Question

answer....it....fast.....

Answer 1

$E=y^2-12y+20$
$\implies E=y^2-(10+2)y+20$
$\implies E=y^2-10y-2y+20$
$\implies E=y(y-10)-2(y-10)$
$\implies E=(y-2)(y-10)$
[This was not required though,but I want to convey that Factorization is not necessary everywhere]
Option (a) is correct which says "E≤0 when 2 ≤y≤10 and E>0 when y <2 or y>10"
You can arrive at this conclusion by trial and error method.
Case 1: let y=4 (under 2≤y≤10)
$\therefore E=(4)^2-12×4+20$
$E=16-48+20$
$E=36-48$
$E=(-12)$
$E<0$
Case 2: let y=1 (under y<2)
$\therefore E=(1)^2-12×1+20$
$E=1-12+20$
$E=9\implies E>0$
Case 3: let y=11 (under y>10)
$\therefore E=(11)^2-11×12+20$
$\implies E=121-132+20$
$\implies E=(-11)+20=9 \implies E>0$
Hence, We arrive at option A.