Let f be a function defined on the interval `[0,2pi]` such that `int_(0)^(x)(f^(')(t)-sin2t)dt=int_(x)^(0)f(t)tantdt` and `f(0)=1`. Then the maximum value of `f(x)`is…………………..

Answer

`1.125`

Explanation

We have
`int_(0)^(x) (f^(')(t)-sin2t)dt=int_(x)^(0)f(t)tantdt`
Differentiate w.r.t. x, we get
`f^(')(x)-sin2x=0-f(x)tanx`
or `y^(')+ytanx=sin2x`, where `y=f(x)`
Above is a linear differential equation.
`I.F. =e^(inttanxdx)=secx`
Therefore, solution is
`ysecx=int2sinxdx`
or `ysecx=-2cosx+c`
or `y=cosx-2cos^(2)x`
`f(0)=1 rArr c=3`
`therefore f(x)=3cosx-2cos^(2)x`
`=-2[cosx-3/4]^(2)+9/8`
`therefore f(x)"max"=9/8`