Roots and coefficients of equations | PRMO 2017 | Question 4

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Try this beautiful problem from the PRMO, 2017 based on Roots and coefficients of equations.

Roots and coefficients of equations - PRMO 2017

Let a,b be integers such that all the roots of the equation $(x^{2}+ax+20)(x^{2}+17x+b)$=0 are negetive integers, find the smallest possible values of a+b.

is 107
is 25
is 840
cannot be determined from the given information

Key Concepts

Polynomials

Roots

Coefficients

Check the Answer

But try the problem first...

Answer: is 25.

Source

Suggested Reading

PRMO, 2017, Question 4

Polynomials by Barbeau

Try with Hints

First hint

$(x^{2}+ax+20)(x^{2}+17x+b)$

where sum of roots $ \lt $ 0 and product $ \gt 0$ for each quadratic equation $x^{2}$+ax+20=0 and

$(x^{2}+17x+b)=0$

$a \gt 0$, $b \gt 0$

now using vieta's formula on each quadratic equation $x^{2}$+ax+20=0 and $(x^{2}+17x+b)=0$, to get possible roots of $x^{2}$+ax+20=0 from product of roots equation $20=(1 \times 20), (2 \times 10), (4 \times 5)$

min a=4+5=9 from all sum of roots possible

Second Hint

again using vieta's formula, to get possible roots of $(x^{2}$+17x+b)=0 from sum of roots equation $17=-(\alpha + \beta) \Rightarrow (\alpha,\beta)=(-1,-16),(-2,-15),$

$(-8,-9)$

min b=(-1)(-16)=16 from all products of roots possible

Final Step

$(a+b)_{min}=a_{min}+b_{min}$=9+16=25.

ISI MStat 2018 PSA Problem 12 | Sequence of positive numbers

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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.

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ISI MStat 2016 Problem 1 | Area bounded by the curves | PSB Sample

This is a beautiful problem from ISI MStat 2016 Problem 1, PSB Sample, based on area bounded by the curves. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem- ISI MStat 2016 Problem 1

In the diagram below, $L(x)$ is a straight line that intersects the graph of a polynomial $P(x)$ of degree 2 at the points $A=(-1,0)$ and $B=(5,12) .$ The area of the shaded region is 36 square units. Obtain the expression for $P(x)$.

Prerequisites

Area bounded by the curve
Polynomials of degree 2
Area of a triangle

Solution

Let $P(x)=ax^2 +bx + c $ as given $P(x)$ is of degree 2 .

Now from the figure we can see that $L(x)$ intersect $P(x)$ at points $A=(-1,0)$ and $B=(5,12) .$

Hence we have $P(-1)=0$ and $P(5)=12$ , which gives ,

$ a-b+c=0 $ ---(1) and $ 25a+5b +c =12 $ ----(2)

Then ,

Fig-1

See from Fig-1 we can say that Area of the shaded region = (Area bounded by the curve P(x) and x-axis )- (Area of the triangle ABC) - (Area bounded by the curve P(x) , x=5 and x=L )

= $ \int^{L}_{-1} P(x)\,dx - \frac{1}{2} \times (5+1) \times 12 -\int^{5}_{L} P(x)\,dx $

=$\int^{L}_{-1} P(x)\,dx - \int^{5}_{L} P(x)\,dx $ -36

=$ \int^{5}_{-1} P(x)\,dx $ -36

Again it is given that area of the shaded region is 36 square units.

So, $ \int^{5}_{-1} P(x)\,dx $ -36 =36 $ \Rightarrow $ $ \int^{5}_{-1} P(x)\,dx $ =$ 2 \times 36 $

$ \int^{5}_{-1} (ax^2+bx+c) \,dx = 2 \times 36 $ . After integration we get ,

$ 7a + 2b +c =12 $ ---(3)

Now we have three equations and three unknows

$ a-b+c=0 $

$ 25a+5b +c =12 $

$ 7a + 2b +c =12 $

Solving this three equations by elimination and substitution we get ,

$ a=-1 , b=6 , c=7 $

Therefore , the expression for $P(x)$ is $ P(x)= -x^2+6x+7 $ .

Previous MStat Posts:

Probability of tossing a coin | AIME I, 2009 | Question 3

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Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.

Probability of tossing a coin - AIME I, 2009 Question 3

A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac{1}{25}$ the probability of five heads and three tails. Let p=$\frac{m}{n}$ where m and n are relatively prime positive integers. Find m+n.

Key Concepts

Probability

Theory of equations

Polynomials

Check the Answer

But try the problem first...

Answer: 11.

Source

Suggested Reading

AIME, 2009

Course in Probability Theory by Kai Lai Chung .

Try with Hints

First hint

here $\frac{8!}{3!5!}p^{3}(1-p)^{5}$=$\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}$

Second hint

then $(1-p)^{2}$=$\frac{1}{25}p^{2}$ then 1-p=$\frac{1}{5}p$

Final Step

then p=$\frac{5}{6}$ then m+n=11

Complex Numbers | AIME I, 2009 | Problem 2

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Try this beautiful problem from AIME, 2009 based on complex numbers.

Complex Numbers - AIME, 2009

There is a complex number z with imaginary part 164 and a positive integer n such that $\frac{z}{z+n}=4i$, Find n.

Key Concepts

Complex Numbers

Theory of equations

Polynomials

Check the Answer

But try the problem first...

Answer: 697.

Source

Suggested Reading

AIME, 2009, Problem 2

Complex Numbers from A to Z by Titu Andreescue .

Try with Hints

First hint

Taking z=a+bi

Second hint

then a+bi=(z+n)4i=-4b+4i(a+n),gives a=-4b b=4(a+n)=4(n-4b)

Final Step

then n=$\frac{b}{4}+4b=\frac{164}{4}+4.164=697$

Combinations | AIME I, 2009 |Problem 9

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Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations

Combinations- AIME, 2009

A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to $9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.

Key Concepts

Combinations

Theory of equations

Polynomials

Check the Answer

But try the problem first...

Answer: 420.

Source

Suggested Reading

AIME I, 2009, Problem 9

Combinatorics by Brualdi .

Try with Hints

First hint

Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35

Second hint

these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.

Final Step

then total number of guesses is 35.12=420

Functions and Equations |Pre-RMO, 2019

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Try this beautiful problem from Pre-RMO, 2019 based on Functions and Equations.

Functions and Equations - PRMO, 2019

Let f(x) = $x^{2}+ax+b$, if for all non zero real x, f(x+$\frac{1}{x})$=f(x)+f($\frac{1}{x}$) and the roots of f(x)=0 are integers, find the value of $a^{2}+b^{2}$.

Key Concepts

Functions

Algebra

Polynomials

Check the Answer

But try the problem first...

Answer: 13.

Source

Suggested Reading

Pre-RMO, 2019

Functional Equation by Venkatchala .

Try with Hints

First hint

f(x+$\frac{1}{x})$=f(x)+f($\frac{1}{x}$)

Second hint

then $(x+\frac{1}{x})^{2}+a(x+\frac{1}{x})+b$=$x^{2}+ax+b+\frac{1}{x^{2}}+\frac{a}{x}+b$ then b=2, product of roots is 2 then roots are (1,2),(-1,-2) and a=3or-3

Final Step

then $a^{2}+b^{2}$=4+9=13