PUNE MCA Exam 2016 Question Paper with answers for online practice

Concept:The sum of cubes $\sum_{r=1}^{10} (x-r)^3$ simplifies to a cubic polynomial; its derivative is always positive, so the cubic is strictly increasing, ensuring exactly one real root.Explanation:Let $S(x) = \sum_{r=1}^{10} (x-r)^3$.Using formulas $\sum_{r=1}^{10} r = 55$, $\sum_{r=1}^{10} r^2 = 385$, $\sum_{r=1}^{10} r^3 = (55)^2 = 3025$, we expand:$S(x) = 10x^3 - 3 \cdot 55 x^2 + 3 \cdot 385 x - 3025 = 10x^3 - 165x^2 + 1155x - 3025$.The derivative $S'(x) = 30x^2 - 330x + 1155$.Its discriminant: $330^2 - 4 \cdot 30 \cdot 1155 = 108900 - 138600 = -29700 < 0$.Since the leading coefficient of $S'(x)$ is positive, $S'(x) > 0$ for all real $x$.Therefore $S(x)$ is strictly increasing, and as $x \to -\infty$, $S(x) \to -\infty$, and as $x \to +\infty$, $S(x) \to +\infty$.Thus the equation $S(x)=0$ has exactly one real root.Answer:1