Question

The number of solutions of the equation 2cos(x/2)=3^x+3^-x

Answer 1

Answer:

Explanation:

We have x3 + 2x2 + 5x + 2cosx = 0 Let f(x) = x3 + 2x2 + 5x + 2cosx ⇒ f'(x) = 3x2 + 4x + 5 – 2sinx Let g(x) = 3x2 + 4x + 5 and h(x) =

2sinx Max value of g(x) is – (16 – 60)/6 = 44/6 = 22/3 and max value of h (x) is 2.

Thus, f'(x) > 0 ⇒ f(x) is strictly increasing function

Also, f(0) = 2 > 0 and f(2π) > 0

Therefore f(x) has no real root.