Question

The equation $8{x^3} - a{x^2} + bx - 1 = 0$ has three real roots in G.P. If ${\lambda _1} \leqslant a \leqslant {\lambda _2}$, then ordered pair $\left( {{\lambda _1},{\lambda _2}} \right)$ can be

1. $\left( { - 2,2} \right)$
2. $\left( {18,12} \right)$
3. $\left( { - 10, - 8} \right)$
4. None of these

Answer 1

Here it is sufficient to find the range of $a$for the required answer. Consider the three real roots in G.P. and then use ${\text{Product of roots = }}-\dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^3}}}$ and ${\text{Sum of roots = }}-\dfrac{{{\text{coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}$ to find the range of $a$. Then, compare the given options with the range of $a$ to find the correct answer.

Complete step-by-step answer:
It is given that these roots are in G.P. Let the common difference of this G.P. be $r$.
Let us consider the roots of the equation be $\dfrac{k}{r}$, $r$ and $kr$ for simplicity.
For a third-degree equation with real roots, it is known that the product of the real roots is equal to the $-\dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^3}}}$.
Here the constant term is $- 1$ , the real roots are $\dfrac{k}{r}$, $r$ and $kr$ and coefficient of ${x^3}$ is 8.
Therefore
$\dfrac{k}{r} \times r \times kr = - \dfrac{{\left( { - 1} \right)}}{8}$
${k^3} = \dfrac{1}{8}$
$k = \dfrac{1}{2}$
Thus the three roots become
$\dfrac{1}{{2r}}$, $\dfrac{1}{2}$ and $\dfrac{r}{2}$.
Also, it is known that the sum of the real roots of the third-degree polynomial is equal to the $-\dfrac{{{\text{coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}$.
Here the coefficient of ${x^2}$ is $- a$, the real roots are $\dfrac{1}{{2r}}$, $\dfrac{1}{2}$,$\dfrac{r}{2}$ and coefficient of ${x^3}$ is 8.
Therefore ,
$\dfrac{1}{{2r}} + \dfrac{1}{2} + \dfrac{r}{2} = - \dfrac{{\left( { - a} \right)}}{8}$
Multiplying the equation throughout with 8 we get
$\dfrac{4}{r} + 4 + 4r = a \\\ a = 4 + 4\left( {\dfrac{1}{r} + r} \right) \\\$
Here the range of a can be defined by defining the scope of the function $\dfrac{1}{r} + r$.
For $r > 0$, the function $\dfrac{1}{r} + r$ ranges from 2 to $\infty$ with its minimum value, 2 at $r = 1$.
And the range of $a$for $r > 0$is therefore defined by
$a = 4 + 4\left( {\dfrac{1}{r} + r} \right)$
For $r > 0$,
Min value of $a = 4 + 4\left( 2 \right)$
$a = 12$
For $r > 0$, $a \in \left( {12,\infty } \right)$
For $r < 0$, the function $\dfrac{1}{r} + r$ ranges from $- \infty$ to –2 with its minimum value (–2) occuring at $r = - 1$.
And the range of $a$ for $r < 0$ is therefore defined by
$a = 4 + 4\left( {\dfrac{1}{r} + r} \right)$
For $r < 0$,
Min value of $a = 4 + 4\left( { - 2} \right)$
$a = - 4$
For $r < 0$, $a \in \left( { - \infty , - 4} \right)$.
On combining the range,
$a \in \left( { - \infty , - 4} \right) \cup \left( {12,\infty } \right)$
Comparing the range of with the given options, we can see that the option 3 matches the range.
Hence, option C is the correct answer.

Note: While taking the roots in G.P., choose numbers such as $\dfrac{k}{r}$, $r$ and $kr$ to avoid tricky calculations. Formulate the equations using the formulas ${\text{Product of roots = }}-\dfrac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^3}}}$ and ${\text{Sum of roots = }}-\dfrac{{{\text{coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}$.