At `x=(5pi)/6,\ \ f(x)=2sin3x+3cos3x` is

Options

A. maximum
B. minimum
C. zero
D. neither maximum nor minimum

Answer

D

Explanation

We have, `f(x) = 2sin3x + 3cosx`
`therefore f^(')(x)=2.cos3x.3+3(-1sin3x).3`
`rArr f^(')(x)=6cos3x-9sin3x`
Now, `f^(')(x)=-18sin3x-27cos3x`…………………(i)
`=-9(2sin3x + 3cos3x)`
`therefore f^(')(5pi)/(6)=6cos(3.(5pi)/6)-9sin(3.(5pi)/6)`
`=6cos(5pi)/2-9sin(5pi)/2`
`=6 cos (2pi+pi/2)-sin(2pi+pi/2)`
`=0-9 ne 0`
So, `x=(5pi)/(6)` cannot be point of maxima or minima.
Hence, f(x) at `x=(5pi)/6` is neither maximum nor minimum.