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Let f(x)=sec^(-1)[1+cos^(2)x], where [.]...

Let `f(x)=sec^(-1)[1+cos^(2)x],` where [.] denotes the greatest integer function. Then the

A

` [ 1, 2 ]`

B

`[ 0, 2] `

C

`{sec ^(-1) 1, sec ^(-1) 2} `

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C

Clearly, `f(x)` is defined for all `x in R` for any ` x in R`we have
`0 le cos^(2) x le 1`
`rArr 1 le 1 + cos ^(2) x le 2`
`rArr [1 + cos^(2) x] = {{:(,1,"for" x in R -{ pi : n in Z}),(,2,"for"x = pi"," n in Z):}`
`rArr f(x) =sec^(-1) [ 1+ cos^(2)]={{:(sec^(-1)1,"for" x in R - {n pi : n in Z}),(sec^(-1)2,"for" x = pi n in Z):}`
Hence, range `(f) = {sec^(-1), sec^(-1)}`
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