Let us define the function `f(x)=x^(2)+x+1`
Statement I The equation `sin x= f(x)` has no solution.
Statement II The curve `y=sinx` and `y=f(x)` do not intersect each other when graph is oberved.

Answer

To solve the problem, we need to analyze the function \( f(x) = x^2 + x + 1 \) and the equation \( \sin x = f(x) \). We will evaluate the statements provided in the question.

### Step 1: Analyze the function \( f(x) \)
The function given is:
\[ f(x) = x^2 + x + 1 \]
This is a quadratic function, and we can determine its minimum value by completing the square.

### Step 2: Complete the square for \( f(x) \)
To complete the square:
\[ f(x) = x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]
From this form, we can see that the minimum value of \( f(x) \) occurs at \( x = -\frac{1}{2} \) and is:
\[ f\left(-\frac{1}{2}\right) = \frac{3}{4} \]

### Step 3: Determine the range of \( f(x) \)
Since \( f(x) \) is a parabola that opens upwards, the minimum value is \( \frac{3}{4} \). Therefore, the range of \( f(x) \) is:
\[ f(x) \geq \frac{3}{4} \]

### Step 4: Analyze the function \( \sin x \)
The function \( \sin x \) oscillates between -1 and 1. Thus, the range of \( \sin x \) is:
\[ -1 \leq \sin x \leq 1 \]

### Step 5: Compare the ranges of \( \sin x \) and \( f(x) \)
From the analysis, we have:
- The minimum value of \( f(x) \) is \( \frac{3}{4} \).
- The maximum value of \( \sin x \) is 1.

Since the minimum value of \( f(x) \) (which is \( \frac{3}{4} \)) is greater than the maximum value of \( \sin x \) (which is 1), it follows that:
\[ f(x) > \sin x \quad \text{for all } x \]

### Step 6: Conclusion about the statements
1. **Statement I:** The equation \( \sin x = f(x) \) has no solution. This is true because \( \sin x \) cannot reach the values of \( f(x) \).
2. **Statement II:** The curves \( y = \sin x \) and \( y = f(x) \) do not intersect. This is also true for the same reason.

Thus, both statements are true, and Statement II correctly explains Statement I.

### Final Answer
Both Statement I and Statement II are true, and Statement II is a correct explanation for Statement I.

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