Question

If for real values of \( x \), \( \cos \theta = x + \frac{1}{x} \), then x is

• A. \( \theta \) is an obtuse angle
• B. No value of \( \theta \) is possible
• C. \( \theta \) is a right angle
• D. \( \theta \) is an acute angle

Hint:

When solving for values of \( \theta \) based on trigonometric equations, always check the range of the function (in this case, \( \cos \theta \)) and compare it with the range of the given expression.

Answer 1

The Correct Option is B

Step 1: Understand the given equation.
We are given:
\[ \cos \theta = x + \frac{1}{x} \] where \( x \) is a real number.

Step 2: Use the identity of cosine.
We know that the range of \( \cos \theta \) is between -1 and 1, i.e.,
\[ -1 \leq \cos \theta \leq 1 \]
Thus, the expression \( x + \frac{1}{x} \) must also lie between -1 and 1 for real values of \( x \).

Step 3: Analyze the expression \( x + \frac{1}{x} \).
Let’s find the minimum and maximum values of \( x + \frac{1}{x} \). Consider the function:
\[ f(x) = x + \frac{1}{x} \]
Differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = 1 - \frac{1}{x^2} \]
Set \( f'(x) = 0 \) to find the critical points:
\[ 1 - \frac{1}{x^2} = 0 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1 \]

Step 4: Find the minimum and maximum values.
Now, evaluate \( f(x) \) at \( x = 1 \) and \( x = -1 \):
- At \( x = 1 \), \( f(1) = 1 + \frac{1}{1} = 2 \)
- At \( x = -1 \), \( f(-1) = -1 + \frac{1}{-1} = -2 \)
Since \( x + \frac{1}{x} \) has a minimum value of -2 and a maximum value of 2, it can never be between -1 and 1.

Step 5: Conclusion.
Since \( x + \frac{1}{x} \) cannot take any value within the range of -1 to 1, no real value of \( \theta \) satisfies the equation \( \cos \theta = x + \frac{1}{x} \).
Final Answer.
Therefore, the correct answer is option (B), "No value of \( \theta \) is possible."