The range of `f(x)=sec^(-1)((log)_3tanx+(log)_(tanx)3)` is (a) `[pi/3,pi/2]uu[pi/2,(2pi)/3]` (b) `[0,pi/2]` (c) `((2pi)/3,pi)` (d) none of these

Options

A. `[(pi)/(3),(pi)/(2)) cup ((pi)/(2),(2pi)/(3)]`
B. `[0,(pi)/(2))`
C. `((2pi)/(3),pi]`
D. None of these

Answer

A

Explanation

`f(x)=sec^(-1)(log_(3)tanx+log_(tanx)3).`
`=sec^(-1)(log_(3)tanx+(1)/(log_(3)tanx))`
Now, for `log_(3)tanx in (-oo,oo) " or " log_(3)tanx in R`
Also, ` x+(1)/(x) le -2 " or " x +(1)/(x) ge 2`
i.e., ` log_(3)tan x+(1)/(log_(3) tanx) le -2`
or `log_(3)tan x+(1)/(log_(3) tanx) ge 2`
i.e., `sec^(-1)(log_(3)tan x+(1)/(log_(3) tanx)) le sec^(-1)(-2)`
or `sec^(-1) (log_(3)tan x+(1)/(log_(3) tanx)) ge sec^(-1) 2`
i.e., `f(x) le (2pi)/(3) " or " f(x) ge (pi)/(3)`
i.e., `f(x) in [(pi)/(3),(pi)/(2)) cup ((pi)/(2),(2pi)/(3)]`