`sin x + cos x = y ^(2) -y + a` has no value of x for any y if 'a' belongs to

Options

A. `(0,sqrt3)`
B. `(-sqrt3, 0)`
C. `(-oo, -sqrt3)`
D. `(sqrt3, oo)`

Answer

To solve the equation \( \sin x + \cos x = y^2 - y + a \) and determine the values of \( a \) for which there is no value of \( x \) for any \( y \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin x + \cos x = y^2 - y + a \]
We know that \( \sin x + \cos x \) can be expressed in a different form.

### Step 2: Use the identity for \( \sin x + \cos x \)
Using the identity:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
This means that the maximum value of \( \sin x + \cos x \) is \( \sqrt{2} \) and the minimum value is \( -\sqrt{2} \).

### Step 3: Establish the range of \( \sin x + \cos x \)
Thus, we have:
\[ -\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2} \]
This indicates that for the equation to have no solution for any \( y \), the right-hand side \( y^2 - y + a \) must fall outside this range.

### Step 4: Analyze the quadratic expression
The quadratic expression \( y^2 - y + a \) opens upwards (since the coefficient of \( y^2 \) is positive). The vertex of this quadratic is given by:
\[ y = -\frac{b}{2a} = \frac{1}{2} \]
Substituting \( y = \frac{1}{2} \) into the quadratic gives us the minimum value:
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + a = \frac{1}{4} - \frac{1}{2} + a = a - \frac{1}{4} \]

### Step 5: Set the conditions for no solution
For there to be no value of \( x \) for any \( y \), we need:
\[ a - \frac{1}{4} > \sqrt{2} \quad \text{or} \quad a - \frac{1}{4} < -\sqrt{2} \]
This leads to two inequalities:
1. \( a > \sqrt{2} + \frac{1}{4} \)
2. \( a < -\sqrt{2} + \frac{1}{4} \)

### Step 6: Conclusion
Thus, the values of \( a \) for which there is no value of \( x \) for any \( y \) are:
\[ a < -\sqrt{2} + \frac{1}{4} \quad \text{or} \quad a > \sqrt{2} + \frac{1}{4} \]