Let f: R → R where $f(x): \frac{x^2+4x+7}{x^2+x+1}$. Is f(x) one-one?
Answer & explanation
Correct answer: option 3
We have $f(x)=\frac{x^2+4x+7}{x^2+x+1}=1+\frac{3(x+2)}{x^2+x+1}$
Let $f(x_1) = f(x_2)$
$⇒1+\frac{3(x_1+2)}{x_1^2+x_1+1}=1+\frac{3(x_2+2)}{x_2^2+x_2+1}$
$⇒x_1x_1^2+x_1x_2+x_1+2x_2^2+2x_2+2$
$x_1^2x_2+x_1x_2+x_2+2x_1^2+2x_1+2$
$⇒(x_1-x_2)(2x_1+2x_2+x_1x_2+1)=0$
Let us consider $2x_1 + 2x_2 +x_1x_2 +1=0$
$⇒x_2=-\frac{1+2x_1}{2+x_1}$
This relation is satisfied by infinite number of pairs $(x_1,x_2)$, where $x_1≠x_2$, e.g., $(0, -1/2), (1, -1)$ etc.
Hence f(x) is many-one.