Quadratic Equation and Inequalities

$ 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

$n(S \cap T) = 3$

$\alpha + \beta = \sqrt 2 $, $\alpha \beta = \sqrt 6 $

${1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}$

$ = 2 + {{2 - 2\sqrt 6 } \over 6} = - a$

$\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\beta ^2}}} + 1} \right) = 1 + {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} + {1 \over {{\alpha ^2}{\beta ^2}}}$

$ = {7 \over 6} + {{2 - 2\sqrt 6 } \over 6} = b$

$ \Rightarrow a + b = {{ - 5} \over 6}$

So, equation is ${x^2} + {{17x} \over 6} + {7 \over 6} = 0$

OR $6{x^2} + 17x + 7 = 0$

Both roots of equation are $-$ve and distinct

${3^{\sqrt {{{\log }_3}5} }} - {5^{\sqrt {{{\log }_5}3} }} = {3^{\sqrt {{{\log }_3}5} }} - {\left( {{3^{{{\log }_3}5}}} \right)^{\sqrt {{{\log }_5}3} }}$

${3^{{{\left( {{{\log }_3}5} \right)}^{{1 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = 0$

Note : In the given equation 'x' is missing.

So

$\alpha + \beta + {1 \over \alpha } + {1 \over \beta } = (\alpha + \beta ) + {{\alpha + \beta } \over {\alpha \beta }}$

$ = 5 - {5 \over 3} = {{10} \over 3}$

$\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = 2 + \alpha \beta + {1 \over {\alpha \beta }} = 2 - 3 - {1 \over 3} = {{ - 4} \over 3}$

So Equation must be option (B).

$\alpha + \beta = a - 3,\,\alpha \beta = 1 - 2a$

$ \Rightarrow {\alpha ^2} + {\beta ^2} = {(a - 3)^2} - 2(1 - 2a)$

$ = {a^2} - 6a + 9 - 2 + 4a$

$ = {a^2} - 2a + 7$

$ = {(a - 1)^2} + 6$

So, ${\alpha ^2} + {\beta ^2} \ge 6$

When, ${x^5} = 1$

then ${x^5} - 1 = 0$

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

Given, ${x^4} + {x^3} + {x^2} + x + 1 = 0$ has roots $\alpha$, $\beta$, $\gamma$ and 8.

$\therefore$ Roots of ${x^5} - 1 = 0$ are 1, $\alpha$, $\beta$, $\gamma$ and 8.

We know, Sum of p^th power of n^th roots of unity = 0. (If p is not multiple of n) or n (If p is multiple of n)

$\therefore$ Here, Sum of p^th power of n^th roots of unity

$ = {1^p} + {\alpha ^p} + {\beta ^p} + {\gamma ^p} + {8^p} = \left\{ {\matrix{ 0 & ; & {\mathrm{If\,p\,is\,not\,multiple\,of\,5}} \cr 5 & ; & {\mathrm{If\,p\,is\,multiple\,of\,5}} \cr } } \right.$

Here, $p = 2021$, which is not multiple of 5.

$\therefore$ ${1^{2021}} + {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {8^{2021}} = 0$

$ \Rightarrow {\alpha ^{2021}} + {\beta ^{2021}} + {\gamma ^{2021}} + {8^{2021}} = - 1$

Given,

${{(x + 2)({x^2} + 3x + 5)} \over { - 2 + 3x - {x^2}}} \ge 0$

${x^2} + 3x + 5$ is a quadratic equation

$a = 1 > 0$ and $D = {( - 3)^2} - 4\,.\,1\,.\,5 = - 11 < 0$

$\therefore$ ${x^2} + 3x + 5 > 0$ (always)

So, we can ignore this quadratic term

${{(x + 2)} \over { - 2 + 3x - {x^2}}} \ge 0$

$ \Rightarrow {{x + 2} \over { - ({x^2} - 3x + 2)}} \ge 0$

$ \Rightarrow {{x + 2} \over {{x^2} - 3x + 2}} \le 0$

$ \Rightarrow {{x + 2} \over {{x^2} - 2x - x + 2}} \le 0$

$ \Rightarrow {{x + 2} \over {(x - 1)(x - 2)}} \le 0$

$\therefore$ $x \in ( - \alpha , - 2] \cup (1,2)$

$\therefore$ ${S_1} = ( - \alpha , - 2] \cup (1,2)$

Now,

${3^{2x}} - {3^{x + 1}} - {3^{x + 2}} + 27 \le 0$

$ \Rightarrow {({3^x})^2} - 3\,.\,{3^x} - {3^2}\,.\,{3^x} + 27 \le 0$

Let ${3^x} = t$

$ \Rightarrow {t^2} - 3\,.\,t - {3^2}\,.\,t + 27 \le 0$

$ \Rightarrow t(t - 3) - 9(t - 3) \le 0$

$ \Rightarrow (t - 3)(t - 9) \le 0$

$\therefore$ $3 \le t \le 9$

$ \Rightarrow {3^1} \le {3^x} \le {3^2}$

$ \Rightarrow 1 \le x \le 2$

$\therefore$ $x \in [1,2]$

$\therefore$ ${S_2} = [1,2]$

$\therefore$ ${S_1} \cup {S_2} = ( - \alpha ,2] \cup (1,2) \cup [1,2]$

$ = ( - \alpha ,2] \cup [1,2]$

Given quadratic equation,

$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$

Let, a and b are the roots of the equation,

$\therefore$ $a + b = - {{\alpha - 6} \over 3}$

and $ab = {{\alpha + 3} \over 3}$

For real roots,

$D \ge 0$

$ \Rightarrow {(\alpha - 6)^2} - 4\,.\,3\,.\,(\alpha + 9) \ge 0$

$ \Rightarrow {\alpha ^2} - 12\alpha + 36 - 12\alpha - 36 \ge 0$

$ \Rightarrow {\alpha ^2} - 24\alpha \ge 0$

$ \Rightarrow \alpha (\alpha - 24) \ge 0$

$\therefore$ $\alpha > 24$ or $\alpha < 0$

$\therefore$ Real roots of the equation possible for $\alpha > 24$ or $\alpha < 0$.

Now, sum of square of roots

$ = {a^2} + {b^2}$

$ = {(a + b)^2} - 2ab$

$ = {{{{(\alpha - 6)}^2}} \over 9} - 2\,.\,{{(\alpha + 3)} \over 3}$

$ = {{{\alpha ^2} - 12\alpha + 36 - 6\alpha - 18} \over 9}$

$ = {{{\alpha ^2} - 18\alpha + 18} \over 9} = f(x)$

$\therefore$ Sum of square of roots are minimum when ${a^2} + {b^2} = $ minimum.

$\therefore$ $f{(\alpha )_{\min }} = {{{\alpha ^2} - 18\alpha + 18} \over 9}$

Value of quadratic equation ${\alpha ^2} - 18\alpha + 18$ is minimum at $\alpha = - {b \over {2a}} = - {{( - 18)} \over {2\,.\,1}} = 9$

But for real roots $\alpha$ should be less than 0 or greater than 24.

So, there is no value of $\alpha$ in the range $\alpha > 24 \cup \alpha < 0$ where sum of squares of two real roots is minimum.

$\therefore$ S is an empty set.

Given, $\alpha$ is a root of the equation 1 + x² + x⁴ = 0

$\therefore$ $\alpha$ will satisfy the equation.

$\therefore$ 1 + $\alpha$² + $\alpha$⁴ = 0

${\alpha ^2} = {{ - 1 \pm \sqrt {1 - 4} } \over 2}$

$ = {{ - 1 \pm \sqrt 3 i} \over 2}$

$\therefore$ ${\alpha ^2} = \omega \,ar\,{\omega ^2}$

Now,

${\alpha ^{1011}} + {\alpha ^{2022}} - {\alpha ^{3033}}$

$ = \alpha \,.\,{({\alpha ^2})^{505}} + {({\alpha ^2})^{1011}} - \alpha \,.\,{({\alpha ^2})^{1516}}$

$ = \alpha {(\omega )^{505}} + {(\omega )^{1011}} - \alpha \,.\,{(\omega )^{1516}}$

$ = \alpha \,.\,{({\omega ^3})^{168}}\,.\,\omega + {({\omega ^3})^{337}} - \alpha \,.\,{({\omega ^3})^{505}}\,.\,\omega $

$ = \alpha \,\omega + 1 - \alpha \,\omega $

$ = 1$

$\because$ x = $-$1 be the roots of f(x) = 0

$\therefore$ Let $f(x) = A(x + 1)(x - 1)$ ...... (i)

Now, $f( - 2) + f(3) = 0$

$ \Rightarrow A[ - 1( - 2 - b) + 4(3 - b)] = 0$

$b = {{14} \over 3}$

$\therefore$ Second root of f(x) = 0 will be ${{14} \over 3}$

$\therefore$ Sum of roots $ = {{14} \over 3} - 1 = {{11} \over 3}$

$f(x) = {x^4} - 4x + 1 = 0$

$f'(x) = 4{x^3} - 4$

$ = 4(x - 1)({x^2} + 1 + x)$

$\Rightarrow$ Two solution

A = ($-$3, 1) and B = ($-$ $\infty$, $-$1] $\cup$ [3, $\infty$)

So, A $-$ B = ($-$1, 1)

B $-$ A = ($-$ $\infty$, $-$3] $\cup$ [3, $\infty$) = R $-$ ($-$3, 3)

A $\cap$ B = ($-$3, $-$1]

and A $\cup$ B = ($-$ $\infty$, 1) $\cup$ [3, $\infty$) = R $-$ [1, 3)

$a{x^2} - 2bx + 15 = 0$ has repeated root so ${b^2} = 15a$ and $\alpha = {{15} \over b}$

$\because$ $\alpha$ is a root of ${x^2} - 2bx + 21 = 0$

So ${{225} \over {{b^2}}} = 9 \Rightarrow {b^2} = 25$

Now ${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta = 4{b^2} - 42 = 100 - 42 = 58$

$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$

Let ${e^x} = t$

$\therefore$ $({t^2} - 4)(6{t^2} - 5t + 1) = 0$

$ \Rightarrow ({t^2} - 4)(2t - 1)(3t - 1) = 0$

$\therefore$ t = 2, $-$2, ${1 \over 2}$, ${1 \over 3}$

$\therefore$ ${e^x} = 2 \Rightarrow x = \ln 2$

${e^x} = - 2$ (not possible)

${e^x} = {1 \over 2} \Rightarrow x = - \ln 2$

${e^x} = {1 \over 3} \Rightarrow x = - \ln 3$

$\therefore$ Sum of all real roots

$ = \ln 2 - \ln 2 - \ln 3$

$ = - \ln 3$

Given equation ${x^7} - 7x - 2 = 0$

Let $f(x) = {x^7} - 7x - 2$

$f'(x) = 7{x^6} - 7 = 7({x^6} - 1)$

and $f'(x) = 0 \Rightarrow x = \, + \,1$

and $f( - 1) = - 1 + 7 - 2 = 5 > 0$

$f(1) = 1 - 7 - 2 = - 8 < 0$

So, roughly sketch of f(x) will be

So, number of real roots of f(x) = 0 and 3

$3{x^2} + \lambda x - 1 = 0$

Given, two roots are $\alpha$ and $\beta$.

$\therefore$ Sum of roots $ = \alpha + \beta = {-\lambda \over 3}$

And product of roots $ = \alpha \beta = {-1 \over 3}$

Given that,

Sum of square of reciprocal of roots $\alpha$ and $\beta$ is 15.

$\therefore$ ${1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 15$

$ \Rightarrow {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} = 15$

$ \Rightarrow {{{{(\alpha + \beta )}^2} - 2\alpha \beta } \over {{{(\alpha \beta )}^2}}} = 15$

$ \Rightarrow {{{{{\lambda ^2}} \over 9} + 2 \times {1 \over 3}} \over {{1 \over 9}}} = 15$

$ \Rightarrow {{{{{\lambda ^2} + 6} \over 9}} \over {{1 \over 9}}} = 15$

$ \Rightarrow {\lambda ^2} + 6 = 15$

$ \Rightarrow {\lambda ^2} = 9$

Now, $6{({\alpha ^3} + {\beta ^3})^2}$

$ = 6{\{ (\alpha + \beta )({\alpha ^2} + {\beta ^2} - \alpha \beta )\} ^2}$

$ = 6{(\alpha + \beta )^2}{\left[ {{{(\alpha + \beta )}^2} - 2\alpha \beta - \alpha \beta } \right]^2}$

$ = 6{\left( {{-\lambda \over 3}} \right)^2}{\left[ {{{\left( {{-\lambda \over 3}} \right)}^2} - 3\,.\,{-1 \over 3}} \right]^2}$

$ = 6 \times {{{\lambda ^2}} \over 9} \times \left[ {{{{\lambda ^2}} \over 9} + 1} \right]$

$ = 6 \times {{9} \over 9} \times {\left[ {{{9} \over 9} + 1} \right]^2}$

$ = 6 \times {\left( {{2}} \right)^2}$

$ = {{6 \times 4}} = 24$

Consider the equation x² + ax + b = 0

If has two roots (not necessarily real $\alpha$ & $\beta$)

Either $\alpha$ = $\beta$ or $\alpha$ $\ne$ $\beta$

Case (1) If $\alpha$ = $\beta$, then it is repeated root. Given that $\alpha$² $-$ 2 is also a root

So, $\alpha$ = $\alpha$² $-$ 2 $\Rightarrow$ ($\alpha$ + 1)($\alpha$ $-$ 2) = 0

$\Rightarrow$ $\alpha$ = $-$1 or $\alpha$ = 2

When $\alpha$ = $-$1 then (a, b) = (2, 1)

$\alpha$ = 2 then (a, b) = ($-$4, 4)

Case (2) If $\alpha$ $\ne$ $\beta$

Then

(I) $\alpha$ = $\alpha$² $-$ 2 and $\beta$ = $\beta$² $-$ 2

Hence, (a, b) = ($-$($\alpha$ + $\beta$), $\alpha$$\beta$)

($-$1, $-$2)

(II) $\alpha$ = $\beta$² $-$ 2 and $\beta$ = $\alpha$² $-$ 2

Then $\alpha$ $-$ $\beta$ = $\beta$² $-$ $\alpha$² = ($\beta$ $-$ $\alpha$) ($\beta$ + $\alpha$)

Since $\alpha$ $\ne$ $\beta$ we get $\alpha$ + $\beta$ = $\beta$² + $\alpha$² $-$ 4

$\alpha$ + $\beta$ = ($\alpha$ + $\beta$)² $-$ 2$\alpha$$\beta$ $-$ 4

Thus $-$1 = 1 $-$2 $\alpha$$\beta$ $-$ 4 which implies

$\alpha$$\beta$ = $-$1 Therefore (a, b) = ($-$($\alpha$ + $\beta$), $\alpha$$\beta$)

= (1, $-$1)

(III) $\alpha$ = $\alpha$² $-$ 2 = $\beta$² $-$ 2 and $\alpha$ $\ne$ $\beta$

$\Rightarrow$ $\alpha$ = $-$ $\beta$

Thus $\alpha$ = 2, $\beta$ = $-$2

$\alpha$ = $-$1, $\beta$ = 1

Therefore (a, b) = (0, $-$4) & (0, 1)

(IV) $\beta$ = $\alpha$² $-$ 2 = $\beta$² $-$ 2 and $\alpha$ $\ne$ $\beta$ is same as (III) Therefore we get 6 pairs of (a, b)

Which are (2, 1), ($-$4, 4), ($-$1, $-$2), (1, $-$1), (0, $-$4)

Option (a)

$x + 1 - 2{\log _2}(3 + {2^x}) + 2{\log _4}(10 - {2^{ - x}}) = 0$

${\log _2}({2^{x + 1}}) - {\log _2}{(3 + {2^x})^2} + {\log _2}(10 - {2^{ - x}}) = 0$

$lo{g_2}\left( {{{{2^{x + 1}}.(10 - {2^{ - x}})} \over {{{(3 + {2^x})}^2}}}} \right) = 0$

${{2({{10.2}^{ - x}} - 1)} \over {{{(3 + {2^x})}^2}}} = 1$

$ \Rightarrow {20.2^x} - 2 = 9 + {2^{2x}} + {6.2^x}$

$\therefore$ ${({2^x})^2} - 14({2^x}) + 11 = 0$

Roots are 2^x₁ & 2^x₂

$\therefore$ 2^x₁ . 2^x₂ = 11

x₁ + x₂ = log₂(11)

$\cos ec18^\circ = {1 \over {\sin 18^\circ }} = {4 \over {\sqrt 5 - 1}} = \sqrt 5 + 1$

Let $\cos ec18^\circ = x = \sqrt 5 + 1$

$ \Rightarrow x - 1 = \sqrt 5 $

Squaring both sides, we get

${x^2} - 2x + 1 = 5$

$ \Rightarrow {x^2} - 2x - 4 = 0$

${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$

Let $3{x^2} + 4x + 3 = a$

and $3{x^2} + 4x + 2 = b \Rightarrow b = a - 1$

Given equation becomes

$ \Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0$

$ \Rightarrow a(a - kb) - b(a - kb) = 0$

$ \Rightarrow (a - kb)(a - b) = 0 \Rightarrow a = kb$ or a = b (reject) $\because$ a = kb

$ \Rightarrow 3{x^2} + 4x + 3 = k(3{x^2} + 4x + 2)$

$ \Rightarrow 3(k - 1){x^2} + 4(k - 1)x + (2k - 3) = 0$ for real roots

D $\ge$ 0

$ \Rightarrow 16{(k - 1)^2} - 4(3(k - 1))(2k - 3) \ge 0$

$ \Rightarrow 4(k - 1)\{ 4(k - 1) - 3(2k - 3)\} \ge 0$

$ \Rightarrow 4(k - 1)\{ - 2k + 5\} \ge 0$

$ \Rightarrow - 4(k - 1)\{ 2k - 5\} \ge 0$

$ \Rightarrow (k - 1)(2k - 5) \le 0$



$\therefore$ $k \in \left[ {1,{5 \over 2}} \right]$

$\therefore$ $k \ne 1$

$\therefore$ $k \in \left( {1,{5 \over 2}} \right]$