JEE Advanced Maths: Basic Maths / Inequalities questions with solutions

Q1. For x < 0, find the maximum value of (3x² + 12)/x.

1. -12
2. 12
3. -6
4. -24

Answer: -12

(3x² + 12)/x = 3x + 12/x. For x < 0, let -3x > 0 and -12/x > 0. By AM-GM, (-3x) + (-12/x) >= 2 sqrt((-3x)(-12/x)) = 2 sqrt(36) = 12. So 3x + 12/x <= -12. The maximum is -12, attained when -3x = -12/x, i.e. x² = 4, x = -2.