Let `f(x)={{:(,(1)/(|x|),"if "|x| gt 2,"then "f(x)is),(,a+bx^(2), , "if"|x| le2):}` is differentiable at x=-2 for

Options

A. `a=(3)/(4), b=(1)/(6)`
B. `a=(3)/(4), b=(1)/(16)`
C. `a=-(1)/(4), b=(1)/(16)`
D. `a=(1)/(4), b=-(1)/(16)`

Explanation

To determine the values of \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} \frac{1}{|x|} & \text{if } |x| > 2 \\ a + bx^2 & \text{if } |x| \leq 2 \end{cases} \]

is differentiable at \( x = -2 \), we need to ensure that \( f(x) \) is continuous and differentiable at that point.

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

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

\[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) = f(-2) \]

Calculating \( f(-2) \):

Since \( |-2| \leq 2 \), we use the second case:

\[ f(-2) = a + b(-2)^2 = a + 4b \]

Now, we calculate the left-hand limit:

\[ \lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} (a + bx^2) = a + b(-2)^2 = a + 4b \]

Next, we calculate the right-hand limit:

\[ \lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} \frac{1}{|x|} = \frac{1}{|-2|} = \frac{1}{2} \]

Setting the left-hand limit equal to the right-hand limit for continuity:

\[ a + 4b = \frac{1}{2} \quad \text{(Equation 1)} \]

### Step 2: Check Differentiability at \( x = -2 \)

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

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

Calculating the left-hand derivative:

\[ f'(x) = \frac{d}{dx}(a + bx^2) = 2bx \]

Thus,

\[ \lim_{x \to -2^-} f'(x) = 2b(-2) = -4b \]

Now, calculating the right-hand derivative:

\[ f'(x) = \frac{d}{dx}\left(\frac{1}{|x|}\right) = -\frac{1}{x^2} \]

Thus,

\[ \lim_{x \to -2^+} f'(x) = -\frac{1}{(-2)^2} = -\frac{1}{4} \]

Setting the left-hand derivative equal to the right-hand derivative:

\[ -4b = -\frac{1}{4} \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

From Equation 2:

\[ 4b = \frac{1}{4} \implies b = \frac{1}{16} \]

Substituting \( b \) back into Equation 1:

\[ a + 4\left(\frac{1}{16}\right) = \frac{1}{2} \]

This simplifies to:

\[ a + \frac{1}{4} = \frac{1}{2} \implies a = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \]

### Final Values

Thus, the values of \( a \) and \( b \) that make \( f(x) \) differentiable at \( x = -2 \) are:

\[ a = \frac{1}{4}, \quad b = \frac{1}{16} \]