QJEEAdv14Mae0af: JEE Advanced 2014 Mathematics - Class 11 JEEAdv14Mae0af Solution: Let be given by . Then,

Question

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Let

f:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R

be given by

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

. Then,

Accepted Answer

f:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R

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

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

= {\left[ {\log \left( {{{(\sec x - \tan x)(\sec x + \tan x)} \over {\sec x + \tan x}}} \right)} \right]^3}

= {\left[ {\log \left( {{1 \over {\sec x + \tan x}}} \right)} \right]^3} = {[ - \log (\sec x + \tan x)]^3}

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

\therefore

f is an odd function. (a) is correct and (d) is not correct.

Also,

f'(x) = 3{[\log (\sec x + \tan x)]^2}\,.\,{{\sec x\tan x + {{\sec }^2}x} \over {\sec x + \tan x}}

= 3\sec x{[\log (\sec x + \tan x)]^2} &gt; 0\,\forall x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

\therefore

f is increasing on

\left( { - {\pi \over 2},{\pi \over 2}} \right)

We know that strictly increasing function is one one.

\therefore

f is one one

\therefore

(b) is correct.

(\sec x + \tan x) = \tan \left( {{\pi \over 4} + {\pi \over 2}} \right)

as

x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

, then

0 &lt; \tan \left( {{\pi \over 4} + {\pi \over 2}} \right) &lt; \infty

0 &lt; \sec x + \tan x &lt; \infty

\Rightarrow - \infty &lt; \ln (\sec x + \tan x) &lt; \infty

- \infty &lt; {[\ln (\sec x + \tan x)]^3} &lt; \infty

\Rightarrow - \infty &lt; f(x) &lt; \infty

Range of f(x) is R and thus f(x) is an onto function.

\therefore

(c) is correct.

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