Question

Let $f:(-\frac{\pi }{2},\frac{\pi }{2})\to R$ be given by f(x) = $\left[log\right(secx+tanx){]}^3$. 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

Answer 1

Answer: (A), (B), (C)

The correct options are
A f(x) is an odd function
B f(x) is a one – one function
C f(x) is an onto function

$f\left(x\right)=\left[ln\right(secx+tanx\left){]}^3{f}^{\prime }\right(x)=\frac{3\left[ln\right(secx+tanx\left){]}^2\right(secxtanx+se{c}^{2}x)}{(secx+tanx)}$
$f^{\prime }\left(x\right)=3secx\left[ln\right(secx+tanx){]}^2>0,\forall xϵ(\frac{-\pi }{2},\frac{\pi }{2})$
f(x) is an increasing function.
$\therefore$ f(x) is an one – one function.
$(secx+tanx)=tan(\frac{\pi }{4}+\frac{x}{2}),asxϵ(-\frac{\pi }{2},\frac{\pi }{2}),then0<tan(\frac{\pi }{2}+\frac{x}{2})<\infty 0<secx+tanx<\infty \Rightarrow -\infty <ln(secx+tanx)<\infty -\infty <\left[ln\right(secx+tanx){]}^3<\infty \Rightarrow -\infty <f(x)<\infty$
Range of f(x) is R and thus f(x) is an onto function.
$f(-x)=\left[ln\right(secx-tanx){]}^3={\left[ln\left(\frac{1}{secx+tanx}\right)\right]}^{3}f(-x)=-[ln(secx+tanx){]}^3$
f(x) + f (-x) = 0
$\Rightarrow$ f(x) is an odd function.