If `0lt=xlt=pi/3` then range of `f(x)=sec(pi/6-x)+sec(pi/6+x)` is `(4/(sqrt(3)),oo)` (b) `(4/(sqrt(3)),oo)` `(0,4/(sqrt(3)))` (d) `(0,4/(sqrt(3)))`

Options

A. `((4)/sqrt(3),oo)`
B. `[(4)/sqrt(3),oo]`
C. `[0,(4)/sqrt(3)]`
D. `(0,(4)/sqrt(3))`

Answer

B

Explanation

if a,b `gt0`
Using AM `ge` GM we get
`(1)/(a)+(1)/(b)ge(2)/sqrt(ab)`
`rArr f(x)ge-(2)/(sqrt(cos((pi)/(6)-x)cos((pi)/(6)+x)))`
`=(2)/(sqrt(cos^(2)((pi)/(6))-sin^(2)x))`
`=(2)/(sqrt((3)/(4)-(1-cos2x)/(2)))`
`=(2)/(sqrt((1)/(4)+(cos2x)/(2)))`
Now for `0lexle(pi)/(3),(-1)/(2)lecos 2xle1`
`rArr 0lesqrt((1)/(4)+(cos2x)/(2))lesqrt(3)/(2)`
`rArr f(x)ge(4)/sqrt(3)`
Since f is continous rang of f is `[(4)/sqrt(3),oo]`