Quadratic Equation and Inequalities

The given equation is $\lambda x^2-(\lambda+3) x+3=0$. We can find the roots by factoring:

$ \lambda x^2-\lambda x-3 x+3=0 $

$\Rightarrow $ $ \lambda x(x-1)-3(x-1)=0 $

$\Rightarrow $ $ (x-1)(\lambda x-3)=0 $

The two roots are : $ x=1, \quad x=\frac{3}{1}$

We are given that $\alpha$ and $\beta$ are the roots such that $\alpha<\beta$ and the following condition holds:

$ \frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3} $

Because we don't know if 1 is smaller or larger than $\frac{3}{\lambda}$, we must consider two distinct cases.

Case A : $\alpha=1$ and $\beta=\frac{3}{\lambda}$

Substitute these into the condition $\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}$ :

$ \begin{array}{r} \frac{1}{1}-\frac{1}{3 / \lambda}=\frac{1}{3} \\ 1-\frac{\lambda}{3}=\frac{1}{3} \\ 1-\frac{1}{3}=\frac{\lambda}{3} \\ 1-\frac{1}{3}=\frac{\lambda}{3} \\ \frac{2}{3}=\frac{\lambda}{3} \Longrightarrow \lambda=\mathbf{2} \end{array} $

Check the constraint $\alpha<\beta$ : If $\lambda=2$, then $\beta=3 / 2=1.5$. Since $1<1.5$, this value is valid.

Case B : $\alpha=\frac{3}{\lambda}$ and $\beta=1$

Substitute these into the condition $\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}$ :

$ \begin{aligned} & \frac{1}{3 / \lambda}-\frac{1}{1}=\frac{1}{3} \\ & \frac{\lambda}{3}-1=\frac{1}{3} \\ & \frac{\lambda}{3}=1+\frac{1}{3} \\ & \frac{\lambda}{3}=\frac{4}{3} \Longrightarrow \lambda=4 \end{aligned} $

Check the constraint $\alpha<\beta$ : If $\lambda=4$, then $\alpha=3 / 4=0.75$. Since $0.75<1$, this value is valid.

The possible values for $\lambda$ are 2 and 4 .

$ \text { Sum }=2+4=6 $

The sum of all possible values of $\lambda$ is 6.

$ S=\left\{x^3+a x^2+b x+c: a, b, c \in \mathbb{N} \text { and } a, b, c \leq 20\right\} \text { so } a, b, c \in\{1,2,3, \ldots 20\} $

$x^3+a x^2+b x+c$ is divisible by $x^2+2$

$\therefore $ $ x^3+a x^2+b x+c=\left(x^2+2\right) L(x)$

Let $L(x)=x+n$ because L.H.S is cubic have leading coefficient 1 .

$ \begin{aligned} x^3+a x^2+b x+c & =\left(x^2+2\right)(x+n) \\ & =x^3+x^2 n+2 x+2 n \end{aligned} $

Compare $n=a, b=2,2 n=c \Longrightarrow c=2 a$

so, $c=2 a$ and $b=2$

Finding the Number of Possible Polynomials

• For $b$ : There is only $\mathbf{1}$ choice $(b=2)$, which is within the range $[1,20]$.

• For $a$ and $c$ : Since $c=2 a$, we must ensure $c \leq 20$.

• If $a=1, c=2$

• If $a=2, c=4$

• …

• If $a=10, c=20$
  
  There are 10 possible values for the pair ( $a, c$ ).

Total polynomials $=($ Choices for $b) \times($ Choices for $a, c$ pairs $)=1 \times 10=\mathbf{1 0}$

Given equation

$ x^4-a x^2+9=0--- \text { (1) } $

let $x^2=t$, equation (1) become

$ t^2-a t+9=0--- \text { (2) } $

If $t_1$ & $t_2$ are roots of equation (2) then four distinct roots of equation (1) is $\left\{-\sqrt{t_1},-\sqrt{t_2}, \sqrt{t_1}, \sqrt{t_2}\right\}$

roots of this equation are real and distinct if $\mathrm{D}>0$

$ \begin{aligned} & (-a)^2-4 \times 9>0 \\\\ & \Rightarrow a^2-36>0 \end{aligned} $

so $a \in(-\infty,-6) \cup(6, \infty)$

so smallest positive integral value of $a$ is 7 as $a>6$.

given $x|x+3|+|x-1|-2=0$.

critical points $x=-3,1$.

two critical points means 3 cases.

Case 1 : $ x<-3,|x+3|=-(x+3)$ and $|x-1|=-(x-1)$. equation become

$ \begin{aligned} & -x(x+3)-(x-1)-2=0 \\ & -x^2-3 x-x+1-2=0 \\ & -x^2-4 x-1=0 \Longrightarrow x^2+4 x+1=0 \\ & x=\frac{-4 \pm \sqrt{16-4}}{2}=\frac{-4 \pm 2 \sqrt{3}}{2}=-2 \pm \sqrt{3} \end{aligned} $

since $x<-3$ so $x=-2-\sqrt{3}$ is only solution.

Case 2 : $-3 \leq x \leq 1,|x+3|=x+3$ and $|x-1|=-(x-1)$. equation become

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

since $-3 \leq x \leq 1$ so $x=-1+\sqrt{2},-1-\sqrt{2}$ both accepted.

Case 3 : $x>1,|x+3|=x+3 \text { and }|x-1|=x-1 \text {. } $

equation become

$ x(x+3)+x-1-2=0 . $

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

$x=\frac{-4 \pm \sqrt{16+12}}{2}=\frac{-4 \pm 2 \sqrt{7}}{2}=-2 \pm \sqrt{7}$. Checking positive solution. since $x>1$, so $x=\sqrt{7}-2 \approx 0.645$, so no solution.

so final solution is

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

Number of solution = 3

$ \begin{aligned} & \text { } f(x)=\log _{\left(10 x^2-17 x+7\right)}\left(18 x^2-11 x+1\right) \\ & \log _b a \text { is defined if } a>0 \text { and } b>0 \text { and } b \neq 1 \text {. } \\ & 18 x^2-11 x+1>0 \\ & 18 x^2-9 x-2 x+1>0 \\ & 9 x(2 x-1)-1(2 x-1)>0 \\ & (2 x-1)(9 x-1)>0 \\ & S_1: x \in(-\infty, 1 / 9) \cup(1 / 2, \infty) \text {. } \\ & 10 x^2-17 x+7>0 \\ & 10 x^2-10 x-7 x+7>0 \\ & 10 x(x-1)-7(x-1)>0 \\ & (10 x-7)(x-1)>0 \\ & S_2: x \in(-\infty, 7 / 10) \cup(1, \infty) \text {. } \\ & \text { And } 10 x^2-17 x+7 \neq 1 \end{aligned} $

$ \begin{aligned} & 10 x^2-17 x+6 \neq 0 \\ & 10 x^2-12 x-5 x+6 \neq 0 \\ & 2 x(5 x-6)-1(5 x-6) \neq 0 \\ & (2 x-1)(5 x-6) \neq 0 \\ & S_3: x \neq 1 / 2,6 / 5 \Rightarrow x \neq 0.5,1.2 . \\ & \text { Final solution } x \in\left[S_1 \cap S_2\right]-\{0.5,1.2\} . \\ & x \in[(-\infty, 1 / 9) \cup(1 / 2, \infty)] \cap[(-\infty, 7 / 10) \cup(1, \infty)]-\{0.5,1.2\} . \\ & x \in(-\infty, 1 / 9) \cup(1 / 2,7 / 10) \cup(1, \infty)-\{1.2\} . \end{aligned} $

Since 0.5 is not included, it need not to be subtracted.

Compare final solution with $(-\infty, a) \cup(b, c) \cup(d, \infty)-\{e\}$.

$ \begin{aligned} & a=1 / 9, b=1 / 2, c=7 / 10, d=1 \text { and } e=1.2 . \\ & 90(a+b+c+d+e)=90(1 / 9+1 / 2+7 / 10+1+1.2) \\ & =10+45+63+90+108=316 . \end{aligned} $

$ \begin{aligned} &\begin{aligned} & (-2+\sqrt{3})(|\sqrt{x}-3|)+(x-6 \sqrt{x}+9)-2 \sqrt{3}=0, x \geq 0 \\ & \text { put }|\sqrt{x}-3|=t \\ & \Rightarrow(-2+\sqrt{3}) t+t^2-2 \sqrt{3}=0 \\ & \Rightarrow t(t-2)+\sqrt{3}(t-2)=0 \\ & \Rightarrow(t+\sqrt{3})(t-2)=0 \\ & t+\sqrt{3} \neq 0 \Rightarrow t=2 \\ & \Rightarrow|\sqrt{x}-3|=2 \\ & \sqrt{x}-3= \pm 2 \\ & \sqrt{x}=5,1 \\ & x=25,1 \\ & \Rightarrow \alpha=1, \beta=25 \\ & \Rightarrow \sqrt{\frac{\beta}{\alpha}}+\sqrt{\alpha \beta}=10 \end{aligned}\\ &\text { Option (2) is Correct. } \end{aligned} $

$ \text { Let B's time }=x \text { days } $

$ \begin{aligned} & \Rightarrow \text { As time }=(x-24) \text { day } \\ & \text { Rate of } \mathrm{A}=\frac{1}{x-24} \\ & \text { Rate of } \mathrm{B}=\frac{1}{x} \\ & \frac{1}{x-24}+\frac{1}{x}=\frac{1}{22.5} \\ & \frac{2 x-24}{x(x-24)}=\frac{1}{22.5} \\ & 45 x-540=x^2-24 x \\ & x^2-69 x+540=0 \\ & (x-60)(x-9)=0 \\ & \Rightarrow x=60 \\ & \text { At time }=60-24 \\ & =36 \text { days } \end{aligned} $

$ \begin{aligned} & \frac{1}{2} \leq|\alpha-\beta| \leq \frac{3}{2} \\ & \frac{1}{4} \leq|\alpha-\beta|^2 \leq \frac{9}{4} \\ & \frac{1}{4} \leq(\alpha+\beta)^2-4 \alpha \beta \leq \frac{9}{4} \\ & \frac{1}{4} \leq \frac{25}{9}-4 \times \frac{\lambda}{4} \leq \frac{9}{4} \\ & -\frac{91}{36} \leq-\lambda \leq \frac{-19}{36} \\ & \frac{19}{36} \leq \lambda \leq \frac{91}{36} \\ & \lambda=1,2 \\ & \text { Sum }=3 \end{aligned} $

$ \begin{aligned} &\text { If }\\ &\begin{aligned} & x < -4: \\ & -x(x+4)-3(x+2)+10=0 \\ & x^2+7 x-4=0 \\ & x=\frac{-7 \pm \sqrt{65}}{2} \\ & -4 < x < -2: \\ & x(x+4)-3(x+2)+10=0 \\ & x^2+x+4=0 \\ & \Delta<0 \\ & x > -2 \\ & x(x+4)+3(x+2)+10=0 \\ & x^2+7 x+16=0 \\ & \Delta<0 \end{aligned} \end{aligned} $

$ \begin{aligned} & \because \alpha<1<\beta \\ & f(1)<0 \\ & \Rightarrow 1+2 a+(3 a+10)<0 \\ & \Rightarrow 5 a+11<0 \\ & a<\frac{-11}{5} \\ & \therefore a \in\left(-\infty, \frac{-11}{5}\right) \end{aligned} $

$ \begin{aligned} & |x-1| 2-5|x-1|+6=0 \\ & \text { Let }|x-1|=t \\ & t^2-5+6=0 \\ & (t-3)(t-2)=0 \\ & t=2,3 \\ & |x-1|=2 \text { or }|x-1|=3 \\ & x-1= \pm 2 \text { or } x-1= \pm 3 \\ & \Rightarrow x=3,-1,4,-2 \\ & \Rightarrow \text { Sum }=3-1+4-2 \\ & \quad=4 \end{aligned} $

$\begin{aligned} & |\mathrm{x}-2|^2+2|\mathrm{x}-2|-|\mathrm{x}-2|-2=0 \\ & \Rightarrow(|\mathrm{x}-2|+2)(|\mathrm{x}-2|-1)=0 \\ & \Rightarrow|\mathrm{x}-2|=1 \\ & \Rightarrow \mathrm{x}=2 \pm 1=3,1 \\ & \Rightarrow \text { sum of square of roots }=9+1=10 \\ & \mathrm{x}^2-2|\mathrm{x}-3|-5=0 \\ & \text { Case-I } \mathrm{x}-3 \geq 0 \\ & \Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=0 \\ & \Rightarrow(\mathrm{x}-1)^2=0 \\ & \Rightarrow \mathrm{x}=1 \\ & \text { But } \mathrm{x} \geq 3 \\ & \Rightarrow \mathrm{x} \in \phi \\ & \text { Case-II } \mathrm{x}-3<0 \\ & \mathrm{x}^2+2 \mathrm{x}-11=0, \mathrm{D}>0 \Rightarrow \text { Real & distinct roots } \\ & \mathrm{f}(\mathrm{x})=\mathrm{x}^2+2 \mathrm{x}-11 \\ & \mathrm{f}(3)>0, \frac{-\mathrm{p}}{2 \mathrm{a}}=-1<3 \\ & \Rightarrow \text { both roots }<3, \text { both roots acceptable } \\ & \text { Sum of square of roots }=(\alpha+\beta)^2-2 \alpha \beta \\ & =4+22=26 \\ & \Rightarrow \text { Final sum }=10+26=36 \end{aligned}$

$ \begin{aligned} & \text { (I) } x<2 \\ & -x^2+2 x-3 x+9+1=0 \\ & \Rightarrow x^2+x-10=0 \\ & \Rightarrow x=\frac{-1+\sqrt{41}}{2}, \frac{-1-\sqrt{41}}{2} \\ & \quad \quad \qquad \quad \times \qquad \qquad \sqrt{ } \end{aligned}$

$\begin{aligned} & \text { (II) } 2 \leq x<3 \\ & \Rightarrow \mathrm{x}^2-2 \mathrm{x}-3 \mathrm{x}+9+1=0 \\ & \Rightarrow \mathrm{x}^2-5 \mathrm{x}+10=0 \\ & \mathrm{D}<0 \\ & \text { (III) } \mathrm{x} \geq 3 \\ & \mathrm{x}^2-2 \mathrm{x}+3 \mathrm{x}-9+2=0 \\ & \Rightarrow \mathrm{x}^2+\mathrm{x}-8=0 \end{aligned}$

$\mathrm{x}=\frac{-1+\sqrt{32}}{2}, \frac{-1-\sqrt{32}}{2}$

$\quad \quad \times \qquad \qquad \times$

1 real roots

To find the set of all values of $ p \in \mathbb{R} $ for which both roots of the equation $ x^2-(p+2)x+(2p+9)=0 $ are negative real numbers, follow these steps:

Discriminant Condition:

The equation's discriminant $ D $ must be non-negative for real roots:

$ (p+2)^2 - 4(2p+9) \geq 0 $

Simplifying this:

$ p^2 + 4p + 4 - 8p - 36 \geq 0 \quad \Rightarrow \quad p^2 - 4p - 32 \geq 0 $

This can be factored as:

$ (p-8)(p+4) \geq 0 $

Meaning $ p \in (-\infty, -4] \cup [8, \infty) $. …… (1)

Sum of Roots Condition:

The sum of the roots (which is $ p+2 $) must be negative:

$ p + 2 < 0 \quad \Rightarrow \quad p < -2 $ …… (2)

Product of Roots Condition:

The product of the roots $ (2p + 9) $ must be positive:

$ 2p + 9 > 0 \quad \Rightarrow \quad p > -\frac{9}{2} $ …… (3)

Determine the Valid Interval:

Combine the results from conditions (1), (2), and (3). From conditions (1) and (2), we find $ p < -2 $:

Intersection of $ (-\infty, -4] $ and $ (-\frac{9}{2}, -2) $ gives:

$ p \in \left(-\frac{9}{2}, -4\right] $

Calculate $\beta - 2\alpha$:

With $\alpha = -\frac{9}{2}$ and $\beta = -4$, compute:

$ \beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right) = -4 + 9 = 5 $

Therefore, the difference $\beta - 2\alpha$ is 5.

$\begin{aligned} & x^2+4 x-n=0 \text { has integer roots } \\ & \Rightarrow x=\frac{-4 \pm \sqrt{16+4 n}}{2}=-2 \pm \sqrt{4+n} \end{aligned}$

For $x$ to be integer $4+n$ must be perfect squares

$\begin{aligned} & n \in[20,100] \\ & n+4 \in[24,104]=S \end{aligned}$

$\left\{25,36, \ldots 10^2\right\} \in S \Rightarrow 5^2, 6^2, \ldots 10^2 \Rightarrow 6$ values of $n$

Given the equation $ x(x+2)(12-k) = 2 $, we want it to have equal roots. To achieve this, we need to manipulate it into a quadratic form in terms of $ x $:

$ x^2 + 2x - \frac{2}{12-k} = 0 $

For this quadratic equation to have equal (repeated) roots, the discriminant $ D $ must be zero. The discriminant $ D $ for the equation $ ax^2 + bx + c = 0 $ is given by:

$ D = b^2 - 4ac $

Substituting $ a = 1 $, $ b = 2 $, and $ c = -\frac{2}{12-k} $ into the discriminant formula, we get:

$ 4 - 4\left(-\frac{2}{12-k}\right) = 0 $

Simplifying the expression:

$ 1 + \frac{2}{12-k} = 0 $

Solving for $ k $:

$ \frac{2}{12-k} = -1 \quad \Rightarrow \quad 2 = -(12 - k) \quad \Rightarrow \quad 2 = -12 + k \quad \Rightarrow \quad k = 14 $

Now, consider the point $(k, \frac{k}{2})$, which becomes $(14, 7)$. We need to find its distance from the line $3x + 4y + 5 = 0$. The formula for the distance $ d $ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is:

$ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} $

Substituting $ (x_1, y_1) = (14, 7) $ and the line coefficients $ A = 3 $, $ B = 4 $, $ C = 5 $:

$ d = \frac{|3(14) + 4(7) + 5|}{\sqrt{3^2 + 4^2}} $

Calculating the numerator:

$ 3 \times 14 + 4 \times 7 + 5 = 42 + 28 + 5 = 75 $

And the denominator:

$ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $

Therefore, the distance is:

$ d = \frac{75}{5} = 15 $

Thus, the distance is 15.

$\begin{aligned} & \Rightarrow \quad P^n=\gamma^n+\delta^n \\ & P_{25}-P_{23}=\left(\gamma^{25}-\gamma^{23}\right)+\left(\delta^{25}-\delta^{23}\right) \\ & =\gamma^{23}\left(\gamma^2-1\right)+\delta^{23}\left(\delta^2-1\right) \\ & =\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \delta)=-3\left[\gamma^{24}+\delta^{24}\right] \end{aligned}$

$\begin{aligned} &\Rightarrow \frac{P_{25}-P_{23}}{P_{24}}=(-3)\\ &\text { Similarly, } \end{aligned}$

$\begin{aligned} & \Rightarrow Q_{25}+\sqrt{3} Q_{24}=\left(\alpha^{25}+\sqrt{3} \alpha^{24}\right)+\left(\beta^{25}+\sqrt{3} \beta^{24}\right) \\ & =\alpha^{23}\left(\alpha^2+\sqrt{3} \alpha\right)+\beta^{23}\left(\beta^2+\sqrt{3} \beta\right) \\ & =\alpha^{23}(16)+16 \beta^{23} \\ & \Rightarrow \frac{Q_{25}+\sqrt{3} Q_{24}}{2 \cdot Q_{23}}=\frac{16\left(\alpha^{23}+\beta^{23}\right)}{2\left(\alpha^{23}+\beta^{23}\right)}=8 \\ & \Rightarrow \frac{Q_{25}+\sqrt{3} Q_{24}}{2 Q_{23}}+\frac{\left(P_{25}-P_{23}\right)}{P_{24}}=8+(-3)=5 \end{aligned}$

Given:

$ P_{10} = 123 $

$ P_9 = 76 $

$ P_8 = 47 $

$ P_1 = 1 $

We know that:

$ P_n = \alpha^n + \beta^n $

According to Newton’s identities, we have the relation:

$ P_{10} = P_9 + P_8 $

This implies:

$ P_{10} - P_9 - P_8 = 0 $

From $ P_1 = 1 $, it follows that:

$ \alpha + \beta = 1 $

$ \alpha \beta = 1 $

Now, we are tasked with finding the quadratic equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. For such a quadratic equation:

Using the relationship between roots and coefficients, the equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:

$ x^2 - \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)x + \frac{1}{\alpha \beta} = 0 $

Substitute the known sum and product of $\alpha$ and $\beta$:

$ x^2 - \left(\frac{\alpha + \beta}{\alpha \beta}\right)x + \frac{1}{\alpha \beta} = 0 $

Given that $\alpha + \beta = 1$ and $\alpha \beta = 1$, we can simplify:

$ x^2 - \left(\frac{1}{1}\right)x + \frac{1}{1} = 0 $

This simplifies to:

$ x^2 + x - 1 = 0 $

Thus, the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:

$ x^2 + x - 1 = 0 $

$\begin{aligned} & (a-5)^2-8(15-3 a)<0 \\ & a^2+14 a+25-120<0 \\ & a^2+14 a-95<0 \\ & (a+19)(a-5)<0 \\ & a \in(-19,5) \\ & \therefore-19< x<5 \\ & \therefore \sum_{x \in X} x^2=\left(1^2+2^2+\ldots .+4^2\right)+\left(1^2+2^2+\ldots+18^2\right) \\ & =\frac{4 \times 5 \times 9}{6}+\frac{18 \times 19 \times 37}{6} \\ & =30+2109 \\ & =2139 \end{aligned}$

Consider $\frac{1}{\sqrt{\mathrm{x}}}=\alpha \quad \mathrm{x}>0$

$\begin{aligned} & \left(9 \alpha^2-9 \alpha+2\right)\left(2 \alpha^2-7 \alpha+3\right)=0 \\ & (3 \alpha-2)(3 \alpha-1)(\alpha-3)(2 \alpha-1)=0 \\ & \alpha=\frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 3 \\ & x=9,4, \frac{9}{4}, \frac{1}{9} \end{aligned}$

So, no. of solutions $=4$

as $f(x)$ is a polynomial of degree two let it be

$f(x)=a x^2+b x+c \quad(a \neq 0)$

on satisfying given conditions we get

$C=1 \& a= \pm 1$

hence $f(x)=1 \pm x^2$

also range $\in(-\infty, 1]$ hence

$\begin{gathered} f(x)=1-x^2 \\ \text { now } f(k)=-2 k \end{gathered}$

$1-\mathrm{k}^2=-2 \mathrm{k} \rightarrow \mathrm{k}^2-2 \mathrm{k}-1=0$

let roots of this equation be $\alpha \& \beta$ then $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$

$=4-2(-1)=6$

To find the sum of the squares of all the roots of the equation $ x^2 + |2x - 3| - 4 = 0 $:

Case I: $ x \geq \frac{3}{2} $

For $ x \geq \frac{3}{2} $, the expression $ |2x - 3| $ becomes $ 2x - 3 $. Thus, the equation becomes:

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

Simplifying gives:

$ x^2 + 2x - 7 = 0 $

Solving this quadratic equation, we find:

$ x = 2\sqrt{2} - 1 $

Case II: $ x < \frac{3}{2} $

For $ x < \frac{3}{2} $, the expression $ |2x - 3| $ becomes $ -(2x - 3) = -2x + 3 $. The equation therefore becomes:

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

Simplifying gives:

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

Solving this quadratic equation, we obtain:

$ x = 1 - \sqrt{2} $

Sum of the Squares of the Roots

The sum of the squares of the roots is:

$ (2\sqrt{2} - 1)^2 + (1 - \sqrt{2})^2 $

Calculating each term:

$(2\sqrt{2} - 1)^2 = (2\sqrt{2})^2 - 2 \cdot 2\sqrt{2} \cdot 1 + 1^2 = 8 - 4\sqrt{2} + 1 = 9 - 4\sqrt{2}$

$(1 - \sqrt{2})^2 = 1^2 - 2 \cdot 1 \cdot \sqrt{2} + (\sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$

Adding these results:

$ (9 - 4\sqrt{2}) + (3 - 2\sqrt{2}) = 12 - 6\sqrt{2} $

This simplifies to $ 6(2 - \sqrt{2}) $.

Thus, the sum of the squares of the roots is $ 6(2 - \sqrt{2}) $.

Only 2 solutions.

To solve the given equation, start by rewriting the expression:

$ (x^2 - 9x + 11)^2 - (x - 4)(x - 5) = 3 $

First, simplify the second part of the expression:

$ (x - 4)(x - 5) = x^2 - 9x + 20 $

Now the equation becomes:

$ (x^2 - 9x + 11)^2 - (x^2 - 9x + 20) = 3 $

Introduce a substitution for simplification:

$ \text{Let } t = x^2 - 9x $

Thus, the equation transforms to:

$ t^2 + 22t + 121 - t - 20 - 3 = 0 $

Simplify further:

$ t^2 + 21t + 98 = 0 $

Factor the quadratic:

$ (t + 14)(t + 7) = 0 $

This gives:

$ t = -7 \quad \text{or} \quad t = -14 $

Address each case where $ t = x^2 - 9x $:

$ x^2 - 9x = -7 $

$ x^2 - 9x + 7 = 0 $

Solving this quadratic equation, we find the roots:

$ x = \frac{9 \pm \sqrt{81 - 4 \times 7}}{2} = \frac{9 \pm \sqrt{53}}{2} $

$ x^2 - 9x = -14 $

$ x^2 - 9x + 14 = 0 $

Solving this quadratic equation:

$ x = \frac{9 \pm \sqrt{81 - 4 \times 14}}{2} = \frac{9 \pm \sqrt{25}}{2} $

$ x = \frac{9 \pm 5}{2} = 7 \quad \text{or} \quad x = 2 $

The rational roots from the second equation are 7 and 2. Thus, the product of all the rational roots is:

$ 7 \times 2 = 14 $

To find the sum of the fourth powers of the roots $\alpha_\theta$ and $\beta_\theta$ of the quadratic equation $2x^2 + (\cos \theta)x - 1 = 0$, we start analyzing the expression $\alpha_\theta^4 + \beta_\theta^4$.

The equation can be rewritten with its roots using:

$ \alpha + \beta = -\frac{\cos \theta}{2}, \quad \alpha \beta = -\frac{1}{2} $

We need to calculate $\alpha^2 + \beta^2$ and $\alpha^2 \beta^2$:

$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{\cos \theta}{2}\right)^2 - 2\left(-\frac{1}{2}\right) = \frac{\cos^2 \theta}{4} + 1 $

$ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} $

Substitute these into:

$ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 = \left(\frac{\cos^2 \theta}{4} + 1\right)^2 - \frac{1}{2} $

Maximize and minimize $\left(\frac{\cos^2 \theta}{4} + 1\right)^2$:

Zero of $\cos\theta$ leads to:

$ \left(\frac{0^2}{4} + 1\right)^2 = 1 $

Max value $\cos^2 \theta = 1$:

$ \left(\frac{1}{4} + 1\right)^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16} $

Substitute back:

$ \text{Max: } \frac{25}{16} - \frac{1}{2} = \frac{25}{16} - \frac{8}{16} = \frac{17}{16} $

$ \text{Min: } 1 - \frac{1}{2} = \frac{1}{2} $

Finally, compute $16(M + m)$:

$ 16\left(\frac{17}{16} + \frac{1}{2}\right) = 16\left(\frac{17}{16} + \frac{8}{16}\right) = 16 \times \frac{25}{16} = 25 $

$\begin{aligned} & x^2-\sqrt{2} x-\sqrt{3}=0 \\ & P_n=\alpha^n-\beta^n \end{aligned}$

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

Using Newton's theorem

$\begin{aligned} & P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\ & \text { Put } n=10 \\ & P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0 \end{aligned}$

$P_{12}=\sqrt{2} P_{11}+\sqrt{3} P_{10}$

Put $n=9$

$\begin{aligned} & P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_9=0 \\ & P_{11}=\sqrt{2} P_{10}+\sqrt{3} P_9 \\ & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12} \end{aligned}$

Put the value of $P_{12}$ & $P_{12}$ in above equation.

$\begin{aligned} = & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10)\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right)-11\left(\sqrt{2} P_{11}+\sqrt{3} P_{10}\right) \\ = & 11 \sqrt{3} P_{10}-10 \sqrt{2} P_{10}+22 P_{10}+10 \sqrt{2} P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2} P_{11}-11 \sqrt{3} P_{10} \\ = & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2}\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right) \\ = & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-22 P_{10}-11 \sqrt{6} P_9 \\ = & 10 \sqrt{3} P_9 \end{aligned}$

$\begin{aligned} & x^2+2 \sqrt{2 x}-1=0 \\ & \alpha+\beta=-2 \sqrt{2} \text { and } \alpha \beta=-1 \\ & \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\ & =8+2=10 \\ & \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha \beta)^2 \\ & =100-2=98 \\ & \alpha^6+\beta^6=\left(\alpha^2+\beta^2\right)^3-3 \alpha^2 \beta^2\left(\alpha^2+\beta^2\right) \\ & =1000-3(10) \\ & =970 \\ & \therefore \quad \frac{1}{10}\left(\alpha^6+\beta^6\right)=97 \end{aligned}$

Equation whose roots are $\alpha^4+\beta^4$ and $\frac{1}{10}\left(\alpha^6+\beta^6\right)$ is

$\begin{aligned} & x^2-(98+97) x+98 \times 97=0 \\ & x^2-195 x+9506=0 \end{aligned}$

First, let's start by substituting $y = (8)^x$ in the given equation. By substituting, the equation $8^{2x} - 16 \cdot 8^x + 48 = 0$ will be transformed into

$ y^2 - 16y + 48 = 0 $

Now, we have a quadratic equation in $y$. To find the roots of this quadratic equation, we can use the quadratic formula:

$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

In this case, $a = 1$, $b = -16$, and $c = 48$. Substituting these values into the formula, we get:

$ y = \frac{16 \pm \sqrt{256 - 192}}{2} $

$ y = \frac{16 \pm \sqrt{64}}{2} $

$ y = \frac{16 \pm 8}{2} $

Solving for the two possible values of $y$, we have:

$ y = \frac{16 + 8}{2} = 12 $

$ y = \frac{16 - 8}{2} = 4 $

Now, recall that we substituted $y = (8)^x$. So, we need to solve for $x$ when $y = 12$ and $y = 4$:

$ 8^x = 12 $

$ x = \log_8(12) $

$ 8^x = 4 $

$ x = \log_8(4) $

Therefore, the solutions for $x$ are $\log_8(12)$ and $\log_8(4)$. The sum of these solutions is:

$ \log_8(12) + \log_8(4) $

Using the logarithmic property that $\log_b(m) + \log_b(n) = \log_b(m \cdot n)$, we get:

$ \log_8(12) + \log_8(4) = \log_8(12 \cdot 4) $

$ = \log_8(48) $

Now, we note that:

$ 48 = 8 \cdot 6 $

Thus,

$ \log_8(48) = \log_8(8 \cdot 6) = \log_8(8) + \log_8(6) = 1 + \log_8(6) $

Therefore, the sum of all the solutions of the equation is:

Option A $1 + \log_8(6)$.

$\begin{aligned} & x^2-\left(t^2-5 t+6\right) x+1=0 \\ & \therefore a_{2025}-\left(t^2-5 t+6\right) a_{2024}+a_{2023}=0 \\ & \Rightarrow \frac{a_{2025}+a_{2023}}{a_{2024}}=t^2-5 t+6 \\ & =\left(t+\frac{5}{2}\right)^2+\left(\frac{-1}{4}\right) \\ & \text { Minimum value }=\frac{-1}{4} \end{aligned}$

Given that the roots of the quadratic equation are $2$ and $6$, we can use Vieta's formulas which relate the coefficients of the polynomial to sums and products of its roots.

The given quadratic equation is:

$a x^2 + b x + 1 = 0$

By Vieta's formulas, the sum of the roots is:

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

So:

$8 = -\frac{b}{a} \Rightarrow b = -8a$

And the product of the roots is:

$2 \times 6 = \frac{1}{a} \Rightarrow 12 = \frac{1}{a} \Rightarrow a = \frac{1}{12}$

Therefore, $b = -8a = -8 \left( \frac{1}{12} \right) = -\frac{2}{3}$.

Given the roots of the new quadratic equation are:

$\frac{1}{2a+b}$ and $\frac{1}{6a+b}$

We know $a = \frac{1}{12}$ and $b = -\frac{2}{3}$, so:

$2a + b = 2 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{6} - \frac{2}{3} = \frac{1 - 4}{6} = -\frac{3}{6} = -\frac{1}{2}$

and:

$6a + b = 6 \left(\frac{1}{12}\right) - \frac{2}{3} = \frac{1}{2} - \frac{2}{3} = \frac{3 - 4}{6} = -\frac{1}{6}$

Thus, the roots of the new quadratic equation are:

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

and:

$\frac{1}{-\frac{1}{6}} = -6$

The new quadratic equation with roots $-2$ and $-6$ can be formulated as:

$x^2 - (\text{sum of roots}) x + (\text{product of roots}) = 0$

The sum of the roots is:

$-2 + (-6) = -8$

The product of the roots is:

$(-2) \times (-6) = 12$

Thus, the quadratic equation becomes:

$x^2 - (-8)x + 12 = x^2 + 8x + 12 = 0$

Hence, the correct option is:

Option A

$x^2 + 8 x + 12 = 0$

Given : $p x^2+q x-r=0$

Let $p=\frac{a}{r_1}, q=a, r=a r_1$

$\begin{aligned} & \text { and } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \\\\ & \Rightarrow \frac{-\frac{q}{p}}{-\frac{r}{p}}=\frac{3}{4} \\\\ & \Rightarrow \frac{q}{r}=\frac{3}{4} \\\\ & \Rightarrow \frac{1}{r_1}=\frac{3}{4} \\\\ & \Rightarrow r_1=\frac{4}{3}\end{aligned}$

$\begin{aligned}(\alpha-\beta)^2 & =(\alpha+\beta)^2-4 \alpha \beta \\\\ & =\left(\frac{-q}{p}\right)^2-4\left(\frac{-r}{p}\right) \\\\ & =\frac{q^2}{p^2}+\frac{4 r}{p} \\\\ & =r_1^2+4 r_1^2=5 r_1^2 \\\\ & =5\left(\frac{4}{3}\right)^2=\frac{80}{9}\end{aligned}$

Notice that $(\sqrt{3} + \sqrt{2})$ and $(\sqrt{3} - \sqrt{2})$ are reciprocals of each other because :

• $(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1$

Using the Reciprocal Property :

• This means $(\sqrt{3} - \sqrt{2})^x = \frac{1}{(\sqrt{3} + \sqrt{2})^x} $

Let $a = (\sqrt{3} + \sqrt{2})^x $. The equation given in the problem becomes :

• $a + \frac{1}{a} = 10$

Simplifying

Multiplying both sides by *a*, we get a quadratic equation :

• $a^2 + 1 = 10a$


• $a^2 - 10a + 1 = 0$

Solving the Quadratic

Using the quadratic formula, we find :

• $a = \frac{10 \pm \sqrt{96}}{2} = 5 \pm 2\sqrt{6}$

Possible values of x

Since $a = (\sqrt{3} + \sqrt{2})^x $, we have two cases :

1. $(\sqrt{3} + \sqrt{2})^x = 5 + 2\sqrt{6}$ = $(\sqrt{3} + \sqrt{2})^2$. There is one real solution for x in this case which is x = 2.


2. $(\sqrt{3} + \sqrt{2})^x = 5 - 2\sqrt{6}$ = $(\sqrt{3} - \sqrt{2})^2= \frac{1}{(\sqrt{3} + \sqrt{2})^2} $ = $(\sqrt{3} + \sqrt{2})^{-2}$. There is one real solution for x in this case which is x = - 2.

Conclusion

There are two real solutions for *x*. Therefore, the number of elements in the set S is 2. This corresponds with option (C).

$x^2-8 x+32>0 \forall x \in R$ as discriminant of this quadratic is $64-4 \times 32<0$

$ \Rightarrow a x^2+2(a+1) x+9 a+4<0 \forall x \in R $

$\Rightarrow$ Only possible when $a<0$ and $D<0$

$\Rightarrow$ Since $S$ is set of positive values of $a \Rightarrow S$ is a null set

$ \Rightarrow n(S)=0 $

$\begin{aligned} & x^2-x-1=0 \\ & S_n=2023 \alpha^n+2024 \beta^n \\ & S_{n-1}+S_{n-2}=2023 \alpha^{n-1}+2024 \beta^{n-1}+2023 \alpha^{n-2}+2024 \beta^{n-2} \\ & =2023 \alpha^{n-2}[1+\alpha]+2024 \beta^{n-2}[1+\beta] \\ & =2023 \alpha^{n-2}\left[\alpha^2\right]+2024 \beta^{n-2}\left[\beta^2\right] \\ & =2023 \alpha^n+2024 \beta^n \\ & S_{n-1}+S_{n-2}=S_n \\ & P_{u t} n=12 \\ & S_{11}+S_{10}=S_{12} \end{aligned}$

$ \begin{aligned} & x|x|-5|x+2|+6=0 \\\\ & \text { Case-I : } \\\\ & \text { When } x<-2 \text { then } \\\\ & -x^2+5(x+2)+6=0 \\\\ & \Rightarrow x^2-5 x-16=0 \\\\ & \Rightarrow x=\frac{5 \pm \sqrt{25+64}}{2} \\\\ & \therefore x=\frac{5-\sqrt{89}}{2} \text { is accepted } \end{aligned} $

Case-II :

$ \begin{aligned} & \text { When } -2 \leq x < 0 \text { then } \\\\ & -x^2-5(x+2)+6=0 \\\\ \Rightarrow & x^2+5 x+4=0 \\\\ \Rightarrow & (x+1)(x+4)=0 \\\\ & x=-1 \text { is accepted } \end{aligned} $

Case-III :

$ \begin{aligned} & \text { When } x \geq 0 \text { then } \\\\ & x^2-5(x+2)+6=0 \\\\ \Rightarrow & x^2-5 x-4=0 \\\\ & x=\frac{5 \pm \sqrt{25+16}}{2} \\\\ & =\frac{5 \pm \sqrt{41}}{2} \\\\ & x=\frac{5 - \sqrt{41}}{2} \text { is accepted } \\\\ \therefore & 3 \text { real roots are possible. } \end{aligned} $

1. Find the roots of the quadratic equation:

   The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. For the quadratic equation $x^2 - \sqrt{2}x + 2 = 0$, we have $a = 1, b = -\sqrt{2}, c = 2$. Plugging these into the quadratic formula gives:
   
   $x = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2 - 4(1)(2)}}{2(1)} = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm \sqrt{6}i}{2}.$
   
   So, we have two roots, $\alpha$ and $\beta$, which are:
   
   $\alpha = \frac{\sqrt{2} + \sqrt{6}i}{2},$
   
   $\beta = \frac{\sqrt{2} - \sqrt{6}i}{2}.$

2. Express the roots in exponential form:

   We can express complex numbers in the form $re^{i\theta}$. For $\alpha$ and $\beta$, we find the magnitude $r = \sqrt{2}$ and the arguments $\theta = \frac{\pi}{3}, -\frac{\pi}{3}$ respectively. So, we have:
   
   $\alpha = \sqrt{2}e^{i\frac{\pi}{3}},$
   
   $\beta = \sqrt{2}e^{-i\frac{\pi}{3}}.$

3. Calculate the 14th power of the roots:

   To find $\alpha^{14}$ and $\beta^{14}$, we use the property of exponents which says that $(a^m)^n = a^{mn}$. So, we have:
   
   $\alpha^{14} = (\sqrt{2}e^{i\frac{\pi}{3}})^{14} = 2^7e^{i\frac{14\pi}{3}} = 128e^{i\frac{2\pi}{3}},$
   
   $\beta^{14} = (\sqrt{2}e^{-i\frac{\pi}{3}})^{14} = 2^7e^{-i\frac{14\pi}{3}} = 128e^{-i\frac{2\pi}{3}}.$

4. Add the 14th powers of the roots:

   We want to find the real part of $\alpha^{14} + \beta^{14}$. To do this, we use the property that $e^{ix} = \cos(x) + i\sin(x)$. We have:
   
   $\alpha^{14} + \beta^{14} = 128e^{i\frac{2\pi}{3}} + 128e^{-i\frac{2\pi}{3}} = 128(2)\cos\left(\frac{2\pi}{3}\right) = -128.$

$ x|x-1|+|x+2|+a=0 $

Case I : If $x<-2$ then

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

$ a=x^2+2 $

$y=x^2+2$ is decreasing $\forall x \in(-\infty,-2)$

Case II : If $-2 \leq x<1$ then

$ \begin{aligned} & -x^2+x+x+2+a=0 \\\\ & a=x^2-2 x-2 \end{aligned} $

$y=x^2-2 x-2$ is decreasing $\forall x \in[-2,1)$.

Case III: If $x \geq 1$ then

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

$y=-\left(x^2+2\right)$ is decreasing $\forall x \in[1, \infty)$

$\therefore$ Exactly one real root $\forall x \in R$

Given quadratic equation: $x^{2}+\sqrt{6} x+3=0$

We can find the roots using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = \sqrt{6}$, and $c = 3$.

Substituting these values into the quadratic formula, we have:

$x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)}$

Simplifying further :

$x = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2}$

$x = \frac{-\sqrt{6} \pm \sqrt{-6}}{2}$

Since the discriminant is negative, the roots are complex numbers. We can express them using the imaginary unit $i$ :

$x = \frac{-\sqrt{6} \pm \sqrt{6}i}{2}$

$ \Rightarrow $ $x = \frac{-1}{2}\sqrt{6} \pm \frac{1}{2}\sqrt{6}i$

$ \therefore $ $\alpha, \beta=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}$.

The required expression can be rewritten in terms of the argument of the exponential form of the roots, which simplifies the calculation:

$ \begin{aligned} & =\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned} $

Here, the exponential power of $\sqrt{3}$ in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by $(\sqrt{3})^{8} = 81$ to simplify:

$ \begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned} $

Since the cosine function has a period of $2\pi$, we can reduce the arguments of the cosine function in the numerator and denominator.

We have $69\pi/4 = \pi/4 + 17\pi = \pi/4$,

$42\pi/4 = 2\pi/4 + 10\pi = \pi/2$,

$45\pi/4 = \pi/4 + 11\pi = \pi/4$, and

$30\pi/4 = 2\pi/4 + 7\pi = \pi/2$.

Therefore, the required expression simplifies to :

$ \begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{\pi}{2}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{\pi}{2}\right)} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}+(\sqrt{3})^{6}\cdot 0}{(\sqrt{3})^{7}\sqrt{2}+(\sqrt{3})^{2}\cdot 0} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}}{(\sqrt{3})^{7}\sqrt{2}} \\\\ & = (\sqrt{3})^{15-7} \\\\ & = (\sqrt{3})^{8} \\\\ & = 81. \end{aligned} $

Given cubic equation is :

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

$\because \alpha, \beta, \gamma$ are the roots of above equation.

And $\beta \gamma=1=-\alpha$

$ \begin{aligned} & \text { So, product of roots }=-c \\\\ & \Rightarrow \alpha \beta \gamma=-c \\\\ & \Rightarrow(-1)(1)=-c \\\\ & \Rightarrow c=1 \end{aligned} $

Since, $\alpha=-1$ is the root. So,

$ \begin{aligned} & \Rightarrow-1-b+c=0 \\\\ & \Rightarrow c-b=1 \\\\ & \Rightarrow 1-b=1 \Rightarrow b=0 \end{aligned} $

The given equation becomes $x^3+1=0$

So, roots are $-1,-\omega,-\omega^2$

$ \begin{aligned} & \therefore b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3 \\\\ & =0+2-3(-1)^3-6(-\omega)^3-8\left(-\omega^2\right)^3 \\\\ & =2+3+6 \omega^3+8 \omega^6 \\\\ & =5+6+8=19 \end{aligned} $

Concept :

For a cubic equation, $a x^3+b x^2+c x+d=0$

Sum of roots $=\frac{-b}{a}$

Product of roots taken two at a time $=\frac{c}{a}$

Product of roots $=\frac{-d}{a}$

We have,

$ \begin{aligned} & A=\{x \in R:[x+3]+[x+4] \leq 3\} \\\\ & \text { Here, }[x+3]+[x+4] \leq 3 \\\\ & \Rightarrow [x]+3+[x]+4 \leq 3 \\\\ & (\because[x+n]=[x]+n, n \in I) \\\\ & \Rightarrow 2[x]+4 \leq 0 \Rightarrow[x] \leq-2 \\\\ & \Rightarrow x \in(-\infty,-1) \\\\ & A \equiv(-\infty,-1) ...........(i) \end{aligned} $

Also,

$ B=\left\{x \in R: 3^x\left(\sum\limits_{r=1}^{\infty} \frac{3}{10^r}\right)^{x-3}<3^{-3 x}\right\} $

Here,

$ \begin{aligned} & 3^x\left(\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\ldots\right)^{x-3}<3^{-3 x} \\\\ \Rightarrow & 3^x\left(\frac{\frac{3}{10}}{1-\frac{1}{10}}\right)^{x-3}<3^{-3 x} \\\\ \Rightarrow & 3^x\left(\frac{1}{3}\right)^{x-3}<3^{-3 x} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & 3^{x-x+3}<3^{-3 x} \Rightarrow 3^3<3^{-3 x} \\\\ \Rightarrow & -3 x>3 \Rightarrow x<-1 \\\\ \Rightarrow & x \in(-\infty,-1) \\\\ \Rightarrow & B \equiv(-\infty,-1) ...........(ii) \end{array} $

From equations (i) and (ii), we get

$ A=B $

$ \begin{aligned} & \text { We have, }\left|x^2-8 x+15\right|-2 x+7=0 \\\\ & \Rightarrow |(x-3)(x-5)|-2 x+7=0 \end{aligned} $



Now, when $x \leq 3$ or $x \geq 5$, then

$ \begin{array}{ccrl} & x^2-8 x+15-2 x+7 =0 \\\\ &\Rightarrow x^2-10 x+22=0 \\\\ &\Rightarrow x^2-10 x+25-3=0 \\\\ &\Rightarrow (x-5)^2-3=0 \\\\ &\Rightarrow (x-5)= \pm \sqrt{3} \\\\ &\Rightarrow x=5+\sqrt{3}, 5-\sqrt{3} \end{array} $

Since, $x \leq 3$ or $x \geq 5$

$ \therefore x=5+\sqrt{3} $

When, $3 < x < 5$, then

$ \begin{array}{rlrl} & -\left(x^2-8 x+15\right)-2 x+7 =0 \\\\ & \Rightarrow -x^2+8 x-15-2 x+7 =0 \\\\ & \Rightarrow -x^2+6 x-8 =0 \\\\ & \Rightarrow x^2-6 x+8 =0 \\\\ & \Rightarrow (x-4)(x-2) =0 \Rightarrow x=2,4 \end{array} $

Since, $3 < x < 5$

$ \therefore x=4 $

$ \therefore \text { Sum of roots }=(5+\sqrt{3})+4=9+\sqrt{3} $

$ \begin{aligned} & f(1)>0 \Rightarrow k>6 \\\\ & f(2)<0 \Rightarrow k<8 \\\\ & f(3)>0 \Rightarrow k>6 \\\\ & k \in(6,8) \end{aligned} $

Only 1 integral value of $k$ is 7.

Let $(\sqrt{3}+\sqrt{2})^{x^{2}-4}=t$

$ \begin{aligned} & t+\frac{1}{t}=10 \\\\ \Rightarrow & t^{2}-10 t+1=0 \\\\ \Rightarrow & t=\frac{10 \pm \sqrt{100-4}}{2}=5 \pm 2 \sqrt{6} \end{aligned} $

Case-I

$ \begin{aligned} & t=5+2 \sqrt{6} = (\sqrt{3}+\sqrt{2})^{2} \\\\ \Rightarrow & (\sqrt{3}+\sqrt{2})^{x^{2}-4}=(\sqrt{3}+\sqrt{2})^{2} \\\\ \Rightarrow & x^{2}-4=2 \Rightarrow x^{2}=6 \Rightarrow x=\pm \sqrt{6} \end{aligned} $

Case-II :

$t=5-2 \sqrt{6}$ = $(\sqrt{3}-\sqrt{2})^{2}$

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

$\Rightarrow\left((\sqrt{3}-\sqrt{2})^{-1}\right)^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}$

$\Rightarrow 4-x^{2}=2$

$\Rightarrow x^{2}=2$

$\Rightarrow x=\pm \sqrt{2}$

$e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0$

Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}$

Now, $\mathrm{t}^{4}+8 \mathrm{t}^{3}+13 \mathrm{t}^{2}-8 \mathrm{t}+1=0$

Dividing equation by $\mathrm{t}^{2}$

$ \begin{aligned} & t^{2}+8 t+13-\frac{8}{t}+\frac{1}{t^{2}}=0 \\\\ & t^{2}+\frac{1}{t^{2}}+8\left(t-\frac{1}{t}\right)+13=0 \\\\ & \left(t-\frac{1}{t}\right)^{2}+2+8\left(t-\frac{1}{t}\right)+13=0 \end{aligned} $

Let $\mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{z}$

$ z^{2}+8 z+15=0 $

$ \begin{aligned} & (z+3)(z+5)=0 \end{aligned} $

$ z=-3 \text { or } z=-5 $

So, $\mathrm{t}-\frac{1}{\mathrm{t}}=-3$ or $\mathrm{t}-\frac{1}{\mathrm{t}}=-5$

$t^{2}+3 t-1=0$ or $t^{2}+5 t-1=0$

$\mathrm{t}=\frac{-3 \pm \sqrt{13}}{2}$ or $\mathrm{t}=\frac{-5 \pm \sqrt{29}}{2}$

as $t=e^{x}$ so $t$ must be positive,

$ t=\frac{\sqrt{13}-3}{2} \text { or } \frac{\sqrt{29}-5}{2} $

So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$

Hence two solutions and both are negative.

$\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}$

$=\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)}$

$\Rightarrow \sqrt{\mathrm{x}-3}=0 \Rightarrow \mathrm{x}=3$ which is in domain

or

$\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+3}=\sqrt{4 \mathrm{x}-2}$

$2 \sqrt{(x-1)(x+3)}=2 x-4$

$x^{2}+2 x-3=x^{2}-4 x+4$

$6 \mathrm{x}=7$

$\mathrm{x}=7 / 6$ (rejected)

$14 x^{2}-31 x+3 \lambda=0$

$ \begin{aligned} & \alpha+\beta=\frac{31}{14} \ldots .(1) \text { and } \alpha \beta=\frac{3 \lambda}{14}\quad...(2) \\\\ & 35 x^{2}-53 x+4 \lambda=0 \\\\ & \alpha+\gamma=\frac{53}{35} \ldots(3) \text { and } \alpha \gamma=\frac{4 \lambda}{35} \quad\ldots(4) \\\\ & \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma \end{aligned} $

(1) $-(3) \Rightarrow \beta-\gamma=\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}$

$\frac{15}{8} \gamma-\gamma=\frac{7}{10} \Rightarrow \gamma=\frac{4}{5}$

$\Rightarrow \beta=\frac{15}{8} \times \frac{4}{5}=\frac{3}{2}$

$\Rightarrow \alpha=\frac{31}{14}-\beta=\frac{31}{14}-\frac{3}{2}=\frac{5}{7}$

$\Rightarrow \lambda=\frac{14}{3} \alpha \beta=\frac{14}{3} \times \frac{5}{7} \times \frac{3}{2}=5$

so, sum of roots $\frac{3 \alpha}{\beta}+\frac{4 \alpha}{\gamma}=\left(\frac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right)$

$ \begin{aligned} & =\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma} \\\\ & =\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5 \end{aligned} $

Product of roots

$ =\frac{3 \alpha}{\beta} \times \frac{4 \alpha}{\gamma}=\frac{12 \alpha^{2}}{\beta \gamma}=\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \frac{4}{5}}=\frac{250}{49} $

So, required equation is $x^{2}-5 x+\frac{250}{49}=0$

$\Rightarrow 49 \mathrm{x}^{2}-245 \mathrm{x}+250=0$

$3\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)+5=0$

$3\left[\left(x+\frac{1}{x}\right)^{2}-2\right]-2\left(x+\frac{1}{x}\right)+5=0$

Put $x+\frac{1}{x}=t \Rightarrow t \in(-\infty,-2] \cup[2, \infty)$

$ \begin{aligned} & 3 t^{2}-2 t-1=0 \\\\ & 3 t^{2}-3 t+t-1=0 \\\\ & \Rightarrow 3 t(t-1)+1(t-1)=0 \Rightarrow t=1,=-\frac{1}{3} \\\\ & \Rightarrow \quad t=1,-\frac{1}{3} \\\\ & \because t \in(-\infty,-2] \cup[2, \infty) \end{aligned} $

No real value of $t \Rightarrow$ no real value of $x$.

${x^2} - 4x + [x] + 3 = x[x]$

$ \Rightarrow {x^2} - 4x + [x] + 3 - x[x] = 0$

$ \Rightarrow (x - 1)(x - 3) - [x](x - 1) = 0$

$ \Rightarrow (x - 1)(x - [x] - 3) = 0$

$\therefore$ $x = 1$

or

$x - [x] - 3 = 0$

$ \Rightarrow \{ x\} - 3 = 0$ [As $\{ x\} = x - [x]$]

$ \Rightarrow \{ x\} = 3$

But we know, $0 < \{ x\} < 1$

$\therefore$ $\{ x\} \ne 3$

$\therefore$ $x$ has only one solution in ($-\infty,\infty$) which is $x = 1$.

${1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {40 - a}} + {1 \over {40 - a}} - {1 \over {60 - a}}\, + \,....\, + \,{1 \over {180 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$

$ \Rightarrow {1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$

$ \Rightarrow {1 \over {20}}\left( {{{180} \over {(20 - a)(200 - a)}}} \right) = {1 \over {256}}$

$ \Rightarrow (20 - a)(200 - a) = 9.256$

OR ${a^2} - 220a + 1696 = 0$

$ \Rightarrow a = 212,\,8$

$|{x^2}| - 7|x| + 9 \le 0$

$ \Rightarrow |x| \in \left[ {{{7 - \sqrt {13} } \over 2},{{7 + \sqrt {13} } \over 2}} \right]$

As $x \in Z$

So, x can be $ \pm \,2, \pm \,3, \pm \,4, \pm \,5$

Out of these values of x,

$x = 3, - 4, - 5$

satisfy S as well

$n(S \cap T) = 3$