Let `alpha (x) = f(x) -f (2x) and beta (x) =f (x) -f (4x) and alpha '(1) =5 alpha'(2) =7` then find the vlaue of `beta'(1)-10`

Answer

To solve the problem, we need to find the value of \( \beta'(1) - 10 \) given the functions \( \alpha(x) = f(x) - f(2x) \) and \( \beta(x) = f(x) - f(4x) \), along with the derivatives \( \alpha'(1) = 5 \) and \( \alpha'(2) = 7 \).

### Step 1: Differentiate \( \alpha(x) \)
We start by differentiating \( \alpha(x) \):
\[ \alpha'(x) = f'(x) - f'(2x) \cdot 2 \]
This is because we apply the chain rule to differentiate \( f(2x) \).

### Step 2: Evaluate \( \alpha'(1) \)
Now we substitute \( x = 1 \) into the derivative:
\[ \alpha'(1) = f'(1) - 2f'(2) \]
Given that \( \alpha'(1) = 5 \), we can write:
\[ f'(1) - 2f'(2) = 5 \quad \text{(Equation 1)} \]

### Step 3: Evaluate \( \alpha'(2) \)
Next, we substitute \( x = 2 \) into the derivative:
\[ \alpha'(2) = f'(2) - 2f'(4) \]
Given that \( \alpha'(2) = 7 \), we can write:
\[ f'(2) - 2f'(4) = 7 \quad \text{(Equation 2)} \]

### Step 4: Solve for \( f'(2) \)
From Equation 1, we can express \( f'(1) \):
\[ f'(1) = 5 + 2f'(2) \]

### Step 5: Substitute into Equation 2
Now we substitute \( f'(1) \) into Equation 2:
\[ f'(2) - 2f'(4) = 7 \]
We can rearrange this to express \( f'(4) \):
\[ f'(4) = \frac{f'(2) - 7}{2} \quad \text{(Equation 3)} \]

### Step 6: Differentiate \( \beta(x) \)
Now we differentiate \( \beta(x) \):
\[ \beta'(x) = f'(x) - f'(4x) \cdot 4 \]

### Step 7: Evaluate \( \beta'(1) \)
Substituting \( x = 1 \):
\[ \beta'(1) = f'(1) - 4f'(4) \]

### Step 8: Substitute \( f'(4) \)
Using Equation 3, we substitute \( f'(4) \) into the expression for \( \beta'(1) \):
\[ \beta'(1) = f'(1) - 4 \left(\frac{f'(2) - 7}{2}\right) \]
This simplifies to:
\[ \beta'(1) = f'(1) - 2(f'(2) - 7) \]
\[ \beta'(1) = f'(1) - 2f'(2) + 14 \]

### Step 9: Substitute \( f'(1) \)
Now we substitute \( f'(1) \) from Equation 1:
\[ \beta'(1) = (5 + 2f'(2)) - 2f'(2) + 14 \]
This simplifies to:
\[ \beta'(1) = 5 + 14 = 19 \]

### Step 10: Find \( \beta'(1) - 10 \)
Finally, we calculate \( \beta'(1) - 10 \):
\[ \beta'(1) - 10 = 19 - 10 = 9 \]

Thus, the final answer is:
\[ \boxed{9} \]