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

Answer: (A)

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.