Let `f(x)={(1/(|x|), for |x| ge 1),(ax^2+b,for |x| < 1))` . If `f(x)` is continuous and differentiable at any point, then

Options

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

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable at the points where the definition of the function changes, specifically at \( x = -1 \) and \( x = 1 \).

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

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

For \( f(x) \) to be continuous at \( x = 1 \), we need:
\[ \lim_{x \to 1^-} f(x) = f(1) \]

Calculating \( f(1) \):
\[ f(1) = \frac{1}{|1|} = 1 \]

Now, calculate \( \lim_{x \to 1^-} f(x) \):
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (ax^2 + b) = a(1)^2 + b = a + b \]

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

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

For \( f(x) \) to be continuous at \( x = -1 \), we need:
\[ \lim_{x \to -1^-} f(x) = f(-1) \]

Calculating \( f(-1) \):
\[ f(-1) = \frac{1}{|-1|} = 1 \]

Now, calculate \( \lim_{x \to -1^+} f(x) \):
\[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (ax^2 + b) = a(-1)^2 + b = a + b \]

Setting these equal for continuity:
\[ a + b = 1 \quad \text{(This is the same as Equation 1)} \]

### Step 3: Check Differentiability at \( x = 1 \)

For \( f(x) \) to be differentiable at \( x = 1 \), we need:
\[ \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x) \]

Calculating the left-hand derivative:
\[ f'(x) = \frac{d}{dx}(ax^2 + b) = 2ax \]
Thus,
\[ \lim_{x \to 1^-} f'(x) = 2a(1) = 2a \]

Calculating the right-hand derivative:
\[ f'(x) = \frac{d}{dx}\left(\frac{1}{|x|}\right) = -\frac{1}{x^2} \]
Thus,
\[ \lim_{x \to 1^+} f'(x) = -\frac{1}{(1)^2} = -1 \]

Setting these equal for differentiability:
\[ 2a = -1 \quad \Rightarrow \quad a = -\frac{1}{2} \quad \text{(Equation 2)} \]

### Step 4: Solve for \( b \)

Substituting \( a = -\frac{1}{2} \) into Equation 1:
\[ -\frac{1}{2} + b = 1 \quad \Rightarrow \quad b = 1 + \frac{1}{2} = \frac{3}{2} \]

### Final Values

Thus, we have:
\[ a = -\frac{1}{2}, \quad b = \frac{3}{2} \]

### Conclusion

The values of \( a \) and \( b \) that make \( f(x) \) continuous and differentiable at all points are:
\[ a = -\frac{1}{2}, \quad b = \frac{3}{2} \]