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 .
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 .