Functions
Invertibility
MMTS_Full_Test_01
Grade 12
Question:
The function $f:\mathbb{R}-\{-1,1\}\to\mathbb{R}$ defined by $f(x)=\dfrac{x}{1-x^2}$ is
injective but not surjective
surjective but not injective
injective and surjective
neither injective nor surjective
Step-by-Step Solution
Key Concept: Check if $f$ is one-one and onto
$f(2)=2/3$, $f(-2)=-2/3$: both values covered. $f'(x)=\frac{1+x^2}{(1-x^2)^2}>0$ on each component but not globally injective on full domain: $f(x)=f(-x)$ is not possible since $f$ is odd, but check $f(x_1)=f(x_2)$: from $x_1(1-x_2^2)=x_2(1-x_1^2)\Rightarrow x_1-x_2=x_1x_2(x_1-x_2)$, so $x_1=x_2$ or $x_1x_2=1$. So not injective. Range: all of $\mathbb{R}$. Surjective but not injective.
Correct Answer: 2