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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  has no real root so discriminant is  so  and ‘s are positive and decreasing so  . 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.

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