If f : `R rarr R`, f(x) `= sin^(2) x + cos^(2) x`, then f is

Options

A. One - one but not onto
B. Onto but not one-one
C. Neither one-one nor onto
D. Both one-one and onto

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^2 x + \cos^2 x \).

### Step 1: Simplify the function
We know from trigonometric identities that:
\[ \sin^2 x + \cos^2 x = 1 \]
Thus, we can rewrite the function as:
\[ f(x) = 1 \]

### Step 2: Determine if the function is one-one (injective)
A function is one-one if every element in the domain maps to a unique element in the codomain. In this case, since \( f(x) = 1 \) for all \( x \), it means that multiple inputs (e.g., \( x_1, x_2, \ldots \)) can produce the same output (which is always 1). Therefore, the function is not one-one.

### Step 3: Determine if the function is onto (surjective)
A function is onto if every element in the codomain has a pre-image in the domain. The codomain of the function \( f \) is \( \mathbb{R} \) (the set of all real numbers). However, the range of \( f(x) \) is just the single value 1, which means there are many elements in \( \mathbb{R} \) that do not have a corresponding input in the domain that maps to them. Therefore, the function is not onto.

### Conclusion
Since the function is neither one-one nor onto, we conclude that:
\[ \text{The function } f \text{ is neither one-one nor onto.} \]

### Final Answer
The correct option is: Neither one-one nor onto.

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