Let f : ( - 2 , 2 ) R be given by f ( x ) = ( log ( sec x + tan x ) ) 3 . Then [2014]

Question

Let $f : (- \frac{π}{2}, \frac{π}{2}) \to R$ be given by $f (x) = (\log {(\sec x + \tan x))}^{3} .$ Then [2014]

Smart Achievers · Accepted Answer

(1, 2, 3)

Given: $f : (- \frac{π}{2}, \frac{π}{2}) \to R$ is given by

$f (x) = (\log {(\sec x + \tan x))}^{3}$

$f (- x) = (\log {(\sec x - \tan x))}^{3}$

$= {[\log (\frac{(\sec x - \tan x) (\sec x + \tan x)}{\sec x + \tan x})]}^{3}$

$= {[\log (\frac{1}{\sec x + \tan x})]}^{3} = {[- \log (\sec x + \tan x)]}^{3}$

$= - {[\log (\sec x + \tan x)]}^{3} = - f (x)$

$∴$ $f (x) is an odd function.$

$∴$ option (1) is correct and (4) is not correct.

Now, $f' (x) = 3 {[\log (\sec x + \tan x)]}^{2} \cdot \frac{\sec x \tan x + \sec^{2} x}{\sec x + \tan x}$

$= 3 \sec x {[\log (\sec x + \tan x)]}^{2} > 0, ∀ x \in (- \frac{π}{2}, \frac{π}{2})$

$∴$ $f (x) is increasing on (- \frac{π}{2}, \frac{π}{2})$

We know that a strictly increasing function is one-one.

$∴$ $f$ is one-one, hence (2) is the correct option.

Also, $\lim_{x \to {\frac{π}{2}}^{-}} {[\log (\sec x + \tan x)]}^{3} = \infty$

and $\lim_{x \to {\frac{π}{2}}^{+}} {[\log (\sec x + \tan x)]}^{3} = - \infty$

$∴$ $Range of f = (- \infty, \infty) = R = domain$

$∴$ $f is an onto function.$

$∴$ option (3) is correct.

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