The greatest value of the function `f(x)=(sin2x)/(sin(x+pi/4))` on the interval `(0,pi/2)i s` `1/(sqrt(2))` (b) `sqrt(2)` (c) 1 (d) `-sqrt(2)`

Options

A. `1/(sqrt(2))`
B. `sqrt(2)`
C. 1
D. `-sqrt(2)`

Answer

3

Explanation

`f(x) =((sin x+cos x)^(2)-1)/(1)sqrt(2)(sinx+cosx)=sqrt(2)(t^(2)-1)/(t)`
`=phi(t) = sqrt(2) (t-(1)/(t))`
Where `t=g(x) = sinx +cos x =sqrt(2 cos (pi)/(4)+x)x in [0,pi//2]`
`therefore t in [1,sqrt(2)]`
Now `phi(t) =sqrt(2)(t-(1)/(t))` where `t in [1,sqrt(2)]`
`therefore t in [sqrt(2)]`
Now `phi(t) =sqrt(2)(t-(1)/(t))` where `t in [1,sqrt(2)]`
Therefore `phi` is increasing
Hence `phi(t)` is greatest at the endpoint of interval `[1,sqrt(2)]`
`t=sqrt(2)therefore`
`f(X)=phi(t)=sqrt[sqrt(2)-(1)/sqrt(2)]=1`