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What is the maximum value of (x^2+2x+3)/(x^2+2x+2)?

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by Max@Math Revolution » Fri Apr 26, 2019 5:42 am

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Answers

A

B

C

D

E

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[GMAT math practice question]

What is the maximum value of (x^2+2x+3)/(x^2+2x+2)?

A. 0
B. 1
C. 2
D. 3
E. 5
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by Max@Math Revolution » Mon Apr 29, 2019 5:15 pm
=>

(x^2+2x+3)/(x^2+2x+2)
=(x^2+2x+2+1)/(x^2+2x+2)
= (x^2+2x+2)/(x^2+2x+2) + 1/(x^2+2x+2)
= 1 + 1/(x^2+2x+2)

x^2+2x+2
=x^2+2x+1+1
=(x+1)^2+1 ≥ 1, with equality when x = -1.
Thus, 1/(x^2+2x+2) ≤ 1 and 1 + 1/(x^2+2x+2) ≤ 2, with equality when x = -1.
Therefore, the maximum value is 2.

Therefore, the answer is C.
Answer: C
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by deloitte247 » Tue Apr 30, 2019 1:44 pm
$$\frac{x^2+2x+3}{x^2+2x+2}=\frac{x^2+2x+1+2}{x^2+2x+1+1}$$
$$where\ x^2+2x+1\ can\ be\ simplified\ as\ \left(x+1\right)^2$$
$$Therefore,\ \frac{\left(x+1\right)^2+2}{\left(x+1\right)^2+1}$$
let (x+1)=a
$$\frac{a^2+2}{a^2+1}$$
$$a^2\ge0$$
$$Since\ a^2\ cannot\ be\ negative$$
$$Therefore,\ 2\ge\frac{a^2+2}{a^2+1}\ge1$$
Hence, the maximum value of the expression = 2
Answer = option c
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