If `a, b in (0, 2)` and the equation `(2x^2+5)/2 = x - 2cos(ax + b)` has at least one solution then

Options

A. 1
B. 2
C. e
D. `pi`

Answer

To solve the equation

\[ \frac{2x^2 + 5}{2} = x - 2\cos(ax + b) \]

we need to analyze it under the condition that \( a, b \in (0, 2) \) and determine the conditions under which this equation has at least one solution.

### Step 1: Rearranging the Equation
First, we can rearrange the equation to isolate the cosine term:

\[ 2\cos(ax + b) = x - \frac{2x^2 + 5}{2} \]

This simplifies to:

\[ 2\cos(ax + b) = x - x^2 - \frac{5}{2} \]

### Step 2: Expressing the Right Side
Now, we rewrite the right side:

\[ 2\cos(ax + b) = -x^2 + x - \frac{5}{2} \]

Let’s denote the right side as \( f(x) = -x^2 + x - \frac{5}{2} \).

### Step 3: Analyzing the Function \( f(x) \)
The function \( f(x) \) is a quadratic function that opens downwards (since the coefficient of \( x^2 \) is negative). To find the maximum value of this function, we can use the vertex formula for a quadratic function \( ax^2 + bx + c \), where the x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \).

Here, \( a = -1 \) and \( b = 1 \):

\[ x = -\frac{1}{2 \cdot -1} = \frac{1}{2} \]

### Step 4: Finding the Maximum Value of \( f(x) \)
Now we substitute \( x = \frac{1}{2} \) back into \( f(x) \):

\[ f\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 + \frac{1}{2} - \frac{5}{2} \]

Calculating this gives:

\[ f\left(\frac{1}{2}\right) = -\frac{1}{4} + \frac{1}{2} - \frac{5}{2} = -\frac{1}{4} + \frac{2}{4} - \frac{10}{4} = -\frac{9}{4} \]

### Step 5: Range of \( 2\cos(ax + b) \)
The range of \( 2\cos(ax + b) \) is from -2 to 2. For the equation to have at least one solution, the maximum value of \( f(x) \) must be at least -2:

\[ -\frac{9}{4} \geq -2 \]

This is not true, thus we need to check if the minimum value of \( f(x) \) can be less than or equal to 2.

### Step 6: Finding Minimum Value of \( f(x) \)
The minimum value of \( f(x) \) occurs at the endpoints of the interval where \( a, b \in (0, 2) \). We can evaluate \( f(x) \) at \( x = 0 \) and \( x = 2 \):

1. For \( x = 0 \):
\[ f(0) = -0^2 + 0 - \frac{5}{2} = -\frac{5}{2} \]

2. For \( x = 2 \):
\[ f(2) = -2^2 + 2 - \frac{5}{2} = -4 + 2 - \frac{5}{2} = -2 - \frac{5}{2} = -\frac{9}{2} \]

### Conclusion
Since the maximum value of \( f(x) \) is -2 and the minimum value is -\(\frac{9}{2}\), the equation \( 2\cos(ax + b) \) can equal \( f(x) \) at least once if the range of \( f(x) \) intersects with the range of \( 2\cos(ax + b) \).

Thus, for the equation to have at least one solution, we need to ensure that:

\[ -\frac{9}{4} \leq 2 \quad \text{and} \quad -2 \leq -\frac{9}{4} \]

This leads to the conclusion that the values of \( a \) and \( b \) must be chosen such that the maximum of the quadratic function does not exceed the maximum of the cosine function.