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Question: Real number $x_1, x_2, x_3$ in A.P. satisfying the equation $x^3 - x^2 + \alpha x + \beta = 0$ then ...

Real number x1,x2,x3x_1, x_2, x_3 in A.P. satisfying the equation x3−x2+αx+β=0x^3 - x^2 + \alpha x + \beta = 0 then select correct alternative(s)

A

α≤13\alpha \leq \frac{1}{3}

B

β≥−127\beta \geq -\frac{1}{27}

C

α>13\alpha > \frac{1}{3}

D

β<−127\beta < -\frac{1}{27}

Answer

α≤13,β≥−127\alpha \leq \frac{1}{3}, \beta \geq -\frac{1}{27}

Explanation

Solution

Let the roots of the equation x3−x2+αx+β=0x^3 - x^2 + \alpha x + \beta = 0 be x1,x2,x3x_1, x_2, x_3. Since x1,x2,x3x_1, x_2, x_3 are in A.P. and are real numbers, let them be a−d,a,a+da-d, a, a+d where a,d∈Ra, d \in \mathbb{R}.

From Vieta's formulas:

  1. Sum of roots: (a−d)+a+(a+d)=−(−1)/1(a-d) + a + (a+d) = -(-1)/1 3a=1  ⟹  a=133a = 1 \implies a = \frac{1}{3}. So, the roots are 13−d,13,13+d\frac{1}{3}-d, \frac{1}{3}, \frac{1}{3}+d.

  2. Since x=13x = \frac{1}{3} is a root, it must satisfy the equation: (13)3−(13)2+α(13)+β=0(\frac{1}{3})^3 - (\frac{1}{3})^2 + \alpha(\frac{1}{3}) + \beta = 0 127−19+α3+β=0\frac{1}{27} - \frac{1}{9} + \frac{\alpha}{3} + \beta = 0 −227+α3+β=0(∗)-\frac{2}{27} + \frac{\alpha}{3} + \beta = 0 \quad (*)

  3. Sum of products of roots taken two at a time: (13−d)(13)+(13)(13+d)+(13−d)(13+d)=α(\frac{1}{3}-d)(\frac{1}{3}) + (\frac{1}{3})(\frac{1}{3}+d) + (\frac{1}{3}-d)(\frac{1}{3}+d) = \alpha (19−d3)+(19+d3)+(19−d2)=α(\frac{1}{9} - \frac{d}{3}) + (\frac{1}{9} + \frac{d}{3}) + (\frac{1}{9} - d^2) = \alpha 39−d2=α  ⟹  13−d2=α\frac{3}{9} - d^2 = \alpha \implies \frac{1}{3} - d^2 = \alpha. Since dd is a real number, d2≥0d^2 \ge 0. Therefore, α=13−d2≤13\alpha = \frac{1}{3} - d^2 \le \frac{1}{3}.

  4. Product of roots: (13−d)(13)(13+d)=−β(\frac{1}{3}-d)(\frac{1}{3})(\frac{1}{3}+d) = -\beta 13(19−d2)=−β\frac{1}{3}(\frac{1}{9} - d^2) = -\beta. Substitute d2=13−αd^2 = \frac{1}{3} - \alpha (from step 3) into this equation: 13(19−(13−α))=−β\frac{1}{3}(\frac{1}{9} - (\frac{1}{3} - \alpha)) = -\beta 13(19−13+α)=−β\frac{1}{3}(\frac{1}{9} - \frac{1}{3} + \alpha) = -\beta 13(−29+α)=−β\frac{1}{3}(-\frac{2}{9} + \alpha) = -\beta −227+α3=−β  ⟹  β=227−α3-\frac{2}{27} + \frac{\alpha}{3} = -\beta \implies \beta = \frac{2}{27} - \frac{\alpha}{3}. This is consistent with equation (∗)(*).

    To find the range of β\beta using d2d^2: From 13(19−d2)=−β\frac{1}{3}(\frac{1}{9} - d^2) = -\beta, we have β=−127+d23\beta = -\frac{1}{27} + \frac{d^2}{3}. Since d2≥0d^2 \ge 0, it follows that d23≥0\frac{d^2}{3} \ge 0. Therefore, β=−127+d23≥−127\beta = -\frac{1}{27} + \frac{d^2}{3} \ge -\frac{1}{27}.

Based on the derived conditions:

  • α≤13\alpha \le \frac{1}{3} is correct.
  • β≥−127\beta \ge -\frac{1}{27} is correct.