If the function $$f(x) = \begin{cases} k_1(x - \pi)^2 - 1, & x \leq \pi \\ k_2 \cos x, & x > \pi \end{cases}$$ is twice differentiable, then the ordered pair $$(k_1, k_2)$$ is equal to:

For the given piece-wise definition we have

$$f(x)=\begin{cases} k_1\,(x-\pi)^2-1,& x\le \pi\\[4pt] k_2\cos x,& x>\pi \end{cases}$$

To be twice differentiable at the joining point $$x=\pi$$, the function itself, its first derivative and its second derivative must all be continuous there.

1. Continuity of the function at $$x=\pi$$

Left-hand value:

$$f(\pi^-)=k_1\,(\pi-\pi)^2-1=k_1\cdot 0-1=-1$$

Right-hand value:

$$f(\pi^+)=k_2\cos\pi=k_2(-1)=-k_2$$

For continuity we equate the two:

$$-1=-k_2\Longrightarrow k_2=1$$

2. Continuity of the first derivative at $$x=\pi$$

First we write the standard differentiation formulas we will use:

• For $$x^n$$ we use $$\dfrac{d}{dx}\bigl(x^n\bigr)=n\,x^{\,n-1}.$$

• For $$\cos x$$ we use $$\dfrac{d}{dx}\bigl(\cos x\bigr)=-\sin x.$$

Left-hand derivative:

$$f'(x)=\dfrac{d}{dx}\bigl(k_1(x-\pi)^2-1\bigr)=k_1\cdot 2(x-\pi)=2k_1(x-\pi)$$

Therefore

$$f'(\pi^-)=2k_1(\pi-\pi)=0$$

Right-hand derivative (using $$k_2=1$$ already obtained):

$$f'(x)=\dfrac{d}{dx}\bigl(k_2\cos x\bigr)=-k_2\sin x=-\sin x$$

Hence

$$f'(\pi^+)=-(\sin\pi)=0$$

Because both one-sided derivatives equal $$0$$, the first derivative is continuous automatically for every $$k_1$$ when $$k_2=1$$. So we proceed to the second derivative.

3. Continuity of the second derivative at $$x=\pi$$

The second derivative on the left side is obtained by differentiating $$f'(x)=2k_1(x-\pi)$$ once more:

$$f''(x)=\dfrac{d}{dx}\bigl(2k_1(x-\pi)\bigr)=2k_1$$

Thus

$$f''(\pi^-)=2k_1$$

On the right side we differentiate $$f'(x)=-\sin x$$ again, recalling the formula $$\dfrac{d}{dx}(\sin x)=\cos x$$ and carrying the minus sign:

$$f''(x)=\dfrac{d}{dx}\bigl(-\sin x\bigr)=-\cos x$$

So

$$f''(\pi^+)=-(\cos\pi)= -(-1)=1$$

For the second derivative to be continuous we need

$$f''(\pi^-)=f''(\pi^+) \Longrightarrow 2k_1=1$$

Solving this simple linear equation gives

$$k_1=\dfrac{1}{2}$$

Combining the two conditions, we finally obtain

$$\bigl(k_1,k_2\bigr)=\left(\dfrac12,\,1\right)$$

Hence, the correct answer is Option A.