Quadratic Equations

Given, equation

$ x^3+4 x^2-9 x-36=0 $

Factors of 36 are

$ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18 $

and $\pm 36$

$x=3$ satisfies this equation, so one of the factor is $(x-3)$, then

$ \begin{aligned} & \quad \quad x^3+4 x^2-9 x-36=0 \\ & \Rightarrow \quad(x-3)\left(x^2+7 x+12\right)=0 \\ & \Rightarrow \quad(x-3)(x+3)(x+4)=0 \\ & \Rightarrow \quad x=3,-3,-4 \\ & \because \quad \alpha<\beta<\gamma \Rightarrow-4<-3<3 \\ & \therefore \alpha=-4, \beta=-3 \text { and } \gamma=3 \\ & \therefore \alpha+2 \beta+3 \gamma=-1 \end{aligned} $

Given equation

$ x^4-4 x^3-7 x^2+22 x+24=0 $

Let $a, b, c$ and $d$ are roots of the equation, such that

$ a+b=c+d \text {, then by relation } $

between roots and coefficients

$ \begin{aligned} & a+b+c+d=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow \quad & a+b=2, c+d=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

and $a b+a c+a d+b c+b d+c d=-7$

$ \begin{array}{r} a b+a(c+d)+b(c+d)+c d=-7 \\ a b+c d+(a+b)(c+d)=-7 \\ a b+c d+4=-7 \\ a b+c d=-11 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{array} $

Now, $(a+b+c+d)^3=(a+b)^3+(c+d)^3$

$ \begin{aligned} +3(a+b)^2(c+d)+3(a+b)(c+d)^2 & \\ (a+b+c+d)^3= & a^3+b^3+c^3+d^3 \\ + & 3 a b(a+b)+3 c d(c+d) \\ + & 3(a+b)(c+d)(a+b+c+d) \end{aligned} $

$ \begin{aligned} &\text { Using Eqs. (i), (ii) and (iii), we get }\\ &\begin{aligned} & 4^3=a^3+b^3+c^3+d^3+6 a b+6 c d+3(2)(2)(4) \\ & 64=a^3+b^3+c^3+d^3-66+48 \\ & \Rightarrow a^3+b^3+c^3+d^3=82 \end{aligned} \end{aligned} $

Given that

$ \begin{aligned} & \quad x^2-1 \leq 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \Rightarrow \quad x^2 \leq 1 \\ & \Rightarrow \quad-1 \leq x \leq 1 \\ & \text { and also } x^2-x-2 \geq 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \Rightarrow(x-2)(x+1) \geq 0 \\ & \Rightarrow x \in(-\infty,-1] \cup[2, \infty) \end{aligned} $

Thus, the set of all value of $x$ which satisfy both the inequalities $x^2-1 \leq 0$ and $x^2-x-2 \geq 0$ simultaneously is $\{-1\}$.

The given equations are

$ x^2-6 x+a=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$and\,\, x^2-c x+6=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Since, $\alpha$ is a common roots in both the equations

Thus, the other roots of first and second equations are $4 \beta$ and $3 \beta$.

Hence, $(\alpha, 4 \beta)$ and $(\alpha, 3 \beta)$ are the roots of first and second equations.

Thus, $4 \alpha \beta=a$ and $3 \alpha \beta=6$

$ \alpha \beta=\frac{a}{4} \Rightarrow a=8 $

So, from Eq. (i), we get

$ x^2-6 x+8=0 \Rightarrow x=2 \text { and } 4 $

If $x=2$ is a common roots, then

From Eq. (ii), we get $4-2 c+6=0$

$ \begin{aligned} & \Rightarrow c=5 \\ & \therefore x^2-5 x+6=0 \Rightarrow x=3,2 \end{aligned} $

Thus, their common root is 2 .

To find $\alpha^{15} + \beta^{15}$ for the roots $\alpha$ and $\beta$ of the equation $x^2 + 2x + 2 = 0$, let's solve the quadratic equation:

$ x^2 + 2x + 2 = 0 $

First, we complete the square:

$ (x + 1)^2 = -1 $

This implies:

$ x + 1 = \pm i $

Thus, the roots are:

$ x = i - 1 \quad \text{or} \quad x = -i - 1 $

Assign $\alpha = i - 1$ and $\beta = -i - 1$.

Next, calculate $\alpha^{15} + \beta^{15}$:

$ \alpha^2 = i^2 - 2i + 1 = -1 - 2i + 1 = -2i $

$ \beta^2 = (-i)^2 - 2(-i) + 1 = -1 + 2i + 1 = 2i $

Now express $\alpha^{15} + \beta^{15}$ as:

$ \alpha^{15} + \beta^{15} = (\alpha^2)^7 \cdot \alpha + (\beta^2)^7 \cdot \beta $

Substitute $\alpha^2$ and $\beta^2$:

$ = (-2i)^7 \cdot (i-1) + (2i)^7 \cdot (-i-1) $

Simplify:

Calculate $(-2i)^7$:

$ (-2i)^7 = (-2)^7 \cdot i^7 = -128 \cdot (-i) = 128i $

Calculate $(2i)^7$:

$ (2i)^7 = 2^7 \cdot i^7 = 128 \cdot (-i) = -128i $

Substitute back:

$ = 128i(i-1) - 128i(-i-1) $

Simplify the expression:

$ = 128i \cdot i - 128i \cdot 1 - 128i \cdot (-i) - 128i \cdot (-1) $

$ = 128(-1) - 128i + 128 - 128i $

Combine terms:

$ = 128(-1 + 1) + 128i - 128i $

$ = 256(-1) = -256 $

Thus, $\alpha^{15} + \beta^{15} = -256$.

Consider the equation:

$ x^4 - 2a x^3 + 4b x^2 + 8a x + 16 = 0 \quad \quad \quad \quad \quad (i) $

Let the roots of equation (i) be $\alpha$, $\beta$, $\gamma$, and $\delta$.

The product of the roots for polynomial equation (i) is given by:

$ \alpha \beta \gamma \delta = \frac{\text{Constant term}}{\text{Coefficient of } x^4} = \frac{16}{1} = 16 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (ii) $

Suppose we have a new equation whose roots are $P$ times the roots of equation (i), thus the roots become $P \alpha$, $P \beta$, $P \gamma$, and $P \delta$.

Now, the product of these new roots is:

$ P \alpha \cdot P \beta \cdot P \gamma \cdot P \delta = P^4 \times \alpha \beta \gamma \delta = 1 $

So,

$ P^4 = \frac{1}{\alpha \beta \gamma \delta} $

Substituting the value from equation (ii), we get:

$ P^4 = \frac{1}{16} $

Thus,

$ |P| = \frac{1}{(16)^{\frac{1}{4}}} = \frac{1}{2} $

Therefore, the absolute value of $P$ is $\frac{1}{2}$.

Given the quadratic equation $4x^2 - 2x + (k - 4) = 0$, one of the roots is the reciprocal of the other. To find the value of $k$, we first compare the equation to the standard form:

$ ax^2 + bx + c = 0 $

Here, we identify:

$ a = 4, \quad b = -2, \quad c = k - 4 $

Since one root is the reciprocal of the other, let $\alpha$ and $\beta$ be the roots. The relationship between the roots is:

$ \alpha = \frac{1}{\beta} \quad \Rightarrow \quad \alpha \beta = 1 $

The product of the roots of a quadratic equation is given by:

$ \alpha \cdot \beta = \frac{c}{a} $

Substituting the known values:

$ 1 = \frac{k - 4}{4} $

To solve for $k$, we set:

$ k - 4 = 4 $

Thus, solving for $k$, we find:

$ k = 4 + 4 = 8 $

Therefore, the value of $k$ is $8$.

$\begin{aligned} & n(S)={ }^9 C_1 \times{ }^8 C_1=72 \\ & x^2+b x+c>0 \Rightarrow b^2-4 c<0\end{aligned}$

$\begin{aligned} & \therefore n(E)=29 \\ & \therefore \text { Required probability }=\frac{29}{72}\end{aligned}$

Given, $(x-2)$ is a common factor of equation $\left(x^2+a x+b\right)$ and

$ \begin{aligned} & \quad\left(x^2+\alpha+d\right) . \\ & \therefore x^2+a x+b=(x-2)(x-\alpha) \\ & x^2+a x+b=x^2-(2+\alpha) x+2 \alpha \\ & \therefore 2 \alpha=b ; \quad \therefore \alpha=\frac{b}{2} \\ & \text { and }-2-\alpha=a \\ & \quad-2-\frac{b}{2}=a \end{aligned} $

$\begin{aligned} x^2+\alpha+d & =(x-2)(x-\beta) \\ x^2+\alpha+d & =x^2-(2+\beta) x+2 \beta \\ d & =2 \beta \\ \therefore \quad \frac{d}{2} & =\beta \\ \therefore \quad-2-\beta & =c \\ \therefore \quad \frac{b-d}{c-a} & =\frac{2 \alpha-2 \beta}{-2-\beta+2+\alpha} \\ & =\frac{2(\alpha-\beta)}{(\alpha-\beta)}=2\end{aligned}$

Given,

$ \begin{aligned} & e^{4 t}-10 e^{3 t}+29 e^{2 t}-20 e^t+4=0 \\ & \text { Let } \quad e^t=x \\ & \Rightarrow x^4-10 x^3+29 x^2-20 x+4=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad x^2-10 x+29-\frac{20}{x}+\frac{4}{x^2}=0 \\ & \Rightarrow x^2+\left(\frac{2}{x}\right)^2-10\left(x+\frac{2}{x}\right)+29=0 \\ & \Rightarrow\left(x+\frac{2}{x}\right)^2-10\left(x+\frac{2}{x}\right)+25=0 \\ & \text { Let } \\ & \therefore \quad x+\frac{2}{x}=P \\ & \therefore \quad P^2-10 P+25=0 \\ & \Rightarrow \quad(P-5)^2=0 \\ & \Rightarrow \quad P \end{aligned} \begin{aligned} & =5 \end{aligned} $

Putting the value of $P$ in Eq. (ii), we get

$ \begin{aligned} x+\frac{2}{x} & =5 \\ \Rightarrow \quad x^2-5 x+2 & =0 \\ \Rightarrow \quad x & =\frac{5 \pm \sqrt{25-8}}{2} \\ \therefore \quad x & =\frac{5+\sqrt{17}}{2}, \frac{5-\sqrt{17}}{2} \\ e^{t_1} & =\frac{5+\sqrt{17}}{2}, \\ e^{t_2} & =\frac{5-\sqrt{17}}{2} \\ e^{t_1} \cdot e^{t_2} & =\frac{25-17}{4}=\frac{8}{4}=2 \\ e^{t_1}+t_2 & =2 \\ t_1+t_2 & =\log _e 2 \end{aligned} $

$\because$ There are 4 roots and 2 of them are repeated.

$ \Rightarrow t_1+t_2+t_3+t_4=2 \log _e 2 $

Statement $\mathrm{I}|x|^2-4|x|+3<0$

For $|x|=$ Positive,

$ \begin{array}{rlrl} & x^2-4 x+3 <0 \\ \Rightarrow & x^2-x-3 x+3 <0 \\ \Rightarrow & x(x-1)-3(x-1) <0 \\ \Rightarrow & (x-1)(x-3) <0 \end{array} $

$ \begin{aligned} &x \in(1,3)\\ &\text { For }|x|=\text { Negative }\\ &\begin{aligned} & x^2+4 x+3<0 \\ \Rightarrow & x^2+x+3 x+3<0 \\ \Rightarrow & x(x+1)+3(x+1)<0 \\ \Rightarrow & (x+1)(x+3)<0 \end{aligned} \end{aligned} $

$ x \in(-3,-1) $

Therefore, $x \in(-3,-1) \cup(1,3)$ Statement I is false.

$ \begin{aligned} &\text { Statement II } x^2-8 x+15>0\\ &\begin{array}{lr} \Rightarrow & x^2-3 x-5 x+15>0 \\ \Rightarrow & x(x-3)-5(x-3)>0 \\ \Rightarrow & (x-3)(x-9)>0 \end{array} \end{aligned} $

$ \begin{aligned} &\Rightarrow x>5 \text { or } x<3\\ &\text { Statement II is true. } \end{aligned} $

$6 x-x^2+12$ attains its extreme value $\beta$ at $x=\alpha$

$ \begin{array}{ll} \text { Let } & y=-x^2+6 x+12 \\ \Rightarrow & y=-\left[x^2-6 x-12\right] \\ \Rightarrow & y=-\left[x^2-6 x+9-9-12\right] \\ \Rightarrow & y=-\left[(x-3)^2-21\right] \\ \Rightarrow & y=21-(x-3)^2 \\ \Rightarrow & y-21=-(x-3)^2 \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad(x-3)^2=-(y-21) \\ & \Rightarrow \quad(x-h)^2=4 p(y-k) \end{aligned}\\ &\text { Vertex, }(h, k)=(3,21) \end{aligned} $

$ \begin{array}{ll} \text { At } & x=3, y \rightarrow y_{\text {max }} \\ \therefore & \alpha=3 \text { and } \beta=21 \\ \Rightarrow & \beta=7 \times 3=7 \times \alpha=7 \alpha \end{array} $

$ \begin{aligned} & \text { } x^3-2 x-25 \lambda=0 \\ & \Rightarrow \quad \frac{x^3-2 x}{25}=\lambda \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 3 x^3-8 x-\frac{175}{3} \lambda=0 \\ & \Rightarrow \quad \frac{9 x^3-24 x}{175}=\lambda \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On equating Eqs. (i) and (ii),

$ \frac{x^3-2 x}{25}=\frac{9 x^3-24 x}{175} $

$ \begin{array}{ll} \Rightarrow & 7 x^3-14 x=9 x^3-24 x \\ \Rightarrow & 10 x=2 x^3 \Rightarrow x^3-5 x=0 \\ \Rightarrow & x\left(x^2-5\right)=0 \\ \Rightarrow & x=0, x= \pm \sqrt{5} \end{array} $

For $x=0$

$\Rightarrow \lambda=0$ [From Eqs. (i) and (ii)]

But $\lambda>0$    (given)

For $x=\sqrt{5}$

$ 5 \sqrt{5}-2 \sqrt{5}-25 \lambda=0 \text { [from Eq. (i) }] $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad \frac{3 \sqrt{5}}{25}=\lambda \\ & \Rightarrow \quad \lambda=\frac{3}{5 \sqrt{5}} \quad(\lambda>0) \\ & \text { At } x=-\sqrt{5} \\ & -5 \sqrt{5}+2 \sqrt{5}-25 \lambda=0 \\ & \Rightarrow \quad \lambda=\frac{-3 \sqrt{5}}{25} \quad(\lambda<0) \end{aligned}\\ &\text { Therefore, } \lambda=\frac{3}{5 \sqrt{5}} \end{aligned} $

$ \text { } \begin{aligned} & x^3-7 p x^2+5 q x-6 r=0 \\ & \alpha+\beta+\gamma=-\frac{(-7 p)}{1}=7 p \end{aligned} $

Sum of two roots is zero.

$ \begin{array}{llrl} \therefore & \text { Let } \alpha+\beta=0 & \\ \text { Then, } & \gamma=7 p & \\ & \alpha \cdot \beta+\beta \cdot \gamma+\gamma \cdot \alpha =5 q \\ \Rightarrow & \alpha \beta+\gamma(\alpha+\beta) =5 q \\ \Rightarrow & \alpha \cdot \beta+\gamma \cdot 0 =5 q \\ \Rightarrow & \alpha \cdot \beta =5 q \end{array} $

Now, $\alpha \cdot \beta \cdot \gamma=\frac{-(-6 r)}{1}=6 r$

$ \begin{aligned} \Rightarrow & & 5 q \cdot 7 p & =6 r \\ \Rightarrow & & 5 q & =6 r / 7 p \end{aligned} $

$3 p^2 x^3+p x^2+q x+3=0$

$ \begin{aligned} & \quad \quad p=1 \text { and } q=-7 \\ & \Rightarrow \quad 3 x^3+x^2-7 x+3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

On putting $x=1$

$ 3 \times 1^3+1^2-7 \times 1+3=7-7=0 $

$\therefore(x-1)$ is one factor in the Eq. (i),

∴ Eq. (i) can be written as

$ (x-1)\left(3 x^2+4 x-3\right)=0 $

$3 x^2+4 x-3=0$ gives the irrational roots.

$ \begin{aligned} \Rightarrow \alpha+\beta & =\frac{-4}{3}, \alpha \cdot \beta=\frac{-3}{3}=-1 \\ |\alpha-\beta| & =\sqrt{\left(\alpha+\beta^2\right)-4 \alpha \cdot \beta} \\ & =\sqrt{\left(\frac{-4}{3}\right)^2-4(-1)} \\ & =\sqrt{\frac{16}{9}+4} \\ & =\sqrt{\frac{52}{9}}=\frac{2 \sqrt{13}}{3} \end{aligned} $

The transformed equation is

$ 8 x^3-54 x-78=0 $

Root of cubic equation $f(x)=0$ are diminished by $\frac{-3}{2}$.

Since, we don't have the original equation but the transformed equation. Therefore, we will verify it through the option.

On taking option (a), we have

$ \begin{aligned} & f(x)=2 x^3-9 x^2-6 \\ & f(x)=2 x^3-9 x^2+0 x-6 \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \therefore & A_0 y^3+A_1 y^2+A_2 y+A_3=0 \\ & 2 y^3+0 y^2+\left(\frac{-27}{2}\right) y-\frac{39}{2}=0 \\ \Rightarrow & 2 y^3-\frac{27 y}{2}-\frac{39}{2}=0 \\ \Rightarrow & 8 y^3-54 y-78=0 \\ \text { or } & 8 x^3-54 x-78=0 \end{array}\\ &\text { which is the transformed equation. } \end{aligned} $

Given, equation $x^2+b x+c=0$ has roots $\alpha$ and $\beta$, then

$ \alpha+\beta=-b, \alpha \beta=c $

Given, $\alpha^2+\beta^2=5 \Rightarrow(\alpha+\beta)^2-2 \alpha \beta=5$

$ \Rightarrow \quad b^2-2 c=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, $\alpha^3+\beta^3=9$

$ \begin{array}{lr} \Rightarrow(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right) =9 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, -b(5-c)=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eq. (i), $c=\frac{b^2-5}{2}$ on putting in Eq. (ii),

$ \begin{array}{rlrl} & -b\left(5-\frac{b^2-5}{2}\right) =9 \\ \Rightarrow & -b\left(\frac{15-b^2}{2}\right) =9 \\ & \Rightarrow b^3-15 b-18 & =0 \end{array} $

Since, $b=-3$ is root of the above equation.

$ \begin{aligned} & \therefore \quad(b+3)\left(b^2-3 b-6\right)=0 \\ & \quad b=-3, c=\frac{b^2-5}{2}=\frac{9-5}{2}=2 \\ & \therefore \quad b+c=-1 \end{aligned} $

Let

$ \begin{aligned} & y=\frac{x^2-x+2}{x^2+x-2} ; \forall x \in R-\{-2,1\} \\ \Rightarrow \quad & y x^2+y x-2 y=x^2-x+2 \\ \Rightarrow & (y-1) x^2+(y+1) x-2(y+1)=0 \end{aligned} $

For $x \in R$,

$ \begin{aligned} & D \geq 0 \\ & \Rightarrow(y+1)^2-4(y-1)[-2(y+1)] \geq 0 \\ & \Rightarrow \quad(y+1)[(y+1)+8(y-1)] \geq 0 \\ & \Rightarrow \quad(y+1)(9 y-7) \geq 0 \end{aligned} $

$ \therefore y \in(-\infty,-1] \cup\left[\frac{7}{9}, \infty\right) $

Given, $\alpha, \beta$ and $\gamma$ are the roots of the equation,

$ x^3-9 x^2+23 x-15=0 $

Then, $\quad \alpha+\beta+\gamma=9$

$ \quad \begin{aligned} \Rightarrow \quad \alpha \beta+\beta \gamma+\gamma \alpha & =23 \\ \alpha \beta \gamma & =15 \end{aligned} $

Since,

$ \begin{array}{r} \alpha^3+\beta^3+\gamma^3-3 \alpha \beta \gamma=(\alpha+\beta+\gamma) \\ \left\{\alpha^2+\beta^2+\gamma^2-(\alpha \beta+\beta \gamma+\gamma \alpha)\right\} \\ \Rightarrow \quad \alpha^3+\beta^3+\gamma^3=3 \alpha \beta \gamma \\ +(\alpha+\beta+\gamma)\left[(\alpha+\beta+\gamma)^2\right. \\ -3(\alpha \beta+\beta \gamma+\gamma \alpha)] \\ =(3 \times 15)+9\left\{(9)^2-3 \times 23\right\} \\ =45+9 \times(81-69)=153 \end{array} $

Given, $\alpha, \beta$ and $2 \beta$ are the real roots of the equation $x^3-9 x^2+k=0$ where $k \in R-\{0\}$.

Then, $\alpha+\beta+2 \beta=9$

$ \begin{array}{ll} \Rightarrow & \alpha+3 \beta=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } & \alpha \cdot \beta \cdot 2 \beta=-k \\ \Rightarrow & 2 \alpha \beta^2=-k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Product of two roots

$ \begin{array}{r} \alpha \beta+\beta \cdot 2 \beta+2 \beta \cdot \alpha=0 \\ \Rightarrow \quad 3 \alpha \beta+2 \beta^2=0 \\ \Rightarrow \quad \beta(3 \alpha+2 \beta)=0 \\ \beta \neq 0 \text { because } k \neq 0 \text { so, } \\ \quad 3 \alpha+2 \beta=0 &\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

From Eqs. (i) and (iii),

$ 7 \beta=27 \Rightarrow 14 \beta=54 $

Given equation,

$ \begin{array}{r} x^5-3 x^4+5 x^3-5 x^2+3 x-1=0 \\ \Rightarrow(x-1)\left(x^4-2 x^3+3 x^2-2 x+1\right)=0 \end{array} $

[ $\because$ ] is a root of given

$ \begin{array}{r} \Rightarrow(x-1) x^2\left(x^2-2 x+3-\frac{2}{x}+\frac{1}{x^2}\right)=0 \\ \Rightarrow(x-1) x^2\left(x^2+\frac{1}{x^2}-2\left(x+\frac{1}{x}\right)+3\right) =0 \\ (x-1) x^2\left\{\left(x+\frac{1}{x}\right)^2-2-2\left(x+\frac{1}{x}\right)+3\right\}=0 \\ =(x-1) x^2\left\{\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)+1\right\} =0 \end{array} $

$ \begin{aligned} & \Rightarrow \quad(x-1) x^2\left(x+\frac{1}{x}-1\right)^2=0 \\ & \Rightarrow \quad(x-1)\left(x^2-x+1\right)^2=0 \end{aligned} $

Now, $x-1=0 \Rightarrow x=1$

$ \begin{aligned}and\,\,\, x^2-x+1 & =0 \\\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x & =\frac{1 \pm 2}{2}=\frac{3}{2},-\frac{1}{2} \end{aligned} $

So, sum of distinct roots

$ =1+\frac{3}{2}-\frac{1}{2}=1+1=2 $

Here, $\left(x^4+1\right)=\frac{1}{a}(x+1)^4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Given equation is a reciprocal equation. Then, replace $x$ by $1 / x ; x \neq 0$

$ \begin{aligned} & \left(\frac{1}{x^4}+1\right) =\frac{1}{a}\left(\frac{1}{x}+1\right)^4 \\ \Rightarrow & \left(\frac{1+x^4}{x^4}\right) =\frac{1}{a} \frac{(1+x)^4}{x^4} \\ \Rightarrow & \left(1+x^4\right) =\frac{1}{a}(1+x)^4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

At $x=0, a=1$

As $x \neq 0, a \in R-\{1\}$.

$ \text { } \begin{aligned} f(x) & =A x^2+B x \\ g(x) & =L x^2+M x+N \\ f(x)-g(x) & =A x^2+B x-L x^2-M x-N \\ & =(A-L) x^2+(B-M) x-N \end{aligned} $

$ \begin{aligned} & \text { } \quad \begin{aligned}Let\,\,\, & A-L=P \\ & B-M=Q \\ & f(x)-g(x)=P x^2+Q x-N \\ & f(2)-g(2)=1 \\ & 4 P+2 Q-N=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} $

Also, $f(3-g(3)=4$

$ 9 P+3 Q-N=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Also, $f(4)-g(4)=9$

$ 16 P+4 Q-N=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Now, Eq. (i) $\times 4$-Eq. (iii), we get

$ 4 Q-3 N=-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Eq. (i) $\times 9-$ Eq. (ii) $\times 4$, we get

$ 6 Q-5 N=-7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

Eq. (iv) $\times 3$-Eq. (v) $\times 2$, we get

$ \begin{array}{rlrl} & -9 N+10 N & =-15+14 \\ & \Rightarrow & N & =-1 \end{array} $

From Eq. (iv),

$ \begin{aligned} & & 4 Q-3(-1) & =-5 \\ \Rightarrow & & 4 Q+3 & =-5 \\ \Rightarrow & & 4 Q & =-8 \\ \Rightarrow & & Q & =-2 \end{aligned} $

From Eq. (i),

$ \begin{array}{rlrl} & & 4 P+2(-2)-(-1) & =1 \\ & & 4 P-4+1 & =1 \\ &&P =1 \\ & \therefore & f(x)-g(x) & =x^2-2 x+1 \\ & & =(x-1)^2 \\ & & f(x)-g(x) & =0 \\ & \Rightarrow & (x-1)^2 & =0 \\ & \Rightarrow & x & =1 \end{array} $

∴ The root of $f(x)-g(x)=0$ is 1 .

$ \begin{aligned} & \text { } f(x)=\frac{2 x-3}{(x-2)(x-3)} \\ & \text { Let } \quad y=\frac{2 x-3}{x^2-5 x+6} \\ & \Rightarrow \quad x^2 y-5 x y+6 y-2 x+3=0 \\ & \Rightarrow \quad x^2 y-x(5 y+2)+6 y+3=0 \\ & \quad D \geq 0 \end{aligned} $

$ \begin{array}{rlrl} & (5 y+2)^2-4 y(6 y+3) \geq 0 \\ \Rightarrow & 25 y^2+4+20 y-24 y^2-12 y \geq 0 \\ \Rightarrow & y^2+8 y+4 \geq 0 \\ \therefore & y^2+8 y+16-12 \geq 0 \\ \Rightarrow & (y+4)^2 \geq 12 \end{array} $

From option (a),

$ (-10+4)^2=36 \geq 12 $

From option (b),

$ (2+4)^2=36 \geq 12 $

From option (c),

$ (1+4)^2=25 \geq 12 $

From option (d),

$ (-2+4)^2=4<12 $

∴ Option (d) is correct.

Given, $\alpha, \beta$ and $\gamma$ are the roots of the equation

$ 5 x^3-3 x^2+2 x-4=0 $

Then, $\alpha+\beta+\gamma=\frac{3}{5}$

$ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{2}{5} $

and   $ \alpha \beta \gamma=\frac{4}{5} $

Now, $\alpha+\beta+\gamma=\frac{3}{5}$

$ \begin{aligned} &\text { On squaring both sides, we get }\\ &\begin{aligned} & (\alpha+\beta+\gamma)^2=\frac{9}{25} \\ \Rightarrow & \alpha^2+\beta^2+\gamma^2 \\ & +2(\alpha \beta+\beta \alpha+\gamma \alpha)=\frac{9}{25} \\ \Rightarrow & \alpha^2+\beta^2+\gamma^2=\frac{9}{25}-\frac{4}{5}=-\frac{11}{25} \\ \therefore & \alpha \beta+\beta \gamma+\gamma \alpha=\frac{2}{5} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { On squaring both sides, we get }\\ &\begin{aligned} & \alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2+2 \alpha \beta^2 \gamma+2 \alpha \beta \gamma^2 +2 \alpha^2 \beta \gamma=\frac{4}{25} \\ & \Rightarrow \Sigma \alpha^2 \beta^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)=4 / 25 \\ & \Rightarrow \Sigma \alpha^2 \beta^2= \frac{4}{25}-2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right)=\frac{4}{25}-\frac{24}{25} \\ &=-\frac{20}{25}=-\frac{4}{5} \end{aligned} \end{aligned} $

Given, $\alpha, \beta$ and $\gamma$ are the roots of equation $x^3+4 x^2-9 x-36=0$ and

$ \begin{aligned} & \alpha+\beta=0 & & \\ \therefore & & \alpha+\beta+\gamma & =-4 \\ \Rightarrow & & \gamma & =-4 \\ & & \alpha \beta+\beta \gamma+\gamma \alpha & =-9 \\ \Rightarrow & & \alpha \beta+\gamma(\alpha+\beta) & =-9 \\ \Rightarrow & & \alpha \beta & =-9 \\ & & \alpha(-\alpha) & =-9 \\ & & & {[\because \alpha+\beta=0] } \end{aligned} $

So, $-\alpha^2=-9 \Rightarrow \alpha^2=9.$

Since $\alpha+\beta=0$, we get $\beta=-\alpha$. Therefore, $\beta^2=\alpha^2=9$.

Also, from above, $\gamma=-4$, so $\gamma^2=16$.

Now substitute these values:

$ \begin{aligned} \alpha^2+2 \beta^2+3 \gamma^2 & =9+2 \times 9+3 \times 16 \\ & =9+18+48 \\ & =75 \end{aligned} $

Given, equation

$ 6 x^6-25 x^5+31 x^4-31 x^2+25 x-6=0 $

which is of reciprocal equation of even degree with opposite sign.

$\therefore x^2-1$ is the root of the equation.

$\therefore x= \pm 1$ are the zeroes of the equation.

$ \begin{aligned} &\Rightarrow 6 x^4-25 x^3+37 x^2-25 x+6=0\\ &\text { Dividing by } x^2 \text {, we get }\\ &\begin{aligned} 6 x^2-25 x+37-\frac{25}{x}+\frac{6}{x^2} & =0 \\ \Rightarrow 6\left(x^2+\frac{1}{x^2}\right)-25\left(x+\frac{1}{x}\right)+37 & =0 \end{aligned} \end{aligned} $

$ \begin{aligned} & 6\left(\left(x+\frac{1}{x}\right)^2-2\right)-25\left(x+\frac{1}{x}\right)+37=0 \\ & \Rightarrow \quad 6\left(x+\frac{1}{x}\right)^2-25\left(x+\frac{1}{x}\right)+25=0 \\ & \therefore \quad 6 t^2-25 t+25=0\left[\text { Let } x+\frac{1}{x}=t\right] \\ & \Rightarrow \quad 6 t^2-15 t-10 t+25=0 \\ & \Rightarrow 3 t(2 t-5)-5(2 t-5)=0 \\ & \Rightarrow t=\frac{5}{3}, t=\frac{5}{2} \\ & \therefore \quad x+\frac{1}{x}=\frac{5}{3} \text { or } x+\frac{1}{x}=\frac{5}{2} \\ & \Rightarrow \quad 3 x^2+3=5 x \text { or } 2 x^2+2=5 x \\ & \Rightarrow \quad 3 x^2-5 x+3=0 \\ & \text { or } \quad 2 x^2-5 x+2=0 \end{aligned} $

The roots of $3 x^2-5 x+3=0$ are not real

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

∴ The rational roots $\frac{1}{2}, 2,-1, \frac{5}{3}, \frac{5}{2}$.

The value of $m=-1$ and $M=2$

$ M-m=2-(-1)=3 $

$\because \alpha$ and $\beta$ are the roots of the equation

$ \begin{aligned} & x^2-2 \sqrt{3} x+4=0 \\ & \therefore \quad \alpha+\beta=2 \sqrt{3} \text { and } \alpha \beta=4 \\ & \text { Now, } \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\ & =12-8=4 \\ & \begin{aligned} \therefore \alpha^6+\beta^6 & =\left(\alpha^2\right)^3+\left(\beta^2\right)^3 \\ & =\left(\alpha^2+\beta^2\right)^3-3 \alpha^2 \beta^2\left(\alpha^2+\beta^2\right) \\ & =(4)^3-3(4)^2(4) \\ {[\because} & \left.a^3+b^3=(a+b)^3-3 a b(a+b)\right] \\ & =4^3(1-3)=64(-2)=-128 \end{aligned} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \text { } c=-15 x(-2)=30 \\ & \text { Now, } x^2+13 x+30=0 \\ & x^2+10 x+3 x+30=0 \\ & x(x+10)+3(x+10)=0 \\ &\begin{aligned} \Rightarrow & \,\,\,\,\,x=-10,-3 \\ \therefore &\,\,\,\,\, |\alpha-\beta|=|-3+10|=7 \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text {(A) } 2 x^2+4 x+5 \\ & 2 x^2+4 x+2+3 \Rightarrow 2(x+1)^2+3 \end{aligned} $

∴ The minimum value of $2(x+1)^2+3\,\,\, is 3 .$

$ \text { } \begin{aligned}(B)\,\,Let\,\, f(x) & =\frac{x^2+4 x+1}{x^2+x+1} \\ & =\frac{x^2+x+1}{x^2+x+1}+\frac{3 x}{x^2+x+1} \\ & =1+\frac{3 x}{x^2+x+1} \end{aligned} $

The maximum value of $\frac{3 x}{x^2+x+1}$ can be 1.

$ \frac{3 x}{x^2+x+1}=\frac{3}{x+\frac{1}{x}+1} \text { and } x+\frac{1}{x} \geq 2 $

Therefore, maximum value of $f(x)$ is 2 .

(C) Given, $1 \leq \frac{3 x^2-5 x+6}{x^2+1} \leq 2$

$ \begin{aligned} & \Rightarrow 1 \leq \frac{3 x^2-5 x+6}{x^2+1} \text { and } \frac{3 x^2-5 x+6}{x^2+1} \leq 2 \\ & \Rightarrow \quad x^2+1 \leq 3 x^2-5 x+5 \\ & \text { and } 3 x^2-5 x+6 \leq 2 x^2+2 \\ & \Rightarrow 2 x^2-5 x+4 \geq 0 \text { and } x^2-5 x+4 \leq 0 \end{aligned} $

The value of $x$ for which

$2 x^2-5 x+4 \geq 0$ are not real.

$\because 2 x^2-5 x+4 \geq 0$ is possible for all real value

$ \begin{aligned} & \Rightarrow \quad x \in R \\ & \text { and } \quad x^2-5 x+4 \leq 0 \\ & \Rightarrow \quad(x-4)(x-1) \leq 0 \Rightarrow x \in[1,4] \end{aligned} $

Now common of $x \in R$ and $x \in[1,4]$

$ \begin{array}{ll} \Rightarrow & x \in[1,4] \\ \therefore & b=4 \end{array} $

(D) $a=1$

Correct order is = IV, III, V, II

$\because \alpha, \beta$ and $\gamma$ are the roots of the equation $5 x^3-2 x-4=0$

$ \therefore \alpha+\beta+\gamma=0, \alpha \beta+\beta \alpha+\gamma \alpha=-\frac{2}{5} $

and $\alpha \beta \gamma=\frac{4}{5}$

If $\alpha+\beta+\gamma=0$, then

$ \alpha^3+\beta^3+\gamma^3=3 \alpha \beta \gamma=3 \times \frac{4}{5}=\frac{12}{5} $

Let $a_1, a_2, a_3, a_4$ and $a_5$ are the roots of

$ x^5-a x^4+b x^3-c x^2+d x-1=0 $

If AM and GM of $a_1, a_2, a_3, a_4$ and $a_5$ are equal, then

$ a_1=a_2=a_3=a_4=a_5=1 $

Sum of roots $={ }^5 C_1=5$

$ \because \quad a=5 $

Sum of roots taken 2 at a time

$ ={ }^5 C_2=10 $

$\therefore \quad b=10$

Sum of roots taken 3 at a time

$ ={ }^5 C_3=10 $

$ \therefore \quad c=10 $

Sum of roots taken 4 at a time $={ }^5 C_4=5$

$ \therefore \quad d=5 $

Hence, $a+b+c+d=5+10+10+5$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =30 $

$ \begin{aligned} P(x): x^{10}-3 x^8+5 x^6 & -5 x^4 +3 x^2-1=0 \end{aligned} $

$ \begin{aligned} & \because P(1) \text { and } P(-1) \text { are zero } \\ & \Rightarrow(x-1) \text { and }(x+1) \text { are factors. } \\ & P(x)=\left(x^2-1\right)\left(x^8-2 x^6+3 x^4-2 x^2+1\right) \\ & \text { Let } g(x)=x^8-2 x^6+3 x^4-2 x^2+1 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =x^2\left(x^6-2 x^4+3 x^2-2\right)+1 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =x^2\left(x^2-1\right)\left(x^4-x^2+2\right)+1 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =x^2\left(x^2-1\right)\left\{\left(x^2-\frac{1}{2}\right)^2+\frac{7}{4}\right\}+1 \end{aligned} $

$g(x) \neq 0$ for any real value.

∴ Total number of non-real roots

$ =10-2=8 $

Condition for a common root

$ \begin{aligned} & \left(c_1 a_2-c_2 a_1\right)^2=\left(b_1 c_2-b_2 c_1\right)\left(a_1 b_2-a_2 b_1\right) \\ & \Rightarrow \quad(3 c-(-5 c))^2=(35 c-3 c)(1-(-7)) \\ & \Rightarrow \quad 64 c^2=32 c \times 8 \\ & \Rightarrow \quad c^2=4 c \\ & \therefore \quad c=4 \,\,\,\,\,\,\,\,[\because c \neq 0]\end{aligned} $

Now, the given expression becomes

$ x^2-3 x+4 $

It discriminant

$ D=9-16=-7<0 $

Coefficient of $x^2=1>0$

Hence, the given expression is positive $\forall x \in R$.

$ \begin{aligned} & \text { } f(x)=\frac{6 x^2-18 x+21}{6 x^2-18 x+17} \\ & =1+\frac{4}{6 x^2-18 x+17} \\ & \text { Now, } 6 x^2-18 x+17 \in\left[\frac{7}{2}, \infty\right) \\ & {\left[\because a x^2+b x+c \in\left[-\frac{D}{4 a}, \infty\right) \text { when } a>0\right]} \\ & \therefore \quad \frac{4}{6 x^2-18 x+17} \in\left(0, \frac{8}{7}\right] \\ & \therefore \quad f(x) \in\left(1, \frac{15}{7}\right] \\ & \therefore \quad m=\frac{15}{7} \text { and } n=1 \\ & \Rightarrow \quad 14 m-7 n=30-7=23 \end{aligned} $

$ \begin{aligned} & \text { } x^3+x^2+x+r=0 \\ & \alpha+\beta+\gamma=-1, \alpha \beta+\beta \gamma+\gamma \alpha=1, \\ & \alpha \beta \gamma=-r \\ & \alpha^3+\beta^3+\gamma^3=5 \\ & \Rightarrow\left[\alpha^3+\beta^3+\gamma^3-3 \alpha \beta \gamma\right]+3 \alpha \beta \gamma=5 \\ & \Rightarrow\left[( \alpha + \beta + \gamma ) \left(\alpha^2+\beta^2+\gamma^2-\alpha \beta\right.\right. \\ & -\beta \gamma-\gamma \alpha)]+3(-r)=5 \\ &\Rightarrow { }{(-1)}{}\left[\left(\alpha^2+\beta^2+\gamma^2+2 \alpha \beta+2 \beta \gamma+2 \gamma \alpha\right)\right. \\ & -3 \alpha \beta-3 \beta \gamma-3 \gamma \alpha]-3 r=5 \\ & \Rightarrow-\left\{(\alpha+\beta+\gamma)^2-3(\alpha \beta+\beta \gamma+\gamma \alpha)\right\} \\ & -3 r=5 \\ & \Rightarrow-\left\{(-1)^2-3(1)\right\}-3 r=5 \Rightarrow 2-3 r=5 \\ & \therefore \quad r=-1 \end{aligned} $

$ \begin{aligned} & \text { } 6 x^6-25 x^5+31 x^4-31 x^2 +25 x-6=0 \\ & \Rightarrow 6\left(x^6-1\right)-25 x\left(x^4-1\right)+31 x^2 \left(x^2-1\right)=0 \end{aligned} $

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

$ \begin{aligned} & \Rightarrow x^2\left(x^2-1\right)\left[6\left(x^2+\frac{1}{x^2}\right)\right. \\ & \left.-25\left(x+\frac{1}{x}\right)+37\right]=0 \\ & \Rightarrow x^2\left(x^2-1\right)\left[6\left\{\left(x+\frac{1}{x}\right)^2-2 x \times \frac{1}{x}\right\}\right. \\ & \left.-25\left(x+\frac{1}{x}\right)+37\right]=0 \\ & \Rightarrow x^2\left(x^2-1\right)\left[6\left(x+\frac{1}{x}\right)^2\right. \\ & \left.-25\left(x+\frac{1}{x}\right)+25\right]=0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow x^2\left(x^2-1\right)\left[\left\{3\left(x+\frac{1}{x}\right)-5\right\}\right. \\ & \left.\quad\left\{2\left(x+\frac{1}{x}\right)-5\right\}\right]=0 \\ & \quad 3\left(x+\frac{1}{x}\right)-5=0 \quad[\therefore x= \pm 1] \\ & \Rightarrow \quad x+\frac{1}{x}=\frac{5}{3} \Rightarrow 3 x^2-5 x+3=0 \\ & \quad x=\frac{5 \pm \sqrt{-11}}{6} ; \\ & 2\left(x+\frac{1}{x}\right)-5=0 \\ & \quad x+\frac{1}{x}=\frac{5}{2} \Rightarrow x=2 \frac{1}{2} \end{aligned}\\ &\text { ∴ Sum of non-real roots }\\ &=\frac{5+\sqrt{-11}}{6}+\frac{5-\sqrt{-11}}{6}=2 \times \frac{5}{6}=\frac{5}{3} \end{aligned} $

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 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} $