+\frac{1}{x} & =\frac{5}{2} \\ & \Rightarrow & x & =2, \frac{1}{2} \\ & & x+\frac{1}{x} & =\frac{5}{3} \text { has no real roots } \\ & \therefore & \alpha+\beta & =\frac{5}{3} \end{aligned} $

We have,

$\sqrt{\frac{5 x}{x-2}}+\sqrt{\frac{x-2}{5 x}}=\frac{29}{10}$

Let $\quad \sqrt{\frac{5 x}{x-2}}=y$

Then, $y+\frac{1}{y}=\frac{29}{10}$

$ \begin{array}{ll} \Rightarrow & \frac{y^2+1}{y}=\frac{29}{10} \\ \Rightarrow & 10 y^2+10=29 y \\ \Rightarrow & 10 y^2-29 y+10=0 \\ \Rightarrow & 10 y^2-25 y-4 y+10=0 \\ \Rightarrow & 5 y(2 y-5)-2(2 y-5)=0 \\ \Rightarrow & (5 y-2)(2 y-5)=0 \end{array} $

$ \begin{aligned} & \Rightarrow \quad y=\frac{2}{5}, \frac{5}{2} \\ & \text { Now, if } y=\frac{2}{5} \Rightarrow \sqrt{\frac{5 x}{x-2}}=\frac{2}{5} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{5 x}{x-2}=\frac{4}{25} \\ \Rightarrow & 125 x=4 x-8 \\ \Rightarrow & 121 x=-8 \\ \Rightarrow & x=-8 / 121 \end{array} $

$ \text { And if } y=\frac{5}{2} \Rightarrow \sqrt{\frac{5 x}{x-2}}=\frac{5}{2} $

$ \begin{array}{ll} \Rightarrow & \frac{5 x}{x-2}=\frac{25}{4} \\ \Rightarrow & \frac{x}{x-2}=\frac{5}{4} \\ \Rightarrow & 5 x-10=4 x \\ \Rightarrow & x=10 \end{array} $

$ \begin{aligned} & \therefore \text { Let } \alpha=10 \text { and } \beta=\frac{-8}{121} \quad(\because \alpha>\beta) \\ & \\ & \quad \begin{aligned} \therefore &\,\,\,\,\,\,\,\,\, \sqrt{\alpha^2-11^4 \beta^2} \\ = & \sqrt{(10)^2-11^4\left(\frac{-8}{121}\right)^2} \\ = & \sqrt{100-64} \\ = & \sqrt{36}=6 \end{aligned} \end{aligned} $

Let $y=\frac{9 \cdot 3^{2 x}+6 \cdot 3^x+4}{9 \cdot 3^{2 x}-6 \cdot 3^x+4}$

Put, $3^x=t$

$ \begin{aligned} & \therefore \quad y=\frac{9 \cdot t^2+6 t+4}{9 t^2-6 t+4} \\ & \Rightarrow y\left(9 t^2-6 t+4\right)=9 t^2+6 t+4 \\ & \Rightarrow 9 t^2(y-1)-6 t(y+1)+(4 y-4)=0 \\ & \Rightarrow \quad(y-1) t^2 \frac{-6}{9}(y+1) t+\frac{4}{9}(y-1)=0 \\ & \Rightarrow(y-1) t^2-\frac{2}{3}(y+1) t+\frac{4}{9}(y-1)=0 \end{aligned} $

$\because t$ is real

$ \therefore D \geq 0 $

$ \begin{array}{lr} \Rightarrow \frac{4}{9}(y+1)^2-4(y-1) \frac{4}{9}(y-1) \geq 0 \\ \Rightarrow \frac{4}{9}(y+1)^2-\frac{16}{9}(y-1)^2 \geq 0 \\ \Rightarrow (y+1)^2-4(y-1)^2 \geq 0 \\ \Rightarrow y^2+1+2 y-4 y^2-4+8 y \geq 0 \\ \Rightarrow -3 y^2+10 y-3 \geq 0 \\ \Rightarrow 3 y^2-10 y+3 \leq 0 \\ \Rightarrow (y-3)(3 y-1) \leq 0 \\ \Rightarrow \frac{1}{3} \leq y \leq 3 \\ \text { Hence, minimum value }=\frac{1}{3} . \end{array} $

Let the roots of the given cubic equation $ x^3 - p x^2 + p x - 1 = 0 $ be $ \alpha, \beta, \gamma $.

From the standard relationships between coefficients and roots, we get:

$ \begin{aligned} \alpha + \beta + \gamma &= p, \\ \alpha\beta + \beta\gamma + \gamma\alpha &= p, \\ \alpha\beta\gamma &= 1. \end{aligned} $

The question says that the new equation, whose roots are the squares of the original roots (i.e., $ \alpha^2, \beta^2, \gamma^2 $), is the same as the given equation itself. Therefore, the symmetric sums of these squared roots must also be:

$ \begin{aligned} \alpha^2 + \beta^2 + \gamma^2 &= p, \\ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 &= p, \\ \alpha^2\beta^2\gamma^2 &= 1. \end{aligned} $

Now, using the identity:

$ (\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha) $

Substitute the known values:

$ p^2 = p + 2p $

Simplify this to get:

$ p^2 = 3p $

Since $ p \neq 0 $, divide both sides by $ p $:

$ p = 3 $

We have,

$ \begin{aligned} & \lambda x^2+13 x+7=0 \\ \therefore \quad & D=(13)^2-4(\lambda)(7)=169-28 \lambda \end{aligned} $

For rational roots, $D=$ perfect square

So, $\lambda \in(-3,7)$ has values $-2,0,6$ so that $D$ become perfect square.

∴ Required sum $=-2+0+6=4$

We have, $f(x)=1-2 x-5 x^2$

$ \begin{aligned} & =-5\left[x^2+\frac{2}{5} x-\frac{1}{5}\right] \\ & =-5\left[\left(x+\frac{1}{5}\right)^2-\frac{1}{5}-\frac{1}{25}\right] \\ & =-5\left(x+\frac{1}{5}\right)^2+1+\frac{1}{5} \\ & =-5\left(x+\frac{1}{5}\right)^2+\frac{6}{5} \\ & \therefore \quad f(x) \leq \frac{6}{5} ; \quad \therefore \quad \alpha=\frac{6}{5} \\ & \text { Again, } g(x)=x^2-2 x+r \\ & \quad=\quad(x-1)^2+r-1 \\ & \therefore \quad g(x) \geq(r-1) \\ & \therefore \quad \beta=r-1 \end{aligned} $

So, $5 \alpha x^2+\beta x+6=6 x^2+(r-1) x+6$

Since, $5 \alpha x^2+\beta x+6>0$ and coefficient of $x^2>0$

$ \begin{array}{lr} \therefore & D<0 \\ \Rightarrow & (r-1)^2-4 \times 6 \times 6<0 \\ \Rightarrow & (r+11)(r-13)<0 \\ \Rightarrow & r \in(-11,13) \end{array} $

$ \begin{aligned} & \text { Let } p(x)=x^4+x^3-4 x^2+x-1 \\ & \quad \begin{array}{l} =(x-1)\left(x^3+2 x^2-2 x-1\right) \\ =(x-1)(x-1)\left(x^2+3 x+1\right) \\ =(x-1)(x-1)\left[x-\left(\frac{-3+\sqrt{5}}{2}\right)\right] \\ \quad\left[x-\left(\frac{-3-\sqrt{5}}{2}\right)\right] \end{array} \end{aligned} $

∴ Roots of the equation, $p(x)=0$ are

$ x=1,1, \frac{-3+\sqrt{5}}{2}, \frac{-3-\sqrt{5}}{2} $

∴ Required ratio

$ \begin{aligned} & =\frac{(1)^2+(1)^2+\left(\frac{-3+\sqrt{5}}{2}\right)^2+\left(\frac{-3-\sqrt{5}}{2}\right)^2}{1 \times\left(\frac{-3+\sqrt{5}}{2}\right)\left(\frac{-3-\sqrt{5}}{2}\right)} \\ & =\frac{1+1+\frac{5+9-6 \sqrt{5}}{4}+\frac{5+9+6 \sqrt{5}}{4}}{\frac{9-5}{4}} \\ & =\frac{8+5+9-6 \sqrt{5}+5+9+6 \sqrt{5}}{4}=\frac{36}{4}=9 \\ & \therefore \text { Ratio }=9: 1 \end{aligned} $

Given,

$\alpha_1, \beta_1, \gamma_1, \delta_1$ are the roots of equation

$ a x^4+b x^3+c x^2+d x+e=0 \text { and } $

$\alpha_2, \beta_2, \gamma_2, \delta_2$ are the roots of equation

$ e x^4+d x^3+c x^2+b x+a=0 $

∴ Clearly,

$\alpha_1, \beta_1, \gamma_1, \delta_1$ are the reciprocal of roots of

$ a x^4+b x^3+c x^2+d x+e=0 $

Also, $\delta_1>\gamma_1>\beta_1>\alpha_1>0$

and $\delta_2>\gamma_2>\beta_2>\alpha_2>0$

$\therefore \delta_1$ and $\alpha_2$ are reciprocal

Similarly $\gamma_1$ and $\beta_2$,

$\gamma_2$ and $\beta_1$ and $\alpha_1$ and $\delta_2$ are reciprocal

$ \begin{aligned} & \therefore \quad \alpha_1 \delta_2=1, \beta_1 \gamma_2=1, \beta_2 \gamma_1=1, \alpha_2 \delta_1=1 \end{aligned} $

$ \begin{aligned} & \alpha_1-\delta_2=2 \\ & \therefore \quad \alpha_1-\frac{1}{\alpha_1}=2 \Rightarrow \alpha_1^2-2 \alpha_1-1=0 \\ & \text { and } \delta_1-\alpha_2=4 \\ & \\ & \delta_1-\frac{1}{\delta_1}=4=\delta_1^2-4 \delta_1-1=0 \end{aligned} $

$\therefore \alpha_1$ and $\delta_1$ are roots of

$ \begin{gathered} a x^4+b x^3+c x^2+d x+e=0 \\ \therefore\left(x^2-2 x-1\right)\left(x^2-4 x-1\right) \\ =a x^4+b x^3+c x^2+d x+e \end{gathered} $

Put $x=1$,

$ \begin{gathered} (1-2-1)(1-4-1)=a+b+c+d+c \\ 8=a+b+c+d+e \end{gathered} $