UPSC NDA 2 2019 Mathematics Solved Paper

Concept:A function is increasing in an interval if its first derivative is positive in that interval.Explanation:Given $F(x)=3x^{4}-20x^{3}-12x^{2}+288x+1$Differentiate: $F'(x)=12x^{3}-60x^{2}-24x+288$Factor: $F'(x)=12(x^{3}-5x^{2}-2x+24)$$=12(x+2)(x-3)(x-4)$Set $F'(x)=0$ to find critical points: $x=-2,\ 3,\ 4$Check sign of $F'(x)$ in each interval:• For $x<-2$: all three factors negative → product negative• For $-2 < x<3$: $(x+2)>0,\ (x-3)<0,\ (x-4)<0$ → product positive• For $3 < x<4$: $(x+2)>0,\ (x-3)>0,\ (x-4)<0$ → product negative• For $x>4$: all three positive → product positiveThus $F'(x)>0$ on $(-2,3)$ and $(4,\infty)$Among the given options, only $(-2,3)$ appears.Answer:Option A: $(-2,3)$