Correct option is (B) \(\frac{4ac-b^2}{4a}\)
\(ax^2+bx+c\) \(=a(x^2+\frac bax+\frac ca)\)
\(=a\left(x^2+2\times x\times\frac b{2a}+(\frac b{2a})^2+\frac ca-(\frac b{2a})^2\right)\)
\(=a\left((x+\frac b{2a})^2+\frac ca-\frac{b^2}{4a^2}\right)\)
\(=a\left((x+\frac b{2a})^2+\frac{4ac-b^2}{4a^2}\right)\)
\(\because\) \((x+\frac b{2a})^2\geq0\)
\(\therefore\) \(ax^2+bx+c\) \(\geq a(\frac{4ac-b^2}{4a^2})\)
\(\Rightarrow\) \(ax^2+bx+c\) \(\geq\frac{4ac-b^2}{4a}\)
Hence, least value of \(ax^2+bx+c\) is \(\frac{4ac-b^2}{4a}.\)
