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Let f: (-(pi)/(2),(pi)/(2)) to R be give...

Let `f: (-(pi)/(2),(pi)/(2)) to R` be given by `f(x)=(log(secx +tanx))^(3)`. Then

A

f(x) is an odd function

B

f(x) is a one-one function

C

f(x) is an onto function

D

f(x) is an even function

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
PLAN
(i) If `f'(x) gt 0, AA x in (a,b)` then f(x) is an increasing function in (a, b) and thus f(x) is one-one function in (a, b).
(ii) If range of f(x) = codomain of f(x), then f(x) is an onto function.
(iii) A function f(x) is said to be an odd function. if
`f(-x)= -f(x), AA x in R,` i.e.
`f(-x) +f(x)=0, AA x in R`
`f(x)=["In"(sec x+tan x)]^(3)`
`f'(x)=(3["In"(sec x+tan x)]^(2)(sec x tan x +sec^(2)x))/((sec x+tan x))`
`f'(x)=3 sec x["In" (sec x +tanx)]^(2) gt 0, AA x in ((-pi)/(2),(pi)/(2))`
f(x) is an increasing function.
`therefore` f(x) is an one-one function.
`(sec x +tan x)=tan((pi)/(4)+(x)/(2)), " as " x in ((-pi)/(2),(pi)/(2)),` then
`0 lt tan ((pi)/(4)+(x)/(2)) lt oo`
` 0 lt sec x + tanx lt oo`
`rArr -oo lt " In " (sec x + tanx) lt oo`
`-oo lt [ "In"(secx+tanx)]^(3) lt oo`
`rArr -oo lt f(x) lt oo`
Range of f(x) is R and thus f(x) is an ont function.
`f(-x)=["In"(secx-tanx)]^(3)=["In"((1)/(secx +tanx)]^(3)`
`f(-x)= -["In" (secx+tanx)]^(3)`
`f(x)+f(-x)=0`
`rArr` f(x) is an odd function.
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