QJEEAdv22Maf0c0: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Maf0c0 Solution: Let Let be the function defined by Then, which of the following …

Given,

$\alpha = \sum\limits_{k = 1}^\alpha {{{\sin }^{2k}}\left( {{\pi \over 6}} \right)}$

$= {\sin ^2}\left( {{\pi \over 6}} \right) + {\sin ^4}\left( {{\pi \over 6}} \right) + {\sin ^6}\left( {{\pi \over 6}} \right)\, + \,......$

This is a infinite G.P.

$\therefore$ $\alpha = {{{{\sin }^2}\left( {{\pi \over 6}} \right)} \over {1 - {{\sin }^2}\left( {{\pi \over 6}} \right)}}$

$= {{{1 \over 4}} \over {1 - {1 \over 4}}}$

$= {1 \over 4} \times {4 \over 3}$

$= {1 \over 3}$

$\therefore$ $g(x) = {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}$

Given $x \in [0,1]$

We know,

$AM \ge GM$

For ${2^{{x \over 3}}}$ and ${2^{{1 \over 3}(1 - x)}}$ both are positive in $x \in [0,1]$

${{{x^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}} \over 2} \ge {\left( {{2^{{x \over 3}}}\,.\,{2^{{1 \over 3}(1 - x)}}} \right)^{{1 \over 2}}}$

$\Rightarrow {{{2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}} \over 2} \ge {\left( {{2^{{1 \over 3}}}} \right)^{{1 \over 2}}}$

$\Rightarrow {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}} \ge 2\,.\,{2^{{1 \over 6}}}$

$\Rightarrow g(x) \ge {2^{{7 \over 6}}}$

We know, $AM = GM$ when terms are equal.

$\therefore$ $g(x) = {2^{{7 \over 6}}}$ when

${2^{{x \over 3}}} = {2^{{1 \over 3}(1 - x)}}$

$\Rightarrow {x \over 3} = {{1 - x} \over 3}$

$\Rightarrow 2x = 1$

$\Rightarrow x = {1 \over 2}$

$\therefore$ At $x = {1 \over 2}$ , ${\left[ {g(x)} \right]_{\min }} = {2^{{7 \over 6}}}$

$\therefore$ Option (A) is correct.

And option (D) is wrong as at only at a single point $x = {1 \over 2},\,g(x)$ is minimum.

Now, $g(x) = {2^{{x \over 3}}} + {2^{{1 \over 3}(1 - x)}}$

$= {2^{{x \over 3}}} + {{{2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}$

$= {2^{{x \over 3}}} + {2^{{1 \over 3}}}\,.\,{2^{ - {x \over 3}}}$

$\therefore$ $g'(x) = {2^{{x \over 3}}}\,.\,\log 2\,.\,{1 \over 3} + {2^{{1 \over 3}}}\,.\,{2^{ - {x \over 3}}}\,.\,\log 2\,.\,\left( { - {1 \over 3}} \right)$

$= {1 \over 3}\log 2\left[ {{2^{{x \over 3}}} - {{{2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right]$

$= {1 \over 3}\log 2\left[ {{{{2^{{{2x} \over 3}}} + {2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right]$

We already found that at $x = {1 \over 2}$ $g(x)$ is minimum.

$\therefore$ $g'(x) = 0$ at $x = {1 \over 2}$

Now, g'(x) > 0 when

{1 \over 3}\log \left[ {{{{2^{{{2x} \over 3}}} - {2^{{1 \over 3}}}} \over {{2^{{x \over 3}}}}}} \right] > 0

\Rightarrow {2^{{{2x} \over 3}}} > {2^{{1 \over 3}}}

\Rightarrow {{2x} \over 3} > {1 \over 3}

\Rightarrow x > {1 \over 2}

Similarly, g'(x) < 0 when x < {1 \over 2}

If we put it in number line we get this

We know $g'(x)$ represent slope of curve $g(x)$ and it is negative when x < {1 \over 2} and positive when x > {1 \over 2} and zero when $x = {1 \over 2}$ .

$\therefore$ Graph of $g(x)$ is

From graph you can see value of $g(x)$ is maximum either at $x = 0$ or $x = 1$ in the range $x \in [0,1]$ .

$\therefore$ $g(0) = {2^0} + {2^{{1 \over 3}}} = 1 + {2^{{1 \over 3}}}$

$g(1) = {2^{{1 \over 3}}} + {2^0} = 1 + {2^{{1 \over 3}}}$

$\therefore$ We get maximum value at $x = 0$ and $x = 1$ both.

$\therefore$ B and C options are correct.