UPTET Paper 2 Math and Science 2018 Solved Paper

Concept:For a quadratic expression $ax^{2} + bx + c$ with $a > 0$, the minimum value is given by $\frac{4ac - b^{2}}{4a}$.Explanation:First, expand the product: $(x - 2)(x - 9) = x^{2} - 11x + 18$.Compare this with $ax^{2} + bx + c$. Here, $a = 1$, $b = -11$, and $c = 18$.Since $a > 0$, the parabola opens upward and has a minimum.Apply the formula: minimum value $= \frac{4(1)(18) - (-11)^{2}}{4(1)}$.Calculate: $4 \times 18 = 72$, $(-11)^{2} = 121$, so numerator $= 72 - 121 = -49$.Denominator is $4$. Thus, minimum value $= \frac{-49}{4}$.This value occurs at $x = -\frac{b}{2a} = \frac{11}{2}$. No need to compute here.Answer:The minimum value is $-\frac{49}{4}$, which corresponds to option D.