Question

The number of the real roots of the equation 2cos(x(x+1)) = 2^x + 2^-x is 

• A. 2
• B. 1
• C. infinite
• D. 0

Answer 1

The Correct Option is B

Solve the equation:

\(2 \cos(x(x+1)) = 2^x + 2^{-x}\)

Step 1: Analyze the right-hand side

The expression \(2^x + 2^{-x}\) is always greater than or equal to 2 for all real values of \(x\). This is due to the AM ≥ GM inequality:

\[\frac{2^x + 2^{-x}}{2} \geq \sqrt{2^x \cdot 2^{-x}} = \sqrt{1} = 1 \Rightarrow 2^x + 2^{-x} \geq 2\]

Therefore, the minimum value of \(2^x + 2^{-x}\) is 2, which occurs at \(x = 0\).

Step 2: Analyze the left-hand side

The function \(\cos(\theta)\) has a range between -1 and 1. Hence,

\[-1 \leq \cos(x(x+1)) \leq 1 \Rightarrow -2 \leq 2 \cos(x(x+1)) \leq 2\]

Therefore, the left-hand side of the equation is always between -2 and 2.

Step 3: Compare both sides

Since \(2^x + 2^{-x} \geq 2\) and \(2 \cos(x(x+1)) \leq 2\), the only way equality holds is when:

• \(2^x + 2^{-x} = 2\) ⇒ This occurs only when \(x = 0\)
• \(\cos(x(x+1)) = 1\) ⇒ So \(x(x+1) = 2\pi n\) for integer \(n\)

At \(x = 0\), both sides evaluate to: 
Left: \(2 \cos(0) = 2\)
Right: \(2^0 + 2^0 = 1 + 1 = 2\)
✅ The equality holds.

Conclusion:

The equation has exactly one real solution: \(x = 0\)

Final Answer: \(\boxed{1}\) real solution