UPSC NDA I 2016 Mathematics Solved Paper

Concept:Local maximum occurs at a critical point where the derivative changes sign from positive to negative.Explanation:First, differentiate $f(x) = (x-1)^{2}(x+1)(x-2)^{3}$ to find critical points.Factor the derivative: $f'(x) = (x-1)(x-2)^{2}(6x^{2} - 5x - 5)$.Critical points: $x = 1$, $x = 2$ (double root), and the roots of $6x^{2} - 5x - 5 = 0$, i.e., $x = \frac{5 \pm \sqrt{145}}{12}$ (approximately $-0.587$ and $1.42$).The term $(x-2)^{2} > 0$ for $x \neq 2$, so sign of $f'$ depends on $(x-1)$ and the quadratic.The quadratic is negative between its two roots and positive outside.Check intervals:$(-\infty, \alpha)$: $f' < 0$ (decreasing).$(\alpha, 1)$: $f' > 0$ (increasing).$(1, \beta)$: $f' < 0$ (decreasing).$(\beta, \infty)$: $f' > 0$ (increasing).At $x = \alpha$ and $x = \beta$, derivative changes from negative to positive → local minima.At $x = 1$, derivative changes from positive to negative → local maximum.At $x = 2$, derivative does not change sign → point of inflection, not an extremum.Answer:There is exactly one point of local maxima. (Option B: One)