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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


But try the problem first…

Answer: is 169.

Source
Suggested Reading

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

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

Check the Answer


But try the problem first…

Answer: is 560.

Source
Suggested Reading

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.

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Algebra Arithmetic Functions I.S.I. and C.M.I. Entrance ISI Entrance Videos

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

Check the Answer


But try the problem first…

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

Source
Suggested Reading

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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Algebra Arithmetic Functional Equations Functions I.S.I. and C.M.I. Entrance ISI Entrance Videos

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

Check the Answer


But try the problem first…

Answer:2

Source
Suggested Reading

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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Algebra Arithmetic Functions I.S.I. and C.M.I. Entrance ISI Entrance Videos

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

Check the Answer


But try the problem first…

Answer:at least 2n

Source
Suggested Reading

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.

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Algebra Arithmetic Functional Equations Functions I.S.I. and C.M.I. Entrance ISI Entrance Videos

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

Check the Answer


But try the problem first…

Answer: 18.

Source
Suggested Reading

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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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

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

Check the Answer


But try the problem first…

Answer: is 502.

Source
Suggested Reading

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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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


But try the problem first…

Answer: is 561.

Source
Suggested Reading

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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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


But try the problem first…

Answer: is 259.

Source
Suggested Reading

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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