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Q. The inequalities $y(-1) \geq-4, y(1) \leq 0$ and $y(3) \geq 5$ are known to hold for $y=a x^2+b x+c$ then the least value of ' $a$ ' is :

Complex Numbers and Quadratic Equations

Solution:

$a-b+c \geq-4$ .....(i)
and a + b + c $\leq 0 \Rightarrow-a-b-c \geq 0$....(ii)
and $9 a +3 b + c \geq 5 $.... (iii)
(i) + (ii) $\Rightarrow-2 b \geq-4$
(iv) ; (ii) + (iii) + (iv) $\Rightarrow 8 a \geq 1 \Rightarrow a \geq 1 / 8$