The number of real solutions of the equation `"sin"e^(x)"cos" e^(x) = 2^(x-2) + 2^(-x-2)`, is

Options

A. 0
B. 1
C. 2
D. infinite

Answer

A

Explanation

We have,
`"sin" e^(x) "cos" e^(x) = 2^(x-2) + 2^(-x-2)`
`rArr "sin" 2e^(x) = (2^(x) + 2^(-x))/(2)`
Now, `(2^(x) + 2^(-x))/(2) ge sqrt(2^(x) xx 2^(-x)) = 1 " " ["using": A.M ge G.M]`
`therefore "sin" 2e^(x) = (2^(x) + 2^(-x))/(2)`
`rArr "sin" 2e^(x) ge 1`
`rArr "sin" (2e^(x)) = 1` and this is true when `2^(x) = 2^(-x)` i.e. x = 0
But, for `x = 0 "sin" (2e^(x)) = "sin" 2 ne 1`.
Hence, the given equation has no solution.