SATHEE: Sequences and Series 2 Question 18

Sequences and Series 2 Question 18

19. The real numbers $x_{1}, x_{2}, x_{3}$ satisfying the equation $x^{3} - x^{2} + β x + γ = 0$ are in AP. Find the intervals in which $β$ and $γ$ lie.

(1996, 3M)

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Answer:

Correct Answer: 19. (0)

Solution:

Since, $x_{1}, x_{2}, x_{3}$ are in an AP. Let $x_{1} = a - d, x_{2} = a$ and $x_{3} = a + d$ and $x_{1}, x_{2}, x_{3}$ be the roots of $x^{3} - x^{2} + β x + γ = 0$

$\begin{array}{lll} ∴ & Σ α & = a - d + a + a + d = 1 \\ \Rightarrow & a & = 1 / 3 \\ Σ α β = (a - d) & a + a (a + d) + (a - d) (a + d) = β \\ and α β γ & = (a - d) a (a + d) = - γ \end{array}$

From Eq. (i),

$3 a = 1 \Rightarrow a = 1 / 3$

From Eq. (ii), $3 a^{2} - d^{2} = β$

$\begin{array}{lcc} \Rightarrow & 3 (1 / 3)^{2} - d^{2} = β & [from Eq. (i)] \\ \Rightarrow & 1 / 3 - β = d^{2} \end{array}$

NOTE In this equation, we have two variables $β$ and $y$ but we have only one equation. So, at first sight it looks that this equation cannot solve but we know that $d^{2} \geq 0, \forall d \in R$ , then $β$ can be solved. This trick is frequently asked in IIT examples.

$\begin{aligned} \Rightarrow \frac{1}{3} - β \geq 0 [∵ d^{2} \geq 0] \\ \Rightarrow β \leq \frac{1}{3} \Rightarrow β \in [- \infty, 1 / 3] \end{aligned}$

From Eq. (iii), $a (a^{2} - d^{2}) = - γ$

$\begin{array}{lc} \Rightarrow & \frac{1}{3} \frac{1}{9} - d^{2} = - γ \Rightarrow \frac{1}{27} - \frac{1}{3} d^{2} = - γ \\ \Rightarrow & γ + \frac{1}{27} = \frac{1}{3} d^{2} \Rightarrow γ + \frac{1}{27} \geq 0 \\ \Rightarrow & γ \in - 1 / 27 \\ \Rightarrow & γ \in - \frac{1}{27}, \infty \end{array}$