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To solve this problem, we need to determine the continuity of each of the three given piecewise functions across their respective domains. Let's examine each function.

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### (i) \( f(x) = \begin{cases} 
x + 2 & \text{if } x \leq 1 \\
3 + 5 \ln x & \text{if } x > 1 
\end{cases} \)

- **For \( x \leq 1 \):** \( f(x) = x + 2 \) is a linear function, which is continuous everywhere.
- **For \( x > 1 \):** \( f(x) = 3 + 5 \ln x \) is continuous wherever \( \ln x \) is defined, i.e., for \( x > 0 \). This ensures continuity for \( x > 1 \).

Now check the continuity at \( x = 1 \):

\[
\lim_{x \to 1^-} f(x) = 1 + 2 = 3
\]
\[
\lim_{x \to 1^+} f(x) = 3 + 5 \ln 1 = 3 + 0 = 3
\]

Since both left and right limits are equal and the function is defined at \( x = 1 \), \( f(x) \) is continuous at \( x = 1 \).

Thus, function **(i) is continuous**.

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### (ii) \( f(x) = \begin{cases} 
7e^x - 4 & \text{if } x < 0 \\
3 \cos x & \text{if } x \geq 0 
\end{cases} \)

- **For \( x < 0 \):** \( f(x) = 7e^x - 4 \) is an exponential function, which is continuous everywhere.
- **For \( x \geq 0 \):** \( f(x) = 3 \cos x \) is also continuous everywhere.

Now check the continuity at \( x = 0 \):

\[
\lim_{x \to 0^-} f(x) = 7e^0 - 4 = 7 - 4 = 3
\]
\[
\lim_{x \to 0^+} f(x) = 3 \cos(0) = 3 \times 1 = 3
\]

Since both left and right limits are equal and the function is defined at \( x = 0 \), \( f(x) \) is continuous at \( x = 0 \).

Thus, function **(ii) is continuous**.

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### (iii) \( f(x) = \begin{cases} 
\frac{x^2 - 49}{x - 7} & \text{if } x \neq 7 \\
12 & \text{if } x = 7 
\end{cases} \)

- **For \( x \neq 7 \):** We can simplify the given expression as follows:
  \[
  f(x) = \frac{x^2 - 49}{x - 7} = \frac{(x - 7)(x + 7)}{x - 7} = x + 7 \quad \text{for } x \neq 7
  \]
  This is a linear function, which is continuous everywhere except possibly at \( x = 7 \).
  
Now check the continuity at \( x = 7 \):

\[
\lim_{x \to 7} f(x) = 7 + 7 = 14
\]
But \( f(7) = 12 \), so there is a jump discontinuity at \( x = 7 \).

Thus, function **(iii) is not continuous**.

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### Conclusion:
- Function (i) is continuous.
- Function (ii) is continuous.
- Function (iii) is not continuous.

Therefore, the correct answer is **(F) (i) and (ii) only**.

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Would you like further details on any of these calculations? Here are 5 related questions for further exploration:

1. What are the conditions for a function to be continuous at a point?
2. How can we handle discontinuities in piecewise functions?
3. Why is logarithmic function \( \ln x \) undefined for \( x \leq 0 \)?
4. What is the significance of removable discontinuities, like in function (iii)?
5. Can a function be continuous but not differentiable? If so, how?

**Tip:** When checking the continuity of piecewise functions, always check if the limits from both sides are equal at the point where the pieces join!

Which of the following functions are continuous on \((-\infty, \infty)\)?

Explore how to determine the continuity of piecewise functions with a detailed step-by-step solution. This problem covers key mathematical concepts, including continuity, piecewise functions, and limits. Suitable for students in Grades 11-12.

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