IF `f(x)=A si n ((pi x)/2)+B and f'(1/2)=sqrt2 and int_0^1 f(x)dx =(2A)/pi` then what is the value of B

Options

A. A) `2/pi`
B. B) `4/pi`
C. C) 0
D. D) 1

Answer

To solve the problem step by step, we will analyze the given function and conditions.

### Step 1: Write down the function and given conditions
We have:
\[ f(x) = A \sin\left(\frac{\pi x}{2}\right) + B \]
Given:
1. \( f'\left(\frac{1}{2}\right) = \sqrt{2} \)
2. \( \int_0^1 f(x) \, dx = \frac{2A}{\pi} \)

### Step 2: Find the derivative of \( f(x) \)
To find \( f'(x) \):
\[ f'(x) = A \cdot \frac{d}{dx}\left(\sin\left(\frac{\pi x}{2}\right)\right) + 0 \]
Using the chain rule:
\[ f'(x) = A \cdot \cos\left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2} \]
Thus:
\[ f'(x) = \frac{A\pi}{2} \cos\left(\frac{\pi x}{2}\right) \]

### Step 3: Evaluate \( f'\left(\frac{1}{2}\right) \)
Now, substituting \( x = \frac{1}{2} \):
\[ f'\left(\frac{1}{2}\right) = \frac{A\pi}{2} \cos\left(\frac{\pi \cdot \frac{1}{2}}{2}\right) = \frac{A\pi}{2} \cos\left(\frac{\pi}{4}\right) \]
Since \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \):
\[ f'\left(\frac{1}{2}\right) = \frac{A\pi}{2} \cdot \frac{\sqrt{2}}{2} = \frac{A\pi\sqrt{2}}{4} \]

### Step 4: Set up the equation from the condition \( f'\left(\frac{1}{2}\right) = \sqrt{2} \)
Setting the expression equal to \( \sqrt{2} \):
\[ \frac{A\pi\sqrt{2}}{4} = \sqrt{2} \]
Dividing both sides by \( \sqrt{2} \):
\[ \frac{A\pi}{4} = 1 \]
Thus:
\[ A\pi = 4 \]
\[ A = \frac{4}{\pi} \]

### Step 5: Evaluate the integral \( \int_0^1 f(x) \, dx \)
Now we need to evaluate the integral:
\[ \int_0^1 f(x) \, dx = \int_0^1 \left(A \sin\left(\frac{\pi x}{2}\right) + B\right) \, dx \]
This can be split into two parts:
\[ = \int_0^1 A \sin\left(\frac{\pi x}{2}\right) \, dx + \int_0^1 B \, dx \]

### Step 6: Calculate \( \int_0^1 \sin\left(\frac{\pi x}{2}\right) \, dx \)
Using the substitution \( u = \frac{\pi x}{2} \), \( du = \frac{\pi}{2}dx \), thus \( dx = \frac{2}{\pi}du \):
When \( x = 0 \), \( u = 0 \) and when \( x = 1 \), \( u = \frac{\pi}{2} \):
\[ \int_0^1 \sin\left(\frac{\pi x}{2}\right) \, dx = \int_0^{\frac{\pi}{2}} \sin(u) \cdot \frac{2}{\pi} \, du \]
\[ = \frac{2}{\pi} \left[-\cos(u)\right]_0^{\frac{\pi}{2}} = \frac{2}{\pi} \left[-\cos\left(\frac{\pi}{2}\right) + \cos(0)\right] = \frac{2}{\pi} (0 + 1) = \frac{2}{\pi} \]

### Step 7: Substitute back into the integral
Now substituting back:
\[ \int_0^1 f(x) \, dx = A \cdot \frac{2}{\pi} + B \cdot 1 = \frac{4}{\pi} \cdot \frac{2}{\pi} + B = \frac{8}{\pi^2} + B \]

### Step 8: Set the integral equal to \( \frac{2A}{\pi} \)
From the given condition:
\[ \frac{8}{\pi^2} + B = \frac{2A}{\pi} \]
Substituting \( A = \frac{4}{\pi} \):
\[ \frac{8}{\pi^2} + B = \frac{2 \cdot \frac{4}{\pi}}{\pi} = \frac{8}{\pi^2} \]

### Step 9: Solve for \( B \)
Thus:
\[ B = \frac{8}{\pi^2} - \frac{8}{\pi^2} = 0 \]

### Conclusion
The value of \( B \) is:
\[ B = 0 \]