Let for function `f (x)= [{:(cos ^(-1)x ,,, -1 le x le 0),( mx +c ,,, 0 lt x le 1):},` Lagrange's mean value theorem is applicable in `[-1,1]` then ordered pair `(m,c )` is:

Options

A. `(1, -(pi)/(2))`
B. `(1, (pi)/(2))`
C. `(-1,- (pi)/(2))`
D. `(-1,(pi)/(2))`

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous and differentiable on the interval \([-1, 1]\). The function is defined as follows:

\[ f(x) = \begin{cases} \cos^{-1}(x) & \text{for } -1 \leq x \leq 0 \\ mx + c & \text{for } 0 < x \leq 1 \end{cases} \]

### Step 1: Check Continuity at \( x = 0 \)

For \( f(x) \) to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

Calculating \( f(0) \):

\[ f(0) = \cos^{-1}(0) = \frac{\pi}{2} \]

Now, we calculate the left-hand limit:

\[ \lim_{x \to 0^-} f(x) = \cos^{-1}(0) = \frac{\pi}{2} \]

And the right-hand limit:

\[ \lim_{x \to 0^+} f(x) = m(0) + c = c \]

Setting these equal for continuity:

\[ c = \frac{\pi}{2} \]

### Step 2: Check Differentiability at \( x = 0 \)

For \( f(x) \) to be differentiable at \( x = 0 \), we need:

\[ f'(0^-) = f'(0^+) \]

Calculating \( f'(0^-) \):

\[ f'(x) = \frac{d}{dx}(\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}} \]

Evaluating at \( x = 0 \):

\[ f'(0^-) = -\frac{1}{\sqrt{1 - 0^2}} = -1 \]

Now calculating \( f'(0^+) \):

\[ f'(x) = \frac{d}{dx}(mx + c) = m \]

Setting the derivatives equal for differentiability:

\[ m = -1 \]

### Step 3: Final Ordered Pair

From the conditions we derived:

1. \( c = \frac{\pi}{2} \)
2. \( m = -1 \)

Thus, the ordered pair \((m, c)\) is:

\[ (m, c) = (-1, \frac{\pi}{2}) \]

### Conclusion

The final answer is:

\[ \text{Ordered pair } (m, c) = (-1, \frac{\pi}{2}) \]