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Question

Let f:(π2, π2)R be given by f(x) = [log(sec x+tan x)]3. Then

A
f(x) is an odd function
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B
f(x) is a one – one function
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C
f(x) is an onto function
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D
f(x) is an even function
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Solution

The correct option is C f(x) is an onto function
f(x)=[ln (sec x+tan x)]3f(x)=3[ln (sec x+tan x)]2(sec x tan x+sec2x)(sec x+tan x)
f(x)=3secx[ln(sec x+tan x)]2>0, x ϵ (π2, π2)
f(x) is an increasing function.
f(x) is an one – one function.
(sec x+tan x)=tan(π4+x2), as x ϵ(π2, π2), then 0<tan (π2+x2)<0<sec x+tan x< <ln(sec x+tan x)<<[ln (sec x+tan x)]3< <f(x)<
Range of f(x) is R and thus f(x) is an onto function.
f(x)=[ln(sec xtan x)]3=[ln(1sec x+tan x)]3f(x)=[ln(sec x+tan x)]3
f(x) + f (-x) = 0
f(x) is an odd function.

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