Quadratic Equations

The roots of the quadratic equation $x^2 - 35x + c = 0$ are given to be in the ratio 2:3.

Let's assume the roots are $2\alpha$ and $3\alpha$.

The sum of the roots is given by the equation:

$ 2\alpha + 3\alpha = 5\alpha = 35 $

Solving for $\alpha$, we get:

$ \alpha = 7 $

The product of the roots can be expressed as:

$ (2\alpha)(3\alpha) = 6\alpha^2 = c $

Substituting the value of $\alpha$, we find:

$ c = 6 \times 7^2 = 6 \times 49 $

Given that $c = 6K$, we equate to find $K$:

$ 6 \times 49 = 6K $

$ K = 49 $

We are given the polynomial equation $x^4-x^3-8x^2+2x+12=0$ with roots $\alpha, \beta, \gamma, $ and $\delta$. It is known that the sum of two of the roots, $\alpha$ and $\beta$, is zero, i.e., $\alpha + \beta = 0$.

From Vieta’s formulas, the sum of all roots is:

$ S_1 = \alpha + \beta + \gamma + \delta = 1 $

Given $\alpha + \beta = 0$, we find:

$ \gamma + \delta = 1 \quad \ldots \text{(ii)} $

Similarly, from Vieta’s formulas, the sum of the products of the roots taken two at a time is:

$ S_2 = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = -8 $

Using $\alpha + \beta = 0$, simplify to:

$ \alpha\beta + \gamma\delta = -8 \quad \ldots \text{(iii)} $

From the sum of the products of the roots taken three at a time, we have:

$ S_3 = \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -2 $

Substitute $(\alpha + \beta) \gamma\delta + \alpha\beta(\gamma + \delta) = -2$ and rearrange:

$ \alpha\beta = -2 \quad \ldots \text{(iv)} $

From Vieta’s formulas, the product of the roots is given by:

$ S_4 = \alpha\beta\gamma\delta = 12 $

Use $\alpha\beta = -2$ to find:

$ \gamma\delta = -6 \quad \ldots \text{(v)} $

Next, solve the system of equations obtained from (ii) and (v):

$\gamma + \delta = 1$

$\gamma \delta = -6$

These are the equations for a quadratic:

$ t^2 - (\gamma + \delta)t + \gamma\delta = 0 \implies t^2 - t - 6 = 0 $

Solving, we find the roots to be:

$ t = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2} $

Thus, $t = 3$ or $t = -2$. Since $\gamma > \delta$, we assign $\gamma = 3$ and $\delta = -2$.

Finally, compute:

$ 3\gamma + 2\delta = 3 \times 3 + 2 \times (-2) = 9 - 4 = 5 $

$ \begin{aligned} &\text { Roots of equation }\\ &\begin{gathered} x^2+a x+2=0 \text { are } \alpha, \beta \\ \therefore \alpha+\beta=-a, \alpha \beta=2 \\ \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=a^2-4 \end{gathered} \end{aligned} $

$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} & \left(\frac{\alpha \beta+1}{\beta}\right)\left(\frac{\alpha \beta+1}{\alpha}\right)\left(\frac{\alpha^2-1}{\alpha}\right)\left(\frac{\beta^2-1}{\beta}\right) \\ & =\frac{(2+1)(2+1)\left[\alpha^2 \beta^2-\left(\alpha^2+\beta^2\right)+1\right]}{(\alpha \beta)^2} \\ & =\frac{9}{(2)^2}\left[4-\left(a^2-4\right)+1\right]=\frac{9}{4}\left(9-a^2\right) \end{aligned} \end{aligned} $

$\left(x^2-7 x+11\right)^{x^2-6 x-7}=1$

Case I

$ \begin{aligned} x^2-7 x+11 & =1 \\ x^2-7 x+10 & =0 \\ x & =2,5 \end{aligned} $

Case II

$ \begin{aligned} x^2-7 x+11 & =-1 \\ x^2-7 x+12 & =0 \\ x & =4,3 \end{aligned} $

and $x^2-6 x-7=$ even

only $x=3$

Case III

$ \begin{aligned} x^2-6 x-7 & =0 \text { and } x^2-7 x+11 \neq 0 \\ (x-7)(x+1) & =0 \\ x & =-1,7 \end{aligned} $

Sum of solutions

$ =2+5+3+(-1)+7=16 $

Given, $x^2+2 p x-2 p+8>0$ for all real values of $x$.

$ \begin{array}{lr} \text { So, }(2 p)^2-4(1)(-2 p+8) < 0 \\ \Rightarrow & 4 p^2+8 p-32 < 0 \\ \Rightarrow & p^2+2 p-8 < 0 \\ \Rightarrow & (p+4)(p-2) < 0 \\ \Rightarrow & -4 < p < 2 \\ \Rightarrow & p \in(-4,2) \end{array} $

The expression $\frac{x+3}{(x-1)(x+2)}$ is not defined only at $x=1$ and $x \neq 2$

Let $y=\frac{x+3}{(x-1)(x+2)}$

$\Rightarrow \quad(x-1)(x+2) y=x+3$

$ \begin{aligned} & \Rightarrow y x^2+y x-2 y-x-3=0 \\ & \Rightarrow y x^2+(y-1) x-(2 y+3)=0 \\ & \Rightarrow x=\frac{-(y-1) \pm \sqrt{\begin{array}{c} (y-1)^2-4(y) \\ (-2 y-3) \end{array}}}{2 y} \end{aligned} $

Now, the range set of $\frac{x+3}{(x-1)(x+2)}$ will be all real values of $y$ except those value for which

$ \begin{array}{rlrl} & (y-1)^2-4 y(-2 y-3) < 0 \\ \Rightarrow & y^2-2 y+1+8 y^2+12 y < 0 \\ \Rightarrow & 9 y^2+10 y+1 < 0 \\ \Rightarrow & (9 y+1)(y+1) < 0 \\ \Rightarrow -1 & < y <-\frac{1}{9} \end{array} $

Thus, range of $\frac{x+3}{(x-1)(x+2)}$ be

$ \begin{aligned} & R-\left(-1,-\frac{1}{9}\right) \\ & \therefore \alpha=-1 \text { and } \beta=-\frac{1}{9} \end{aligned} $

The intercepts of the equation

$ (-1) x+\left(-\frac{1}{9}\right) y+1=0 \text { or } \frac{x}{1}+\frac{y}{9}=1 \text { be } 1 $

and 9.

Thus, the sum of intercept be $1+9=10$.

Given equation is,

$ x^4+x^3-4 x^2+x+1=0 $

On putting $x=y+h$

$ \begin{array}{r} \therefore(y+h)^4+(y+h)^3-4(y+h)^2 \\ +(y+h)+1=0 \\ \Rightarrow\left(y^2+h^2+2 y h\right)^2+y^3+h^3+3 y h^2+3 y^2 h \\ -4 y^2-4 h^2-8 y h+y+h+1=0 \\ \Rightarrow y^4+h^4+4 y^2 h^2+2 y^2 h^2+4 y h^3 \\ +4 y^3 h+y^3+h^3+3 y h^2+3 y^2 h \\ -4 y^2-4 h^2-8 y h+y+h+1=0 \\ \Rightarrow y^4+(4 h+1) y^3+\left(6 h^2+3 h-4\right) y^2 \\ +\left(4 h^3+3 h^2-8 h+1\right) y \\ +h^4+h^3-4 h^2+h+1=0 \end{array} $

Given, transformed equation does not contain $x^2$ term.

$ \begin{aligned} & \therefore \quad\left(6 h^2+3 h-4\right)=0 \\ & \Rightarrow \quad h=\frac{-3 \pm \sqrt{105}}{12} \end{aligned} $

So, $\quad \alpha=\frac{-3+\sqrt{105}}{12}$ and $\beta=\frac{-3-\sqrt{105}}{12}$

$ \begin{aligned} & \text { Now, } 12(\alpha-\beta)^2=12\left[\left(\frac{-3+\sqrt{105}}{12}\right)\right. \\ & \left.-\left(\frac{-3-\sqrt{105}}{12}\right)\right]^2 \\ & =12\left[\frac{-3+\sqrt{105}+3+\sqrt{105}}{12}\right]^2 \\ & =12\left(\frac{2 \sqrt{105}}{12}\right)^2=12 \times \frac{4 \times 105}{12 \times 12}=35 \end{aligned} $

We have, $x^2+a x-c=0$ and $\sin 2 \theta$ and $\cos 2 \theta$ are roots.

$ \begin{aligned} \therefore \quad \sin 2 \theta+\cos 2 \theta & =\frac{-a}{1}=-a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \sin 2 \theta \cos 2 \theta & =\frac{-c}{1}=-c \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Squaring on both sides in Eq. (i), we get Then, $(\sin 2 \theta+\cos 2 \theta)^2=(-a)^2$

$ \begin{gathered} \sin ^2 2 \theta+\cos ^2 2 \theta+2 \sin 2 \theta \cos 2 \theta=a^2 \\ 1+2 \times-c=a^2 \quad \text { [from Eq. (i)] } \\ a^2+2 c-1=0 \end{gathered} $

Given, $a x^2-7 x+c=0$ and $a x^2+5 x-c=0$ have a common roots.

$ \begin{aligned} &c=6 x\\ &\text { }\\ So,\,\,&x=\frac{c}{6}\\ &\begin{aligned} & \therefore \quad a \quad\left(\frac{c^2}{36}\right)-7\left(\frac{c}{6}\right)+c=0 \\ & \Rightarrow \quad a c^2-42 c+36 c=0 \Rightarrow a c^2-6 c=0 \\ & \Rightarrow \quad c=0,(a c-6)=0 \\ & \Rightarrow \quad a=\frac{6}{c}, a c=6 \end{aligned} \end{aligned} $

Now, according to the question, 3 is a root of $a x^2-7 x+c=0$

$ \begin{array}{rlrl} \Rightarrow & 9 a-21+c =0 \\ \Rightarrow \quad 9\left(\frac{6}{c}\right)-21+c & =0 \Rightarrow 54-21 c+c^2=0 \\ \Rightarrow \quad c^2-21 c+54 & =0 \\ c & =18,3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ x & =\frac{c}{6} \Rightarrow x=3, \frac{1}{2} \end{array} $

So, common root $=\frac{1}{2}$

Given, inequality $x^2-7 x+10 \geq 0$ and $2 x+3-x^2>0$

Case $\mathbf{I}(x-5)(x-2) \geq 0$

$ \begin{aligned} & x \in(-\infty, 2] \cup[5, \infty)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \text { Case } \text { II }-x^2+2 x+3>0 \\ & -\left(x^2-2 x-3\right)>0 \\ \Rightarrow & x^2-2 x-3<0 \\ \Rightarrow & x^2-3 x+x-3<0 \\ \Rightarrow & x(x-3)+1(x-3)<0 \\ & (x-3)(x+1)<0 \end{aligned} $

$ \begin{aligned} &x \in(-1,3)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { From Eqs. (i) and (ii), we get }\\ &x \in(-1,2] \end{aligned} $

Let roots of $x^2+3 x-2 k=0$ be $\alpha$ and $\beta$ and roots of $x^2-2 x-7 k=0$ be $\alpha$ and $\gamma$,

$ \begin{aligned} \alpha+\beta & =-3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \alpha+\gamma & =2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \alpha \beta & =-2 k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ \alpha \gamma & =-7 k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Solving Eqs. (i) and (ii), we get

$ \beta-\gamma=-5 $

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

$ \frac{\beta}{\gamma}=\frac{2}{7} \Rightarrow \beta=\frac{2}{7} \gamma $

$ \therefore \quad \frac{2}{7} \gamma-\gamma=-5 \Rightarrow \frac{-5}{7} \gamma=-5 $

$ \begin{aligned} & \Rightarrow \quad \gamma=7 \\ & \because \gamma \text { is a root of } x^2-2 x-7 k=0 \\ & \therefore \quad 7^2-2(7)-7 k=0 \\ & \Rightarrow \quad k=5 \end{aligned} $

Now, roots of $5 x^2+7 x-6=0$

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

Positive root is $x=\frac{3}{5}$.

Let $y=\frac{x^2-2 x+1}{x^2+x-1}$

$ \Rightarrow \quad(y-1) x^2+(y+2) x-y-1=0 $

value does not lie in the given interval, if discriminant $<0$.

$ \begin{array}{lr} \Rightarrow & (y+2)^2+4(y-1)(y+1) < 0 \\ \Rightarrow & y^2+4 y+4+4 y^2-4 < 0 \\ \Rightarrow & 5 y^2+4 y < 0 \Rightarrow y(5 y+4) < 0 \\ \Rightarrow & -\frac{4}{5} < y < 0 \\ \therefore & y \text { does not lie in }\left(-\frac{4}{5}, 0\right) \end{array} $

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

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

Transforming the equation whose roots are $\alpha^2, \beta^2$ and $\gamma^2$.

For this replacing $x$ with $x^{1 / 2}$

$ \begin{array}{ll} & x^{3 / 2}-3 x+3 x^{1 / 2}+1=0 \\ \Rightarrow & \left(x^{3 / 2}+3 x^{1 / 2}\right)^2=(3 x-1)^2 \\ \Rightarrow & x^3+9 x+6 x^2=9 x^2+1-6 x \\ \Rightarrow & x^3-3 x^2+15 x-1=0 \end{array} $

Then, sum of product of roots taken two at a time is $\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2=15$

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} $