The number of solutions of the equation `x^3+2x^2+6x+2cosx=0` where `x in[0,2pi]` is (a) one (b) two (c) three (d) zero

Options

A. one
B. two
C. three
D. zero

Answer

4

Explanation

Let f(x) =`x^(3)+2x^(2)+5x+2cosx`
`therefore f(X)=3x^(2)+4x+5-2sinx`
Now the least value of `3x^(2)+4x+5` is
`-(D)/(4a)=-((4)^(2)-4(3)(5))/(4(3))=11/3` ltbr and the greatest value of 2 sin x is 2 .Therefore `3x^(2)+4x+5gt2 sin x forall x in R`
or `f(X) =3x^(2)+4x+5-2 sinx gt 0 forall x in R`
Thus f(X) is strictly an incresing function
Alos `f(x) =0 and f(2pi)gt0`
Hence the numebr of roots is zero .