Quadratic Equations

$ \begin{aligned} & \alpha+\beta+\gamma=0 \\ & \quad \alpha \beta+\beta \gamma+\alpha \gamma=+\frac{a}{2} \\ & \quad \alpha \beta \gamma=-b \end{aligned} $

Let $y+a=z$

$\Rightarrow z^3+K z^2+L=0$ has roots.

$z_1, z_2$ and $z_3$

$ \begin{array}{lc} \Rightarrow & z_1=3 \beta \gamma, z_2=3 \alpha \gamma, z_3=3 \alpha \beta \\ \Rightarrow & 3(\alpha \beta+\beta \gamma+\alpha \gamma)=-K \\ \Rightarrow & 27(\alpha \beta \gamma)^2=-L \Rightarrow-27 b^2=L \\ \Rightarrow & K=\frac{-3 a}{2} \\ \Rightarrow & \frac{L}{K}=\frac{27 b^2 \times 2}{3 a}=\frac{18 b^2}{a} \end{array} $

$ \begin{aligned} &\text { } f \text { is symmetric about } x=1\\ &\begin{aligned} \Rightarrow & & f^{\prime}(1) & =0 \\ \Rightarrow & & 2+b & =0 \\ \Rightarrow & & b & =-2 \end{aligned} \end{aligned} $

$ \begin{aligned} f(x) & =x^2-2 x+c \\ f(0) & =c \\ f(1) & =c-1 \\ f(-1) & =c+3 \\ \Rightarrow \quad f(-1) & >f(0)>f(1) \end{aligned} $

$ \begin{gathered} \alpha+\beta=-3 \\ \alpha \beta=k \end{gathered} $

$ \begin{aligned} & \Rightarrow \quad\left(\alpha+\frac{1}{\alpha}\right)\left(\beta+\frac{1}{\beta}\right)=\frac{18}{4} \\ & \Rightarrow \quad \alpha \beta+\frac{\alpha^2+\beta^2}{\alpha \beta}+\frac{1}{\alpha \beta}=\frac{9}{2} \\ & \Rightarrow \quad k+\frac{9-2 k}{k}+\frac{1}{k}=\frac{9}{2} \\ & \Rightarrow \quad \frac{k^2-2 k+10}{k}=\frac{9}{2} \\ & \Rightarrow \quad 2 k^2-4 k+20=9 k \\ & \Rightarrow \quad 2 k^2-13 k+20=0 \\ & \Rightarrow k \text { satisfies } 2 x^2-13 x+20=0 \end{aligned} $

$ \begin{array}{ll} \text { Let } & f(x)=a x^2+b x+c \\ & f(-3)=0 \\ \Rightarrow & f(x)=a(x+3)^2 \\ \Rightarrow & f(0)=9 a \Rightarrow a=2 \\ \text { So, } & f(x)=2(x+3)^2 \\ & f(3)=2 \times 6^2=72 \end{array} $

$ \begin{aligned} &\text { Let the integral root be } \gamma \text {. }\\ &\begin{aligned} & \Rightarrow \quad 6 \gamma^3-25 \gamma^2+2 \gamma+8=0 \\ & \Rightarrow \quad 6 \gamma^3-24 \gamma^2-\gamma^2+2 \gamma+8=0 \\ & \Rightarrow \quad 6 \gamma^2(\gamma-4)-\left(\gamma^2-2 \gamma-8\right)=0 \\ & \Rightarrow \quad 6 \gamma^2(\gamma-4)-(\gamma-4)(\gamma+2)=0 \\ & \Rightarrow \quad(\gamma-4)\left(6 \gamma^2-\gamma-2\right)=0 \\ & \Rightarrow \quad(\gamma-4)\left(6 \gamma^2+3 \gamma-4 \gamma-2\right)=0 \\ & \Rightarrow \quad(\gamma-4)(3 \gamma(2 \gamma+1))-2(2 \gamma+1))=0 \\ & \Rightarrow \quad \gamma=4, \beta=\frac{-1}{2}, \quad \alpha=\frac{2}{3} . \\ & \therefore \quad \frac{4}{\alpha}+\frac{1}{\beta}=\frac{4 \times 3}{2}-2=4 \end{aligned} \end{aligned} $

$ \begin{aligned} & \Rightarrow x^5+x^4-13 x^3-13 x^2+36 x+36=0 \\ & \Rightarrow x^4(x+1)-13 x^2(x+1)+36(x+1)=0 \\ & \Rightarrow(x+1)\left(x^4-4 x^2-9 x^2+36\right)=0 \\ & \Rightarrow(x+1)\left(x^2-4\right)\left(x^2-9\right)=0 \\ & x=-1, \pm 2, \pm 3 \\ & \text { So, } \varepsilon=3, \delta=2, \gamma=-1 \\ & \beta=-2 \alpha=-3 \\ & \text { So, } \frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}=\frac{3}{-3}+\frac{2}{-2}+\frac{1}{-1} \\ & \quad=-3 \end{aligned} $

The equation $ax^2+bx+c=0$ has the roots $\tan \theta$ and $\cot \theta$.

For a quadratic equation $ax^2+bx+c=0$, if $p$ and $q$ are roots, then:

Sum of roots: $p+q=-\frac{b}{a}$
Product of roots: $pq=\frac{c}{a}$

Here, the roots are $\tan \theta$ and $\cot \theta$. So, $\tan \theta + \cot \theta = -\frac{b}{a}$ and $\tan \theta \cdot \cot \theta = \frac{c}{a}$.

$\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$

This simplifies to: $\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}$,
because $\sin^2 \theta+\cos^2 \theta = 1$

Now, $\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}$
But from earlier, $\tan \theta + \cot \theta = -\frac{b}{a}$.
So, $\frac{1}{\sin \theta \cos \theta} = -\frac{b}{a}$.

For $\tan \theta \cdot \cot \theta = \frac{c}{a}$:
$\tan \theta \cdot \cot \theta = 1$, so $\frac{c}{a} = 1 \implies c = a$.

Now, take the result $\frac{1}{\sin \theta \cos \theta} = -\frac{b}{a}$.
Flip both sides: $\sin \theta \cos \theta = -\frac{a}{b}$

$2 \sin \theta \cos \theta = 2 \times \left(-\frac{a}{b}\right) = -\frac{2a}{b}$

But $2 \sin \theta \cos \theta = \sin 2\theta$.
So, $\sin 2\theta = -\frac{2a}{b}$.

Since $a = c$, we can also write $\sin 2\theta = -\frac{2c}{b}$.

We have,

$ \begin{array}{ll} & ||2 x-3|-4|=2 \\ \Rightarrow & |2 x-3|-4= \pm 2 \\ \Rightarrow & |2 x-3|= \pm 2+4 \\ \Rightarrow & |2 x-3|=6 \text { or }|2 x-3|=2 \\ \Rightarrow & 2 x-3= \pm 6 \text { or } 2 x-3= \pm 2 \\ \Rightarrow & 2 x= \pm 6+3 \text { or } 2 x= \pm 2+3 \\ \Rightarrow & 2 x=9,-3 \text { or } 2 x=5,1 \\ \Rightarrow & x=\frac{9}{2}, \frac{-3}{2} \text { or } x=\frac{5}{2}, \frac{1}{2} \end{array} $

∴ Sum of roots

$ \begin{aligned} & =\frac{9}{2}-\frac{3}{2}+\frac{5}{2}+\frac{1}{2} \\ & =\frac{9-3+5+1}{2}=\frac{12}{2}=6 \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & 3 x^5-6 x^4+2 x^3+4 x^2-5 x+8 \\ \Rightarrow \quad & 3 x^3\left(x^2-2 x+3\right)-7 x\left(x^2-2 x+3\right) -10\left(x^2-2 x+3\right)-4 x+38 \\ \Rightarrow \quad & \left(x^2-2 x+3\right)\left(3 x^3-7 x-10\right)-4 x+38 \\ \therefore \quad & a=3, b=0, c=-7, d=-10 \\ \quad & p=-4, q=38 \\ \therefore \quad & a b+c d=70 \\ \quad & p+2 q-2=-4+76-2=70 \\ \therefore \quad & a b+c d=p+2 q-2 \end{aligned} \end{aligned} $

$ \begin{aligned} & \alpha \beta=\alpha \delta=1 \\ & 12\left(x^2+\frac{1}{x^2}\right)-56\left(x+\frac{1}{x}\right)+89=0 \\ & 12\left(x+\frac{1}{x}\right)^2-24-56\left(x+\frac{1}{x}\right)+89=0 \\ & 12\left(x+\frac{1}{x}\right)^2-56\left(x+\frac{1}{x}\right)+65=0 \\ & 12 t^2-56 t+65=0, \quad x+\frac{1}{x}=t \\ & t=\frac{56 \pm \sqrt{(56)^2-4(12)(65)}}{2 \times 12} \end{aligned} $

$ \begin{aligned} & x+\frac{1}{x}=\frac{5}{2}, x+\frac{1}{x}=\frac{13}{6} \\ & x=2, \frac{1}{2}, x=\frac{2}{3}, \frac{3}{2} \\ \therefore & \frac{\alpha+\beta}{\gamma+\delta}=\frac{\frac{5}{2}}{\frac{13}{6}}=\frac{15}{13} \end{aligned} $

$ \begin{aligned} & \text { } x^2+p x+2=0 \\ & x^2+x+2 p=0 \end{aligned} $

Let ' $\alpha$ ' be the common root

$ \begin{array}{ll} \therefore & \alpha=2 \\ & 4+2 p+2=0 \\ & 2 p=-6 \\ & p=-3 \end{array} $

∴ Sum of roots of $x^2+2 p x+8=0$ (i.e., $x^2-6 x+8=0$ ) is 6 .

$x^2-5 a x+6 a=0$

∵ Both roots are greater than 2 .

$ \begin{aligned} & D \geq 0, \frac{-b}{2 a}>1, f(0)>0 \\ & (5 a)^2-4(1)(6 a) \geq 0, \frac{5 a}{2}>1,6 a>0 \\ & \Rightarrow a(25 a-24) \geq 0, a>\frac{2}{5}, a>0 \\ & \therefore a \in\left[\frac{24}{25}, \infty\right) \end{aligned} $

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

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

$ \begin{aligned} & =1+2+(\gamma+\delta)^2-2 \gamma \delta \\ & =3+4+4=11 \end{aligned} $

We have, $x^4+4 x^3-16 x-16=0$

$ \begin{aligned} & \Rightarrow x^3(x-2)+6 x^2(x-2)+12 x(x-2) +8(x-2)=0 \\ & \Rightarrow(x-2)\left(x^3+6 x^2+12 x+8\right)=0 \\ & \Rightarrow(x-2)\left[x^2(x+2)+4 x(x+2)+4(x+2)\right]=0 \\ & \Rightarrow(x-2)(x+2)\left(x^2+4 x+4\right)=0 \\ & \Rightarrow(x-2)(x+2)^3=0 \end{aligned} $

∴ Multiply roots is -2 .

(a) $x^2+2 x-3=0 \Rightarrow(x+3)(x-1)=0$

(b) $x^2-3 x+2=0 \Rightarrow(x-2)(x-1)=0$

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

(d) $x^2-4 x+3=0 \Rightarrow(x-3)(x-1)=0$

Given equation is

$ x^2-3 a x+a^2-2 a-k=0 $

For distinct real roots, $D>0$

$ \begin{aligned} & =(-3 a)^2-4 \times 1 \times\left(a^2-2 a-k\right)>0 \\ & =9 a^2-4 a^2+8 a+4 k>0 \\ & =5 a^2+8 a+4 k>0 \end{aligned} $

Since, leading coefficient of $a^2$ is $5>0$

So, $(8)^2-4 \times(5)(4 k)<0=64-80 k<0$

$ \begin{aligned} \Rightarrow \quad 64 & <80 k \\ & =k>\frac{4}{5} \Rightarrow k \in\left(\frac{4}{5}, \infty\right) \end{aligned} $

Let

$ \begin{aligned} & f(x)=x^4-10 x^3+37 x^2-60 x+36=0 \\ & \text { At } x=1, f(1)=1-10+37-60+36 \neq 0 \\ & \text { At } x=2, f(2)=\left(2^4-10\left(2^3+37\left(2^2\right.\right.\right. \\ & -60(2+36=0 \\ & \quad=16-80+148-120+36 \\ & =200-200=0 \end{aligned} $

So, $x=2$ is a factor of $f(x)$

Similarly $x=3$ is a factor of $f(x)$

$ \therefore f(x)=(x-2)^2(x-3)^2 $

$\therefore$ Original roots are $2,2,3,3$

The distinct roots are 2 and 3

There are three possible pairs of distinct roots

(i) Increase the two $2 s$ by 1

(ii) Increase the two $3 s$ by 1

(iii) Increase one 2 and one 3 by 1

Let us assume we choose one root 2 and one root 3 to increase by 1

Since, these distinct roots.

Increase one ' 2 ' by $1: 2+1=3$

Increase one ' 3 ' by $1: 3+1=4$

Other two roots ( 2 and 3 ) are kept fixed

$\therefore$ New sets of roots of he transformed equation is $3,4,2,3$

Let the transformed equation be $g(x)$. its roots are $2,3,3,4$

$ \begin{aligned} \therefore g(x) & =(x-2)(x-3)(x-3)(x-4) \\ & =(x-2)(x-3)^2(x-4) \end{aligned} $

The common roots are value of $x$ for which $f(x)=0$ and $g(x)=0$

$\therefore$ Common roots are $2,3,3$

$\therefore$ Number of common roots $=3$

Given, $\alpha, \beta, \gamma$ are the roots of the equations $x^3-P x^2+Q x-R=0$

$ \therefore \quad \begin{array}{r} \alpha+\beta+\gamma=P \\ \alpha \beta+\beta \gamma+\gamma \alpha=Q \end{array} $

And $\alpha \beta \gamma=R$

Also, $(\alpha-2)^2,(\beta-2)^2,(\gamma-2)^2$ are the roots of equations, $x^3-5 x^2+4 x=0$

Now, $x^2-5 x^2+4 x=0$

$ \begin{array}{lc} \Rightarrow & x\left(x^2-5 x+4\right)=0 \\ \Rightarrow & x(x-1)(x-4)=0 \\ \Rightarrow & x=0,1,4 \\ \therefore & (\alpha-2)^2=0 \Rightarrow \alpha=2 \\ & (\beta-2)^2=1 \Rightarrow \beta-2= \pm 1 \Rightarrow \beta=1,3 \end{array} $

And $(\gamma-2)^2=4 \Rightarrow \gamma-2= \pm 2 \Rightarrow \gamma=0,4$

Now,

$ \begin{aligned} P+Q+R=(\alpha+\beta+\gamma)+(\alpha \beta+\beta \gamma & +\gamma \alpha) +\alpha \beta \gamma \end{aligned} $

Case I when $\alpha=2, \beta=1, \gamma=0$

then,

$ \begin{aligned} P+Q+R & =(2+1+0)+(2+0+0) \\ & =5 \end{aligned} $

Case II when $\alpha=2, \beta=1, \gamma=4$

then,

$ \begin{aligned} P+Q+R=(2+1+4)+(2+4+8) & + (2 \times 1 \times 4)=29 \end{aligned} $

Case III when $\alpha=2, \beta=3, \gamma=0$ then

$ \begin{aligned} P+Q+R & =(2+3+0)+(6+0+0)+0 \\ & =11 \end{aligned} $

Case IV when $\alpha=2, \beta=3, \gamma=4$

then

$ \begin{aligned} P+Q+R & =(2+3+4)+(6+12+8)+24 \\ & =59 \end{aligned} $

Clearly, least value of $P+Q+R$ is 5 .

Given, quadratic equation is

$ a x^2+a x+5=0 $

For a quadratic equation

$ \begin{gathered} A x^2+B x+C=0 \text { to have no real roots, } \\ \Delta=B^2-4 A C<0 \end{gathered} $

So, $\quad \Delta=a^2-4(a)(5)<0$

$ \begin{aligned} & \Rightarrow \quad a^2-20 a<0 \\ & \Rightarrow \quad a(a-20)<0 \end{aligned} $

So, $\quad 0

Thus, $a$ must be an integer.

So, the integral values of $a$ in this range are $1,2,3, \ldots, 19$.

The number of such value is

$ 19-1+1=19 $

Given equation is

$ 32 x^3-48 x^2+22 x-3=0 $

Since, the roots of this equation are in AP .

So, sum of roots $=a-d+a+a+d=3 a$

$ =\frac{-(-48)}{32}=\frac{3}{2} \Rightarrow a=\frac{1}{2} $

Now, sum of the products of roots $=\frac{22}{32}$

$ \Rightarrow(a-d) a+a(a+d)+(a-d)(a+d)=\frac{11}{16} $

$ \begin{aligned} & \Rightarrow a^2-a d+a^2+a d+a^2-d^2=\frac{11}{16} \\ & \Rightarrow 3 a^2-d^2=\frac{11}{16} \Rightarrow 3\left(\frac{1}{2}\right)^2-d^2=\frac{11}{16} \\ & \Rightarrow d^2=\frac{3}{4}-\frac{11}{16}=\frac{12-11}{16}=\frac{1}{16} \\ & \Rightarrow d^2=\frac{1}{16} \end{aligned} $

So, the square of the common difference of the root is $\frac{1}{16}$.

Given, equation is

$ x^4-2 x^3+x^2+4 x-6=0 $

Let the roots of this equation be $\alpha, \beta, \gamma$ and $\delta$.

Sum of the roots

$ =\alpha+\beta+\gamma+\delta=\frac{-(-2)}{1}=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Sum of the products of roots

$ \begin{aligned} & =\alpha \beta+\alpha \gamma+\alpha \delta+\beta \gamma+\beta \delta+\gamma \delta \\ & =\frac{1}{1}=1 \end{aligned} $

Sum of products of roots taken three at a time

$ \begin{aligned} & =\alpha \beta \gamma+\alpha \beta \delta+\alpha \gamma \delta+\beta \gamma \delta \\ & =\frac{-4}{1}=-4 \end{aligned} $

Product of roots $=\alpha \beta \gamma \delta=\frac{-6}{1}=-6$

Since, the sum of two roots be zero, i.e.,

$ \begin{aligned} & & \alpha+\beta & =0 \\ \Rightarrow & & \beta & =-\alpha \end{aligned} $

So, $\quad \gamma+\delta=2$        [from Eq. (i)]

And,

$ \begin{aligned} & \alpha(-\alpha)+\alpha \gamma+\alpha \delta+(-\alpha \gamma)+(-\alpha) \delta+\gamma \delta=1 \\ & \Rightarrow \quad-\alpha^2+\alpha \gamma+\alpha \delta-\alpha \gamma-\alpha \delta+\gamma \delta=1 \\ & \Rightarrow \quad-\alpha^2+\gamma \delta=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

And $\quad \alpha(-\alpha) \gamma \delta=-6$

$ \Rightarrow \quad-\alpha^2 \gamma \delta=-6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

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

$ \begin{aligned} & \gamma \delta=1+\alpha^2 \\ \Rightarrow \quad & \alpha^2\left(1+\alpha^2\right)=6 \\ \Rightarrow \quad & \alpha^4+\alpha^2=6 \end{aligned} $

Let $y=\alpha^2$, then we get

$ \begin{array}{ll} & y^2+y-6=0 \\ \Rightarrow & (y+3)(y-2)=0 \\ \Rightarrow & y=-3 \text { and } y=2 \end{array} $

Since, $\alpha^2$ must be non-negative,

$ y=\alpha^2=2 $

Then, $\gamma \delta=1+\alpha^2=1+2=3$

Now, $(\gamma+\delta)^2=\gamma^2+\delta^2+2 \gamma \delta$

$ \begin{aligned} \Rightarrow \quad & \gamma^2+\delta^2=(\gamma+\delta)^2-2 \gamma \delta \\ & =2^2-2(3)=4-6=-2 \end{aligned} $

$f(x)=1+x^2$ is the polynomial satisfying the given condition

$ \begin{aligned} & f\left(\frac{2}{3}\right)=1+\left(\frac{2}{3}\right)^2=1+\frac{4}{9}=\frac{13}{9} \\\\ & f\left(\frac{3}{2}\right)=1+\left(\frac{3}{2}\right)^2=1+\frac{9}{4}=\frac{13}{4} \end{aligned} $

$ \begin{aligned} & f\left(\frac{2}{3}\right)+f\left(\frac{3}{2}\right)=\frac{13}{9}+\frac{13}{4}=13\left(\frac{1}{9}+\frac{1}{4}\right) \\\\ & \quad=\frac{13 \cdot 13}{36}=\left(\frac{13}{6}\right)^2 \\\\ & \therefore \quad \sqrt{f\left(\frac{2}{3}\right)+f\left(\frac{3}{2}\right)}=\sqrt{\left(\frac{13}{6}\right)^2}=\frac{13}{6} \end{aligned} $

We have $\alpha^2-\alpha+1=0$

$ \begin{array}{ll} \Rightarrow & \alpha+\frac{1}{\alpha}=1 \Rightarrow \alpha^2+\frac{1}{\alpha^2}+2=1 \\\\ \Rightarrow & \alpha^2+\frac{1}{\alpha^2}=-1 \Rightarrow \alpha^4+\frac{1}{\alpha^4}+2=1 \\\\ \Rightarrow & \alpha^4+\frac{1}{\alpha^4}=-1 \Rightarrow\left(\alpha+\frac{1}{\alpha}\right)^3=1 \\\\ \Rightarrow & \alpha^3+\frac{1}{\alpha^3}+3\left(\alpha+\frac{1}{\alpha}\right)=1 \\\\ \Rightarrow & \alpha^3+\frac{1}{\alpha^3}=-2 \\\\ \Rightarrow & 1^3+(-1)^3+(-2)^3+(-1)^3 \\\\ \Rightarrow & 1-1-8-1=-9 \end{array} $

Given the quadratic equation $x^2 + ax + b = 0$, we know from the problem statement:

The sum of the roots, $\alpha + \beta$, is $\frac{1}{2}$.

The sum of the cubes of the roots, $\alpha^3 + \beta^3$, is $\frac{37}{8}$.

From Vieta's formulas:

$\alpha + \beta = -a = \frac{1}{2}$. Thus, $a = -\frac{1}{2}$.

The product of the roots $\alpha \beta = b$.

We are given that $\alpha^3 + \beta^3 = \frac{37}{8}$. Using the identity for the sum of cubes:

$ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) $

Substituting the known values:

$ \alpha^3 + \beta^3 = \left(\frac{1}{2}\right)^3 - 3b \left(\frac{1}{2}\right) $

Substitute the given value for $\alpha^3 + \beta^3$:

$ \frac{37}{8} = \frac{1}{8} - \frac{3b}{2} $

Rearranging gives us:

$ \frac{3b}{2} = \frac{1}{8} - \frac{37}{8} = \frac{-36}{8} = \frac{-9}{2} $

Thus,

$ 3b = -9 \quad \Rightarrow \quad b = -3 $

To find $a - \frac{1}{b}$:

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

Calculate the above expression:

$ a - \frac{1}{b} = \frac{-3}{6} + \frac{2}{6} = \frac{-1}{6} $

$4 x^3-3 x^2+2 x-1=0 \quad \text { [given] } $

$ \begin{array}{r} \alpha+\beta+\gamma=\frac{3}{4} \\\\ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{2} \end{array} $

and $\quad \alpha \beta \gamma=\frac{1}{4}$

We know that

$\alpha^3+\beta^3+\gamma^3-3 \alpha \beta \gamma$

$=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)$

$\Rightarrow \alpha^3+\beta^3+\gamma^3-\frac{3}{4}=\frac{3}{4}\left(-\frac{7}{16}-\frac{1}{2}\right)$.

$\left[\because(\alpha+\beta+\gamma)^2=\left(\alpha^2+\beta^2+\gamma^2\right)\right.+2(\alpha \beta+\beta \gamma+\gamma \alpha)]$

$ \left(\frac{3}{4}\right)^2=\left(\alpha^2+\beta^2+\gamma^2\right)+2 \cdot \frac{1}{2} $

$ \left.\alpha^2+\beta^2+\gamma^2=\frac{9}{16}-1=-\frac{7}{16}\right] $

$ \begin{aligned} \Rightarrow \quad & \alpha^3+\beta^3+\gamma^3=-\frac{3}{4}\left(\frac{7+8}{16}\right)+\frac{3}{4} \\\\ & =-\frac{45}{64}+\frac{3}{4}=\frac{-45+48}{64} \\\\ & \alpha^3+\beta^3+\gamma^3=\frac{3}{64} \end{aligned} $

$16 x^4+16 x^3-4 x-1=0$

$ \begin{aligned} & \Rightarrow \quad\left(16 x^4-1\right)+\left(16 x^3-4 x\right)=0 \\\\ & \Rightarrow \quad\left(4 x^2+1\right)\left(4 x^2-1\right)+4 x\left(4 x^2-1\right)=0 \\\\ & \Rightarrow \quad\left(4 x^2-1\right)\left(4 x^2+1+4 x\right)=0 \\\\ & \Rightarrow \quad\left(4 x^2-1\right)(2 x+1)^2=0 \Rightarrow x= \pm \frac{1}{2},-\frac{1}{2} \\\\ & \begin{aligned} \therefore \alpha= & \frac{1}{2}, \beta=-\frac{1}{2}, \gamma=-\frac{1}{2} \text { and } \delta=-\frac{1}{2} \end{aligned} \\\\ & \begin{aligned} \frac{1}{\alpha}= & 2, \frac{1}{\beta}=-2, \frac{1}{\gamma}=-2 \text { and } \frac{1}{\delta}=-2 \end{aligned} \\\\ & \begin{aligned} \therefore \frac{1}{\alpha^4} & +\frac{1}{\beta^4}+\frac{1}{\gamma^2}+\frac{1}{\delta^4} \\\\ & =2^4+(-2)^4+(-2)^4+(-2)^4 \\\\ & =4 \times 16=64 \end{aligned} \end{aligned} $

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

$ 3^{x} + \frac{3}{3^{x}} - 4 < 0 $

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

Substitute into the equation:

$ t + \frac{3}{t} - 4 < 0 $

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

$ t^2 - 4t + 3 < 0 $

Factoring the quadratic, we get:

$ (t-3)(t-1) < 0 $

The solution to this inequality is:

$ t \in (1, 3) $

Since $t = 3^{x}$, substitute back:

$ 3^{x} \in (1, 3) $

Taking the logarithm with base 3 of both sides gives:

$ x \in \left(\log_3{1}, \log_3{3}\right) $

Since $\log_3{1} = 0$ and $\log_3{3} = 1$, we find:

$ x \in (0, 1) $

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

$x^2 - 4x - 12 \leq 0$

$x^2 - 2x - 15 \geq 0$

Next, we factor each equation:

$ (x-6)(x+2) \leq 0 $

$ (x-5)(x+3) \geq 0 $

Now, we determine the solution sets for each inequality:

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

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

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

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

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

Let $ f(x)=3 x^{3}+b x^{2}+b x+3 $

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

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

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

[ $ \because $ two roots are complex]

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

$ \Rightarrow \quad b^{2}-6 b-27 < 0 $

$ \Rightarrow \quad(b+3)(b-9) < 0 $

$ \Rightarrow \quad-3 < b < 9 $

Hence, $ \mathrm{B} \rightarrow $ II.

(C) We know that one root is $ x=-1 $

(Negative).

Two roots are complex if $ -3 < b < 9 $.

If $ b=-3 $, then $ f(x)=0 $

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

In this case two roots are positive and equal.

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

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

Thus, D $ \rightarrow $ III

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

Given, $ x+\frac{4}{x}=2 \sqrt{3} $

$ x^{2}-2 \sqrt{3} x+4=0 $

Roots, $ x=\sqrt{3} \pm i \Rightarrow \alpha, \beta=\sqrt{3} \pm i $

$ \begin{aligned} & \alpha^{2024}, \beta^{2024}=(\sqrt{3} \pm i)^{2024} \\\\ & =2^{2024}\left[\frac{\sqrt{3}}{2} \pm \frac{1}{2} i\right]^{2024} \\\\ & =2^{2024}\left[\cos \frac{\pi}{6} \pm i \sin \frac{\pi}{6}\right]^{2024} \\\\ & =2^{2024}\left[\cos \frac{2024 \pi}{6} \pm i \sin \frac{2024 \pi}{6}\right] \\\\ \therefore \mid \alpha^{2024} & -\beta^{2024}\left|=2^{2024} \times\left|2 i \sin \left(\frac{2024 \pi}{6}\right)\right|\right. \\\\ & =2^{2024} \times 2 \times\left|\sin \left(337 \pi+\frac{\pi}{3}\right)\right| \\\\ & =2^{2025} \times \frac{\sqrt{3}}{2} \end{aligned} $

Hence, $ \frac{2}{\sqrt{3}}\left|\alpha^{2024}-\beta^{2024}\right|=2^{2025} $

Given, $ 12 x^{1 / 3}-25 x^{1 / 6}+12=0 $

Let $ y=x^{1 / 6} $, then $ y^{2}=x^{1 / 3} $

So, $ 12 y^{2}-25 y+12=0 $

$ \begin{aligned} y & =\frac{4}{3} \text { or } \frac{3}{4} \\\\ x^{1 / 6} & =\frac{4}{3} \text { or } \frac{3}{4} \\\\ x & =\left(\frac{4}{3}\right)^{6} \text { or }\left(\frac{3}{4}\right)^{6} \end{aligned} $

Here, $ \alpha=\left(\frac{4}{3}\right)^{6} $ and $ \beta=\left(\frac{3}{4}\right)^{6} $

$ \sqrt[6]{\frac{\alpha}{\beta}}=\left(\frac{4}{3} \times \frac{4}{3}\right)^{6 \times \frac{1}{6}}=\frac{16}{9} $

Given, $ x^{3}+3 x^{2}-10 x-24=0 $

Factorise

$ \Rightarrow(x+4)(x+2)(x-3)=0 $

Here, $ \alpha=3, \beta=-2 \gamma=-4 $

$ \alpha^{3}+3 \beta^{2}-10 \gamma-24=11 K $

$ 27+3 \times 4-10 \times(-4)-24=11 K $

$ 27+12+40-24=11 K \Rightarrow K=\frac{55}{11}=5 $

Given,

$ 8 x^{3}-42 x^{2}+63 x-27=0 $

Factorise $ (4 x-3)(2 x-3)(x-3)=0 $

Roots are $ x=\frac{3}{4}, \frac{3}{2} $ or 3

So, $ \beta=\frac{3}{4}, \gamma=\frac{3}{2} $ and $ \alpha=3 $

Let $ f(x)=\frac{3}{2} x^{2}+3 x+3 $

Extreme value $ =f\left(\frac{-b}{2 a}\right) $

$ \frac{-b}{2 a}=\frac{-3}{2 \times \frac{3}{2}} \Rightarrow-1 \Rightarrow f(-1)=\frac{3}{2}-3+3=\frac{3}{2} $

Given, $ \frac{2 x^{3}+1}{2 x^{2}-x-6} $

$ \begin{array}{r} \left.2 x^{2}-x-6\right) \frac{}{2 x^{3} \quad \quad \quad.+1}\left(x+\frac{1}{2}\right. \\\\ 2 x^{3}-x^{2}-6 x \\\\ \frac{(-) \quad(+) \quad(+)}{x^{2}+6 x+1} \\\\ x^{2}-\frac{1}{2} x-3 \\\\ \frac{(-)(+) \quad(+)}{\frac{13}{2} x+4} \end{array} $

$ \begin{aligned} \frac{2 x^{3}+1}{2 x^{2}-x-6} & =\left(x+\frac{1}{2}\right)+\frac{\frac{13}{2} x+4}{2 x^{2}-x-6} \\ & =\left(x+\frac{1}{2}\right)+\frac{\frac{13 x}{2}+4}{(2 x+3)(x-2)}\end{aligned} $

Now, $ \frac{\frac{13 x}{2}+4}{(2 x-3)(x+2)}=\frac{A}{2 x+3}+\frac{B}{x-2} $

$ \frac{13 x}{2}+4=B(2 x+3)+A(x-2) $

Put $ x=2 $

$ 17=7 B \Rightarrow B=\frac{17}{7} $

Put $ x=\frac{-3}{2} $

$ \begin{aligned} \frac{-39}{4}+4= & A\left(\frac{-3}{2}-2\right) \\\\ -\frac{23}{4}= & A\left(\frac{-7}{2}\right) \Rightarrow A=\frac{23}{14} \\\\ \therefore \frac{2 x^{3}+1}{2 x^{2}-x-6}= & \left(x+\frac{1}{2}\right) \\\\ & +\frac{23}{14(2 x+3)}+\frac{17}{7(x-2)} \end{aligned} $

Here, $ \quad a=1, b=\frac{1}{2}, p=1, q=3 $

$ \begin{array}{l} A=\frac{17}{7}, B=\frac{23}{14} \\\\ \Rightarrow \quad 51 \mathrm{apB}=51 \times 1 \times 1 \times \frac{23}{14}=\frac{1173}{14} \\\\ \Rightarrow 23 b q A=23 \times \frac{1}{2} \times 3 \times \frac{17}{7}=\frac{1173}{14} \\\\ \therefore \quad 51 a p B=23 b q A \end{array} $

Given $\alpha$ is a root of the equation

$ \begin{aligned} & \frac{x-1}{\sqrt{2 x^2-5 x+2}}=\frac{41}{60}, \text { where }-\frac{1}{2}<\alpha<0 \\\\ \therefore \quad & \frac{\alpha-1}{\sqrt{2 \alpha^2-5 \alpha+2}}=\frac{41}{60} \end{aligned} $

$\begin{aligned} & \Rightarrow \quad \frac{(\alpha-1)^2}{2 \alpha^2-5 \alpha+2}=\left(\frac{41}{60}\right)^2 \\\\ & \Rightarrow 3600\left(\alpha^2-2 \alpha+1\right)=1681\left(2 \alpha^2-5 \alpha+2\right) \\\\ & \Rightarrow 238 \alpha^2+1205 \alpha+238=0 \\\\ & \Rightarrow 238 \alpha^2+1156 \alpha+49 \alpha+238=0 \\\\ & \Rightarrow 34 \alpha(7 \alpha+34)+7(7 \alpha+34)=0 \\\\ & \Rightarrow(34 \alpha+7)(7 \alpha+34)=0 \\\\ & \Rightarrow \alpha=-\frac{7}{34},-\frac{34}{7} \\\\ & \because-\frac{1}{2}<\alpha<0 \\\\ & \therefore \alpha=-\frac{7}{34}\end{aligned}$

Given, roots of the equation

$ x^5+4 x^4-13 x^3-52 x^2+36 x+144=0 $

are $\alpha_1 \beta, \gamma, 2$, and $\varepsilon$, where

$ \alpha<\beta<\gamma<2<\varepsilon $

$ \begin{aligned} & \text { Now, } x^5+4 x^4-13 x^3-52 x^2+36 x+144=0 \\\\ & \Rightarrow \quad(x+4)(x+3)(x+2)(x-2)(x-3)=0 \\\\ & \Rightarrow \quad x=-4,-3,-2,2,3 \\\\ & \therefore \alpha=-4, \beta=-3, \gamma=-2 \text { and } \varepsilon=3 \\\\ & \text { Now, } \alpha+2 \beta+3 \gamma+5 \varepsilon \\\\ & =(-4)+2(-3)+3(-2)+5(3) \\\\ & =-4-6-6+15 \\\\ & =-16+15=-1 \end{aligned} $

We have,

$ \begin{aligned} \qquad 3 x^2+(2 k+1) x-5 k=0 \end{aligned} $

$\because \quad$ roots are real and equal.

$ \begin{aligned} \Rightarrow & D=0 \\\\ \Rightarrow & (2 k+1)^2-4(-5 k) \times 3=0 \\\\ \Rightarrow & 4 k^2+4 k+1+60 k=0 \\\\ \Rightarrow & 4 k^2+64 k+1=0 \\\\ & k=\frac{-64 \pm \sqrt{4096-16}}{2 \times 4} \\\\ & =\frac{-64 \pm \sqrt{4080}}{8}=\frac{-16 \pm \sqrt{255}}{2} \end{aligned} $

$Hence, k=\frac{-16+\sqrt{255}}{2}( \because \frac{1}{2} < k < 0)$

Let $\alpha$ be the common root of

$ \begin{aligned} & 2 x^2+a x-2=0 \text { and } x^2+x+0 \rightarrow 2=0 \\\\ & \Rightarrow 2 \alpha^2+a \alpha-2=0 \quad \ldots \text { (i) } \\\\ & \alpha^2+\alpha+2 a=0 \quad \ldots \text { (ii) } \\\\ & \Rightarrow \frac{\alpha^2}{2 a^2+2}=\frac{\alpha}{-2-4 a}=\frac{1}{2-a} \quad\left[a \neq 2,-\frac{1}{2}\right] \\\\ & \Rightarrow \quad \alpha=\frac{4 a+2}{a-2} \end{aligned} $

So, $\frac{(4 a+2)^2}{(a-2)^2\left(2 a^2+2\right)}=\frac{-1}{a-2}$

$ \begin{gathered} \frac{2(2 a+1)^2}{\left(a^2+1\right)}=2-a \\\\ 8 a^2+2+8 a=2 a^2+2-a^3-a \\\\ a^3+6 a^2+9 a=0 \Rightarrow a\left(a^2+6 a+9\right)=0 \\\\ \Rightarrow a=-3 \quad[\because a \neq 0) \\\\ \therefore a x^2-4 x-2 a=-3 x^2-4 x+6=0 \\\\ 3 x^2+4 x-6=0 \\\\ x=\frac{-4 \pm \sqrt{16+72}}{6} \\\\ =\frac{-2 \pm \sqrt{22}}{3} \end{gathered} $

So, one root of $a x^2-4 x-2 a=0$ is

$ =\frac{-2+\sqrt{22}}{3} . $

We have,

$ \begin{gathered} 2 x^3-3 x^2+5 x-7=0 \\\\ \therefore \quad \alpha+\beta+\gamma=-\left(-\frac{3}{2}\right)=\frac{3}{2} \quad \ldots \text { (i) } \\\\ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{5}{2} \quad \ldots \text { (ii) } \\\\ \alpha \beta \gamma=\frac{7}{2} \quad \ldots \text { (iii) } \end{gathered} $

On squaring both side of Eq. (ii), we get

$ \begin{aligned} & (\alpha \beta+\beta \gamma+\gamma \alpha)^2=\frac{25}{4} \\\\ & \alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2+2 \\\\ & \left(\alpha \beta^2 \gamma+\alpha \beta \gamma^2+\alpha^2 \beta \gamma\right)=\frac{25}{4} \\\\ & \Sigma \alpha^2 \beta^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)=\frac{25}{4} \end{aligned} $

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

$ \begin{aligned} & \Sigma \alpha^2 \beta^2+2\left(\frac{7}{2}\right)\left(\frac{3}{2}\right)=\frac{25}{4} \\\\ & \Sigma \alpha^2 \beta^2=\frac{25}{4}-\frac{21}{2}=-\frac{17}{4} \end{aligned} $

We have,

$ x^4-x^3-16 x^2+4 x+48=0 \quad \ldots \text { (i) } $

By hit and trial, we get, $x^2-4$ is the factor of

$ \begin{gathered} x^4-x^3-16 x^2+4 x+48=0 \\\\ \Rightarrow \quad \frac{x^4-x^3-16 x^2+4 x+48}{x^2-4} \\\\ \Rightarrow x^2-x-12 \\\\ \Rightarrow x^4-x^3-16 x^2+4 x+48=\left(x^2-4\right) \\\\ \left(x^2-x-12\right) \\\\ \Rightarrow(x-2)(x+2)(x-4)(x+3)=0 \end{gathered} $

$\Rightarrow$ Roots of above Eq. (i) are,

$ -3,-2,2,4 $

$ \text { So, } \begin{aligned} \alpha^4+\beta^4+\gamma^4+\delta^4 & =(-3)^4+\left(-2^4\right. \\\\ + & \left(2^4+(4)^4=369\right. \end{aligned} $

$ \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 $\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2=15$