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Question

Let f:(π2,π2)R be given by
f(x)=(log(secx+tanx))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)=(log(secx+tanx))3
f(x)=(log(1cosx+sinxcosx))3

f(x)=(log(1+sinxcosx))3

f(x)=⎜ ⎜ ⎜ ⎜ ⎜log⎜ ⎜ ⎜ ⎜ ⎜(sinx2+cosx2)2cos2x2sin2x2⎟ ⎟ ⎟ ⎟ ⎟⎟ ⎟ ⎟ ⎟ ⎟3

f(x)=⎜ ⎜log⎜ ⎜sinx2+cosx2cosx2sinx2⎟ ⎟⎟ ⎟3

f(x)=⎜ ⎜log⎜ ⎜1+tanx21tanx2⎟ ⎟⎟ ⎟3

f(x)=(log(tan(x2+π4)))3


when x(π2,π2),
tan(x2+π4)(0,)
and from the above graph, clearly y=logx is one one for x(0,) as the graph is continuously increasing.
Hence, f(x) is a one-one function.

As for x(0,), logx(,)
Similarly f(x)(,)
Hence, f(x) is an onto function.
Now,
f(x)=(log(secx+tanx))3
f(x)=(log(secxtanx))3
f(x)=(log(1(secx+tanx)))3
f(x)=(log(secx+tanx))3=f(x)
Hence, f(x) is an odd function.

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