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MathMedAdv 2016 • Paper 1

Roots of Quadratic Equations with Trigonometric Coefficients

Assume −π6<θ<−π12-\frac{\pi}{6} < \theta < -\frac{\pi}{12}. Let α1\alpha_1 and β1\beta_1 represent the roots of x2−2xsec⁡θ+1=0x^2 - 2x \sec \theta + 1 = 0, while α2\alpha_2 and β2\beta_2 are the roots of x2+2xtan⁡θ−1=0x^2 + 2x \tan \theta - 1 = 0. Given that α1>β1\alpha_1 > \beta_1 and α2>β2\alpha_2 > \beta_2, evaluate the sum α1+β2\alpha_1 + \beta_2.

Options

A

2(sec⁡θ−tan⁡θ)2(\sec \theta - \tan \theta)

B

2sec⁡θ2\sec \theta

C

−2tan⁡θ-2\tan \theta

Correct
D

00

Topics & Concepts

Complex Numbers and Quadratic EquationsQuadratic EquationsLocation of RootsTrigonometric Ratios and Identities

Step-by-Step Solution

To find the value of α1+β2\alpha_1 + \beta_2, we need to determine the explicit values of the roots α1\alpha_1 and β2\beta_2 for the given quadratic equations under the constraint −π6<θ<−π12-\frac{\pi}{6} < \theta < -\frac{\pi}{12}.

Step 1: Solve the First Quadratic Equation

The first equation is given by: x2−2xsec⁡θ+1=0x^2 - 2x \sec \theta + 1 = 0

Using the quadratic formula, the roots are: x=2sec⁡θ±4sec⁡2θ−42=sec⁡θ±sec⁡2θ−1=sec⁡θ±∣tan⁡θ∣x = \frac{2\sec\theta \pm \sqrt{4\sec^2\theta - 4}}{2} = \sec\theta \pm \sqrt{\sec^2\theta - 1} = \sec\theta \pm |\tan\theta|

Since θ∈(−π6,−π12)\theta \in \left(-\frac{\pi}{6}, -\frac{\pi}{12}\right), θ\theta lies in the fourth quadrant. Therefore:

  • sec⁡θ>0\sec\theta > 0
  • tan⁡θ<0  ⟹  ∣tan⁡θ∣=−tan⁡θ\tan\theta < 0 \implies |\tan\theta| = -\tan\theta

Substituting ∣tan⁡θ∣=−tan⁡θ|\tan\theta| = -\tan\theta, the two roots are: x=sec⁡θ−tan⁡θandx=sec⁡θ+tan⁡θx = \sec\theta - \tan\theta \quad \text{and} \quad x = \sec\theta + \tan\theta

Since tan⁡θ<0\tan\theta < 0, we have −tan⁡θ>0-\tan\theta > 0, which implies: sec⁡θ−tan⁡θ>sec⁡θ+tan⁡θ\sec\theta - \tan\theta > \sec\theta + \tan\theta

Given that α1>β1\alpha_1 > \beta_1, we identify: α1=sec⁡θ−tan⁡θ\alpha_1 = \sec\theta - \tan\theta β1=sec⁡θ+tan⁡θ\beta_1 = \sec\theta + \tan\theta

Step 2: Solve the Second Quadratic Equation

The second equation is given by: x2+2xtan⁡θ−1=0x^2 + 2x \tan \theta - 1 = 0

Using the quadratic formula, the roots are: x=−2tan⁡θ±4tan⁡2θ+42=−tan⁡θ±tan⁡2θ+1=−tan⁡θ±sec⁡θx = \frac{-2\tan\theta \pm \sqrt{4\tan^2\theta + 4}}{2} = -\tan\theta \pm \sqrt{\tan^2\theta + 1} = -\tan\theta \pm \sec\theta

Since sec⁡θ>0\sec\theta > 0 in the fourth quadrant, the two roots are: x=−tan⁡θ+sec⁡θandx=−tan⁡θ−sec⁡θx = -\tan\theta + \sec\theta \quad \text{and} \quad x = -\tan\theta - \sec\theta

Comparing these two values, since sec⁡θ>0\sec\theta > 0: −tan⁡θ+sec⁡θ>−tan⁡θ−sec⁡θ-\tan\theta + \sec\theta > -\tan\theta - \sec\theta

Given that α2>β2\alpha_2 > \beta_2, we identify: α2=−tan⁡θ+sec⁡θ\alpha_2 = -\tan\theta + \sec\theta β2=−tan⁡θ−sec⁡θ\beta_2 = -\tan\theta - \sec\theta

Step 3: Compute α1+β2\alpha_1 + \beta_2

Now, we calculate the sum of α1\alpha_1 and β2\beta_2: α1+β2=(sec⁡θ−tan⁡θ)+(−tan⁡θ−sec⁡θ)\alpha_1 + \beta_2 = (\sec\theta - \tan\theta) + (-\tan\theta - \sec\theta)

Simplifying the terms: α1+β2=sec⁡θ−sec⁡θ−tan⁡θ−tan⁡θ=−2tan⁡θ\alpha_1 + \beta_2 = \sec\theta - \sec\theta - \tan\theta - \tan\theta = -2\tan\theta

Conclusion

The value of α1+β2\alpha_1 + \beta_2 is equal to −2tan⁡θ-2\tan\theta, which corresponds to Option (C).

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