Discussion

Question:

For all real values of x, the range of the function f(x) = $\frac{x^{2} + 2 x + 4}{2 x^{2} + 4 x + 9}$ is

Explanation:

Given, f(x) = $\frac{x^{2} + 2 x + 4}{2 x^{2} + 4 x + 9}$

⇒ f(x) = $\frac{1}{2} [\frac{2 x^{2} + 4 x + 8}{2 x^{2} + 4 x + 9}]$

⇒ f(x) = $\frac{1}{2} [\frac{2 x^{2} + 4 x + 9 - 1}{2 x^{2} + 4 x + 9}]$

⇒ f(x) = $\frac{1}{2} [1 - \frac{1}{2 x^{2} + 4 x + 9}]$

f(x) will be minimum when (2x² + 4x + 9) is minimum.

Now, 2x² + 4x + 9 will be minimum when x = -(4)/2 × 2 = -1

∴ Minimum value of 2x² + 4x + 9 = 2(-1)² + 4(-1) + 9 = 7

∴ Minimum value of f(x) = $\frac{1}{2} [1 - \frac{1}{7}]$ = $\frac{3}{7}$

f(x) will be maximum when (2x² + 4x + 9) is maximum.

Maximum value of (2x² + 4x + 9) will be ∞.

∴ Maximum value of f(x) = $\frac{1}{2} [1 - \frac{1}{\infty}]$ = $\frac{1}{2}$

∴ Range of f(x) = $[\frac{3}{7}, \frac{1}{2})$

Upper value of ½ is in open bracket as value of (2x² + 4x + 9) will never actually be ∞.

Hence, option (a).

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