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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

Finding smallest positive Integer – AIME I, 1996


Find the smallest positive integer solution to \(tan19x=\frac{cos96+sin96}{cos96-sin96}\).

  • is 107
  • is 159
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Trigonometry

Integers

Check the Answer


Answer: is 159.

AIME I, 1996, Question 10

Plane Trigonometry by Loney

Try with Hints


First hint

\(\frac{cos96+sin96}{cos96-sin96}\)

=\(\frac{sin(90+96)+sin96}{sin(90+96)-sin96}\)

=\(\frac{sin186+sin96}{sin186-sin96}\)

=\(\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}\)

=\(\frac{2sin141cos45}{2cos141sin45}\)

=tan141

Second Hint

here \(tan(180+\theta)\)=\(tan\theta\)

\(\Rightarrow 19x=141+180n\) for some integer n is first equation

Final Step

multiplying equation with 19 gives

\(x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)\) [since 2679 divided by 180 gives remainder 159]

\(\Rightarrow x=159\).

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AIME I Algebra Arithmetic Functional Equations Math Olympiad USA Math Olympiad

Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

Roots of Equation and Vieta’s formula – AIME I, 1996


Suppose that the roots of \(x^{3}+3x^{2}+4x-11=0\) are a,b and c and that the roots of \(x^{3}+rx^{2}+sx+t=0\) are a+b,b+c and c+a, find t.

  • is 107
  • is 23
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Roots of Equation

Vieta s formula

Check the Answer


Answer: is 23.

AIME I, 1996, Question 5

Polynomials by Barbeau

Try with Hints


With Vieta s formula

\(f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0\)

\(\Rightarrow a+b+c=-3\), \(ab+bc+ca=4\) and \(abc=11\)

Let a+b+c=-3=p

here t=-(a+b)(b+c)(c+a)

\(\Rightarrow t=-(p-c)(p-a)(p-b)\)

\(\Rightarrow t=-f(p)=-f(-3)\)

\(t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]\)

=23.

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Calculus Functions I.S.I. and C.M.I. Entrance IIT JAM Statistics ISI M.Stat PSB ISI MSAT Statistics

ISI MStat PSB 2012 Problem 2 | Dealing with Polynomials using Calculus

This is a very beautiful sample problem from ISI MStat PSB 2012 Problem 2 based on calculus . Let’s give it a try !!

Problem– ISI MStat PSB 2012 Problem 2


Let \(f\) be a polynomial. Assume that \( f(0)=1, \lim _{x \rightarrow \infty} f”(x)=4\) and \( f(x) \geq f(1) \) for all \( x \in \mathbb{R} .\) Find \( f(2)\) .

Prerequisites


Limit

Derivative

Polynomials

Solution :

Here given \(f(x) \) is a polynomial and \( \lim _{x \rightarrow \infty} f”(x)=4\)

So, Case 1: If f(x) is a polynomial of degree 1 then f”(x)=0 hence limit can’t be 4.

Case 2: If f(x) is a polynomial of degree 2 ,say \( f(x) = ax^2+bx+c \) then \( f”(x)= 2a \) .Hence taking limit we get \( 2a=4 \Rightarrow a=2 \)

Case 3: If f(x) is a polynomial of degree >2 then \( f”(x) = O(x) \) . So, it tends to infinity or – infinity as x tends to infinity .

Therefore the only case that satisfies the condition is Case 2 .

So , f(x) = \( 2x^2+bx+c \) ,say . Now given that \( f(0)=1 \Rightarrow c=1 \) .

Again , it is given that \( f(x) \geq f(1) \) for all \( x \in \mathbb{R} \) which implies that f(x) has minimum at x=1 .

That is f'(x)=0 at x=1 . Here we have \( f'(x)=4x+b=0 \Rightarrow x=\frac{-b}{4}=1 \Rightarrow b=-4 \)

Thus we get \( f(x)=2x^2-4x+1 \) . Putting x=2 , we get \( f(2)=1 \) .


Food For Thought

Assume f is differentiable on \( (a, b)\) and is continuous on \( [a, b]\) with \( f(a)=f(b)=0\). Prove that for every real \( \lambda\) there is some c in \( (a, b)\) such that \( f'(c)=\lambda f(c) \).


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Function Problem | AIME I, 1988 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on function.

Function Problem – AIME I, 1988


For any positive integer k, let \(f_1(k)\) denote the square of the sum of the digits of k. For \(n \geq 2\), let \(f_n(k)=f_1(f_{n-1}(k))\), find \(f_{1988}(11)\).

  • is 107
  • is 169
  • is 634
  • cannot be determined from the given information

Key Concepts


Functions

Equations

Algebra

Check the Answer


Answer: is 169.

AIME I, 1988, Question 2

Functional Equation by Venkatchala

Try with Hints


First hint

\(f_1(11)=4\)

or, \(f_2(11)=f_1(4)=16\)

or, \(f_3(11)=f_1(16)=49\)

Second Hint

or, \(f_4(11)=f_1(49)=169\)

or, \(f_5(11)=f_1(169)=256\)

or, \(f_6(11)=f_1(256)=169\)

or, \(f_7(11)=f_1(169)=256\)

Final Step

This goes on between two numbers with this pattern, here 1988 is even,

or, \(f_1988(11)=f_4(11)=169\).

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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Remainders and Functions | AIME I, 1994 | Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Remainders and Functions.

Remainders and Functions – AIME I, 1994


The function f has the property that, for each real number x, \(f(x)+f(x-1)=x^{2}\) if f(19)=94, find the remainder when f(94) is divided by 1000.

  • is 107
  • is 561
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Remainder

Functions

Check the Answer


Answer: is 561.

AIME I, 1994, Question 7

Elementary Number Theory by David Burton

Try with Hints


First hint

f(94)=\(94^{2}-f(93)=94^{2}-93^{2}+f(92)\)

=\(94^{2}-93^{2}+92^{2}-f(91)\)

Second Hint

=\((94^{2}-93^{2})+(92^{2}-91^{2})\)

\(+….+(22^{2}-21^{2})+20^{2}-f(19)\)

Final Step

=94+93+…..+21+400-94

=4561

\(\Rightarrow\) remainder =561.

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Algebra Arithmetic Math Olympiad PRMO

Trigonometry & natural numbers | PRMO 2017 | Question 11

Try this beautiful problem from the Pre-RMO, 2017 based on Trigonometry & natural numbers.

Trigonometry & natural numbers – PRMO 2017


Let f(x) =\(sin\frac{x}{3}+cos\frac{3x}{10}\) for all real x, find the least natural number x such that \(f(n\pi+x)=f(x)\) for all real x.

  • is 107
  • is 60
  • is 840
  • cannot be determined from the given information

Key Concepts


Trigonometry

Least natural number

Functions

Check the Answer


Answer: is 60.

PRMO, 2017, Question 11

Plane Trigonometry by Loney

Try with Hints


First hint

here f(x) =\(sin\frac{x}{3}+cos\frac{3x}{10}\)

Second Hint

period of\(sin\frac{x}{3}\) is \(6\pi\)

period of \(cos\frac{3x}{10}\) is \(\frac{20\pi}{3}\)

Final Step

Lcm=\(\frac{60\pi}{3}\) \(\Rightarrow n=60\).

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AIME I Algebra Complex Numbers Functions Math Olympiad USA Math Olympiad

Function of Complex numbers | AIME I, 1999 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Function of Complex Numbers and Integers.

Function of Complex Numbers – AIME I, 1999


Let f(z) =(a+bi)z where a,b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin given that |a+bi|=8 and that \(b^{2}\)=\(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 259
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Integers

Complex Numbers

Check the Answer


Answer: is 259.

AIME I, 1999, Question 9

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

Let z=1+i f(1+i)=(a+bi)(1+i)=(a-b)+(a+b)i The image point must be equidistant from (1,1) and(0,0) then the image point lie on the line with slope -1 and which passes through \((\frac{1}{2},\frac{1}{2})\) that is x+y=1

Second Hint

putting x=(a-b) and y=(a+b) gives 2a=1 and \(a=\frac{1}{2}\)

Final Step

and \((\frac{1}{2})^{2} +b^{2}=8^{2}\) then \(b^{2}=\frac{255}{4}\) then 255+4=259.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Function and symmetry | AIME I, 1984 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1984 based on Function and symmetry.

Function and Symmetry – AIME I 1984


A function f is defined for all real numbers and satisfies f(2+x)=f(2-x) and f(7+x)=f(7-x) for all x. If x=0 is root for f(x)=0, find the least number of roots f(x) =0 must have in the interval \(-1000 \leq x\leq 1000\).

  • is 107
  • is 401
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Symmetry

Number Theory

Check the Answer


Answer: is 401.

AIME I, 1984, Question 12

Elementary Number Theory by David Burton

Try with Hints


First hint

by symmetry with both x=2 and x=7 where x=0 is a root, x=4 and x=14 are also roots

Second Hint

here 0(mod 10) or 4(mod10) are roots there are 201 roots as multiples of 10 and 200 roots as for 4(mod10)

Final Step

Then least number of roots as 401.

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Math Olympiad Singapore Math Olympiad

Functional Equation Problem | SMO, 2013 – Problem19 (Senior Section)

Try this beautiful problem from Singapore Mathematics Olympiad based on Functional Equation.

Problem – Functional Equation (SMO Exam)


Let f and g be functions such that for all real numbers x and y,

\( g (f (x+y)) = f( x ) + (x+y) g (y)\).

Find the value of \( g(0) + g (1) + ……………………+ g (2013) \)

  • 1
  • 3
  • 2
  • 0

Key Concepts


Functional Equation

Funcion

Arbitrary Numbers

Check the Answer


Answer: 0

Singapore Mathemaics Olympiad

Challenges and thrills

Try with Hints


We can start this problem by considering y = -x.

Then \( g (f (0) ) = f (x) \) for all x. This \(f\) is is a constant function ; namely

\( f (x) = c \) for some c.

Try the rest of the sum ……………………………………………………

For all value of x,y we have

\( (x+y) g(y) = g(f(x+y)) – f(x) = g(c) – c = 0 \)

Since x + y is arbitrary , we must have \( g (y) = 0 \) for all y .Hence

\( g (0) + g ( 1 ) + …………………………..+ g(2013) = 0 \) (Answer).

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Functions India Math Olympiad Math Olympiad PRMO

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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