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

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

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## Arranging in column | AIME I, 1990 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Arranging in column.

## Arranging in column – AIME I, 1990

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marks man is to break all the targets according to the following rules

1 ) The marksman first chooses a column from which a target is to be broken,

2 ) the marksman must then break the lowest remaining target in the chosen column. If the rules are followed, in how many different orders can the eight targets be broken?

• is 107
• is 560
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Arrangement

Algebra

AIME I, 1990, Question 8

Combinatorics by Brualdi

## Try with Hints

First hint

Let the columns be labelled A,B and C such that first three targets are A, A and A the next three being B, B and B and the next being C and C in which we consider the string AAABBBCC.

Second Hint

Since the arrangement of the strings is one-one correspondence and onto to the order of shooting for example first A is shot first, second A is shot second, third A is shot third, first B is shot fourth, second B is shot fifth, third B is shot sixth, first C is shot seventh, second C is shot eighth,

or, here arrangement of the strings is bijective to the order of the shots taken

Final Step

the required answer is the number of ways to arrange the letters which is $\frac{8!}{3!3!2!}$=560.

Categories

## Problem on Function | TOMATO BStat Objective 720

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Function.

## Problem on Function (B.Stat Objective Question )

Consider the function f(x)=$tan^{-1}(2tan(\frac{x}{2}))$, where $\frac{-\pi}{2} \leq f(x) \leq \frac{\pi}{2}$ Then

• $\lim\limits_{x \to \pi-0}f(x)=\frac{\pi}{2}$, $\lim\limits_{x \to \pi+0}f(x)=\frac{-\pi}{2}$
• $\lim\limits_{x \to \pi}f(x)=\frac{\pi}{2}$
• $\lim\limits_{x \to \pi-0}f(x)=\frac{-\pi}{2}$, $\lim\limits_{x \to \pi+0}f(x)=\frac{\pi}{2}$
• $\lim\limits_{x \to \pi}f(x)=\frac{-\pi}{2}$

### Key Concepts

Equation

Roots

Algebra

Answer:$\lim\limits_{x \to \pi}f(x)=\frac{\pi}{2}$

B.Stat Objective Problem 720

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

f(x)=$tan^{-1}(2tan{\frac{x}{2}})$

Second Hint

$\lim\limits_{x \to \pi}f(x)$

$=\lim\limits_{x \to \pi}tan^{-1}(2tan{\frac{x}{2}})=\frac{\pi}{2}$

$\lim\limits_{x \to \pi-0}f(x)$

$=\lim\limits_{x \to \pi-0}tan^{-1}(2tan{\frac{x}{2}})=\frac{\pi}{2}$

Final Step

$\lim\limits_{x \to \pi+0}f(x)$

$=\lim\limits_{x \to \pi+0}tan^{-1}(2tan{\frac{x}{2}})=\frac{\pi}{2}$

So $\lim\limits_{x \to \pi}f(x)=\frac{\pi}{2}$

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## Roots of Equation | TOMATO B.Stat Objective 711

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Roots of Equation.

## Roots of Equations (B.Stat Objective Question )

The number of roots of the equation $x^2+sin^2{x}-1$ in the closed interval $[0,\frac{\pi}{2}]$ is

• 0
• 2
• 53361
• 5082

### Key Concepts

Equation

Roots

Algebra

B.Stat Objective Problem 711

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

$x^2+sin^2{x}-1=0$

$\Rightarrow x^{2}=cos^{2}x$

we draw two graphs $y=x^{2} and y=cos^{2}x$

where intersecting point gives solution now we look for intersecting points

Second Hint

we get two intersecting points

Final Step

so number of roots is 2.

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## Periodic Function | TOMATO B.Stat Objective 710

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Periodic Function.

## Periodic Function (B.Stat Objective Question )

If f(x) = $a_0+a_1cosx+a_2cos2x+….+a_ncosnx$ where $a_0,a_1,….,a_n$ are non zero real numbers and $a_n > |a_0|+|a_1|+….+|a_{n-1}|$, then number of roots of f(x)=0 in $0 \leq x \leq 2\pi$, is

• 0
• at least 2n
• 53361
• 5082

### Key Concepts

Periodic

Real Numbers

Inequality

B.Stat Objective Problem 710

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

f is periodic with period $2\pi$

here $0 <|\displaystyle\sum_{k=0}^{n-1}a_kcos(kx)| \leq \displaystyle\sum_{k=0}^{n-1}|a_k|<a_n, x\in$set of reals

Second Hint

for points $x_k=\frac{k\pi}{n}$ $1 \leq k \leq 2n$

$f(x_k)=a_n[cos(k\pi)+\theta], |\theta|<1 for 1 \leq k \leq 2n$

[ since cos $\theta$ is periodic and f(x) is expressed for every point x=$x_k$]

Final Step

f has at least 2n points in such a period interval where f has alternating sign.

Categories

## Equations and Roots | TOMATO B.Stat Objective 123

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Equations and Roots.

## Equation and Roots ( B.Stat Objective Question )

Consider the equation of the form $x^{2}+bx+c=0$. The number of such equations that have real roots and have coefficients b and c in the set {1,2,3,4,5,6}, (b may be equal to c), is

• 1113
• 18
• 53361
• 5082

### Key Concepts

Equation

Integers

Roots

B.Stat Objective Problem 123

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

We know that if a quadratic equations have real roots then it’s discriminant is >=0 so here $b^{2}-4c \geq 0$. Now we will go casewise . First we will choose a particular Value for b then check what are the values of c that satisfies the above inequality.

Second Hint

for b=2, c=1

for b=3, c=1,2

for b=4, c=1,2,3,4

for b=5, c=1,2,3,4,5

for b=6, c=1,2,3,4,5,6

Final Step

we get required number =18.

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## Digits and Order | AIME I, 1992 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Digits and Order.

## Digits and order – AIME I, 1992

A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. Find number of ascending positive integers are there.

• is 107
• is 502
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Digits

Order

AIME I, 1992, Question 2

Elementary Number Theory by David Burton

## Try with Hints

First hint

There are nine digits that we use 1,2,3,4,5,6,7,8,9.

Second Hint

Here each digit may or may not be present.

$\Rightarrow 2^{9}$=512 potential ascending numbers, one for subset of {1,2,3,4,5,6,7,8,9}

Final Step

Subtracting empty set and single digit set

=512-10

=502.

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## 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.

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## 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.