If `f (x)= [{:((sin [x^(2)]pi)/(x ^(2)-3x+8)+ax ^(3)+b,,, 0 le x le 1),( 2 cos pix + tan ^(-1)x ,,, 1 lt x le 2):}` is differentiable in `[0,2]` then: ([.] denotes greatest integer function)

Options

A. `a= 1/3`
B. `a=1/6`
C. `b =(pi)/(4)- (13)/(6)`
D. `b =(pi)/(4) -(7)/(3)`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is both continuous and differentiable over the interval \([0, 2]\). The function is defined piecewise:

1. For \( 0 \leq x \leq 1 \):
\[ f(x) = \frac{\sin(\lfloor x^2 \rfloor \pi)}{x^2 - 3x + 8} + ax^3 + b \]

2. For \( 1 < x \leq 2 \):
\[ f(x) = 2 \cos(\pi x) + \tan^{-1}(x) \]

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

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

Calculating \( f(1^-) \):
\[ f(1^-) = \frac{\sin(\lfloor 1^2 \rfloor \pi)}{1^2 - 3 \cdot 1 + 8} + a \cdot 1^3 + b = \frac{\sin(0)}{1 - 3 + 8} + a + b = 0 + a + b = a + b \]

Calculating \( f(1^+) \):
\[ f(1^+) = 2 \cos(\pi \cdot 1) + \tan^{-1}(1) = 2 \cdot (-1) + \frac{\pi}{4} = -2 + \frac{\pi}{4} \]

Setting these equal for continuity:
\[ a + b = -2 + \frac{\pi}{4} \quad \text{(1)} \]

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

Next, we need to ensure that \( f'(1^-) = f'(1^+) \).

Calculating \( f'(1^-) \):
Using the derivative of the first part:
\[ f'(x) = \frac{d}{dx}\left(\frac{\sin(\lfloor x^2 \rfloor \pi)}{x^2 - 3x + 8}\right) + 3ax^2 \]
For \( 0 \leq x < 1 \), \( \lfloor x^2 \rfloor = 0 \), so:
\[ f'(1^-) = 3a \cdot 1^2 = 3a \]

Calculating \( f'(1^+) \):
Using the derivative of the second part:
\[ f'(x) = -2\pi \sin(\pi x) + \frac{1}{1 + x^2} \]
At \( x = 1 \):
\[ f'(1^+) = -2\pi \sin(\pi) + \frac{1}{1 + 1^2} = 0 + \frac{1}{2} = \frac{1}{2} \]

Setting these equal for differentiability:
\[ 3a = \frac{1}{2} \quad \text{(2)} \]

### Step 3: Solve the Equations

From equation (2):
\[ a = \frac{1}{6} \]

Substituting \( a \) back into equation (1):
\[ \frac{1}{6} + b = -2 + \frac{\pi}{4} \]
\[ b = -2 + \frac{\pi}{4} - \frac{1}{6} \]

To combine the constants:
Convert \(-2\) to a fraction with a common denominator of 12:
\[ -2 = -\frac{24}{12} \]
Convert \(\frac{\pi}{4}\) to a fraction with a common denominator of 12:
\[ \frac{\pi}{4} = \frac{3\pi}{12} \]
Now substituting:
\[ b = -\frac{24}{12} + \frac{3\pi}{12} - \frac{2}{12} = \frac{3\pi - 26}{12} \]

### Final Values

Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{1}{6}, \quad b = \frac{3\pi - 26}{12} \]