Let `f (x)` be function defined on `[0,1] ` such that `f (1)=0` and for any `a in (0,1], int _(0)^(a) f (x) dx - int _(a)^(1) f (x) dx =2 f (a) +3a +b` where b is constant.
The length of the subtangent of the curve ` y= f (x ) at x=1//2` is:

Options

A. `sqrte-1`
B. `(sqrte-1)/(2)`
C. `sqrte+1`
D. `(sqrte+1)/(2)`

Answer

To solve the problem step by step, we will follow the given conditions and use integration properties to find the length of the subtangent of the curve \( y = f(x) \) at \( x = \frac{1}{2} \).

### Step 1: Understanding the given condition
We are given that for any \( a \in (0, 1] \):
\[ \int_0^a f(x) \, dx - \int_a^1 f(x) \, dx = 2f(a) + 3a + b \]
where \( b \) is a constant.

### Step 2: Simplifying the integral expression
Using the property of definite integrals, we can rewrite the left-hand side:
\[ \int_0^1 f(x) \, dx = \int_0^a f(x) \, dx + \int_a^1 f(x) \, dx \]
Thus, we can express the equation as:
\[ \int_0^a f(x) \, dx = \int_0^1 f(x) \, dx - \int_a^1 f(x) \, dx \]
Substituting this into our original equation gives:
\[ \int_0^a f(x) \, dx - \int_a^1 f(x) \, dx = 2f(a) + 3a + b \]

### Step 3: Applying Newton-Leibniz formula
Using the Newton-Leibniz formula, we differentiate both sides with respect to \( a \):
\[ f(a) - (-f(a)) = 2f'(a) + 3 \]
This simplifies to:
\[ 2f(a) = 2f'(a) + 3 \]
Thus, we have:
\[ f(a) = f'(a) + \frac{3}{2} \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ f'(a) = f(a) - \frac{3}{2} \]

### Step 5: Solving the differential equation
This is a first-order linear differential equation. We can solve it using the method of integrating factors. The integrating factor \( e^{\int 1 \, da} = e^a \):
\[ e^a f'(a) - \frac{3}{2} e^a = \frac{d}{da}(e^a f(a)) \]
Integrating both sides:
\[ \int \frac{d}{da}(e^a f(a)) \, da = \int \left( \frac{3}{2} e^a \right) \, da \]
This gives:
\[ e^a f(a) = \frac{3}{2} e^a + C \]
where \( C \) is a constant.

### Step 6: Finding \( f(a) \)
Dividing by \( e^a \):
\[ f(a) = \frac{3}{2} + Ce^{-a} \]

### Step 7: Using the condition \( f(1) = 0 \)
Substituting \( a = 1 \):
\[ 0 = \frac{3}{2} + Ce^{-1} \]
Thus, we find:
\[ C = -\frac{3}{2} e \]
So,
\[ f(a) = \frac{3}{2} - \frac{3}{2} e^{-a} \]

### Step 8: Finding \( f\left(\frac{1}{2}\right) \) and \( f'\left(\frac{1}{2}\right) \)
Now, we need to find \( f\left(\frac{1}{2}\right) \):
\[ f\left(\frac{1}{2}\right) = \frac{3}{2} - \frac{3}{2} e^{-\frac{1}{2}} = \frac{3}{2}(1 - e^{-\frac{1}{2}}) \]
Next, we find \( f'(a) \):
\[ f'(a) = \frac{3}{2} e^{-a} \]
Thus,
\[ f'\left(\frac{1}{2}\right) = \frac{3}{2} e^{-\frac{1}{2}} \]

### Step 9: Finding the length of the subtangent
The length of the subtangent at \( x = \frac{1}{2} \) is given by:
\[ \text{Length of subtangent} = \frac{|f\left(\frac{1}{2}\right)|}{f'\left(\frac{1}{2}\right)} \]
Substituting the values:
\[ \text{Length of subtangent} = \frac{\frac{3}{2}(1 - e^{-\frac{1}{2}})}{\frac{3}{2} e^{-\frac{1}{2}}} = \frac{1 - e^{-\frac{1}{2}}}{e^{-\frac{1}{2}}} \]
This simplifies to:
\[ \text{Length of subtangent} = e^{\frac{1}{2}}(1 - e^{-\frac{1}{2}}) = \sqrt{e}(1 - \frac{1}{\sqrt{e}}) = \sqrt{e} - 1 \]

### Final Answer
The length of the subtangent of the curve \( y = f(x) \) at \( x = \frac{1}{2} \) is:
\[ \sqrt{e} - 1 \]