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 both 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 \( x = 2 \), the left-hand limit as \( x \) approaches 2 must equal the right-hand limit at \( x = 2 \).

**Left-hand limit:**
\[ \lim_{x \to 2^-} f(x) = a \sqrt{2 + 7} = a \sqrt{9} = 3a \]

**Right-hand limit:**
\[ \lim_{x \to 2^+} f(x) = b(2) + 5 = 2b + 5 \]

Setting these equal for continuity:
\[ 3a = 2b + 5 \quad \text{(1)} \]

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

For \( f(x) \) to be differentiable at \( x = 2 \), the derivative from the left must equal the derivative from the right.

**Derivative from the left:**
\[ f'(x) = \frac{d}{dx}(a \sqrt{x + 7}) = a \cdot \frac{1}{2\sqrt{x + 7}} \quad \text{(using chain rule)} \]
Evaluating at \( x = 2 \):
\[ f'(2^-) = a \cdot \frac{1}{2\sqrt{2 + 7}} = a \cdot \frac{1}{2\sqrt{9}} = \frac{a}{6} \]

**Derivative from the right:**
\[ f'(x) = b \quad \text{(since the derivative of } bx + 5 \text{ is } b\text{)} \]
So:
\[ f'(2^+) = b \]

Setting these equal for differentiability:
\[ \frac{a}{6} = b \quad \text{(2)} \]

### Step 3: Solve the System of Equations

Now we have a system of two equations:
1. \( 3a = 2b + 5 \)
2. \( b = \frac{a}{6} \)

Substituting equation (2) into equation (1):
\[ 3a = 2\left(\frac{a}{6}\right) + 5 \]
\[ 3a = \frac{2a}{6} + 5 \]
Multiplying through by 6 to eliminate the fraction:
\[ 18a = 2a + 30 \]
\[ 18a - 2a = 30 \]
\[ 16a = 30 \]
\[ a = \frac{30}{16} = \frac{15}{8} \]

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

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

### Final Answer
Thus, \( 2a + 4b = 5 \).

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