QJEEAdv22Ma2af1: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Ma2af1 Solution: The greatest integer less than or equal to is ___________.

Let $I = \int_1^2 {{{\log }_2}({x^3} + 1)dx + \int_1^{\log _2^9} {{{({2^x} - 1)}^{{1 \over 3}}}dx} }$

Let ${I_1} = \int_1^2 {{{\log }_2}({x^3} + 1)dx}$

and ${I_2} = \int_1^{\log _2^9} {{{({2^x} - 1)}^{{1 \over 3}}}dx}$

Let ${2^x} - 1 = {y^3}$

$\Rightarrow {2^x}\,.\,\log _2^2 = 3{y^2}\,.\,{{dy} \over {dx}}$

$\Rightarrow dx = {{3{y^2}} \over {{2^x}}}dy$

$\Rightarrow dx = {{3{y^2}dy} \over {{y^3} + 1}}$

When $x = 1$ then ${y^3} = {2^1} - 1 = 1 \Rightarrow y = 1$

When $x = \log _2^9$ then ${y^3} = {2^{\log _2^9}} - 1 \Rightarrow {y^3} = 8 \Rightarrow y = 2$

$\therefore$ ${I_2} = \int_1^2 {y\,.\,{{3{y^2}} \over {{y^3} + 1}}dy}$

$= \int_1^2 {{{3{y^3}dy} \over {{y^3} + 1}}}$

Let $y = x$

$\Rightarrow dy = dx$

$\therefore$ ${I_2} = \int_1^2 {{{3{x^3}dx} \over {{x^3} + 1}}}$

Now, ${I_1} = \int_1^2 {{{\log }_2}({x^3} + 1)dx}$

$= \left[ {{{\log }_2}({x^3} + 1)\,.\,x} \right]_1^2 - \int_1^2 {{1 \over {{x^3} + 1}}\,.\,{{3{x^2}} \over {\log _2^2}}\,.\,x\,dx}$

$= \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_2^2 - \int_1^2 {{{3{x^3}dx} \over {{x^3} + 1}}}$

$\therefore$ $I = {I_1} + {I_2}$

$= \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_1^2 - \int_1^2 {{{3{x^3}dx} \over {({x^3} + 1)}} + \int_1^2 {{{3{x^3}dx} \over {({x^3} + 1)}}} }$

$= \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_1^2$

$= 2\log _2^9 - 1\,.\,\log _2^2$

$= 2\log _2^9 - 1$

$= 4\log _2^3 - 1$

$= 4 \times 1.58 - 1$

$= 6.32 - 1$

$= 5.32$

$\therefore$ Greatest integer value of

$I = [5.32] = 5$

Other Method :-

Let $f(x) = {\log _2}({x^3} + 1) = y$

$\Rightarrow {x^3} + 1 = {2^y}$

$\Rightarrow {x^3} = {2^y} - 1$

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

$\therefore$ ${f^{ - 1}}(x) = {\left( {{2^x} - 1} \right)^{{1 \over 3}}}$

And $f(1) = \log _2^{(1 + 1)} = 1 = f(a)$ (Assume)

$f(2) = \log _2^{(8 + 1)} = \log _2^9 = f(b)$

$\therefore$ $I = \int_a^b {f(x)dx + \int_{f(a)}^{f(b)} {{f^{ - 1}}(x)dx} }$

Let ${f^{ - 1}}(x) = t$

$\Rightarrow x = f(t)$

$\Rightarrow dx = f'(t)dt$

When $x = f(a)$ then $f(a) = f(t) \Rightarrow t = a$

When $x = f(b)$ then $f(b) = f(t) \Rightarrow t = b$

$\therefore$ $I = \int_a^b {f(t)dt + \int_a^b {t\,.\,f'(t)dt} }$

$= \int_a^b {\left( {f(t) + t\,.\,f'(t)} \right)dt}$

$= \left[ {t\,.\,f(t)} \right]_a^b$

$= b\,.\,f(b) - a\,.\,f(a)$

Here $b = 2$ , $f(b) = \log _2^9$

and $a = 1$ , $f(a) = 1$

$\therefore$ $I = 2\log _2^9 - 1\,.\,1$

$= 4\,.\,\log _2^3 - 1$

$= 4 \times 1.58 - 1$

$= 6.32 - 1$

$= 5.32$

$\therefore$ $[I] = 5$