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ISI MStat 2018 PSA Problem 12 | Sequence of positive numbers

This is a beautiful problem from ISI MSAT 2018 PSA problem 12 based on Sequence of positive numbers. We provide sequential hints so that you can try .

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


Answer: is (D)

ISI MStat 2018 PSA Problem 12

Introduction to Real Analysis by Bertle Sherbert

Try with Hints


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

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 ?

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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