Problem 12 - A Textbook of Differential Calculus

Question

If \(x\) and \(\alpha\) are real, then the inequation \(\log _{2} x+\log _{x} 2+2 \cos \alpha \leq 0\) (a) has no solution (b) has exactly two solutions (c) is satisfied for any real \(\alpha\) and any real \(x\) in \((0,1)\) (d) is satisfied for any real \(\alpha\) and any real \(x\) in \((1, \infty)\)

Step-by-Step Solution

Answer

(d) The inequation is satisfied for any real \( \alpha \) and real \( x \) in \((1, \infty)\).

Step 1: Understanding the Equation

First, rewrite the inequation \( \log_{2} x + \log_{x} 2 + 2\cos \alpha \leq 0 \) using the property of logarithms that \( \log_{a} b = \frac{1}{\log_{b} a} \). This gives us \( \log_{2} x + \frac{1}{\log_{2} x} + 2\cos \alpha \leq 0 \).

Step 2: Considering the general behavior of the expression

Since \( \cos \alpha \) can take any value between -1 and 1, simplify the problem further by substituting different values for \( \cos \alpha \). Consider first when \( \cos \alpha = 0 \), the expression becomes \( \log_{2} x + \frac{1}{\log_{2} x} \leq 0 \).

Step 3: Finding solutions for \( x \) when \( \cos \alpha = 0 \)

Set \( y = \log_{2} x \), then the expression \( y + \frac{1}{y} \leq 0 \) becomes relevant. This inequality holds when \( y > 0, y < -1 \) or when \( x > 2 \) since \( y \) represents \( \log_{2} x \), this occurs when \( x \in (1, 2) \).

Step 4: Expanding to consider \( \cos \alpha \neq 0\)

For general \( \alpha \), adjust the range for \( x \) such that \( \log_{2} x + \frac{1}{\log_{2} x} \leq -2\cos \alpha \). Evaluate if solutions exist at extremes of \( \cos \alpha \) considering positivity of \( y + \frac{1}{y} \geq 2 \).

Step 5: Conclusion on range of \( x \) and \( \alpha \)

The inequality holds true for all \( x \in (1, \, \infty) \). The behavior of \( \log_{2} x + \frac{1}{\log_{2} x} \) ensures that its minimum value is 2 when \( \log_{2} x = 1 \), meaning any \( x > 1 \) satisfies the inequality regardless of \( \alpha \), hence solutions in option (c).

Key Concepts

Inequality Solutions

Inequality solutions involve finding the set of values that satisfy a given inequality. Here, we have the inequality \( \log_{2} x + \log_{x} 2 + 2\cos \alpha \leq 0 \). The first step is to manipulate this inequality using logarithmic properties to make it easier to analyze. By leveraging the identity \( \log_{a} b = \frac{1}{\log_{b} a} \), we can rewrite the inequality as \( \log_{2} x + \frac{1}{\log_{2} x} + 2\cos \alpha \leq 0 \).

To find the values of \( x \) that meet this condition, we substitute different values for \( \cos \alpha \), ranging from -1 to 1, since \( \alpha \) is a real number.

• When \( \cos \alpha = 0 \), the inequality simplifies to \( \log_{2} x + \frac{1}{\log_{2} x} \leq 0 \).
• Setting \( y = \log_{2} x \), the expression becomes \( y + \frac{1}{y} \leq 0 \).
• This inequality is satisfied when \( y > 0 \) or \( y < -1 \).

Considering these cases, the key interval for \( x \) is determined by the range of possible values of \( y \), ultimately leading us to conclude that any \( x \) in the interval \((1, \infty)\) satisfies the inequality regardless of \( \alpha \).

Logarithmic Functions

Logarithmic functions are central to this problem, especially given the presence of \( \log_{2} x \) and \( \log_{x} 2 \) in the inequality. A logarithmic function is the inverse of an exponential function. Here, the properties of logarithms help us transform and solve the inequality.

Key logarithmic principles include:

• The change of base formula, \( \log_{a} b = \frac{1}{\log_{b} a} \), and
• The property that \( \log_{b}(mn) = \log_{b}(m) + \log_{b}(n) \).

In our case, applying \( \log_{a} b = \frac{1}{\log_{b} a} \), we can express \( \log_{x} 2 \) as \( \frac{1}{\log_{2} x} \), simplifying the analysis of the inequality.

Understanding these properties allows one to analyze \( \log_{2} x + \frac{1}{\log_{2} x} \). By considering the behavior of this function, especially its minimum value, we focus on satisfying \( \log_{2} x + \frac{1}{\log_{2} x} \geq 2 \) when \( y = \log_{2} x \) is positive, giving insight into potential solutions. This method helps solve inequalities involving logarithmic terms by adjusting variables and leveraging logarithmic identities.

Trigonometric Functions

Trigonometric functions like \( \cos \alpha \) appear frequently in calculus and inequalities, influencing the outcomes based on their periodic and bounded nature. The cosine function ranges between -1 and 1, depending on the angle \( \alpha \), affecting our inequality's flexibility.

Analyzing the original problem:

• We focused on \( 2\cos \alpha \), which can vary the inequality's results due to its range \([-2, 2]\).
• When \( \cos \alpha = 0 \), it simplifies the expression to \( \log_{2} x + \frac{1}{\log_{2} x} \leq 0 \).

Considering other values, the maximum and minimum bounds of \( 2\cos \alpha \) are important as they manipulate the inequality on either end, yet ultimately, the general solution lies in understanding \( x \) in terms of logarithmic behavior, ensuring that any solution remains valid under all trigonometric conditions.

Trigonometric analysis coupled with logarithms highlights the interplay between these mathematical concepts. Effective strategy involves considering both parts to determine if values of \( x \) exist such that the original inequality holds true, leading to viable solutions within the allowed bounds.