ISI MStat 2018 PSA Problem 12 | Sequence of positive numbers

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 $<img decoding=$ "> has no real root so discriminant is $<img decoding=$ "> so $<img decoding=$ "> and $<img decoding=$ ">'s are positive and decreasing so $<img decoding=$ "> . 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.

Similar Problems and Solutions

ISI MStat 2018 PSA Problem 12 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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 $<img decoding=$ "> has no real root so discriminant is $<img decoding=$ "> so $<img decoding=$ "> and $<img decoding=$ ">'s are positive and decreasing so $<img decoding=$ "> . 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.

Similar Problems and Solutions

ISI MStat 2018 PSA Problem 12 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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