UPSC CDS 2 2019 Math Paper

Concept:The maximum value of $\frac{1}{\text{quadratic}}$ occurs when the quadratic's value is minimum, provided the quadratic is always positive.Explanation:The denominator is $x^2 + 5x + 10$.It is a quadratic with $a = 1 > 0$, so it opens upward and has a minimum.The minimum value of $ax^2 + bx + c$ is $\frac{4ac - b^2}{4a}$.Substitute $a = 1$, $b = 5$, $c = 10$:Minimum $= \frac{4(1)(10) - (5)^2}{4(1)} = \frac{40 - 25}{4} = \frac{15}{4}$.Since the denominator has a minimum $\frac{15}{4}$, the reciprocal $\frac{1}{x^2 + 5x + 10}$ has a maximum at the same $x$ value.Maximum of the expression $= \frac{1}{\frac{15}{4}} = \frac{4}{15}$.Answer:$\frac{4}{15}$, which corresponds to option D.