Let $f : D \to R$ be defined as $f (x) = \frac{x^{2} + 2 x + a}{x^{2} + 4 x + 3 a}$ , where D and R denote the complete domain of f for corresponding values of ‘a’ and the set of all real numbers respectively. If ‘f’ is surjective mapping, then the value of $\frac{a^{2} - 4 a + 1}{a^{2} - 2 a + 3}$ can be

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$\frac{1}{9}$

$- 1$

$\frac{1}{4}$

$\frac{- 1}{9}$

answer is A, B, C.

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Detailed Solution

$y = \frac{x^{2} + 2 x + a}{x^{2} + 4 x + 3 a}$

After supplication, we get
$a (a - 1) \leq 0$
If $a = 0, 1$ the range of f is not equal to R so $0 < a < 1$

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