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 \( x \in \mathbb{R} \). The function \( f(x) \) is defined as follows:

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

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

Since \( g(x) \) is continuous for all \( x \), \( f(x) \) must also be continuous at the point where the definition changes, which is at \( x = 1 \).

To ensure continuity at \( x = 1 \):

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

Calculating the left-hand limit:

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

Calculating the right-hand limit:

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

Setting these equal for continuity:

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

### Step 2: Ensure Differentiability at \( x = 1 \)

Next, we need to ensure that \( f(x) \) is differentiable at \( x = 1 \). This requires that the left-hand derivative and right-hand derivative at \( x = 1 \) are equal.

Calculating the left-hand derivative:

\[ f'(1^-) = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} \]

For \( 1-h < 1 \):

\[ f(1-h) = a(1-h)^2 + b = a(1 - 2h + h^2) + b = a - 2ah + ah^2 + b \]

Thus,

\[ f(1) = 1 \quad \text{(from the right-hand limit)} \]

Now substituting into the derivative limit:

\[ f'(1^-) = \lim_{h \to 0} \frac{(a - 2ah + ah^2 + b) - 1}{-h} = \lim_{h \to 0} \frac{(a + b - 1) - 2ah + ah^2}{-h} \]

Using \( a + b = 1 \):

\[ f'(1^-) = \lim_{h \to 0} \frac{-2ah + ah^2}{-h} = \lim_{h \to 0} (2a - ah) = 2a \]

Calculating the right-hand derivative:

\[ f'(1^+) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{\frac{1}{|1+h|} - 1}{h} = \lim_{h \to 0} \frac{\frac{1 - |1+h|}{|1+h|}}{h} \]

For small \( h \), \( |1+h| \approx 1 \):

\[ f'(1^+) = \lim_{h \to 0} \frac{-h}{h} = -1 \]

Setting the left-hand and right-hand derivatives equal:

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

### Step 3: Solve the Equations

From equation (2):

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

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

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

### Final Values

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

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