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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 solving polynomials using calculus . Let’s give it a try !!

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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