If `f:R rarr[-1, 1]` be a function defined as `f(x)=sin((x^(2)-8)/(x^(2)+2))`, then f is

Options

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

Answer

To determine whether the function \( f(x) = \sin\left(\frac{x^2 - 8}{x^2 + 2}\right) \) is one-one or onto, we will analyze the inner function \( g(x) = \frac{x^2 - 8}{x^2 + 2} \) first.

### Step 1: Analyze the function \( g(x) \)

We start by rewriting \( g(x) \):

\[ g(x) = \frac{x^2 - 8}{x^2 + 2} \]

### Step 2: Check if \( g(x) \) is one-one

To check if \( g(x) \) is one-one, we will compute its derivative \( g'(x) \):

Using the quotient rule, we have:

\[ g'(x) = \frac{(x^2 + 2)(2x) - (x^2 - 8)(2x)}{(x^2 + 2)^2} \]

Simplifying the numerator:

\[ g'(x) = \frac{2x(x^2 + 2 - x^2 + 8)}{(x^2 + 2)^2} = \frac{2x(10)}{(x^2 + 2)^2} = \frac{20x}{(x^2 + 2)^2} \]

### Step 3: Determine the sign of \( g'(x) \)

The derivative \( g'(x) = \frac{20x}{(x^2 + 2)^2} \) indicates that:

- For \( x > 0 \), \( g'(x) > 0 \) (increasing)
- For \( x < 0 \), \( g'(x) < 0 \) (decreasing)
- At \( x = 0 \), \( g'(0) = 0 \)

Since \( g(x) \) is increasing for \( x > 0 \) and decreasing for \( x < 0 \), it is not a one-one function.

### Step 4: Check if \( g(x) \) is onto

Next, we need to check if \( g(x) \) can take all values in the range of \( [-1, 1] \).

Calculating the limits as \( x \to \infty \) and \( x \to -\infty \):

\[ \lim_{x \to \infty} g(x) = \lim_{x \to -\infty} g(x) = 1 \]

Now, we evaluate \( g(x) \) at specific points:

- At \( x = 0 \):
\[ g(0) = \frac{0^2 - 8}{0^2 + 2} = \frac{-8}{2} = -4 \]

- At \( x = 2 \):
\[ g(2) = \frac{2^2 - 8}{2^2 + 2} = \frac{4 - 8}{4 + 2} = \frac{-4}{6} = -\frac{2}{3} \]

- At \( x = 3 \):
\[ g(3) = \frac{3^2 - 8}{3^2 + 2} = \frac{9 - 8}{9 + 2} = \frac{1}{11} \]

From this analysis, we see that \( g(x) \) does not cover the entire range of \( [-1, 1] \). Thus, \( g(x) \) is not onto.

### Step 5: Conclusion about \( f(x) \)

Since \( g(x) \) is neither one-one nor onto, the function \( f(x) = \sin(g(x)) \) will also not be one-one or onto.

Thus, we conclude:

\[ \text{The function } f(x) \text{ is neither one-one nor onto.} \]