Let the function `f(x)` be defined as below `f(x)=sin^(-1) lambda+x^2` when `0 < x < 1` and `f(x) = 2x` when `x>=1` . `f(x)` can have a minimum at `x=1` then value of `lambda` is

Answer

To solve the problem, we need to determine the value of \( \lambda \) such that the function \( f(x) \) has a minimum at \( x = 1 \). The function is defined as follows:

\[ f(x) = \begin{cases} \sin^{-1} \lambda + x^2 & \text{for } 0 < x < 1 \\ 2x & \text{for } x \geq 1 \end{cases} \]

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

For \( f(x) \) to have a minimum at \( x = 1 \), it must be continuous at that point. This means we need to check if the left-hand limit (LHL) as \( x \) approaches 1 from the left is equal to the right-hand limit (RHL) as \( x \) approaches 1 from the right, and both should equal \( f(1) \).

1. **Calculate LHL as \( x \to 1^- \):**
\[ \lim_{x \to 1^-} f(x) = \sin^{-1} \lambda + 1^2 = \sin^{-1} \lambda + 1 \]

2. **Calculate RHL as \( x \to 1^+ \):**
\[ \lim_{x \to 1^+} f(x) = 2 \cdot 1 = 2 \]

3. **Calculate \( f(1) \):**
\[ f(1) = 2 \cdot 1 = 2 \]

### Step 2: Set the Limits Equal

For continuity at \( x = 1 \):
\[ \sin^{-1} \lambda + 1 = 2 \]

### Step 3: Solve for \( \lambda \)

Now, we can solve for \( \lambda \):
\[ \sin^{-1} \lambda = 2 - 1 \]
\[ \sin^{-1} \lambda = 1 \]

Taking the sine of both sides:
\[ \lambda = \sin(1) \]

### Conclusion

Thus, the value of \( \lambda \) such that \( f(x) \) has a minimum at \( x = 1 \) is:
\[ \lambda = \sin(1) \]