The number of solutions of the equation x3+2x2+5x+2+cosx=0 in 0,2π is :
0
1
2
3
Explanation for correct option
Given, f(x)=x3+2x2+5x+2cosx
Now differentiating with respect to x we get
⇒f'x=3x2+4x+5-2sinx⇒f'x=3x+232+113-2sinx
113-2sinx>0∀x [as -1≤sinθ≤1]
⇒f'x>0∀x
Therefore fx is an increasing function.
f0=2
fx=0 has no solution in 0,2π
Hence, the correct option is A