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

Hint: 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 \[{\text{Sum of roots = }}-\dfrac{{{\text{coefficient of }}{x^2}}}{{{\text{coefficient of }}{x^3}}}\].