The value of a for which the function
`f(x)={{:(tan^(-1)a -3x^2" , " 0ltxlt1),(-6x" , "xge1):}` has a maximum at x=1 , is

Options

A. 0
B. 1
C. 2
D. none of these

Answer

To solve the problem, we need to find the value of \( a \) for which the function

\[ f(x) = \begin{cases} \tan^{-1}(a) - 3x^2 & \text{for } 0 < x < 1 \\ -6x & \text{for } x \geq 1 \end{cases} \]

has a maximum at \( x = 1 \).

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

For \( f(x) \) to have a maximum at \( x = 1 \), it must be continuous at that point. This means that the left-hand limit as \( x \) approaches 1 must equal the right-hand limit at \( x = 1 \).

**Left-hand limit:**
\[ \lim_{x \to 1^-} f(x) = \tan^{-1}(a) - 3(1^2) = \tan^{-1}(a) - 3 \]

**Right-hand limit:**
\[ \lim_{x \to 1^+} f(x) = -6(1) = -6 \]

Setting these equal for continuity:
\[ \tan^{-1}(a) - 3 = -6 \]
\[ \tan^{-1}(a) = -6 + 3 = -3 \]

### Step 2: Solve for \( a \)

To solve for \( a \), we take the tangent of both sides:
\[ a = \tan(-3) \]

### Step 3: Check for Differentiability at \( x = 1 \)

Next, we need to ensure that the function is differentiable at \( x = 1 \). This requires that the left-hand derivative equals the right-hand derivative at \( x = 1 \).

**Left-hand derivative:**
\[ f'(x) = \frac{d}{dx}(\tan^{-1}(a) - 3x^2) = -6x \quad \text{(for } 0 < x < 1\text{)} \]
Evaluating at \( x = 1 \):
\[ f'(1) = -6(1) = -6 \]

**Right-hand derivative:**
\[ f'(x) = -6 \quad \text{(for } x \geq 1\text{)} \]
Evaluating at \( x = 1 \):
\[ f'(1) = -6 \]

Since both derivatives are equal, the function is differentiable at \( x = 1 \).

### Conclusion

The value of \( a \) for which the function has a maximum at \( x = 1 \) is:
\[ a = \tan(-3) \]