Key Concepts and Formulas

• Differentiability at a Point: A function $f(x)$ is differentiable at a point $x=a$ if the limit of the difference quotient exists. For a piecewise function, this implies that the function must be continuous at the point where the definition changes, and the left-hand and right-hand derivatives must be equal at that point.
• Twice Differentiability: For a function to be twice differentiable at a point $x=a$, its first derivative $f'(x)$ must be differentiable at $x=a$. This means $f'(x)$ must be continuous at $x=a$, and the left-hand and right-hand derivatives of $f'(x)$ must be equal at $x=a$. Equivalently, the second derivative $f''(x)$ must exist on both sides of $a$, and the limits of $f''(x)$ as $x$ approaches $a$ from the left and right must be equal.
• Derivatives of Basic Functions:
  • $\frac{d}{dx}(c) = 0$
  • $\frac{d}{dx}(x^n) = nx^{n-1}$
  • $\frac{d}{dx}(\cos x) = -\sin x$

Step-by-Step Solution

1. State the given function: We are given the piecewise function: f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right. We are told that $f(x)$ is twice differentiable.

2. Calculate the first derivative, $f'(x)$: To find $f'(x)$, we differentiate each piece of the function with respect to $x$:

   • For $x < \pi$, $f(x) = k_1(x - \pi)^2 - 1$. Differentiating, we get $\frac{d}{dx}(k_1(x - \pi)^2 - 1) = k_1 \cdot 2(x - \pi) \cdot 1 - 0 = 2k_1(x - \pi)$.
   • For $x > \pi$, $f(x) = k_2\cos x$. Differentiating, we get $\frac{d}{dx}(k_2\cos x) = k_2(-\sin x) = -k_2\sin x$. So, the first derivative is: f'\left( x \right) = \left\{ {\matrix{ {2{k_1}\left( {x - \pi } \right),} & {x < \pi } \cr { - {k_2}\sin x,} & {x > \pi } \cr } } \right. Since $f(x)$ is differentiable, it must be continuous at $x=\pi$. This means $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi)$. $f(\pi) = k_1(\pi - \pi)^2 - 1 = -1$. $\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} k_2\cos x = k_2\cos \pi = -k_2$. Therefore, $-1 = -k_2$, which implies $k_2 = 1$. Also, for $f'(x)$ to be differentiable at $x=\pi$, it must be continuous at $x=\pi$. This means $\lim_{x \to \pi^-} f'(x) = \lim_{x \to \pi^+} f'(x)$. $\lim_{x \to \pi^-} f'(x) = \lim_{x \to \pi^-} 2k_1(x - \pi) = 2k_1(\pi - \pi) = 0$. $\lim_{x \to \pi^+} f'(x) = \lim_{x \to \pi^+} -k_2\sin x = -k_2\sin \pi = -k_2 \cdot 0 = 0$. This condition $0=0$ is always satisfied and does not give us any new information about $k_1$ or $k_2$.

3. Calculate the second derivative, $f''(x)$: To find $f''(x)$, we differentiate $f'(x)$ with respect to $x$:

   • For $x < \pi$, $f'(x) = 2k_1(x - \pi)$. Differentiating, we get $\frac{d}{dx}(2k_1(x - \pi)) = 2k_1 \cdot 1 = 2k_1$.
   • For $x > \pi$, $f'(x) = -k_2\sin x$. Differentiating, we get $\frac{d}{dx}(-k_2\sin x) = -k_2\cos x$. So, the second derivative is: f''\left( x \right) = \left\{ {\matrix{ {2{k_1},} & {x < \pi } \cr { - {k_2}\cos x,} & {x > \pi } \cr } } \right.

4. Apply the condition for twice differentiability at $x=\pi$: For $f(x)$ to be twice differentiable at $x=\pi$, the first derivative $f'(x)$ must be differentiable at $x=\pi$. This means that the second derivative $f''(x)$ must exist on both sides of $\pi$, and the limit of $f''(x)$ as $x$ approaches $\pi$ from the left must be equal to the limit of $f''(x)$ as $x$ approaches $\pi$ from the right. We need to evaluate the limits of $f''(x)$ as $x \to \pi^-$ and $x \to \pi^+$:

   • $\lim_{x \to \pi^-} f''(x) = \lim_{x \to \pi^-} 2k_1 = 2k_1$.
   • $\lim_{x \to \pi^+} f''(x) = \lim_{x \to \pi^+} (-k_2\cos x) = -k_2\cos \pi = -k_2(-1) = k_2$.

   For $f''(x)$ to be continuous at $x=\pi$, these two limits must be equal: $2k_1 = k_2$

5. Combine the conditions and solve for $(k_1, k_2)$: From step 2, we found that $k_2 = 1$ for the function to be continuous. From step 4, we found the condition $2k_1 = k_2$ for the function to be twice differentiable. Substituting $k_2 = 1$ into the second equation: $2k_1 = 1$ $k_1 = \frac{1}{2}$

   Therefore, the ordered pair is $(k_1, k_2) = \left(\frac{1}{2}, 1\right)$.

Common Mistakes & Tips

• Confusing Differentiability and Continuity: Remember that differentiability at a point implies continuity at that point. However, continuity does not guarantee differentiability. For piecewise functions, ensure both continuity and the equality of derivative limits are checked at the junction point.
• Incorrectly Differentiating Piecewise Functions: When differentiating, apply the chain rule correctly for each piece. The derivative of $(x-\pi)^2$ is $2(x-\pi)$, not just $2x$.
• Forgetting the Second Derivative Condition: The problem states the function is twice differentiable. This means the first derivative must be differentiable, which implies the second derivative must be continuous at the junction point.

Summary

The problem requires finding the values of $k_1$ and $k_2$ such that the given piecewise function is twice differentiable. This involves ensuring continuity of the function at $x=\pi$ and continuity of its first derivative at $x=\pi$. We first calculated the first derivative and used the continuity condition of the original function at $x=\pi$ to find $k_2=1$. Then, we calculated the second derivative and used the continuity condition of the first derivative (which means the limits of the second derivative from both sides must be equal) at $x=\pi$ to establish a relationship between $k_1$ and $k_2$. Finally, by combining these conditions, we found $k_1 = \frac{1}{2}$ and $k_2 = 1$.

The final answer is \boxed{\left( {{1 \over 2},1} \right)}. This corresponds to option (D).