Quadratic Equation and Inequalities

$\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$