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Let f(x) = (x² + 3x + a)/(x² + x + 1) with f: R -> R. Find the value(s) of the parameter 'a' for which f is one-one (injective).

1. a = 2
2. a = 1
3. a = 3
4. No value of a makes it one-one

Correct answer: No value of a makes it one-one

Solution

The denominator x² + x + 1 is always positive, so f is defined and continuous on all of R. Since both x² terms dominate, f(x) -> 1 as x -> +/- infinity. A continuous function with equal limits at both infinities must turn around, so it cannot be strictly monotonic, hence never one-one for any a.

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