Assertion (A) : The number of real solutions of the equation `sin x = x^(2) + 3x + 4` is zero
Reason (R): `-1 le sin x le 1`, `AA x in R`

Options

A. Both A, R are true and R explain Assertion
B. Both A, R are true but R does't explain A
C. A is true R is false
D. A is false R is true

Answer

To solve the problem, we need to analyze the equation given in the assertion:

**Assertion (A)**: The number of real solutions of the equation \( \sin x = x^2 + 3x + 4 \) is zero.

**Reason (R)**: \( -1 \leq \sin x \leq 1 \) for all \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the Range of \( \sin x \)**:
- The sine function oscillates between -1 and 1 for all real values of \( x \).
- Therefore, we can conclude that \( \sin x \) can only take values in the interval \([-1, 1]\).

2. **Analyzing the Quadratic Function**:
- The right-hand side of the equation is \( x^2 + 3x + 4 \).
- This is a quadratic function, and we need to determine its minimum value to see if it can fall within the range of \( \sin x \).

3. **Finding the Minimum Value of the Quadratic**:
- The general form of a quadratic equation \( ax^2 + bx + c \) has its vertex (minimum or maximum) at \( x = -\frac{b}{2a} \).
- Here, \( a = 1 \) and \( b = 3 \), so the vertex is at:
\[ x = -\frac{3}{2 \cdot 1} = -\frac{3}{2} \]
- Now, we substitute \( x = -\frac{3}{2} \) into the quadratic to find its minimum value:
\[ f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) + 4 \]
\[ = \frac{9}{4} - \frac{9}{2} + 4 = \frac{9}{4} - \frac{18}{4} + \frac{16}{4} = \frac{7}{4} \]

4. **Comparing the Minimum Value with the Range of \( \sin x \)**:
- The minimum value of \( x^2 + 3x + 4 \) is \( \frac{7}{4} \), which is approximately \( 1.75 \).
- Since \( \frac{7}{4} > 1 \), the quadratic function \( x^2 + 3x + 4 \) is always greater than 1 for all \( x \).

5. **Conclusion**:
- Since \( \sin x \) can only take values between -1 and 1, and \( x^2 + 3x + 4 \) is always greater than 1, there are no real values of \( x \) that can satisfy the equation \( \sin x = x^2 + 3x + 4 \).
- Therefore, the assertion is true: the number of real solutions is zero.

### Final Answer:
- **Assertion (A)** is true.
- **Reason (R)** is true and correctly explains the assertion.