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Q. Let $f ( x )= ax ^2+ bx + c ( a < b )$ and $f ( x ) \geq 0 \forall x \in R$. Find the minimum value of $\frac{ a + b + c }{ b - a }$.

Relations and Functions - Part 2

Solution:

$ f (1)= a + b + c $
$f(-2)=4 a-2 b+c $
$\text { hence } f (1)- f (-2)=3( b - a ) $
$E=\frac{a+b+c}{b-a}=\frac{3 f(1)}{f(1)-f(-2)}=\frac{3}{1-\frac{f(-2)}{f(1)}} $
$\text { Hence } E_{\min .} \text { occurs when } f(-2)=0$
$\text { Hence } E_{\min .}=3$