# 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

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 $$17=-(\alpha + \beta) \Rightarrow (\alpha,\beta)=(-1,-16),(-2,-15),$$

$$(-8,-9)$$

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

Discriminant

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.

# 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 ,

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 $$P(x)= -x^2+6x+7$$ .

# 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.

• 10
• 20
• 30
• 11

### Key Concepts

Probability

Theory of equations

Polynomials

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.

• 101
• 201
• 301
• 697

### Key Concepts

Complex Numbers

Theory of equations

Polynomials

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.

• 110
• 420
• 430
• 111

### Key Concepts

Combinations

Theory of equations

Polynomials

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 $a^{2}+b^{2}$.

• 10
• 20
• 30
• 13

### Key Concepts

Functions

Algebra

Polynomials

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