    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)2cos2x2−sin2x2⎞⎟ ⎟ ⎟ ⎟ ⎟⎠⎞⎟ ⎟ ⎟ ⎟ ⎟⎠3 f(x)=⎛⎜ ⎜⎝log⎛⎜ ⎜⎝sinx2+cosx2cosx2−sinx2⎞⎟ ⎟⎠⎞⎟ ⎟⎠3 f(x)=⎛⎜ ⎜⎝log⎛⎜ ⎜⎝1+tanx21−tanx2⎞⎟ ⎟⎠⎞⎟ ⎟⎠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(secx−tanx))3 ⇒f(−x)=(log(1(secx+tanx)))3 f(−x)=−(log(secx+tanx))3=−f(x) Hence, f(x) is an odd function.  Suggest Corrections  0      Similar questions  Explore more