If the function $f(x) = \begin{cases} \frac{2}{x} \left\{ \sin (k_1 + 1) x + \sin (k_2 - 1) x \right\}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x > 0 \end{cases}$ is continuous at $x = 0$, then $k_1^2 + k_2^2$ is

Step 1: Apply Continuity Condition

For a function to be continuous at a point $x=0$, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal. We are given $f(0) = 4$. Therefore, we need to find the left-hand limit and the right-hand limit and set them equal to 4.

Step 2: Evaluate Left-Hand Limit

We evaluate the left-hand limit using the standard limit $\lim_{x \to 0} \frac{\sin ax}{x} = a$. Applying this, we get $2((k_1+1) + (k_2-1))$, which simplifies to $2(k_1+k_2)$. Since the function is continuous, this limit must be equal to $f(0) = 4$.

Step 3: Set Left-Hand Limit Equal to f(0)

From the previous step, we found the left-hand limit to be $2(k_1 + k_2)$. Equating this to $f(0) = 4$, we get the equation $2(k_1 + k_2) = 4$, which simplifies to $k_1 + k_2 = 2$. This is our first equation relating $k_1$ and $k_2$.

Step 4: Evaluate Right-Hand Limit

We evaluate the right-hand limit. We use the property of logarithms $\log_e(A/B) = \log_e A - \log_e B$. Then, we apply the standard limit $\lim_{x \to 0} \frac{\log_e(1+ax)}{x} = a$. To use this, we rewrite $\log_e(2+kx)$ as $\log_e(2(1+kx/2)) = \log_e 2 + \log_e(1+kx/2)$. The $\log_e 2$ terms cancel out, leaving us with $2 \left( \frac{k_1}{2} - \frac{k_2}{2} \right)$, which simplifies to $k_1 - k_2$.

Step 5: Set Right-Hand Limit Equal to f(0)

Since the function is continuous, the right-hand limit must also be equal to $f(0) = 4$. Therefore, we have the second equation: $k_1 - k_2 = 4$.

Step 6: Solve System of Equations

Now we have a system of two linear equations: $k_1 + k_2 = 2$ and $k_1 - k_2 = 4$. Adding these two equations eliminates $k_2$, giving $2k_1 = 6$, so $k_1 = 3$. Substituting $k_1 = 3$ into the first equation gives $3 + k_2 = 2$, which means $k_2 = -1$.

Step 7: Calculate $k_1^2 + k_2^2$

Finally, we substitute the values of $k_1 = 3$ and $k_2 = -1$ into the expression $k_1^2 + k_2^2$. This gives $3^2 + (-1)^2 = 9 + 1 = 10$.

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