JEE Challenger
More from Complex Numbers and Quadratic Equations
MathMedAdv 2015 • Paper 2

Find Subsets of Alpha Values for Quadratic Equation Roots Difference

Let SS denote the set containing all non-zero real parameters α\alpha such that the quadratic equation αx2−x+α=0\alpha x^2 - x + \alpha = 0 yields two distinct real solutions x1x_1 and x2x_2 fulfilling the condition ∣x1−x2∣<1|x_1 - x_2| < 1. Which among the following intervals represents a subset of SS?

Options

A

(−12,−15)\left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right)

Correct
B

(−15,0)\left(-\frac{1}{\sqrt{5}}, 0\right)

C

(0,15)\left(0, \frac{1}{\sqrt{5}}\right)

D

(15,12)\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)

Correct

Topics & Concepts

Complex Numbers and Quadratic EquationsQuadratic EquationsLocation of Roots

Step-by-Step Solution

To find the set SS of all non-zero real parameters α\alpha for which the given quadratic equation has two distinct real roots satisfying the given condition, we analyze the constraints step-by-step.

The given quadratic equation is: αx2−x+α=0,where α≠0\alpha x^2 - x + \alpha = 0, \quad \text{where } \alpha \neq 0

Step 1: Condition for two distinct real roots

For the quadratic equation to have two distinct real roots x1x_1 and x2x_2, its discriminant DD must be strictly positive: D=(−1)2−4(α)(α)>0D = (-1)^2 - 4(\alpha)(\alpha) > 0 1−4α2>01 - 4\alpha^2 > 0 4α2<1  ⟹  α2<144\alpha^2 < 1 \implies \alpha^2 < \frac{1}{4}

Since α\alpha is a non-zero real number: α∈(−12,0)∪(0,12)— (1)\alpha \in \left(-\frac{1}{2}, 0\right) \cup \left(0, \frac{1}{2}\right) \quad \text{--- (1)}

Step 2: Condition for ∣x1−x2∣<1|x_1 - x_2| < 1

The sum and product of the roots are given by Vieta's formulas: x1+x2=1αandx1x2=1x_1 + x_2 = \frac{1}{\alpha} \quad \text{and} \quad x_1 x_2 = 1

The magnitude of the difference between the roots is expressed as: ∣x1−x2∣=(x1+x2)2−4x1x2=1α2−4=1−4α2∣α∣|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1 x_2} = \sqrt{\frac{1}{\alpha^2} - 4} = \frac{\sqrt{1 - 4\alpha^2}}{|\alpha|}

We are given that ∣x1−x2∣<1|x_1 - x_2| < 1: 1−4α2∣α∣<1\frac{\sqrt{1 - 4\alpha^2}}{|\alpha|} < 1

Since both sides are positive, we can square both sides: 1−4α2α2<1\frac{1 - 4\alpha^2}{\alpha^2} < 1 1−4α2<α21 - 4\alpha^2 < \alpha^2 5α2>1  ⟹  α2>15— (2)5\alpha^2 > 1 \implies \alpha^2 > \frac{1}{5} \quad \text{--- (2)}

Step 3: Determining the Set SS

Combining conditions (1) and (2): 15<α2<14\frac{1}{5} < \alpha^2 < \frac{1}{4}

Taking the square root gives: α∈(−12,−15)∪(15,12)\alpha \in \left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right) \cup \left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)

Thus, the set SS is: S=(−12,−15)∪(15,12)S = \left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right) \cup \left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)

Conclusion

Comparing this set SS with the given options:

  • Option (A): (−12,−15)\left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right) is a subset of SS.
  • Option (D): (15,12)\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right) is a subset of SS.

Thus, the correct options are A and D.

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