ISI MStat PSB 2012 Problem 2 | Dealing with Polynomials using Calculus
Join Trial or Access Free ResourcesJoin Trial or Access Free ResourcesThis is a very beautiful sample problem from ISI MStat PSB 2012 Problem 2 based on calculus . Let's give it a try !!
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
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 \( f(2)=1 \) .
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 \( f'(c)=\lambda f(c) \).
