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 at \( x = 3 \). The function is defined as:

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

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

For \( f(x) \) to be continuous at \( x = 3 \), we must have:

\[ \lim_{x \to 3^-} f(x) = f(3) = \lim_{x \to 3^+} f(x) \]

Calculating \( f(3) \):

\[ f(3) = 2(3^2) + 3 = 2(9) + 3 = 18 + 3 = 21 \]

Calculating \( \lim_{x \to 3^-} f(x) \):

\[ \lim_{x \to 3^-} f(x) = a(3^2) + b(3) + 1 = 9a + 3b + 1 \]

Setting the limits equal for continuity:

\[ 9a + 3b + 1 = 21 \]

This simplifies to:

\[ 9a + 3b = 20 \quad \text{(Equation 1)} \]

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

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

\[ \lim_{x \to 3^-} f'(x) = \lim_{x \to 3^+} f'(x) \]

Calculating the derivatives:

For \( x < 3 \):

\[ f'(x) = 2ax + b \]

For \( x \geq 3 \):

\[ f'(x) = 4x \]

Calculating \( \lim_{x \to 3^-} f'(x) \):

\[ \lim_{x \to 3^-} f'(x) = 2a(3) + b = 6a + b \]

Calculating \( \lim_{x \to 3^+} f'(x) \):

\[ \lim_{x \to 3^+} f'(x) = 4(3) = 12 \]

Setting the derivatives equal for differentiability:

\[ 6a + b = 12 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations

Now we have a system of 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:

\[ 9a + 3(12 - 6a) = 20 \]

Expanding and simplifying:

\[ 9a + 36 - 18a = 20 \]
\[ -9a + 36 = 20 \]
\[ -9a = 20 - 36 \]
\[ -9a = -16 \]
\[ a = \frac{16}{9} \]

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

\[ b = 12 - 6\left(\frac{16}{9}\right) \]
\[ b = 12 - \frac{96}{9} \]
\[ b = \frac{108}{9} - \frac{96}{9} = \frac{12}{9} = \frac{4}{3} \]

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

Now we can find \( \frac{a}{b^2} \):

\[ b^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]

Thus,

\[ \frac{a}{b^2} = \frac{\frac{16}{9}}{\frac{16}{9}} = 1 \]

### Final Answer

Therefore, the value of \( \frac{a}{b^2} \) is:

\[ \boxed{1} \]