UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The product of the first, fourth, and fifth terms of an AP is expressed as a function of the common difference $d$.Maximizing this product requires finding the value of $d$ that makes the first derivative zero and checking the second derivative for a maximum.Explanation:Given: sixth term of the AP is 2.Let the first term be $a$ and common difference $d$.Then $a + 5d = 2$ $\Rightarrow$ $a = 2 - 5d$.The product $P$ of first, fourth, and fifth terms:$P = a \cdot (a + 3d) \cdot (a + 4d)$.Substitute $a = 2 - 5d$:$P = (2-5d)(2-5d+3d)(2-5d+4d) = (2-5d)(2-2d)(2-d)$.Simplify: $P = 2(-5d^3 + 17d^2 - 16d + 4)$.Set $f(d) = 2(-5d^3 + 17d^2 - 16d + 4)$.First derivative: $f'(d) = 2(-15d^2 + 34d - 16)$.Second derivative: $f''(d) = 2(-30d + 34)$.For maxima/minima, $f'(d) = 0$:$2(-15d^2 + 34d - 16) = 0$ $\Rightarrow$ $15d^2 - 34d + 16 = 0$.Factor: $(5d - 8)(3d - 2) = 0$ $\Rightarrow$ $d = \frac{8}{5}$ or $d = \frac{2}{3}$.Check second derivative:At $d = \frac{8}{5}$: $f''\left(\frac{8}{5}\right) = 2\left(-30 \times \frac{8}{5} + 34\right) = 2(-48 + 34) = -28 < 0$ $\Rightarrow$ maximum.At $d = \frac{2}{3}$: $f''\left(\frac{2}{3}\right) = 2\left(-30 \times \frac{2}{3} + 34\right) = 2(-20 + 34) = 28 > 0$ $\Rightarrow$ minimum.Thus, the product is greatest when $d = \frac{8}{5}$.Answer:A. $\frac{8}{5}$