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