If `f(x)={{:(,5 tan^(-1)alpha-3x^(2),0 lt x lt 1),(,5x-3,x ge 1):}` then find the least integer value of `alpha` for which f(x) has minima at x=1.

Answer

To solve the problem, we need to analyze the piecewise function given and determine the conditions under which it has a minimum at \( x = 1 \).

### Step 1: Define the function
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} 5 \tan^{-1}(\alpha) - 3x^2 & \text{for } 0 < x < 1 \\ 5x - 3 & \text{for } x \geq 1 \end{cases} \]

### Step 2: Calculate \( f(1) \)
To find the minimum at \( x = 1 \), we first need to calculate \( f(1) \):
\[ f(1) = 5(1) - 3 = 5 - 3 = 2 \]

### Step 3: Analyze the behavior of \( f(x) \) as \( x \) approaches 1 from the left
We need to ensure that the function is greater than or equal to \( f(1) \) when approaching from the left:
\[ \lim_{x \to 1^-} f(x) = 5 \tan^{-1}(\alpha) - 3(1)^2 = 5 \tan^{-1}(\alpha) - 3 \]
We require:
\[ 5 \tan^{-1}(\alpha) - 3 \geq 2 \]
This simplifies to:
\[ 5 \tan^{-1}(\alpha) \geq 5 \]
\[ \tan^{-1}(\alpha) \geq 1 \]

### Step 4: Solve for \( \alpha \)
Taking the tangent of both sides, we have:
\[ \alpha \geq \tan(1) \]
The value of \( \tan(1) \) (in radians) is approximately \( 1.5574 \).

### Step 5: Find the least integer value of \( \alpha \)
Since we need the least integer value of \( \alpha \) that satisfies this inequality:
\[ \alpha \geq 1.5574 \]
The least integer value satisfying this condition is:
\[ \alpha = 2 \]

### Final Answer
The least integer value of \( \alpha \) for which \( f(x) \) has a minimum at \( x = 1 \) is:
\[ \boxed{2} \]