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Question: If $x_1, x_2$ and $x_3$ are the positive roots of the equation $x^3 - 6x^2 + 3px - 2p = 0, p \in R$ ...

If x1,x2x_1, x_2 and x3x_3 are the positive roots of the equation x3−6x2+3px−2p=0,p∈Rx^3 - 6x^2 + 3px - 2p = 0, p \in R then the value of sin⁡−1(1x1+1x2)+cos⁡−1(1x2+1x3)−tan⁡−1(1x3+1x1)\sin^{-1}(\frac{1}{x_1} + \frac{1}{x_2}) + \cos^{-1}(\frac{1}{x_2} + \frac{1}{x_3}) - \tan^{-1}(\frac{1}{x_3} + \frac{1}{x_1}) is greater than or equal to

A

π6\frac{\pi}{6}

B

π4\frac{\pi}{4}

C

π3\frac{\pi}{3}

D

π2\frac{\pi}{2}

Answer

π4\frac{\pi}{4}

Explanation

Solution

The given cubic equation is x3−6x2+3px−2p=0x^3 - 6x^2 + 3px - 2p = 0. Let x1,x2,x3x_1, x_2, x_3 be the positive roots of this equation.

From Vieta's formulas:

  1. x1+x2+x3=6x_1 + x_2 + x_3 = 6
  2. x1x2+x2x3+x3x1=3px_1x_2 + x_2x_3 + x_3x_1 = 3p
  3. x1x2x3=2px_1x_2x_3 = 2p

Since x1,x2,x3x_1, x_2, x_3 are positive, x1x2x3=2p>0x_1x_2x_3 = 2p > 0, thus, p>0p > 0.

Now, consider the sum of the reciprocals of the roots: 1x1+1x2+1x3=x2x3+x1x3+x1x2x1x2x3=3p2p=32\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{x_2x_3 + x_1x_3 + x_1x_2}{x_1x_2x_3} = \frac{3p}{2p} = \frac{3}{2}.

Let y1=1x1y_1 = \frac{1}{x_1}, y2=1x2y_2 = \frac{1}{x_2}, y3=1x3y_3 = \frac{1}{x_3}. Then y1,y2,y3y_1, y_2, y_3 are positive, and y1+y2+y3=32y_1+y_2+y_3 = \frac{3}{2}.

The expression to evaluate is E=sin⁡−1(1x1+1x2)+cos⁡−1(1x2+1x3)−tan⁡−1(1x3+1x1)E = \sin^{-1}(\frac{1}{x_1} + \frac{1}{x_2}) + \cos^{-1}(\frac{1}{x_2} + \frac{1}{x_3}) - \tan^{-1}(\frac{1}{x_3} + \frac{1}{x_1}).

Using y1+y2+y3=32y_1+y_2+y_3 = \frac{3}{2}: E=sin⁡−1(32−y3)+cos⁡−1(32−y1)−tan⁡−1(32−y2)E = \sin^{-1}(\frac{3}{2} - y_3) + \cos^{-1}(\frac{3}{2} - y_1) - \tan^{-1}(\frac{3}{2} - y_2).

For the cubic equation to have three real roots, its discriminant must be non-negative. The discriminant Δ\Delta is given by: Δ=−108p3+864p2−1728p≥0\Delta = -108p^3 + 864p^2 - 1728p \ge 0. This simplifies to p(p−4)2≤0p(p-4)^2 \le 0.

Since p>0p > 0 and (p−4)2≥0(p-4)^2 \ge 0, the inequality can only hold if (p−4)2=0(p-4)^2 = 0. This implies p=4p=4.

Substitute p=4p=4 into the original equation: x3−6x2+12x−8=0x^3 - 6x^2 + 12x - 8 = 0, which simplifies to (x−2)3=0(x-2)^3 = 0. So, x1=x2=x3=2x_1=x_2=x_3=2.

Since x1=x2=x3=2x_1=x_2=x_3=2, we have y1=y2=y3=12y_1 = y_2 = y_3 = \frac{1}{2}.

Now substitute these values back into the expression for EE: E=sin⁡−1(12+12)+cos⁡−1(12+12)−tan⁡−1(12+12)=sin⁡−1(1)+cos⁡−1(1)−tan⁡−1(1)=π2+0−π4=π4E = \sin^{-1}(\frac{1}{2}+\frac{1}{2}) + \cos^{-1}(\frac{1}{2}+\frac{1}{2}) - \tan^{-1}(\frac{1}{2}+\frac{1}{2}) = \sin^{-1}(1) + \cos^{-1}(1) - \tan^{-1}(1) = \frac{\pi}{2} + 0 - \frac{\pi}{4} = \frac{\pi}{4}.

Thus, the value of the expression is π4\frac{\pi}{4}.