Let f(x) = ax2 – b|x|, where a and b are constants. Then at x = 0, f(x) is
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).
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