If range of `f(x)=cosx`, `x in ((-pi)/(3),(pi)/(6))` is (a, b) then :

Options

A. `a+b=(3)/(2)`
B. `b-a=1-(sqrt(3))/(2)`
C. `a^(2)+b^(2)=(5)/(6)`
D. `a^(2)+b^(2)=(7)/(4)`

Answer

To find the range of the function \( f(x) = \cos x \) for \( x \) in the interval \( \left(-\frac{\pi}{3}, \frac{\pi}{6}\right) \), we will follow these steps:

### Step 1: Identify the endpoints of the interval
We need to evaluate the function \( f(x) = \cos x \) at the endpoints of the interval \( x = -\frac{\pi}{3} \) and \( x = \frac{\pi}{6} \).

### Step 2: Calculate \( f\left(-\frac{\pi}{3}\right) \)
Using the cosine function:
\[ f\left(-\frac{\pi}{3}\right) = \cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

### Step 3: Calculate \( f\left(\frac{\pi}{6}\right) \)
Now, we calculate the value at the other endpoint:
\[ f\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]

### Step 4: Determine the maximum and minimum values
Now we have the values:
- At \( x = -\frac{\pi}{3} \), \( f(-\frac{\pi}{3}) = \frac{1}{2} \)
- At \( x = \frac{\pi}{6} \), \( f(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \)

### Step 5: Identify the range
The range of \( f(x) \) over the interval \( \left(-\frac{\pi}{3}, \frac{\pi}{6}\right) \) is from the minimum value \( \frac{1}{2} \) to the maximum value \( \frac{\sqrt{3}}{2} \).

Thus, the range is:
\[ \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \]

### Step 6: Identify \( a \) and \( b \)
Here, \( a = \frac{1}{2} \) and \( b = \frac{\sqrt{3}}{2} \).

### Step 7: Calculate \( a + b \) and \( a^2 + b^2 \)
1. \( a + b = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \)
2. \( a^2 + b^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1 \)

### Final Answer
The range of \( f(x) \) is \( \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \).

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