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05 Jan 2023, 02:30
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C
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:
65% (01:54) correct
35%
(01:49)
wrong
based on 382
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What is the maximum value of \(\frac{(x^2+2x+3)}{(x^2+2x+2)}\)?
A. 0
B. 1
C. 2
D. 3
E. 5
A. 0
B. 1
C. 2
D. 3
E. 5
05 Jan 2023, 03:39
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Bunuel
\(\frac{(x^2+2x+3)}{(x^2+2x+2)}\)
\(\frac{(x^2+2x+2+1)}{(x^2+2x+2)}\)
\(\frac{(x^2+2x+2)}{(x^2+2x+2)}\) + \(\frac{1}{(x^2+2x+2)}\)
\(1\) + \(\frac{1}{(x^2+2x+2)}\)
As we want the maximum value, \(\frac{1}{(x^2+2x+2)}\) should be maximum. Hence \(x^2+2x+2\) should be minimum.
A quadratic function of the form \(ax^2 + bx + c\) has its minima at \(\frac{-b}{2a}\)
Hence the minima of \(x^2+2x+2\) is at x = \(\frac{-2}{2}\) = -1
Value at minima = \(\frac{1}{((-1)^2+2(-1)+2)}\) = 1
Maximum value = 1+1 = 2
Option C
14 Jul 2024, 05:23
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Here's an easier approach:
If you notice, the numerator and denominator would be consecutive numbers. Numerator would always be 1 more than the denominator. Therefore, the expression could be \(\frac{2}{1}, \frac{3}{2}, \frac{4}{3}\), and so on. And at \(\frac{2}{1} = 2\) is the maximum value.
If you notice, the numerator and denominator would be consecutive numbers. Numerator would always be 1 more than the denominator. Therefore, the expression could be \(\frac{2}{1}, \frac{3}{2}, \frac{4}{3}\), and so on. And at \(\frac{2}{1} = 2\) is the maximum value.
