The number of real solution of the equation sin(e^x) = 2^x + 2^-x
Answers
Answered by
81
we all know exponential function is always positive
here , 2⁻ˣ and 2ˣ both are exponential function. so, both are positive.
we also know, all positive term follow
AM≥ GM
so,
(2⁻ˣ + 2ˣ)/2 ≥√(2⁻ˣ.2ˣ)
(2⁻ˣ + 2ˣ) ≥ 2
hence sin(eˣ) = (2⁻ˣ + 2ˣ) ≥2
but we know maximum value of sine function is 1
hence sin(eˣ) ≠ 2
hence no any solution is possible here.
no of solution = 0
here , 2⁻ˣ and 2ˣ both are exponential function. so, both are positive.
we also know, all positive term follow
AM≥ GM
so,
(2⁻ˣ + 2ˣ)/2 ≥√(2⁻ˣ.2ˣ)
(2⁻ˣ + 2ˣ) ≥ 2
hence sin(eˣ) = (2⁻ˣ + 2ˣ) ≥2
but we know maximum value of sine function is 1
hence sin(eˣ) ≠ 2
hence no any solution is possible here.
no of solution = 0
Answered by
3
Answer:
no solution
Step-by-step explanation:
use the concept
A.M. ≥ G.M.
2ˣ +2-ˣ
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