Roots and coefficients of equations | PRMO 2017 | Question 4

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


Answer: is 25.

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.

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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 $P_n$ 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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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 \(P(x)\).

ISI MStat 2016 Problem 1 figure

Prerequisites

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 ,

ISI MStat 2016 Problem 1 graph
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

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

Check the Answer


Answer: 11.

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

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Complex Numbers | AIME I, 2009 | Problem 2

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

Check the Answer


Answer: 697.

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

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Combinations | AIME I, 2009 |Problem 9

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

Check the Answer


Answer: 420.

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

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Functions and Equations |Pre-RMO, 2019

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

Check the Answer


Answer: 13.

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

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