If `f(x)={{:(2x^(2)+3,,,xge3),(ax^(2)+bx+1,,,xle3):}` is differentiable everywhere, then `(a)/(b^(2))` is equal to

Options

A. 5
B. `(7)/(3)`
C. 1
D. `(16)/(9)`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable everywhere, particularly at the point \( x = 3 \).

### Step 1: Ensure Continuity at \( x = 3 \)

The function is defined as:
\[ f(x) = \begin{cases} 2x^2 + 3 & \text{if } x \geq 3 \\ ax^2 + bx + 1 & \text{if } x < 3 \end{cases} \]

For \( f(x) \) to be continuous at \( x = 3 \), we need:
\[ f(3^-) = f(3^+) \]

Calculating \( f(3^-) \):
\[ f(3^-) = a(3^2) + b(3) + 1 = 9a + 3b + 1 \]

Calculating \( f(3^+) \):
\[ f(3^+) = 2(3^2) + 3 = 18 + 3 = 21 \]

Setting these equal for continuity:
\[ 9a + 3b + 1 = 21 \]
\[ 9a + 3b = 20 \quad \text{(Equation 1)} \]

### Step 2: Ensure Differentiability at \( x = 3 \)

Next, we need to ensure that the derivatives from both sides at \( x = 3 \) are equal.

Calculating \( f'(x) \) for \( x < 3 \):
\[ f'(x) = 2ax + b \]
Thus,
\[ f'(3^-) = 2a(3) + b = 6a + b \]

Calculating \( f'(x) \) for \( x \geq 3 \):
\[ f'(x) = 4x \]
Thus,
\[ f'(3^+) = 4(3) = 12 \]

Setting these equal for differentiability:
\[ 6a + b = 12 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations

We now have two equations:
1. \( 9a + 3b = 20 \)
2. \( 6a + b = 12 \)

From Equation 2, we can express \( b \) in terms of \( a \):
\[ b = 12 - 6a \]

Substituting this expression for \( b \) into Equation 1:
\[ 9a + 3(12 - 6a) = 20 \]
\[ 9a + 36 - 18a = 20 \]
\[ -9a + 36 = 20 \]
\[ -9a = 20 - 36 \]
\[ -9a = -16 \]
\[ a = \frac{16}{9} \]

Now substituting \( a \) back into the expression for \( b \):
\[ b = 12 - 6\left(\frac{16}{9}\right) \]
\[ b = 12 - \frac{96}{9} \]
\[ b = \frac{108}{9} - \frac{96}{9} = \frac{12}{9} = \frac{4}{3} \]

### Step 4: Calculate \( \frac{a}{b^2} \)

Now we can calculate \( \frac{a}{b^2} \):
\[ b^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
\[ \frac{a}{b^2} = \frac{\frac{16}{9}}{\frac{16}{9}} = 1 \]

Thus, the final answer is:
\[ \frac{a}{b^2} = 1 \]