Let `f :R to R ` is defined by `f(x)={{:((x+1)^(3),,x le1),(ln x+(b^(2)-3b+10),, x gt1):}` If f (x) is invertible, then the set of all values of 'b' is :

Options

A. `{1,2}`
B. `phi`
C. `{2,5}`
D. None of these

Answer

To determine the set of all values of \( b \) for which the function \( f(x) \) is invertible, we need to analyze the function defined as:

\[ f(x) = \begin{cases} (x + 1)^3 & \text{if } x \leq 1 \\ \ln x + (b^2 - 3b + 10) & \text{if } x > 1 \end{cases} \]

### Step 1: Check if \( f(x) \) is one-to-one (1-1)

1. **For \( x \leq 1 \)**: The function \( f(x) = (x + 1)^3 \) is a cubic function. Cubic functions are always one-to-one over the real numbers since they are strictly increasing or decreasing. Therefore, \( f(x) \) is one-to-one for \( x \leq 1 \).

2. **For \( x > 1 \)**: The function \( f(x) = \ln x + (b^2 - 3b + 10) \) is the sum of a logarithmic function and a constant. The logarithmic function \( \ln x \) is also one-to-one and strictly increasing for \( x > 0 \). Thus, \( f(x) \) is one-to-one for \( x > 1 \).

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

For \( f(x) \) to be invertible, it must also be continuous. We need to check the left-hand limit and right-hand limit at \( x = 1 \):

- **Left-hand limit** as \( x \) approaches 1:
\[ \lim_{x \to 1^-} f(x) = (1 + 1)^3 = 2^3 = 8 \]

- **Right-hand limit** as \( x \) approaches 1:
\[ \lim_{x \to 1^+} f(x) = \ln(1) + (b^2 - 3b + 10) = 0 + (b^2 - 3b + 10) = b^2 - 3b + 10 \]

### Step 3: Set the limits equal to ensure continuity

For \( f(x) \) to be continuous at \( x = 1 \):
\[ 8 = b^2 - 3b + 10 \]

Rearranging gives:
\[ b^2 - 3b + 10 - 8 = 0 \implies b^2 - 3b + 2 = 0 \]

### Step 4: Solve the quadratic equation

Factoring the quadratic:
\[ b^2 - 3b + 2 = (b - 1)(b - 2) = 0 \]

This gives us the solutions:
\[ b - 1 = 0 \implies b = 1 \]
\[ b - 2 = 0 \implies b = 2 \]

### Conclusion

The set of all values of \( b \) for which the function \( f(x) \) is invertible is:
\[ \{1, 2\} \]