This is a problem from ISI MStat 2018 PSA Problem 12 based on Sequence of positive numbers

Sequence of positive numbers – ISI MStat Year 2018 PSA Question 12


Let \(a_n \) ,\( n \ge 1\) be a sequence of positive numbers such that \(a_{n+1} \leq a_{n}\) for all n, and \(\lim {n \rightarrow \infty} a{n}=a .\) Let \(p_{n}(x)\) be the polynomial \( p_{n}(x)=x^{2}+a_{n} x+1\) and suppose \(p_{n}(x)\) has no real roots for every n . Let \(\alpha\) and \(\beta\) be the roots of the polynomial \(p(x)=x^{2}+a x+1 .\) What can you say about \( (\alpha, \beta) \)?

  • (A) \( \alpha=\beta, \alpha\) and \(\beta\) are not real
  • (B) \( \alpha=\beta, \alpha\) and \(\beta\) are real.
  • (C) \(\alpha \neq \beta, \alpha\) and \(\beta\) are real.
  • (D) \(\alpha \neq \beta, \alpha\) and \(\beta\) are not real

Key Concepts


Sequence

Quadratic equation

Discriminant

Check the Answer


But try the problem first…

Answer: is (D)

Source
Suggested Reading

ISI MStat 2018 PSA Problem 12

Introduction to Real Analysis by Bertle Sherbert

Try with Hints


First hint

Write the discriminant. Use the properties of the sequence \( a_n \) .

Second Hint

Note that as $P_n$ has no real root so discriminant is $(a_n)^2-4<0$ so $|a_n|<2$ and $a_n$‘s are positive and decreasing so $0\leq a_n<2$ . So , what can we say about a ?

Final Step

Therefore we can say that \( 0 \le a < 2 \) hence discriminant of P  , \(a^2-4 \) must be strictly negative so option D.

ISI MStat 2018 PSA Problem 12
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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