Quadratic Equation and Inequalities

Given quadratic equation

${x^2} - x - 1 = 0$ having roots $\alpha $ and $\beta $, ($\alpha $ > $\beta $)

So, $\alpha = {{1 + \sqrt 5 } \over 2}$ and $\beta = {{1 - \sqrt 5 } \over 2}$

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

$ \because $ ${a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1$

So, ${a_{n + 1}} = {{{\alpha ^{n + 1}} - {\beta ^{n + 1}}} \over {\alpha - \beta }}$

${\alpha ^n} + {\alpha ^{n - 1}}\beta + {\alpha ^{n - 2}}{\beta ^2} + ... + \alpha {\beta ^{n - 1}} + {\beta ^n}$

${\alpha ^n} - {\alpha ^{n - 2}} - {\alpha ^{n - 3}}\beta - ... - {\beta ^{n - 2}} + {\beta ^n}$

[as $\alpha $$\beta $ = $ - $1]

= ${\alpha ^n} + {\beta ^n} - ({\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}})$

= ${\alpha ^n} + {\beta ^n} - {a_{n - 1}}$

$\left[ {as\,{a_{n - 1}} = {{{\alpha ^{n - 1}} - {\beta ^{n - 1}}} \over {\alpha - \beta }} = {\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}}} \right]$

$ \Rightarrow $ ${a_{n + 1}} + {a_{n - 1}} = {\alpha ^n} + {\beta ^n} = {b_n},\,\forall n \ge 1$

So, option (b) is correct.

Now, $\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} + {\beta ^n}} \over {{{10}^n}}}} $

[as, b_n = $\alpha $ⁿ + $\beta $ⁿ]

= $\sum\limits_{n = 1}^\infty {{{\left( {{\alpha \over {10}}} \right)}^n}} + \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} $

$ \because $ $\left[ {\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]$

$ = {{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} + {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}} = {\alpha \over {10 - \alpha }} + {\beta \over {10 - \beta }}$

$ = {{10\alpha - \alpha \beta + 10\beta - \alpha \beta } \over {(10 - \alpha )(10 - \beta )}}$

$ = {{10(\alpha + \beta ) - 2\alpha \beta } \over {100 - 10(\alpha + \beta ) + \alpha \beta }}$

$ = {{10(1) - 2( - 1)} \over {100 - 10(1) - 1}}$

[as $\alpha $ + $\beta $ = 1 and $\alpha $$\beta $ = $ - $1]

$ = {{12} \over {89}}$

So, option (a) is not correct.

$ \because $ ${\alpha ^2} = \alpha + 1$ and ${\beta ^2} = \beta + 1$

$ \Rightarrow $${\alpha ^{n + 2}} = {\alpha ^{n + 1}} + {\alpha ^n}$ and ${\beta ^{n + 2}} = {\beta ^{n + 1}} + {\beta ^n}$

$ \Rightarrow $ $({\alpha ^{n + 2}} + {\beta ^{n + 2}}) = ({\alpha ^{n + 1}} + {\beta ^{n + 1}}) + ({\alpha ^n} + {\beta ^n})$

$ \Rightarrow {a_{n + 2}} = {a_{n + 1}} + {a_n}$

Similarly, ${a_{n + 1}} = {a_n} + {a_{n - 1}}$

${a_n} = {a_{n - 1}} + {a_{n - 2}}$

..............

............

${a_3} = {a_2} + {a_1}$

On adding, we get

${a_{n + 2}} = ({a_n} + {a_{n - 1}} + {a_{n - 2}} + ... + {a_2} + {a_1}) + {a_2}$

$ \because $ $\left[ {{a_2} = {{{\alpha ^2} - {\beta ^2}} \over {\alpha - \beta }} = \alpha + \beta = 1} \right]$

So, ${a_{n + 2}} - 1 = {a_1} + {a_2} + {a_3} + ...... + {a_n}$

So, option (c) is also correct.

And, now $\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} - {\beta ^n}} \over {(\alpha - \beta ){{10}^n}}}} $

$ = {1 \over {\alpha - \beta }}\left[ {\sum\limits_{n = 1}^\infty {{{\left( {{a \over {10}}} \right)}^n}} - \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} } \right]$

$ = {1 \over {\alpha - \beta }}\left[ {{{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} - {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}}} \right]$,

$\left[ {as\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]$

$ = {{10(\alpha - \beta )} \over {(\alpha - \beta )[100 - 10(\alpha + \beta ) + \alpha \beta ]}}$

$ = {{10} \over {100 - 10 - 1}} = {{10} \over {89}}$

Hence, options (b), (c) and (d) are correct.

Given, x₁ and x₂ are roots of

$\alpha {x^2} - x + \alpha = 0$

$\therefore$ ${x_1} + {x_2} = {1 \over \alpha }$ and ${x_1}{x_2} = 1$

Also, $\left| {{x_1} - {x_2}} \right| < 1$

$ \Rightarrow {\left| {{x_1} - {x_2}} \right|^2} < 1 \Rightarrow {({x_1} - {x_2})^2} < 1$

or, ${({x_1} + {x_2})^2} - 4{x_1}{x_2} < 1$

$ \Rightarrow {1 \over {{\alpha ^2}}} - 4 < 1$ or ${1 \over {{\alpha ^2}}} < 5$

$ \Rightarrow 5{\alpha ^2} - 1 > 0$

or, $(\sqrt 5 \alpha - 1)(\sqrt 5 \alpha + 1) > 0$

$\therefore$ $\alpha \in \left( { - \infty , - {1 \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},\infty } \right)$ .....(i)

Also, $D > 0$

$ \Rightarrow 1 - 4{\alpha ^2} > 0$ or $\alpha \in \left( { - {1 \over 2},{1 \over 2}} \right)$ ...... (ii)

$\alpha \in \left( { - {1 \over 2},{{ - 1} \over {\sqrt 5 }}} \right) \cup \left( {{1 \over {\sqrt 5 }},{1 \over 2}} \right)$

Given, $3^x=4^{x-1}$

Taking $\log$ on both side with base 3

$\begin{aligned} & \Rightarrow \log _3 3^x=\log _3\left(4^{x-1}\right) \\ & \Rightarrow x \log _3 3=(x-1) \log _3 4 \\ & \Rightarrow \quad x .1=(x-1) \log _3 4 \\ & \Rightarrow \quad x=\frac{\log _3 4}{\log _3 4-1} \\ & \Rightarrow \quad x=\frac{1}{1-\frac{1}{\log _3 4}} \\ \end{aligned}$

$\text { According to base changing rule } \frac{1}{\log _3 4}=\log _4 3$

$\begin{array}{ll} \therefore & x=\frac{1}{1-\log _4 3} \quad \text{... (i)}\\ \Rightarrow & x=\frac{1}{1-\log _{2^2} 3} \\ \Rightarrow & x=\frac{1}{1-\frac{1}{2} \log _2 3} \quad\left[\log _{a^m} x=\frac{1}{m} \log _a x\right] \\ \Rightarrow & x=\frac{2}{2-\log _2 3} \quad \text{... (ii)} \end{array}$

$\begin{aligned} & \Rightarrow \quad x=\frac{2}{2-\frac{1}{\log _3 2}} \\ & \Rightarrow \quad x=\frac{2 \log _3 2}{2 \log _3 2-1} \quad \text{... (iii)} \end{aligned}$

From equation (i), (ii) and (iii), it is clear that option (A), (B) and (C) are correct.

Hints:

(i) Base changing Rule $\log _b^a=\frac{1}{\log _a^b}$

(ii) $\log _a x^n=n \log _a x$

(iii) $\log _{a^m} x=\frac{1}{m} \log _a x$