Question

The equation for $8{x^3} - a{x^2} + bx - 1 = 0$ 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
A.$\left( { - 2,2} \right)$
B.$(24,29)$
C.$( - 10, - 8)$
D.None of these

Answer 1

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

Complete step-by-step answer:
It is given these roots are in G.P. let the common difference of this G.P. be $r$
Let us consider the roots of the equation to be $\dfrac{k}{r}$ $k$ 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 $- \dfrac{{cons\tan term}}{{coefficient{\text{ of }}{{\text{x}}^3}}}$
Here the constant term is -1, the real roots are $\dfrac{k}{r}$, $k$ , $kr$ and coefficient of ${x^3}$ is $8$
Therefore
$\ \dfrac{k}{r} \times k \times kr = - \dfrac{{ - 1}}{8} \\\ \Rightarrow {k^3} = \dfrac{1}{8} \\\ \Rightarrow 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
$- \dfrac{{{\text{coefficient of }}{{\text{x}}^2}}}{{coefficient{\text{ of }}{{\text{x}}^3}}}$
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,
$\dfrac{1}{{2r}} + \dfrac{1}{2} + \dfrac{r}{2} = - \dfrac{{( - a)}}{8}$
Multiplying the equation throughout with 8 we get
$\ \Rightarrow \dfrac{4}{r} + 4 + 4r = a \\\ \Rightarrow a = 4 + 4(\dfrac{1}{r} + r) \\\ \$
Here the range of a can is defined by defining the scope of the function $\dfrac{1}{r} + r$.
For $r \succ 0$
Min value of $\ \Rightarrow a = 4 + 4(2) \\\ \Rightarrow a = 12 \\\ \$
For $r \succ 0,a \in (12,\infty )$
For $r \prec 0,$the function $\dfrac{1}{r} + r$ ranges from $- \infty$ to $- 2$with its maximum value $( - 2)$occurring at $r = - 1$.
And the range of a for $r \prec 0$ is therefore defined by
$a = 4 + 4(-2)$
For $r \prec 0,a \in ( - \infty , - 4)$
On combining the range,
$a \in ( - \infty , - 4)\infty \cup (12,\infty )$
Comparing the range of with the given option,
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 $\dfrac{k}{r},r,kr$ to avoid tricky calculations. Formulate the equation using the formulas
Product of roots $= - \dfrac{{{\text{constant term}}}}{{coefficient{\text{ of }}{{\text{x}}^3}}}$
Sum of roots $= - \dfrac{{coefficient{\text{ of }}{{\text{x}}^2}}}{{{\text{coefficient of }}{{\text{x}}^3}}}$.