Please submit your concern

Explanation:

f(x) = ax² – b|x|

x² and |x| both are positive. Let x ≠ 0.

At x = 0, f(0) = 0

Consider the following cases:

1. a > 0, b > 0
∴ f(x) > 0, when ax² > b|x|
∴ f(x) < 0, when  ax² < b|x|
f(x) is neither maximised or minimised when x = 0.

2. a > 0, b < 0
∴ f(x) = ax² + |b||x| > 0
Thus f(x) is greater than 0 when x ≠ 0.
​​​​​​​∴ f(x) is minimised at x = 0 whenever a > 0, b < 0.

Hence, option (d).