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 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 \( x = 1 \), we need to follow these steps:

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

For \( f(x) \) to have a maximum at \( x = 1 \), it must be continuous at that point. This means:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]

Calculating the left-hand limit:

\[ \lim_{x \to 1^-} f(x) = \tan^{-1}(a) - 3(1^2) = \tan^{-1}(a) - 3 \]

Calculating the right-hand limit:

\[ \lim_{x \to 1^+} f(x) = -6(1) = -6 \]

Setting these equal for continuity:

\[ \tan^{-1}(a) - 3 = -6 \]

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

Rearranging the equation gives:

\[ \tan^{-1}(a) = -6 + 3 = -3 \]

Now, we need to find \( a \) such that:

\[ a = \tan(-3) \]

### Step 3: Check the Derivative at \( x = 1 \)

Next, we need to check the derivative of \( f(x) \) at \( x = 1 \). The derivative from the left is:

\[ f'(x) = \frac{d}{dx}(\tan^{-1}(a) - 3x^2) = -6x \]

Calculating the left-hand derivative at \( x = 1 \):

\[ f'(1) = -6(1) = -6 \]

The derivative from the right is:

\[ f'(x) = -6 \quad \text{(for } x \geq 1\text{)} \]

Calculating the right-hand derivative at \( x = 1 \):

\[ f'(1) = -6 \]

### Step 4: Check the Condition for Maximum

For \( f(x) \) to have a maximum at \( x = 1 \), the derivative must equal zero:

\[ f'(1) = -6 \neq 0 \]

This means that there is no value of \( a \) such that \( f(x) \) has a maximum at \( x = 1 \).

### Conclusion

Thus, the value of \( a \) for which the function has a maximum at \( x = 1 \) is:

**None of the options provided.**