Categories

## 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

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]

$\Rightarrow x=159$.

Categories

## 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

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.

Categories

## 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 $f(2)$ .

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 $f'(c)=\lambda f(c)$.

Categories

## 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

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, $f_1988(11)=f_4(11)=169$.

Categories

## 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

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.

Categories

## 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

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=$\frac{60\pi}{3}$ $\Rightarrow n=60$.

Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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).

Categories

## 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

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