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Inequality

The minimum p...

Question

The minimum positive value of $\frac{x^{2} + 2 x + 4}{x + 2}$ , is?

A

0

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B

1

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C

2

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D

3

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Solution

The correct option is A $0$
Let $f (x) = \frac{x^{2} + 2 x + 4}{x + 2}$
$f^{1} (x) = 0$ for max/min
So, $f^{1} (x) = \frac{(x + 2) (2 x + 2) - (x^{2} + 2 x + 4) (1)}{{(x + 2)}^{2}}$
$f^{1} (x) = \frac{{2 x}^{2} + 4 x + 2 x + 4 - x^{2} - 2 x - 4}{{(x + 2)}^{2}}$
$f^{1} (x) = \frac{x^{2} + 4 x}{{(x + 2)}^{2}} = 0 \Rightarrow x = 0, x = - 4$
$f (0) = \frac{4}{2} = 2$
$f (- 4) = \frac{{(- 4)}^{2} - 8 + 4}{- 4 + 2} = - 6$
So, Minimum positive value for $f (x)$ is $2$ but since range of $f (x)$ is $(- \infty, \infty)$ .
So, We can say that
Minimum positive value $= 0$

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