11sorry & thanks aditya.
please check it
i have to determine a parabola passing through (1,1)
it's eq should be (x-1)2=a(y-1)
now it also pass through (11,181)
so 100=a.180
so 5=9a
so the eq is 9(x-1)2=5(y-1)
\hspace{-16}$If $\bf{p(x)}$ is a Quadratic Polynomial with Real Coeff. satisfying\\\\ $\bf{x^2-2x+2\leq p(x)\leq 2x^2-4x+3\;\forall x\in \mathbb{R}}$ and $\bf{p(11)=181}$\\\\ Then $\bf{p(2012)=}$
-
UP 0 DOWN 0 0 3
3 Answers
1111x2-2x+2=2x2-4x+3 at x=1
so, p(x)must pass through the points-(1,1) &(11,181)
we get p(x)=x2+6x-6
262@sougata - you are missing something .
p(x) = ax2 +bx+c
three unknowns and two equations . you cant solve .
Hint : check for derivative for the polynomials at x=1
