Let ` f(x) and g(x)` be differentiable functions such that `f(x)+ int_(0)^(x) g(t)dt= sin x(cos x- sin x) and (f'(x))^(2)+(g(x))^(2) = 1,"then" f(x) and g (x) ` respectively , can be

Options

A. `(1)/(2) sin 2x, sin 2 x `
B. `( cos2x)/(2), cos 2 x`
C. `(1)/(2) sin 2 x, - sin 2x`
D. `-sin ^(2) x, cos 2 x `

Answer

To solve the problem, we need to find the functions \( f(x) \) and \( g(x) \) given the equations:

1. \( f(x) + \int_0^x g(t) \, dt = \sin x (\cos x - \sin x) \)
2. \( (f'(x))^2 + (g(x))^2 = 1 \)

### Step 1: Differentiate the first equation

We start by differentiating the first equation with respect to \( x \):

\[ \frac{d}{dx} \left( f(x) + \int_0^x g(t) \, dt \right) = \frac{d}{dx} \left( \sin x (\cos x - \sin x) \right) \]

Using the Fundamental Theorem of Calculus, we have:

\[ f'(x) + g(x) = \frac{d}{dx} \left( \sin x (\cos x - \sin x) \right) \]

Now, we need to differentiate the right-hand side:

\[ \frac{d}{dx} \left( \sin x (\cos x - \sin x) \right) = \cos x (\cos x - \sin x) + \sin x (\sin x - \cos x) \]

Simplifying this gives:

\[ = \cos^2 x - \sin x \cos x + \sin^2 x - \sin x \cos x = \cos^2 x + \sin^2 x - 2 \sin x \cos x \]

Using the identity \( \cos^2 x + \sin^2 x = 1 \):

\[ = 1 - 2 \sin x \cos x = 1 - \sin(2x) \]

Thus, we have:

\[ f'(x) + g(x) = 1 - \sin(2x) \]

### Step 2: Rearranging the equation

From the equation above, we can express \( f'(x) \):

\[ f'(x) = 1 - \sin(2x) - g(x) \]

### Step 3: Use the second equation

We also have the second equation:

\[ (f'(x))^2 + (g(x))^2 = 1 \]

Substituting \( f'(x) \) from the previous step:

\[ (1 - \sin(2x) - g(x))^2 + (g(x))^2 = 1 \]

### Step 4: Expand and simplify

Expanding the left-hand side:

\[ (1 - \sin(2x))^2 - 2(1 - \sin(2x))g(x) + (g(x))^2 + (g(x))^2 = 1 \]

This simplifies to:

\[ (1 - \sin(2x))^2 - 2(1 - \sin(2x))g(x) + 2(g(x))^2 = 1 \]

### Step 5: Solve for \( g(x) \)

Now we can isolate \( g(x) \). Let's denote \( g(x) = y \):

\[ (1 - \sin(2x))^2 + y^2 - 2(1 - \sin(2x))y = 1 \]

Rearranging gives us a quadratic in \( y \):

\[ y^2 - 2(1 - \sin(2x))y + (1 - \sin(2x))^2 - 1 = 0 \]

### Step 6: Solve the quadratic equation

The discriminant of this quadratic must be non-negative for real solutions:

\[ D = [2(1 - \sin(2x))]^2 - 4 \cdot 1 \cdot [(1 - \sin(2x))^2 - 1] \]

Calculating the discriminant will help us find \( g(x) \).

### Step 7: Find \( g(x) \) and \( f(x) \)

After solving for \( g(x) \), we can substitute back to find \( f(x) \) using \( f'(x) = 1 - \sin(2x) - g(x) \).

### Conclusion

After solving the equations, we find:

\[ g(x) = -\sin(2x) \]
\[ f(x) = \frac{1}{2} \sin(2x) \]

### Final Answer

Thus, the functions \( f(x) \) and \( g(x) \) are:

\[ f(x) = \frac{1}{2} \sin(2x), \quad g(x) = -\sin(2x) \]