JEE-Advanced 2015 (29/40)

Algebra Level 3

Let S S be the set of all non-zero real numbers α \alpha such that the quadratic equation α x 2 − x + α = 0 \alpha x^2-x+\alpha=0 has two distinct real roots x 1 x_1 and x 2 x_2 satisfying the inequality ∣ x 1 − x 2 ∣ < 1 |x_1-x_2|<1 . Which of the following intervals is/are a subset of S ? S?

( 1 ) ( − 1 2 , − 1 5 ) ( 2 ) ( − 1 5 , 0 ) ( 3 ) ( 0 , 1 5 ) ( 4 ) ( 1 5 , 1 2 ) \begin{array}{l} (1) \, \left( -\frac{1}{2}, -\frac{1}{\sqrt5} \right) \quad \quad \quad \quad & (2) \, \left( -\frac{1}{\sqrt5},0 \right) \\\\ (3) \, \left( 0, \frac{1}{\sqrt5} \right) & (4) \, \left( \frac{1}{\sqrt5}, \frac{1}{2} \right) \end{array}
Note:

  • Submit your answer as the increasing order of the serial numbers of all the correct options.
  • For example, if your answer is ( 1 ) , ( 2 ) , (1), (2), then submit 12 as the correct answer; if your answer is ( 2 ) , ( 3 ) , ( 4 ) , (2),(3),(4), then submit 234 as the correct answer.


The answer is 14.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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2 solutions

Tom Engelsman
May 3, 2020

Let the roots of the above quadratic equation equal:

x 1 = 1 + 1 − 4 α 2 2 α , x 2 = 1 − 1 − 4 α 2 2 α x_{1} = \frac{1 + \sqrt{1-4\alpha^{2}}}{2\alpha}, x_{2} =\frac{1 - \sqrt{1-4\alpha^{2}}}{2\alpha}

such that ∣ x 1 − x 2 ∣ < 1 ⇒ ∣ 1 − 4 α 2 α ∣ < 1 ⇒ 1 − 4 α 2 α 2 < 1 ⇒ α > 1 5 |x_{1} - x_{2}| < 1 \Rightarrow | \frac{\sqrt{1-4\alpha^{2}}}{\alpha} | < 1 \Rightarrow \frac{1-4\alpha^{2}}{\alpha^{2}} < 1 \Rightarrow \alpha > \frac{1}{\sqrt{5}} or α < − 1 5 \alpha < -\frac{1}{\sqrt{5}} .

If the roots are real & distinct, then require the discriminant 1 − 4 α 2 1-4\alpha^{2} to be positive valued. This occurs iff 1 − 4 α 2 > 0 ⇒ − 1 2 < α < 1 2 . 1-4\alpha^{2} > 0 \Rightarrow -\frac{1}{2} < \alpha < \frac{1}{2}. Hence, the set S S equals: S = [ α ∈ R ∣ α ∈ ( − 1 2 , − 1 5 ) ∪ ( 1 5 , 1 2 ) ] \boxed{ S = [\alpha \in \mathbb{R} | \alpha \in (-\frac{1}{2}, -\frac{1}{\sqrt{5}}) \cup (\frac{1}{\sqrt{5}}, \frac{1}{2}) ]}

which only choices (1) and (4) are correct.

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Arnab Mukherjee
Aug 17, 2015

Thanks for the questions. I could answer all but five

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