If function `f(x)={(k_1(x-pi)^2-1,xlepi),(k_2cosx,xgtpi):}` is twice differentiable in ordered pair `(k_1,k_2)`. Find this ordered pair.

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is twice differentiable at the point \( x = \pi \). The function is defined as follows:

\[ f(x) = \begin{cases} k_1 (x - \pi)^2 - 1 & \text{if } x \leq \pi \\ k_2 \cos x & \text{if } x > \pi \end{cases} \]

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

For \( f(x) \) to be continuous at \( x = \pi \), the left-hand limit must equal the right-hand limit at that point:

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

Calculating the left-hand limit:

\[ f(\pi) = k_1 (\pi - \pi)^2 - 1 = -1 \]

Calculating the right-hand limit:

\[ \lim_{x \to \pi^+} f(x) = k_2 \cos(\pi) = k_2 (-1) = -k_2 \]

Setting these equal for continuity:

\[ -1 = -k_2 \implies k_2 = 1 \]

### Step 2: Ensure Differentiability at \( x = \pi \)

For \( f(x) \) to be differentiable at \( x = \pi \), the derivatives from both sides must equal at that point:

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

Calculating the left-hand derivative:

\[ f'(x) = \frac{d}{dx}[k_1 (x - \pi)^2 - 1] = 2k_1 (x - \pi) \]
Thus,
\[ f'(\pi^-) = 2k_1 (\pi - \pi) = 0 \]

Calculating the right-hand derivative:

\[ f'(x) = \frac{d}{dx}[k_2 \cos x] = -k_2 \sin x \]
Thus,
\[ f'(\pi^+) = -k_2 \sin(\pi) = 0 \]

Setting these equal for differentiability:

\[ 0 = 0 \quad \text{(which is always true)} \]

### Step 3: Ensure Twice Differentiability at \( x = \pi \)

For \( f(x) \) to be twice differentiable at \( x = \pi \), the second derivatives from both sides must equal at that point:

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

Calculating the second derivative from the left:

\[ f''(x) = \frac{d}{dx}[2k_1 (x - \pi)] = 2k_1 \]
Thus,
\[ f''(\pi^-) = 2k_1 \]

Calculating the second derivative from the right:

\[ f''(x) = \frac{d}{dx}[-k_2 \sin x] = -k_2 \cos x \]
Thus,
\[ f''(\pi^+) = -k_2 \cos(\pi) = -k_2 (-1) = k_2 \]

Setting these equal for twice differentiability:

\[ 2k_1 = k_2 \]

### Step 4: Substitute \( k_2 \)

From Step 1, we found \( k_2 = 1 \). Substituting this into the equation:

\[ 2k_1 = 1 \implies k_1 = \frac{1}{2} \]

### Conclusion

The ordered pair \( (k_1, k_2) \) is:

\[ \left( \frac{1}{2}, 1 \right) \]