Consider a function `g(x)=f(x-2),AA x in R`, where `f(x)={{:((1)/(|x|),":",|x|ge1),(ax^(2)+b,":",|x|lt1):}`. If `g(x)` is continuous as well as differentiable for all x, then

Options

A. `a=(-1)/(2),b=(3)/(2)`
B. `a=(1)/(2), b=(3)/(2)`
C. `a=(-1)/(2), b=(-3)/(2)`
D. None of these

Answer

To solve the problem, we need to ensure that the function \( g(x) = f(x-2) \) is continuous and differentiable for all \( x \). The function \( f(x) \) is defined piecewise as follows:

\[ f(x) = \begin{cases} \frac{1}{|x|} & \text{if } |x| \ge 1 \\ ax^2 + b & \text{if } |x| < 1 \end{cases} \]

### Step 1: Determine the intervals for \( g(x) \)

Since \( g(x) = f(x-2) \), we need to analyze the behavior of \( f(x) \) at \( x-2 \).

- For \( |x-2| \ge 1 \): This occurs when \( x \le 1 \) or \( x \ge 3 \).
- For \( |x-2| < 1 \): This occurs when \( 1 < x < 3 \).

Thus, we can write \( g(x) \) as:

\[ g(x) = \begin{cases} \frac{1}{|x-2|} & \text{if } x \le 1 \text{ or } x \ge 3 \\ a(x-2)^2 + b & \text{if } 1 < x < 3 \end{cases} \]

### Step 2: Check continuity at \( x = 3 \)

For \( g(x) \) to be continuous at \( x = 3 \), we need:

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

\[ \lim_{x \to 3^+} g(x) = \frac{1}{|3-2|} = 1 \]

Setting these equal for continuity:

\[ a + b = 1 \quad \text{(1)} \]

### Step 3: Check differentiability at \( x = 3 \)

For \( g(x) \) to be differentiable at \( x = 3 \), we need:

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

Calculating the left-hand derivative:

\[ g'(x) = \frac{d}{dx}(a(x-2)^2 + b) = 2a(x-2) \]
So,
\[ \lim_{x \to 3^-} g'(x) = 2a(3-2) = 2a \]

Calculating the right-hand derivative:

\[ g'(x) = \frac{d}{dx}\left(\frac{1}{|x-2|}\right) = -\frac{1}{(x-2)^2} \]
So,
\[ \lim_{x \to 3^+} g'(x) = -\frac{1}{(3-2)^2} = -1 \]

Setting these equal for differentiability:

\[ 2a = -1 \quad \text{(2)} \]

### Step 4: Solve the equations

From equation (2):

\[ a = -\frac{1}{2} \]

Substituting \( a \) into equation (1):

\[ -\frac{1}{2} + b = 1 \]

Solving for \( b \):

\[ b = 1 + \frac{1}{2} = \frac{3}{2} \]

### Final Result

Thus, the values of \( a \) and \( b \) are:

\[ a = -\frac{1}{2}, \quad b = \frac{3}{2} \]