UPSC NDA I 2015 Mathematics Paper

Concept:A function's critical points are found by setting its first derivative to zero. The second derivative test helps determine if a critical point is a local minimum or maximum. The sign of the first derivative in an interval decides if the function is increasing or decreasing there.Explanation:Consider the function $f(x) = 0.75x^4 - x^3 - 9x^2 + 7$.First derivative: $f'(x) = 3x^3 - 3x^2 - 18x = 3x(x^2 - x - 6) = 3x(x-3)(x+2)$.Set $f'(x)=0$ gives critical points $x=-2$, $x=0$, $x=3$.For statement 1: Second derivative $f''(x)=9x^2-6x-18$.$f''(-2) = 9(4) -6(-2)-18 = 36+12-18=30 > 0$, so local minimum at $x=-2$.$f''(3) = 9(9) -6(3)-18 = 81-18-18=45 > 0$, so local minimum at $x=3$.Thus statement 1 is correct.For statement 2: On the interval $(-2,0)$, take a test point like $x=-1$.$f'(-1) = 3(-1)^3 -3(-1)^2 -18(-1) = -3-3+18=12 > 0$.Since $f'(x)>0$ throughout $(-2,0)$ (as $f'$ changes sign only at critical points), the function is increasing on $(-2,0)$.Thus statement 2 is also correct.Answer:Both statements 1 and 2 are correct. Hence, option C is the correct answer (both 1 and 2).