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Question: The difference of maximum & minimum value of \(\frac{x^{2} + 4x + 9}{x^{2} + 9}\) is...

The difference of maximum & minimum value of

x2+4x+9x2+9\frac{x^{2} + 4x + 9}{x^{2} + 9} is

A

1/3

B

2/3

C

–2/3

D

4/3

Answer

4/3

Explanation

Solution

Let x2+4x+9x2+9\frac{x^{2} + 4x + 9}{x^{2} + 9} = y

⇒ (y – 1) x2 – 4x + 9 (y – 1) = 0

For real value of x, D ≥ 0

(5 – 3y) (3y – 1) ≥ 0 ⇒ 13≤y≤53\frac{1}{3} \leq y \leq \frac{5}{3}

∴ Difference = 53−13=43\frac{5}{3} - \frac{1}{3} = \frac{4}{3}