1.The maximum possible area bounded by the parabola y=x2+x+10y = x^2 + x + 10y=x2+x+10 and a chord of the parabola of length 111 isa.112\frac{1}{12}121b.16\frac{1}{6}61c.13\frac{1}{3}31d.12\frac{1}{2}21Login to continueOnly logged in users canattempt or see the solution.
