Find whether f(x) is one-one or many-one & into or onto if f:D->R where D is its domain. `f(x)=3/4x^2-cospix`

Explanation

To determine whether the function \( f(x) = \frac{3}{4}x^2 - \cos(\pi x) \) is one-one or many-one, and whether it is into or onto, we will follow these steps:

### Step 1: Determine if the function is one-one or many-one

**Solution:**
To check if \( f(x) \) is one-one, we can evaluate \( f(-x) \) and see if it equals \( f(x) \).

\[ f(-x) = \frac{3}{4}(-x)^2 - \cos(\pi(-x)) = \frac{3}{4}x^2 - \cos(-\pi x) \]
Since \( \cos(-\theta) = \cos(\theta) \), we have:
\[ f(-x) = \frac{3}{4}x^2 - \cos(\pi x) = f(x) \]

Since \( f(x) = f(-x) \), this indicates that the function is even. An even function is not one-one because it maps both \( x \) and \( -x \) to the same value.

**Conclusion:** The function \( f(x) \) is many-one.

### Step 2: Determine if the function is into or onto

**Solution:**
Next, we need to find the range of the function. Since \( f(x) \) is an even function, we know that its range cannot cover all real numbers \( \mathbb{R} \).

To find the minimum value of the function, we can analyze the two components:
1. The term \( \frac{3}{4}x^2 \) is always non-negative and reaches its minimum value of 0 when \( x = 0 \).
2. The term \( -\cos(\pi x) \) oscillates between -1 and 1.

At \( x = 0 \):
\[ f(0) = \frac{3}{4}(0)^2 - \cos(0) = 0 - 1 = -1 \]

As \( x \) increases or decreases, \( \frac{3}{4}x^2 \) increases, and \( -\cos(\pi x) \) oscillates. Therefore, the function will have values greater than -1, but will not reach all real numbers.

Since the codomain of \( f(x) \) is \( \mathbb{R} \) and the range does not equal \( \mathbb{R} \), we conclude that the function is into.

**Conclusion:** The function \( f(x) \) is into.

### Final Summary:
- The function \( f(x) = \frac{3}{4}x^2 - \cos(\pi x) \) is many-one.
- The function \( f(x) \) is into.

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