If function `f(x)={{:(asqrt(x+7),,,0lexlt2),(bx+1,,,xge2):}` is differentiable `a ge 0`, then the `2a+4b` is equal to

Options

A. 1
B. 5
C. 4
D. 9

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is both continuous and differentiable at the point where the definition of the function changes, which is at \( x = 2 \).

### Step 1: Define the function
The function is defined as:
\[ f(x) = \begin{cases} a \sqrt{x + 7} & \text{for } 0 \leq x < 2 \\ bx + 1 & \text{for } x \geq 2 \end{cases} \]

### Step 2: Ensure continuity at \( x = 2 \)
For \( f(x) \) to be continuous at \( x = 2 \), the left-hand limit (as \( x \) approaches 2 from the left) must equal the right-hand limit (as \( x \) approaches 2 from the right). Therefore, we set:
\[ f(2^-) = f(2^+) \]
Calculating \( f(2^-) \):
\[ f(2^-) = a \sqrt{2 + 7} = a \sqrt{9} = 3a \]
Calculating \( f(2^+) \):
\[ f(2^+) = b(2) + 1 = 2b + 1 \]
Setting these equal gives us our first equation:
\[ 3a = 2b + 1 \quad \text{(Equation 1)} \]

### Step 3: Ensure differentiability at \( x = 2 \)
For \( f(x) \) to be differentiable at \( x = 2 \), the derivatives from the left and right must also be equal:
\[ f'(2^-) = f'(2^+) \]
Calculating \( f'(2^-) \):
\[ f'(x) = \frac{d}{dx}(a \sqrt{x + 7}) = \frac{a}{2\sqrt{x + 7}} \]
Thus,
\[ f'(2^-) = \frac{a}{2\sqrt{2 + 7}} = \frac{a}{2\sqrt{9}} = \frac{a}{6} \]
Calculating \( f'(2^+) \):
\[ f'(x) = b \quad \text{(since the derivative of } bx + 1 \text{ is } b) \]
Setting these equal gives us our second equation:
\[ \frac{a}{6} = b \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 2 into Equation 1
From Equation 2, we have \( a = 6b \). Substituting this into Equation 1:
\[ 3(6b) = 2b + 1 \]
This simplifies to:
\[ 18b = 2b + 1 \]
Subtracting \( 2b \) from both sides:
\[ 16b = 1 \]
Thus,
\[ b = \frac{1}{16} \]

### Step 5: Find \( a \)
Substituting \( b \) back into Equation 2:
\[ a = 6b = 6 \left(\frac{1}{16}\right) = \frac{6}{16} = \frac{3}{8} \]

### Step 6: Calculate \( 2a + 4b \)
Now we can find \( 2a + 4b \):
\[ 2a + 4b = 2\left(\frac{3}{8}\right) + 4\left(\frac{1}{16}\right) \]
Calculating each term:
\[ 2a = \frac{6}{8} = \frac{3}{4} \]
\[ 4b = \frac{4}{16} = \frac{1}{4} \]
Adding these:
\[ 2a + 4b = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1 \]

### Final Answer:
Thus, the value of \( 2a + 4b \) is:
\[ \boxed{1} \]