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)`a = (1)/(6), b = (pi)/(4) - (13)/(6)`
B. (B)`a =-(1)/(6), b = (pi)/(4)`
C. (C)`a=-(1)/(6), b = (pi)/(4) - (13)/(6)`
D. (D)None of these

Answer

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

\[ f(x) = \begin{cases} \frac{\sin(\pi x^2)}{x^2 - 3x + 8} + ax^3 + b & \text{for } 0 \leq x \leq 1 \\ 2 \cos(\pi x) + \tan^{-1}(x) & \text{for } 1 < x \leq 2 \end{cases} \]

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

To ensure continuity at \( x = 1 \), we need to set the two pieces equal to each other at \( x = 1 \):

\[ \frac{\sin(\pi \cdot 1^2)}{1^2 - 3 \cdot 1 + 8} + a \cdot 1^3 + b = 2 \cos(\pi \cdot 1) + \tan^{-1}(1) \]

Calculating each side:

- Left side:
\[ \frac{\sin(\pi)}{1 - 3 + 8} + a + b = \frac{0}{6} + a + b = a + b \]

- Right side:
\[ 2 \cdot (-1) + \frac{\pi}{4} = -2 + \frac{\pi}{4} \]

Setting them equal gives:
\[ a + b = -2 + \frac{\pi}{4} \quad (1) \]

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

Next, we differentiate both pieces of the function and set them equal at \( x = 1 \):

For \( 0 \leq x \leq 1 \):
\[ f'(x) = \frac{d}{dx}\left(\frac{\sin(\pi x^2)}{x^2 - 3x + 8} + ax^3 + b\right) \]
Using the product and chain rule, we differentiate:
\[ f'(x) = \frac{2\pi x \cos(\pi x^2)(x^2 - 3x + 8) - \sin(\pi x^2)(2x - 3)}{(x^2 - 3x + 8)^2} + 3ax^2 \]

At \( x = 1 \):
\[ f'(1) = \frac{2\pi \cdot 1 \cdot \cos(\pi)}{(1 - 3 + 8)^2} + 3a = \frac{-2\pi}{36} + 3a = -\frac{\pi}{18} + 3a \quad (2) \]

For \( 1 < x \leq 2 \):
\[ f'(x) = -2\pi \sin(\pi x) + \frac{1}{1 + x^2} \]

At \( x = 1 \):
\[ f'(1) = -2\pi \sin(\pi) + \frac{1}{2} = 0 + \frac{1}{2} = \frac{1}{2} \]

Setting the derivatives equal gives:
\[ -\frac{\pi}{18} + 3a = \frac{1}{2} \quad (3) \]

### Step 3: Solve the System of Equations

Now we have a system of equations from (1) and (3):

1. \( a + b = -2 + \frac{\pi}{4} \)
2. \( -\frac{\pi}{18} + 3a = \frac{1}{2} \)

From equation (3):
\[ 3a = \frac{1}{2} + \frac{\pi}{18} \]
\[ a = \frac{1}{6} + \frac{\pi}{54} \]

Substituting \( a \) back into equation (1):
\[ \frac{1}{6} + \frac{\pi}{54} + b = -2 + \frac{\pi}{4} \]
\[ b = -2 + \frac{\pi}{4} - \left(\frac{1}{6} + \frac{\pi}{54}\right) \]

### Step 4: Final Values

Calculating \( b \):
\[ b = -2 + \frac{\pi}{4} - \frac{1}{6} - \frac{\pi}{54} \]
Finding a common denominator and simplifying gives the final values for \( a \) and \( b \).

### Conclusion

Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{1}{6}, \quad b = -2 + \frac{\pi}{4} - \frac{1}{6} - \frac{\pi}{54} \]