Let `f` be a differentiable function satisfying `int_(0)^(f(x))f^(-1)(t)dt-int_(0)^(x)(cost-f(t)dt=0` and `f((pi)/2)=2/(pi)`
The value of `lim_(x to 0)(cosx)/(f(x))` is equal to where [.] denotes greatest integer function

Options

A. `0`
B. `1`
C. `1//2`
D. `2`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Given the integral equation
We start with the equation:
\[ \int_{0}^{f(x)} f^{-1}(t) \, dt - \int_{0}^{x} (\cos t - f(t)) \, dt = 0 \]

### Step 2: Differentiate both sides with respect to \(x\)
Using the Fundamental Theorem of Calculus and the chain rule, we differentiate both sides:
\[ \frac{d}{dx} \left( \int_{0}^{f(x)} f^{-1}(t) \, dt \right) - \frac{d}{dx} \left( \int_{0}^{x} (\cos t - f(t)) \, dt \right) = 0 \]
This gives us:
\[ f^{-1}(f(x)) \cdot f'(x) - (\cos x - f(x)) = 0 \]
Since \(f^{-1}(f(x)) = x\), we simplify:
\[ x f'(x) - \cos x + f(x) = 0 \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ x f'(x) + f(x) = \cos x \]

### Step 4: Recognizing the derivative
We can rewrite the equation as:
\[ \frac{d}{dx}(x f(x)) = \cos x \]

### Step 5: Integrate both sides
Integrating both sides with respect to \(x\):
\[ \int \frac{d}{dx}(x f(x)) \, dx = \int \cos x \, dx \]
This results in:
\[ x f(x) = \sin x + C \]
where \(C\) is a constant.

### Step 6: Solve for \(f(x)\)
From the equation, we can express \(f(x)\):
\[ f(x) = \frac{\sin x + C}{x} \]

### Step 7: Use the condition \(f\left(\frac{\pi}{2}\right) = \frac{2}{\pi}\)
Substituting \(x = \frac{\pi}{2}\):
\[ f\left(\frac{\pi}{2}\right) = \frac{\sin\left(\frac{\pi}{2}\right) + C}{\frac{\pi}{2}} = \frac{1 + C}{\frac{\pi}{2}} = \frac{2(1 + C)}{\pi} \]
Setting this equal to \(\frac{2}{\pi}\):
\[ \frac{2(1 + C)}{\pi} = \frac{2}{\pi} \]
This implies:
\[ 1 + C = 1 \implies C = 0 \]

### Step 8: Final expression for \(f(x)\)
Thus, we have:
\[ f(x) = \frac{\sin x}{x} \]

### Step 9: Find the limit
Now, we need to evaluate:
\[ \lim_{x \to 0} \frac{\cos x}{f(x)} = \lim_{x \to 0} \frac{\cos x}{\frac{\sin x}{x}} = \lim_{x \to 0} \frac{x \cos x}{\sin x} \]
Using the limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\):
\[ \lim_{x \to 0} \frac{x \cos x}{\sin x} = \lim_{x \to 0} \cos x \cdot \frac{x}{\sin x} = 1 \cdot 1 = 1 \]

### Step 10: Apply the greatest integer function
Finally, since the limit evaluates to \(1\), we apply the greatest integer function:
\[ \lfloor 1 \rfloor = 1 \]

### Final Answer
The value of \(\lim_{x \to 0} \frac{\cos x}{f(x)}\) is:
\[ \boxed{1} \]