Let f(x)=ax2−b|x|, where a and b are constants. Then at x = 0, f(x) is
A
Maximized whenever a > 0, b > 0
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B
Maximized whenever a > 0, b < 0
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C
Minimized whenever a > 0, b > 0
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D
Minimized whenever a > 0, b < 0
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Solution
The correct option is D Minimized whenever a > 0, b < 0 f(x)=ax2−bx. In this function, x2 and x are always positve. The value thus depends on a and b. f(1) = a - b. Using different options, we find that a - b will be positive if a > 0 and b < 0. The minimum value of a positive function is 0. Hence (d) is correct option.