Question

“The inequality `2^sintheta + 2^costheta ≥ 2^(1/sqrt(2))` holds for all real values of θ” 

विकल्प

• True

• False

Answer 1

This statement is True.

Explanation:

Since `2sin^theta` and `2^costheta` are positive real numbers, so A.M. (Arithmetic Mean) of these two numbers is greater or equal to their G.M. (Geometric Mean)

Hence `(2^sintheta + 2^costheta)/2 ≥ sqrt(2^sintheta xx 2^costheta)`

= `sqrt(2^(sintheta + costheta))`

`≥ 2 ^((sintheta + costheta)/2) = 2^(1/sqrt(2)(1/sqrt(2) sintheta + 1/sqrt(2) cos theta))`

`≥ 2^(1/sqrt(2) sin(pi/4 + theta))`

Since, `-1 ≤ sin(pi/4 + theta) ≤ 1`

We have `(2^sintheta + 2^costheta)/2 ≥ 2^((-1)/sqrt(2))`

⇒ `2^sintheta + 2^costheta ≥ 2^(1 - 1/sqrt(2))`