#### MATH-TRICK-14

**Basic knowledge of Algebra(Polynomials):-**

If we are given that f'(x)=constant then the polynomial is of the form f(x)=ax+b .

If we are given that f''(x)=constant then the polynomial is of the form f(x)=ax²+bx+c .

**Problem:- Let f(0)=1, Lt{f"(x)}=4, where x->∞ and f(x)≥**

**f(1) . Let f(x) is a polynomial for all x in R. Find f(2) .**

__Sol:-__

**Since f"(x)=4, then f(x)=2x²+bx+c .**

So, here f(0)=1 implies c=1 .

Now, f(1)=3+b.

Since f(x)≥ f(1), so f'(1)=0 implying 4+b=0. So, b=-4 .

Therefore, f(x)=2x²-4x+1

Hence f(2)=1 .