Question

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

• A. 1
• B. 3
• C. 0
• D. infinite

Answer 1

The Correct Option is A

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 need to analyze and solve the given equation for \( x \).

The equation we need to solve is: 

\(2 \cos \left( \frac{x_2 + x}{6} \right) = 4^x + 4^{-x}\)

To simplify, let's analyze both sides of the equation:

The left-hand side, \( 2 \cos \left( \frac{x_2 + x}{6} \right) \), is bound between -2 and 2, because the range of the cosine function is \([-1, 1]\).

The right-hand side, \( 4^x + 4^{-x} \), using the property of exponents and recognizing this as an exponential function, is always greater than or equal to 2, since:

\(4^x + 4^{-x} = 2 + 2 \left( \frac{4^x - 4^{-x}}{2} \right)^2 \geq 2\) using the identity \( a + \frac{1}{a} \geq 2 \) for \( a > 0 \).

The only possible value for \( 4^x + 4^{-x} \) which matches the maximum possible value of \( 2 \cos \left( \frac{x_2 + x}{6} \right) \) is 2. This implies:

\( 4^x + 4^{-x} = 2 \)

This equality holds only when \( 4^x = 4^{-x} = 1 \). Therefore, \( x \) must satisfy:

\(4^x = 1 \Rightarrow 4^x = 4^0 \Rightarrow x = 0\).

Let's verify: For \( x = 0 \), the original condition becomes:

\( 2 \cos \left( \frac{x_2 + 0}{6} \right) = 1 + 1 = 2 \)

At \( x = 0 \), the equation holds since \( 2 \cos(0) = 2 \).

Therefore, the only solution is \( x = 0 \).

The number of elements in the set \( S \) is thus 1.

Answer 2

S={x∈R:2cos⁡\((\frac{x_2+x}{6})=4^x+4^{-x}\)}
L.H.S. is less than or equal to 2 and RHS is greater than or equal to 2.
So equality holds only if LHS = RHS = 2
R.H.S. is 2 when x = 0
and at x = 0, LHS is also 2.
So, only one solution exist.
Therefore, the correct option is (A): 1.