Let f(x) be the function given by
`f(x) = 3x^5-5x^3+21x+3sinx+cosx+5.` Then ,

Options

A. f(x) is increasing on R and f(x) =0 has exactaly one negative root
B. f(x) is increasing on R and f(x) =0 has excatly one positive root
C. f(x) is an increasing and f(x) =0 has excatly one negative root
D. f(x) is an increasing and f(x) = 0 has excatly one positive root

Answer

A, C

Explanation

We have
`f(x)=3x^5-5x^3+21x+3sinx+4cosx+5`
`rArr f'(x) = 15x^4-15x^2+21+3cosx-4sinx`
`rArrf'(x)=phi(x)+psi(x), " where "`
`phi(x)=15x^4-15x^2+21 " and " phi(x) = 3cosx-4sinx`
`phi(x) =15x^4-15x^2+21`
`rArr phi(x) 15(x^4-x^2+7/5)`
`rArr phi(x) = 15[(x^2-1/2)^2+(23)/(25)]`
`rArr phi(x) 15(x^2-1/2)^2+(69)/4ge (69)/4 " for all " x in Rand psi(x) =3cosx-4sinx`
`rArr-5le psi (x) le 5 " for all " x in R`
`f'(x) ge (69)/4-5gt0 " for all " x in R`
`rArr` f(x) is increasing on R
We observe that `f(0)= 9 gt 0 and f (oo)= -oo)`
Therfore y=f(x) cuts x-axis excatly ones
Hence f(x) = 0 has excatly one negative root