Nipungautam23 wrote:

If function f(x) = x^2 + 2bx + 4, then for any b, f(x) is the least when x equals

A. -b – √(b² – 4)

B. -2

C. 0

D. -b

E. b – 4

Source: Scoregetter Gmat Test

Explanation attached however I couldn't understand, Please help.

f(x) is a quadratic in x. We know the minimum value of ax^2 + bx + c is at x = -b/2a (draw the graph if you are not sure how we arrive at this - it's good to remember this though).

So minimum value of f(x) is at x = -2b/2*1 = -b

Or

Put b = 0. f(x) = x^2 + 4. This will be minimum when x^2 = 0 since x^2 can anyway not be negative. So x = 0. Two options give 0: (C) and (D)

Now notice that when b is a large negative value such as -100, the minimum value of f(x) will not be at x = 0.

For example, f(x) = x^2 - 200x + 4

If x = 0, f(x) = 4 but if x = 1, f(x) = - 195

So answer must be option (D).

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Karishma

Veritas Prep GMAT Instructor

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