If the fucntion `f(x)={{:(asqrt(x+7),,,0lexlt2),(bx+5,,,x ge2):}` is differentiable `AA x ge0`, then `2a+4b` is equal to

Options

A. `(240)/(16)`
B. 5
C. `(85)/(16)`
D. `(250)/(16)`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at \( x = 2 \). The function is defined as follows:

\[ f(x) = \begin{cases} a\sqrt{x + 7} & \text{for } 0 \leq x < 2 \\ bx + 5 & \text{for } x \geq 2 \end{cases} \]

### Step 1: Ensure Continuity at \( x = 2 \)

For \( f(x) \) to be continuous at \( x = 2 \), we need:

\[ f(2^-) = f(2^+) \]

Calculating \( f(2^-) \):

\[ f(2^-) = a\sqrt{2 + 7} = a\sqrt{9} = 3a \]

Calculating \( f(2^+) \):

\[ f(2^+) = b(2) + 5 = 2b + 5 \]

Setting these equal for continuity:

\[ 3a = 2b + 5 \quad \text{(Equation 1)} \]

### Step 2: Ensure Differentiability at \( x = 2 \)

For \( f(x) \) to be differentiable at \( x = 2 \), we need:

\[ f'(2^-) = f'(2^+) \]

Calculating \( f'(2^-) \):

Using the derivative of \( f(x) = a\sqrt{x + 7} \):

\[ f'(x) = \frac{a}{2\sqrt{x + 7}} \]

Thus,

\[ f'(2^-) = \frac{a}{2\sqrt{2 + 7}} = \frac{a}{2\sqrt{9}} = \frac{a}{6} \]

Calculating \( f'(2^+) \):

\[ f'(x) = b \quad \text{(since the derivative of } bx + 5 \text{ is } b\text{)} \]

Setting these equal for differentiability:

\[ \frac{a}{6} = b \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

Now we have two equations:

1. \( 3a = 2b + 5 \)
2. \( b = \frac{a}{6} \)

Substituting Equation 2 into Equation 1:

\[ 3a = 2\left(\frac{a}{6}\right) + 5 \]

This simplifies to:

\[ 3a = \frac{2a}{6} + 5 \]

Multiplying through by 6 to eliminate the fraction:

\[ 18a = 2a + 30 \]

Rearranging gives:

\[ 18a - 2a = 30 \implies 16a = 30 \implies a = \frac{30}{16} = \frac{15}{8} \]

Now substituting \( a \) back into Equation 2 to find \( b \):

\[ b = \frac{a}{6} = \frac{\frac{15}{8}}{6} = \frac{15}{48} = \frac{5}{16} \]

### Step 4: Calculate \( 2a + 4b \)

Now we can calculate \( 2a + 4b \):

\[ 2a + 4b = 2\left(\frac{15}{8}\right) + 4\left(\frac{5}{16}\right) \]

Calculating each term:

\[ 2a = \frac{30}{8} = \frac{15}{4} \]
\[ 4b = \frac{20}{16} = \frac{5}{4} \]

Adding these together:

\[ 2a + 4b = \frac{15}{4} + \frac{5}{4} = \frac{20}{4} = 5 \]

Thus, the final answer is:

\[ \boxed{5} \]