Question

Let $f: R \rightarrow R$ be function defined by $f(x) = \frac{x^2-8}{x^2+2}$, then $f$ is

• A. one-one but not onto
• B. one-one and onto
• C. onto but not one-one
• D. neither one-one nor onto

Answer 1

Answer: (D)

D

Answer 2

Let the function be $f: R \rightarrow R$ defined by $f(x) = \frac{x^2-8}{x^2+2}$.

To check if $f$ is one-one (injective), we examine if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Let $f(x_1) = f(x_2)$. $\frac{x_1^2-8}{x_1^2+2} = \frac{x_2^2-8}{x_2^2+2}$

$(x_1^2-8)(x_2^2+2) = (x_2^2-8)(x_1^2+2)$

$x_1^2 x_2^2 + 2x_1^2 - 8x_2^2 - 16 = x_1^2 x_2^2 + 2x_2^2 - 8x_1^2 - 16$

$2x_1^2 - 8x_2^2 = 2x_2^2 - 8x_1^2$

$10x_1^2 = 10x_2^2$

$x_1^2 = x_2^2$

$x_1 = \pm x_2$.

Since $f(x_1) = f(x_2)$ does not imply $x_1 = x_2$ (it can be $x_1 = -x_2$), the function is not one-one. For example, $f(2) = \frac{2^2-8}{2^2+2} = \frac{4-8}{4+2} = \frac{-4}{6} = -\frac{2}{3}$ and $f(-2) = \frac{(-2)^2-8}{(-2)^2+2} = \frac{4-8}{4+2} = \frac{-4}{6} = -\frac{2}{3}$. Since $f(2) = f(-2)$ but $2 \neq -2$, $f$ is not one-one.

To check if $f$ is onto (surjective), we find the range of the function and compare it with the codomain $R$. Let $y$ be in the range of $f$. Then there exists $x \in R$ such that $f(x) = y$.

$y = \frac{x^2-8}{x^2+2}$

$y(x^2+2) = x^2-8$

$yx^2 + 2y = x^2 - 8$

$2y + 8 = x^2 - yx^2$

$2y + 8 = x^2(1-y)$

If $1-y = 0$, i.e., $y=1$, the equation becomes $2(1)+8 = x^2(0)$, which is $10 = 0$. This is a contradiction, so $y=1$ is not in the range of $f$. If $1-y \neq 0$, i.e., $y \neq 1$, we can write $x^2 = \frac{2y+8}{1-y}$.

For $x$ to be a real number, $x^2$ must be non-negative. So, we require $\frac{2y+8}{1-y} \ge 0$.

We analyze the sign of the expression $\frac{2y+8}{1-y}$. The critical points are $y=-4$ (where $2y+8=0$) and $y=1$ (where $1-y=0$).

• If $y < -4$, e.g., $y=-5$, $\frac{2(-5)+8}{1-(-5)} = \frac{-2}{6} < 0$.

• If $-4 \le y < 1$, e.g., $y=0$, $\frac{2(0)+8}{1-0} = \frac{8}{1} \ge 0$.

• If $y > 1$, e.g., $y=2$, $\frac{2(2)+8}{1-2} = \frac{12}{-1} < 0$.

So, $\frac{2y+8}{1-y} \ge 0$ when $-4 \le y < 1$. The range of the function is $[-4, 1)$.

The codomain of the function is given as $R$. Since the range $[-4, 1)$ is a proper subset of the codomain $R$, the function is not onto. For example, any real number $y \ge 1$ or $y < -4$ is in the codomain but not in the range.

Since the function is neither one-one nor onto, the correct option is D.

Explanation:

1. The function $f(x) = \frac{x^2-8}{x^2+2}$ is not one-one because $f(x) = f(-x)$ for all $x \in R$, and $x \neq -x$ for $x \neq 0$.

2. To find the range, let $y = f(x)$. Solving for $x^2$, we get $x^2 = \frac{2y+8}{1-y}$. For $x$ to be real, $x^2 \ge 0$, which requires $\frac{2y+8}{1-y} \ge 0$. This inequality holds for $-4 \le y < 1$. The range is $[-4, 1)$.

3. The codomain is $R$. Since the range $[-4, 1)$ is not equal to the codomain $R$, the function is not onto.

4. Therefore, the function is neither one-one nor onto.