Let f and g be two differentiable functins such that:
`f (x)=g '(1) sin x+ (g'' (2) -1) x`
`g (x) = x^(2) -f'((pi)/(2)) x+ f'(-(pi)/(2))`
If `phi (x) =f ^(-1) (x)` then `phi'((pi)/(2) +1)` equals to :

Options

A. `(pi)/(2) +1`
B. `pi/2`
C. 1
D. 0

Answer

To solve the problem, we need to follow a series of steps to find \(\phi'(\frac{\pi}{2} + 1)\) where \(\phi(x) = f^{-1}(x)\).

### Step 1: Differentiate \(f(x)\)
Given:
\[ f(x) = g'(1) \sin x + (g''(2) - 1)x \]
We differentiate \(f(x)\) with respect to \(x\):
\[ f'(x) = g'(1) \cos x + (g''(2) - 1) \]

### Step 2: Differentiate \(g(x)\)
Given:
\[ g(x) = x^2 - f'(\frac{\pi}{2}) x + f'(-\frac{\pi}{2}) \]
We differentiate \(g(x)\) with respect to \(x\):
\[ g'(x) = 2x - f'(\frac{\pi}{2}) \]

### Step 3: Find \(g''(x)\)
Differentiate \(g'(x)\):
\[ g''(x) = 2 \]
Thus, \(g''(2) = 2\).

### Step 4: Substitute \(g''(2)\) into \(f(x)\)
From the expression for \(f(x)\):
\[ f(x) = g'(1) \sin x + (2 - 1)x = g'(1) \sin x + x \]

### Step 5: Find \(f'(\frac{\pi}{2})\)
Substituting \(x = \frac{\pi}{2}\) into \(f'(x)\):
\[ f'(\frac{\pi}{2}) = g'(1) \cos(\frac{\pi}{2}) + (2 - 1) = g'(1) \cdot 0 + 1 = 1 \]

### Step 6: Find \(f'(-\frac{\pi}{2})\)
Substituting \(x = -\frac{\pi}{2}\) into \(f'(x)\):
\[ f'(-\frac{\pi}{2}) = g'(1) \cos(-\frac{\pi}{2}) + (2 - 1) = g'(1) \cdot 0 + 1 = 1 \]

### Step 7: Find \(g'(1)\)
Using \(f'(\frac{\pi}{2}) = 1\):
\[ g'(1) = 2 - f'(\frac{\pi}{2}) = 2 - 1 = 1 \]

### Step 8: Find \(f(x)\) and \(g(x)\)
Substituting \(g'(1) = 1\) into \(f(x)\):
\[ f(x) = 1 \cdot \sin x + x = \sin x + x \]
Substituting \(f'(\frac{\pi}{2})\) and \(f'(-\frac{\pi}{2})\) into \(g(x)\):
\[ g(x) = x^2 - 1 \cdot x + 1 = x^2 - x + 1 \]

### Step 9: Set up the equation for \(\phi(x)\)
We have:
\[ x = f(y) = \sin y + y \]
Differentiating both sides with respect to \(x\):
\[ 1 = \cos y \cdot \frac{dy}{dx} + \frac{dy}{dx} \]
This simplifies to:
\[ 1 = (1 + \cos y) \frac{dy}{dx} \]
Thus,
\[ \frac{dy}{dx} = \frac{1}{1 + \cos y} \]

### Step 10: Find \(\phi'(\frac{\pi}{2} + 1)\)
Set \(x = \frac{\pi}{2} + 1\):
\[ \sin y + y = \frac{\pi}{2} + 1 \]
Assuming \(y = \frac{\pi}{2}\):
\[ \sin(\frac{\pi}{2}) + \frac{\pi}{2} = 1 + \frac{\pi}{2} = \frac{\pi}{2} + 1 \]
Thus, \(y = \frac{\pi}{2}\).

Substituting \(y = \frac{\pi}{2}\) into \(\frac{dy}{dx}\):
\[ \phi'(\frac{\pi}{2} + 1) = \frac{1}{1 + \cos(\frac{\pi}{2})} = \frac{1}{1 + 0} = 1 \]

### Final Answer
\[ \phi'(\frac{\pi}{2} + 1) = 1 \]