    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 functionf(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 x−tan 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.  Suggest Corrections  0      Similar questions  Related Videos   Types of Functions
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