The number of solutions of the equation `sinx=x^(2)+x+1` is-

Options

A. 0
B. 1
C. 2
D. none

Answer

To find the number of solutions of the equation \( \sin x = x^2 + x + 1 \), we will analyze both sides of the equation step by step.

### Step 1: Understand the Functions
The left side of the equation is \( \sin x \), which oscillates between -1 and 1 for all values of \( x \). The right side is a quadratic function \( f(x) = x^2 + x + 1 \).

**Hint:** Remember that \( \sin x \) is bounded between -1 and 1, while a quadratic function can take any value depending on \( x \).

### Step 2: Analyze the Quadratic Function
Let's analyze the quadratic function \( f(x) = x^2 + x + 1 \):
- The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \), where \( a = 1 \) and \( b = 1 \).
- Thus, \( x = -\frac{1}{2} \).
- Now, substitute \( x = -\frac{1}{2} \) into the function to find the minimum value:
\[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4} \]
- The minimum value of \( f(x) \) is \( \frac{3}{4} \).

**Hint:** The minimum value of the quadratic function is crucial to determine if it intersects with \( \sin x \).

### Step 3: Compare the Ranges
Since \( \sin x \) ranges from -1 to 1 and \( f(x) \) has a minimum value of \( \frac{3}{4} \), we can conclude:
- \( f(x) \) is always greater than or equal to \( \frac{3}{4} \) and never reaches or goes below 0.
- Therefore, \( f(x) \) does not intersect with \( \sin x \) because \( \sin x \) can take values below \( \frac{3}{4} \).

**Hint:** Check the ranges of both functions to see if they can intersect.

### Step 4: Conclusion
Since the minimum value of \( f(x) \) is \( \frac{3}{4} \) and \( \sin x \) can take values below \( \frac{3}{4} \), there are no points where \( \sin x = f(x) \).

Thus, the number of solutions to the equation \( \sin x = x^2 + x + 1 \) is **0**.

**Final Answer:** The number of solutions is **0**.