If ` f:R to R` be the function defined by `f(x) = sin(3x+2) AA x in R.` Then, f is invertible.

Explanation

To determine whether the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \sin(3x + 2) \) is invertible, we need to check if it is a one-to-one (injective) and onto (surjective) function.

### Step 1: Check if \( f(x) \) is one-to-one (injective)

A function is one-to-one if \( f(a) = f(b) \) implies \( a = b \).

Assume \( f(a) = f(b) \):
\[ \sin(3a + 2) = \sin(3b + 2) \]
This implies:
\[ 3a + 2 = 3b + 2 + 2k\pi \quad \text{or} \quad 3a + 2 = \pi - (3b + 2) + 2k\pi \quad \text{for some integer } k \]
From the first equation:
\[ 3a = 3b + 2k\pi \implies a = b + \frac{2k\pi}{3} \]
This shows that \( a \) can be different from \( b \) if \( k \neq 0 \), indicating that \( f \) is not one-to-one.

### Step 2: Check if \( f(x) \) is onto (surjective)

A function is onto if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

The range of \( \sin(3x + 2) \) is limited to the interval \([-1, 1]\). Therefore, \( f(x) \) cannot cover all real numbers \( \mathbb{R} \).

### Conclusion

Since \( f(x) \) is neither one-to-one nor onto, we conclude that the function \( f(x) = \sin(3x + 2) \) is not invertible.