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Updated on: 03 Mar 2024, 14:50
Originally posted by Nipungautam23 on 01 Nov 2014, 17:36.
Last edited by Bunuel on 03 Mar 2024, 14:50, edited 2 times in total.
Last edited by Bunuel on 03 Mar 2024, 14:50, edited 2 times in total.
Renamed the topic and edited the question.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
57% (01:33) correct
43%
(02:11)
wrong
based on 668
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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
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.
GM.png [ 22.95 KiB | Viewed 11293 times ]
Explanation attached however I couldn't understand, Please help.
Attachment:
GM.png [ 22.95 KiB | Viewed 11293 times ]
07 Jan 2015, 22:38
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Nipungautam23
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).
02 Nov 2014, 07:21
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Nipungautam23
\(x^2 + 2bx + 4 = (x^2+ 2bx + b^2) + 4 - b^2 = (x + b)^2 + (4 - b^2)=(nonnegative \ value) + (constant)\). To minimize this expression we should minimize (x + b)^2. The least value of it is 0,, which is obtained when x = -b.
Answer: D.
