Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if the following three conditions are met:
  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists. This means $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.
  3. $\lim_{x \to c} f(x) = f(c)$.
• Standard Limits:
  • $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$
  • $\lim_{x \to 0} \frac{\ln(1+ax)}{x} = a$

Step-by-Step Solution

The problem states that the function $f(x)$ is continuous at $x=0$. This means that the left-hand limit, the right-hand limit, and the function value at $x=0$ must all be equal. We are given that $f(0) = 4$. Therefore, we must have: $\lim_{x \to 0^-} f(x) = 4$ and $\lim_{x \to 0^+} f(x) = 4$

Step 1: Evaluate the left-hand limit. For $x < 0$, $f(x) = \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}$. We need to find $\lim_{x \to 0^-} \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}$. We can rewrite this limit by separating the terms and using the standard limit $\lim_{y \to 0} \frac{\sin(ay)}{y} = a$. $\lim_{x \to 0^-} \frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\} = 2 \lim_{x \to 0^-} \left\{\frac{\sin \left(k_1+1\right) x}{x}+\frac{\sin \left(k_2-1\right) x}{x}\right\}$ Applying the standard limit formula, we get: $= 2 \left\{ (k_1+1) + (k_2-1) \right\}$ $= 2 (k_1 + k_2)$ Since the function is continuous at $x=0$, this left-hand limit must be equal to $f(0) = 4$. $2(k_1 + k_2) = 4$ $k_1 + k_2 = 2 \quad (*)$

Step 2: Evaluate the right-hand limit. For $x > 0$, $f(x) = \frac{2}{x} \log_e\left(\frac{2+k_1 x}{2+k_2 x}\right)$. We need to find $\lim_{x \to 0^+} \frac{2}{x} \log_e\left(\frac{2+k_1 x}{2+k_2 x}\right)$. We can rewrite the argument of the logarithm: $\frac{2+k_1 x}{2+k_2 x} = \frac{2\left(1+\frac{k_1}{2} x\right)}{2\left(1+\frac{k_2}{2} x\right)} = \frac{1+\frac{k_1}{2} x}{1+\frac{k_2}{2} x}$ So, the limit becomes: $\lim_{x \to 0^+} \frac{2}{x} \log_e\left(\frac{1+\frac{k_1}{2} x}{1+\frac{k_2}{2} x}\right)$ Using the property of logarithms, $\log_e(a/b) = \log_e(a) - \log_e(b)$: $= 2 \lim_{x \to 0^+} \frac{1}{x} \left\{ \log_e\left(1+\frac{k_1}{2} x\right) - \log_e\left(1+\frac{k_2}{2} x\right) \right\}$ $= 2 \left\{ \lim_{x \to 0^+} \frac{\log_e\left(1+\frac{k_1}{2} x\right)}{x} - \lim_{x \to 0^+} \frac{\log_e\left(1+\frac{k_2}{2} x\right)}{x} \right\}$ We use the standard limit $\lim_{y \to 0} \frac{\ln(1+ay)}{y} = a$. $= 2 \left\{ \frac{k_1}{2} - \frac{k_2}{2} \right\}$ $= 2 \cdot \frac{1}{2} (k_1 - k_2)$ $= k_1 - k_2$ Since the function is continuous at $x=0$, this right-hand limit must be equal to $f(0) = 4$. $k_1 - k_2 = 4 \quad (**)$

Step 3: Solve the system of equations for $k_1$ and $k_2$. We have two equations:

1. $k_1 + k_2 = 2$
2. $k_1 - k_2 = 4$

Adding equation (1) and equation (2): $(k_1 + k_2) + (k_1 - k_2) = 2 + 4$ $2k_1 = 6$ $k_1 = 3$ Substituting $k_1 = 3$ into equation (1): $3 + k_2 = 2$ $k_2 = 2 - 3$ $k_2 = -1$

Step 4: Calculate $k_1^2 + k_2^2$. Now that we have the values of $k_1$ and $k_2$, we can find $k_1^2 + k_2^2$. $k_1^2 + k_2^2 = (3)^2 + (-1)^2$ $= 9 + 1$ $= 10$

Common Mistakes & Tips

• Incorrect application of standard limits: Ensure you are using the correct form of the standard limits, especially when the argument of $\sin$ or $\ln$ is not simply $x$. For example, $\sin(ax)$ requires the denominator to be $x$ and the limit to be $a$, not $ax$.
• Algebraic errors with logarithms: Be careful when simplifying logarithmic expressions. Using properties like $\log(a/b) = \log(a) - \log(b)$ and $\log(1+y) \approx y$ for small $y$ are crucial.
• Mistakes in solving simultaneous equations: Double-check your arithmetic when solving the system of linear equations for $k_1$ and $k_2$.

Summary

For the function to be continuous at $x=0$, the left-hand limit, the right-hand limit, and the function value at $x=0$ must be equal. We used the given function value $f(0)=4$ and evaluated the left-hand limit using the standard limit for sine functions, which yielded the equation $k_1 + k_2 = 2$. We then evaluated the right-hand limit using the standard limit for logarithmic functions, resulting in the equation $k_1 - k_2 = 4$. Solving these two simultaneous equations, we found $k_1=3$ and $k_2=-1$. Finally, we calculated $k_1^2 + k_2^2 = 3^2 + (-1)^2 = 9 + 1 = 10$.

The final answer is $\boxed{10}$.