The number of elements in the set S= {x∈R : 2cos⁡\((\frac{x_2+x}{6})\)=\(4^x+4^{-x}\)}

Question

Let the sets A and B denote the domain and range respectively of the function \(f(x)=\frac{1}{\sqrt[x]-x}\) where [x] denotes the smallest integer greater than or equal to x . Then among the statements
(S1) : A ∩ B = (1, ∞) – N and
(S2) : A ∪ B = (1, ∞)

Answer 1

To find the number of elements in the set \(S = \{x \in \mathbb{R} : 2\cos\left(\frac{x_2 + x}{6}\right) = 4^x + 4^{-x}\}\), we will evaluate the given equation:

The equation is: \[ 2\cos\left(\frac{x_2 + x}{6}\right) = 4^x + 4^{-x} \]

The right side of the equation, \(4^x + 4^{-x}\), can be simplified using the identity:

\[ 4^x + 4^{-x} = 2\left(2^x + 2^{-x}\right) \]

Therefore, the equation becomes:

\[ 2\cos\left(\frac{x_2 + x}{6}\right) = 2(2^x + 2^{-x}) \]

Dividing both sides by 2, we get:

\[ \cos\left(\frac{x_2 + x}{6}\right) = 2^x + 2^{-x} \]

Now, let's analyze the range of the functions:

The maximum value of \(\cos\left(\frac{x_2 + x}{6}\right)\) is 1.
However, \(2^x + 2^{-x}\) achieves its minimum value of 2 when \(x = 0\). For any \(x \neq 0\), this value is greater than 2 due to its symmetrical behavior around 0 and the fact that both \(2^x\) and \(2^{-x}\) are exponential functions.

Hence, the only possibility for equality is when:

\[ \cos\left(\frac{x_2 + x}{6}\right) = 1 \quad \text{and} \quad 2^x + 2^{-x} = 2 \]

This occurs when \(x = 0\) because only at \(x = 0\), \(2^x + 2^{-x} = 2\) and the cosine function \(\cos\left(\frac{x_2 + 0}{6}\right) = 1\), which is its maximum value.

Therefore, the number of elements in set \(S\) is 1, as only \(x = 0\) satisfies this equation.

The correct answer is 1.

Answer 2