If the function `f(x)={{:(asqrt(x+7),":",0lexlt9),(bx+5,":",xge9):}` is differentiable for `xge0`, then the value of `5a+6b` is equal to

Options

A. `(240)/(23)`
B. 10
C. `(80)/(23)`
D. `(250)/(23)`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is both continuous and differentiable at \( x = 9 \). The function is defined as follows:

\[ f(x) = \begin{cases} a \sqrt{x + 7} & \text{for } 0 \leq x < 9 \\ bx + 5 & \text{for } x \geq 9 \end{cases} \]

### Step 1: Ensure Continuity at \( x = 9 \)

For \( f(x) \) to be continuous at \( x = 9 \), the left-hand limit as \( x \) approaches 9 must equal the right-hand limit at \( x = 9 \):

\[ f(9^-) = f(9^+) \]

Calculating \( f(9^-) \):

\[ f(9^-) = a \sqrt{9 + 7} = a \sqrt{16} = 4a \]

Calculating \( f(9^+) \):

\[ f(9^+) = b(9) + 5 = 9b + 5 \]

Setting these equal gives us our first equation:

\[ 4a = 9b + 5 \quad \text{(Equation 1)} \]

### Step 2: Ensure Differentiability at \( x = 9 \)

For \( f(x) \) to be differentiable at \( x = 9 \), the derivative from the left must equal the derivative from the right:

\[ f'(9^-) = f'(9^+) \]

Calculating \( f'(x) \) for \( 0 \leq x < 9 \):

\[ f'(x) = \frac{d}{dx}(a \sqrt{x + 7}) = a \cdot \frac{1}{2\sqrt{x + 7}} \cdot 1 = \frac{a}{2\sqrt{x + 7}} \]

Evaluating at \( x = 9 \):

\[ f'(9^-) = \frac{a}{2\sqrt{16}} = \frac{a}{8} \]

Calculating \( f'(x) \) for \( x \geq 9 \):

\[ f'(x) = b \]

Setting these equal gives us our second equation:

\[ \frac{a}{8} = b \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations

From Equation 2, we can express \( a \) in terms of \( b \):

\[ a = 8b \]

Substituting this into Equation 1:

\[ 4(8b) = 9b + 5 \]

This simplifies to:

\[ 32b = 9b + 5 \]

Subtracting \( 9b \) from both sides:

\[ 32b - 9b = 5 \]

\[ 23b = 5 \]

Thus,

\[ b = \frac{5}{23} \]

Now substituting back to find \( a \):

\[ a = 8b = 8 \left(\frac{5}{23}\right) = \frac{40}{23} \]

### Step 4: Calculate \( 5a + 6b \)

Now we can calculate \( 5a + 6b \):

\[ 5a + 6b = 5\left(\frac{40}{23}\right) + 6\left(\frac{5}{23}\right) \]

Calculating each term:

\[ = \frac{200}{23} + \frac{30}{23} = \frac{230}{23} = 10 \]

### Final Answer

Thus, the value of \( 5a + 6b \) is:

\[ \boxed{10} \]