Functions Ques 7
- Let $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow R$ be given by $f(x)=[\log (\sec x+\tan x)]^3$. 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
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Answer:
Correct Answer: 7.(a,b,c)
Solution: PLAN
(i) If $f^{\prime}(x)>0, \forall x \in(a, b)$, 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
$f(-x)=-f(x), \forall x \in R$, i.e.
$ \begin{aligned} & f(-x)+f(x)=0, \forall x \in R \\ & f(x)=[\ln (\sec x+\tan x)]^3 \\ & f^{\prime}(x)=\frac{3[\ln (\sec x+\tan x)]^2\left(\sec x \tan x+\sec ^2 x\right)}{(\sec x+\tan x)} \\ & f^{\prime}(x)=3 \sec x[\ln (\sec x+\tan x)]^2>0, \forall x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right) \end{aligned} $
$f(x)$ is an increasing function.
$\therefore f(x)$ is an one-one function.
$(\sec x+\tan x)=\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)$, as $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then
$ \begin{aligned} & 0<\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)<\infty \\ & 0<\sec x+\tan x<\infty \\ & \Rightarrow \quad-\infty<\ln (\sec x+\tan x)<\infty \\ & -\infty<[\ln (\sec x+\tan x)]^3<\infty \\ & \Rightarrow \quad-\infty<f(x)<\infty \\ \end{aligned} $
Range of $f(x)$ is $R$ and thus $f(x)$ is an ont function.
$ f(-x)=[\ln (\sec x-\tan x)]^3=\left[\ln \left(\frac{1}{\sec x+\tan x}\right)\right]^3 $
$ \begin{array}{r} f(-x)=-[\ln (\sec x+\tan x)]^3 \\ f(x)+f(-x)=0 \end{array} $
$\Rightarrow f(x)$ is an odd function.
BETA
