Sequences and Series 2 Question 18
19. The real numbers $x _1, x _2, x _3$ satisfying the equation $x^{3}-x^{2}+\beta x+\gamma=0$ are in AP. Find the intervals in which $\beta$ and $\gamma$ 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 $x^{3}-x^{2}+\beta x+\gamma=0$
$$ \begin{array}{lll} \therefore & \Sigma \alpha & =a-d+a+a+d=1 \\ \Rightarrow & a & =1 / 3 \\ & \Sigma \alpha \beta=(a-d) & a+a(a+d)+(a-d)(a+d)=\beta \\ & \text { and } \quad \alpha \beta \gamma & =(a-d) a(a+d)=-\gamma \end{array} $$
From Eq. (i),
$$ 3 a=1 \Rightarrow a=1 / 3 $$
From Eq. (ii), $3 a^{2}-d^{2}=\beta$
$$ \begin{array}{lcc} \Rightarrow & 3(1 / 3)^{2}-d^{2}=\beta & \text { [from Eq. (i)] } \\ \Rightarrow & 1 / 3-\beta=d^{2} \end{array} $$
NOTE In this equation, we have two variables $\beta$ 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 $\beta$ can be solved. This trick is frequently asked in IIT examples.
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{3}-\beta \geq 0 \quad\left[\because d^{2} \geq 0\right] \\ & \Rightarrow \quad \beta \leq \frac{1}{3} \Rightarrow \beta \in[-\infty, 1 / 3] \end{aligned} $$
From Eq. (iii), $\quad a\left(a^{2}-d^{2}\right)=-\gamma$
$$ \begin{array}{lc} \Rightarrow & \frac{1}{3} \frac{1}{9}-d^{2}=-\gamma \Rightarrow \frac{1}{27}-\frac{1}{3} d^{2}=-\gamma \\ \Rightarrow & \gamma+\frac{1}{27}=\frac{1}{3} d^{2} \Rightarrow \gamma+\frac{1}{27} \geq 0 \\ \Rightarrow & \gamma \in-1 / 27 \\ \Rightarrow & \gamma \in-\frac{1}{27}, \infty \end{array} $$
Hence, $\beta \in(-\infty, 1 / 3]$ and $\gamma \in[-1 / 27, \infty)$
