Let `f (x)= [{:(x ^(2)+a,0 le x lt 1),( 2x+b,1le x le 2):}and g (x)=[{:(3x+b,0 le x lt 1),(x ^(3), 1 le x le 2):}`
If derivative of `f(x)` w.r.t. `g (x ) at x =1` exists and is equal to `lamda,` then which of the followig is/are correct?

Options

A. `a+b=-3`
B. `a-b=1`
C. `(ab)/(lamda)=3`
D. `(-b)/(lamda)=3`

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) and ensure they are continuous and differentiable at the point \( x = 1 \). Then, we will compute the derivative of \( f(x) \) with respect to \( g(x) \) at \( x = 1 \) and find the values of \( a \) and \( b \).

### Step-by-Step Solution:

1. **Define the Functions**:
\[ f(x) = \begin{cases} x^2 + a & \text{for } 0 \leq x < 1 \\ 2x + b & \text{for } 1 \leq x \leq 2 \end{cases} \]
\[ g(x) = \begin{cases} 3x + b & \text{for } 0 \leq x < 1 \\ x^3 & \text{for } 1 \leq x \leq 2 \end{cases} \]

2. **Check Continuity at \( x = 1 \) for \( f(x) \)**:
- Left-hand limit as \( x \to 1^- \):
\[ f(1^-) = 1^2 + a = 1 + a \]
- Right-hand limit as \( x \to 1^+ \):
\[ f(1^+) = 2(1) + b = 2 + b \]
- Set the limits equal for continuity:
\[ 1 + a = 2 + b \quad \text{(Equation 1)} \]

3. **Check Continuity at \( x = 1 \) for \( g(x) \)**:
- Left-hand limit as \( x \to 1^- \):
\[ g(1^-) = 3(1) + b = 3 + b \]
- Right-hand limit as \( x \to 1^+ \):
\[ g(1^+) = 1^3 = 1 \]
- Set the limits equal for continuity:
\[ 3 + b = 1 \quad \text{(Equation 2)} \]

4. **Solve Equation 2 for \( b \)**:
\[ b = 1 - 3 = -2 \]

5. **Substitute \( b \) into Equation 1 to find \( a \)**:
\[ 1 + a = 2 - 2 \implies 1 + a = 0 \implies a = -1 \]

6. **Find the Derivative of \( f(x) \) and \( g(x) \) at \( x = 1 \)**:
- Derivative of \( f(x) \):
\[ f'(x) = \begin{cases} 2x & \text{for } 0 \leq x < 1 \\ 2 & \text{for } 1 \leq x \leq 2 \end{cases} \]
At \( x = 1 \):
\[ f'(1) = 2 \]

- Derivative of \( g(x) \):
\[ g'(x) = \begin{cases} 3 & \text{for } 0 \leq x < 1 \\ 3x^2 & \text{for } 1 \leq x \leq 2 \end{cases} \]
At \( x = 1 \):
\[ g'(1) = 3(1^2) = 3 \]

7. **Compute the Derivative of \( f \) with respect to \( g \)**:
\[ \frac{df}{dg} = \frac{f'(1)}{g'(1)} = \frac{2}{3} \]
Thus, \( \lambda = \frac{2}{3} \).

8. **Check the Options**:
- \( a + b = -1 - 2 = -3 \) (Correct)
- \( a - b = -1 + 2 = 1 \) (Correct)
- \( \frac{ab}{\lambda} = \frac{(-1)(-2)}{\frac{2}{3}} = \frac{2}{\frac{2}{3}} = 3 \) (Correct)
- \( -\frac{b}{\lambda} = -\frac{-2}{\frac{2}{3}} = \frac{2 \cdot 3}{2} = 3 \) (Correct)

### Conclusion:
All options are correct.