To determine which of the given statements must be true, we need to analyze each inequality based on the fundamental condition that \(x > y > 1\).

Natural Logarithm Inequality: \(\ln x > \ln y\)

Let's consider the natural logarithm function, denoted as \(f(t) = \ln(t)\). This function is well-defined for all positive values of \(t\) (i.e., \(t > 0\)). A crucial characteristic of the natural logarithm function is that it is a strictly increasing function over its entire domain. This property means that if we have two numbers, say \(a\) and \(b\), such that \(a > b\), then it necessarily follows that \(\ln(a) > \ln(b)\).

Given the initial condition \(x > y > 1\), it clearly implies that both \(x\) and \(y\) are positive numbers, and \(x\) is strictly greater than \(y\). Because the natural logarithm function is an increasing function, if \(x > y\), then the inequality \(\ln x > \ln y\) must hold true.

• Statement (i): \(\ln x > \ln y\) is TRUE.

Exponential Function Inequality: \(e^x > e^y\)

Next, let's examine the exponential function, represented as \(f(t) = e^t\). This function is defined for all real numbers \(t\), meaning its domain covers all possible real values. Similar to the natural logarithm, the exponential function is also a strictly increasing function over its entire domain. This property dictates that if \(a > b\), then it must be true that \(e^a > e^b\).

Considering the given condition \(x > y > 1\), it means that \(x\) and \(y\) are real numbers, and \(x\) is strictly greater than \(y\). Since the exponential function is an increasing function, if \(x > y\), then the inequality \(e^x > e^y\) must hold true.

• Statement (ii): \(e^x > e^y\) is TRUE.

Quadratic Inequality: \(y^2 > x^2\)

Now, let's analyze the inequality involving squares, \(y^2 > x^2\). We are provided with the condition \(x > y > 1\). This means that both \(x\) and \(y\) are positive numbers. For any positive numbers, if one number is greater than another, say \(x > y\), then squaring both numbers preserves the direction of the inequality, resulting in \(x^2 > y^2\).

Let's use a straightforward example to illustrate this point:

In this specific example, \(x = 3\) and \(y = 2\) perfectly satisfy the condition \(x > y > 1\). We calculated that \(x^2 = 9\) and \(y^2 = 4\). It is evident that \(4\) is not greater than \(9\), which means \(y^2 \not> x^2\). This contradicts the statement.

• Statement (iii): \(y^2 > x^2\) is FALSE.

Cosine Function Inequality: \(\cos x > \cos y\)

Finally, let's evaluate the trigonometric inequality, \(\cos x > \cos y\). The cosine function is a periodic function that oscillates between \(-1\) and \(1\). It is neither strictly increasing nor strictly decreasing over large intervals. The given condition \(x > y > 1\) implies that \(x\) and \(y\) are real numbers greater than 1 radian (which is approximately \(57.3^\circ\)).

Let's consider an example where the statement does not hold true:

In this example, \(x = \pi\) and \(y = \frac{\pi}{2}\) fulfill the condition \(x > y > 1\). However, we find that \(\cos x = -1\) and \(\cos y = 0\). Since \(-1\) is not greater than \(0\), it means \(\cos x \not> \cos y\), proving this statement to be false.

Another example to confirm:

Here, \(x=1.2\) and \(y=1.1\) satisfy \(x > y > 1\). However, \(\cos x \approx 0.3624\) and \(\cos y \approx 0.4536\). Clearly, \(\cos x \not> \cos y\) because \(0.3624\) is not greater than \(0.4536\). In fact, in this specific interval, the cosine function is decreasing, leading to \(\cos y > \cos x\).

• Statement (iv): \(\cos x > \cos y\) is FALSE.

Summary of True Statements

Based on our comprehensive analysis of each statement:

• Statement (i): \(\ln x > \ln y\) is TRUE.
• Statement (ii): \(e^x > e^y\) is TRUE.
• Statement (iii): \(y^2 > x^2\) is FALSE.
• Statement (iv): \(\cos x > \cos y\) is FALSE.

Therefore, only statements (i) and (ii) are necessarily true given the condition \(x > y > 1\).