Test of Mathematics Solution Subjective 76 - Range of a Rational Polynomial
This is a Test of Mathematics Solution Subjective 76 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
Problem
Find the set of all values of $ {m}$ such that $ {\displaystyle {y} = {\frac{x^2-x}{1-mx}}}$ can take all real values.
Solution
$ {\displaystyle {y} = {\frac{x^2-x}{1-mx}}}$
$ {\Leftrightarrow}$ $ {\displaystyle{y - myx = x^2 - x}}$
$ {\Leftrightarrow}$ $ {x^2 + x(my - 1) - y = 0}$
$ {\Leftrightarrow}$ $ {\displaystyle{x} = {\frac{1 - my {\pm} {\sqrt{(my-1)^2+4y}}}{2}}}$
Now $ {y}$ takes all real values if discriminant $ {(my-1)^2 + 4y}$ is allways $ {> 0}$.
So now we have to find the all values of $ {m}$ such that $ {(my-1)^2 + 4y}$ $ {> 0}$ for all $ {y}$ $ {\in}$ |R.
$ {\Leftrightarrow}$ $ {m^2 y^2 - 2my + 1 + 4y > 0}$
$ {\Leftrightarrow}$ $ {m^2 y^2 + y(4 - 2m) + 1 > 0}$ ...(i)
Now this is a equation of upside open parabola. If the discriminant is $ {\le{0}}$ of equation (i) then $ {(my - 1)^2 + 4y}$ will always positive.
$ {\Rightarrow}$ $ {\displaystyle{(4 - 2m)^2 - 4m^2} {\le {0}}}$
$ {\Rightarrow}$ $ {{-16m + 16} {\le {0}}}$
$ {\Rightarrow}$ $ {-m + 1 {\le{0}}}$
$ {\Rightarrow}$ $ {1 {\le{m}}}$
Conclusion:
If $ {1 {\le{m}}}$, then $ {\displaystyle{y = {\frac{x^2 - x}{1 - mx}}}}$ can take all the values as $ {x}$ varies over |R.
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 \(tan19x=\frac{cos96+sin96}{cos96-sin96}\).
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 \( f'(c)=\lambda f(c) \).
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
\(f_1(11)=4\)
or, \(f_2(11)=f_1(4)=16\)
or, \(f_3(11)=f_1(16)=49\)
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, \(f_7(11)=f_1(169)=256\)
This goes on between two numbers with this pattern, here 1988 is even,
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
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
putting x=(a-b) and y=(a+b) gives 2a=1 and \(a=\frac{1}{2}\)
and \((\frac{1}{2})^{2} +b^{2}=8^{2}\) then \(b^{2}=\frac{255}{4}\) then 255+4=259.
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
by symmetry with both x=2 and x=7 where x=0 is a root, x=4 and x=14 are also roots
here 0(mod 10) or 4(mod10) are roots there are 201 roots as multiples of 10 and 200 roots as for 4(mod10)
