If the function `f(x)={{:((1+cospix)/(pi^(2)(x+1)^(2)),",",xlt-1),(ax+b,",",-1lexle1"is continuous" AA x "in R, then the value of 3a-b is-"),((x-1)/(lnsqrt(x)),",",xgt1):}`

Options

A. 3
B. 1
C. `-1`
D. `-3`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at the points \( x = -1 \) and \( x = 1 \). The function is defined piecewise, so we will analyze the limits from both sides at these points.

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

1. **Left-hand limit as \( x \to -1^- \)**:
\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{1 - \cos(\pi x)}{\pi^2 (x + 1)^2} \]
Substituting \( x = -1 \) gives:
\[ \frac{1 - \cos(-\pi)}{\pi^2 (0)^2} = \frac{1 - (-1)}{0} = \frac{2}{0} \text{ (undefined)} \]
Since we get \( 0/0 \), we apply L'Hôpital's Rule.

2. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}(1 - \cos(\pi x)) = \pi \sin(\pi x) \]
\[ \text{Denominator: } \frac{d}{dx}(\pi^2 (x + 1)^2) = \pi^2 \cdot 2(x + 1) \]
So we have:
\[ \lim_{x \to -1^-} \frac{\pi \sin(\pi x)}{\pi^2 \cdot 2(x + 1)} = \lim_{x \to -1} \frac{\sin(\pi x)}{2\pi (x + 1)} \]
Substituting \( x = -1 \) again gives \( 0/0 \), so we apply L'Hôpital's Rule again.

3. **Differentiate again**:
\[ \text{Numerator: } \frac{d}{dx}(\sin(\pi x)) = \pi \cos(\pi x) \]
\[ \text{Denominator: } \frac{d}{dx}(2\pi (x + 1)) = 2\pi \]
Now we have:
\[ \lim_{x \to -1} \frac{\pi \cos(\pi x)}{2\pi} = \frac{\cos(-\pi)}{2} = \frac{-1}{2} \]

4. **Right-hand limit as \( x \to -1^+ \)**:
\[ \lim_{x \to -1^+} f(x) = A(-1) + B = -A + B \]

5. **Set limits equal for continuity**:
\[ -A + B = -\frac{1}{2} \quad \text{(Equation 1)} \]

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

1. **Left-hand limit as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = A(1) + B = A + B \]

2. **Right-hand limit as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1} \frac{x - 1}{\ln(\sqrt{x})} \]
This also gives \( 0/0 \), so we apply L'Hôpital's Rule:
\[ \text{Numerator: } 1 \]
\[ \text{Denominator: } \frac{1}{2\sqrt{x}} \text{ (derivative of } \ln(\sqrt{x}) \text{)} \]
Thus:
\[ \lim_{x \to 1} \frac{1}{\frac{1}{2}} = 2 \]

3. **Set limits equal for continuity**:
\[ A + B = 2 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations

From Equation 1:
\[ -B + A = \frac{1}{2} \quad \text{(Rearranging gives } A = B + \frac{1}{2}\text{)} \]

Substituting into Equation 2:
\[ (B + \frac{1}{2}) + B = 2 \]
\[ 2B + \frac{1}{2} = 2 \Rightarrow 2B = \frac{3}{2} \Rightarrow B = \frac{3}{4} \]

Substituting \( B \) back to find \( A \):
\[ A = \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} \]

### Step 4: Calculate \( 3a - b \)

Now we need to find \( 3A - B \):
\[ 3A - B = 3 \left(\frac{5}{4}\right) - \frac{3}{4} = \frac{15}{4} - \frac{3}{4} = \frac{12}{4} = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]