If `[f((sqrt(3))/2)]` is [.] denotes the greatest integer function) 4 (b) 5 (c) 6 (d) `-7`

Options

A. `4`
B. `5`
C. `6`
D. `-7`

Explanation

To solve the problem, we need to find the value of \( f\left(\frac{\sqrt{3}}{2}\right) \) where \( f(x) \) is defined by the integral

\[ f(x) = \int_{\cos x}^1 t^2 f(t) \, dt \]

Given that \( f(x) = 1 - \cos x \).

### Step-by-Step Solution

1. **Set Up the Integral**:
We start with the equation:
\[ f(x) = \int_{\cos x}^1 t^2 f(t) \, dt \]
We also know that \( f(x) = 1 - \cos x \).

2. **Differentiate Both Sides**:
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx} f(x) = \frac{d}{dx} \left( \int_{\cos x}^1 t^2 f(t) \, dt \right) \]

By the Leibniz rule for differentiation under the integral sign, we have:
\[ \frac{d}{dx} \int_{\cos x}^1 t^2 f(t) \, dt = -\sin x \cdot \cos^2 x f(\cos x) \]

Thus, we obtain:
\[ f'(\cos x) \cdot (-\sin x) = -\sin x + \sin x \]

3. **Simplify the Equation**:
This simplifies to:
\[ f'(\cos x) \cdot (-\sin x) = \sin x \]

Therefore, we can write:
\[ f'(\cos x) = \frac{1}{\cos^2 x} \]

4. **Find \( f(\cos x) \)**:
We have established that:
\[ f(\cos x) = \frac{1}{\cos^2 x} \]

5. **Substitute \( x \)**:
Now we need to find \( f\left(\frac{\sqrt{3}}{2}\right) \). We set \( \cos x = \frac{\sqrt{3}}{2} \), which gives us \( x = \frac{\pi}{6} \).

6. **Calculate \( f\left(\frac{\sqrt{3}}{2}\right) \)**:
Substitute \( \cos x = \frac{\sqrt{3}}{2} \) into the equation:
\[ f\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

7. **Evaluate the Greatest Integer Function**:
Now we evaluate the greatest integer function:
\[ \left\lfloor f\left(\frac{\sqrt{3}}{2}\right) \right\rfloor = \left\lfloor \frac{4}{3} \right\rfloor = 1 \]

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