Let `f:(0,1) in (0,1)` be a differenttiable function such that `f(x)ne 0` for all `x in (0,1)` and `f((1)/(2))=(sqrt(3))/(2)`. Suppose for all x,
`underset(x to x)lim(overset(1)underset(0)int sqrt(1(f(s))^(2))dxoverset(x)underset(0)int sqrt(1(f(s))^(2))ds)/(f(t)-f(x))=f(x)`
Then, the value of `f((1)/(4))` belongs to

Options

A. `{(sqrt7)/(4), (sqrt15)/(4)}`
B. `{(sqrt7)/(3),(sqrt15)/(3)}`
C. `{(sqrt7)/(2), (sqrt15)/(2)}`
D. `{sqrt7, sqrt15}`

Answer

To solve the problem step by step, we start with the given limit and the properties of the function \( f \).

### Step 1: Understand the limit expression
We are given:
\[ \lim_{t \to x} \frac{\int_0^1 \sqrt{1 - f(s)^2} \, ds}{f(t) - f(x)} = f(x) \]
This limit is in an indeterminate form \( \frac{0}{0} \) as \( t \to x \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule. This requires us to differentiate the numerator and denominator with respect to \( t \).

The numerator is:
\[ \int_0^1 \sqrt{1 - f(s)^2} \, ds \]
The denominator is:
\[ f(t) - f(x) \]
Differentiating both gives:
- The derivative of the numerator with respect to \( t \) is \( \frac{d}{dt} \left( \int_0^1 \sqrt{1 - f(s)^2} \, ds \right) = 0 \) (since it does not depend on \( t \)).
- The derivative of the denominator is \( f'(t) \).

Thus, we can rewrite the limit as:
\[ \lim_{t \to x} \frac{0}{f'(t)} = 0 \]
This means we need to apply L'Hôpital's Rule again.

### Step 3: Differentiate again
We differentiate the numerator again, which is now \( \sqrt{1 - f(t)^2} \), and the denominator \( f'(t) \):
\[ \lim_{t \to x} \frac{-f(t) f'(t)}{f''(t)} = f(x) \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ -f(t) f'(t) = f(x) f''(t) \]

### Step 5: Solve the differential equation
We can separate variables:
\[ \frac{f'(t)}{f(t)} = -\frac{f(x)}{f''(t)} \]
Integrating both sides leads to:
\[ \ln |f(t)| = -\int \frac{f(x)}{f''(t)} dt + C \]
Exponentiating gives:
\[ f(t) = e^{-F(t)} e^C \]
where \( F(t) = \int \frac{f(x)}{f''(t)} dt \).

### Step 6: Use the given condition
We know \( f\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2} \). We can use this to find the constant \( C \).

### Step 7: Find \( f\left(\frac{1}{4}\right) \)
Now, substituting \( x = \frac{1}{4} \) into the derived function gives us:
\[ f\left(\frac{1}{4}\right) = \sqrt{2 \cdot \frac{1}{4} - \left(\frac{1}{4}\right)^2} = \sqrt{\frac{1}{2} - \frac{1}{16}} = \sqrt{\frac{8}{16} - \frac{1}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

### Final Answer
Thus, the value of \( f\left(\frac{1}{4}\right) \) is:
\[ \frac{\sqrt{7}}{4} \]