QJEEM22Ma56db: JEE Main 2022 Mathematics - Class 12 JEEM22Ma56db Solution: Let . If , then the value of 150 is equal to ___________.

Given,

$f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx}$

$f'(t) = {e^{{t^3}}}\left( {{{{t^8}} \over {{{({t^6} + 2{t^3} + 2)}^2}}}} \right)$

$\therefore$ $f'(1) = {e^1}\left( {{1 \over {{{(1 + 2 + 2)}^2}}}} \right)$

$= {e \over {{5^2}}}$

Now, $f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^2} + 2)}^2}}}} \right)dx}$

Let ${x^3} = z \Rightarrow 3{x^2}dx = dz$

$= \int\limits_0^{{t^3}} {{e^z}\left( {{{{x^6}\,.\,{x^2}dx} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)}$

$= {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2}dz} \over {{{({z^2} + 2z + 2)}^2}}}} \right)}$

$= {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2} + 2z + 2 - 2z - 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)} dz$

$= {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2} + 2z + 2} \over {{{({z^2} + 2z + 2)}^2}}} - {{2z + 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)dz}$

$= {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{1 \over {{z^2} + 2z + 2}} - {{2z + 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)dz}$

$= {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {f(z) + f'(z)} \right)dz}$

$= {1 \over 3}\left[ {{e^z}f(z)} \right]_0^{{t^3}}$

$= {1 \over 3}\left[ {{e^z} \times {1 \over {{z^2} + 2z + 2}}} \right]_0^{{t^3}}$

$= {1 \over 3}\left[ {{e^{{t^3}}} \times {1 \over {{t^6} + 2{t^3} + 2}} - {1 \over 2}} \right]$

$\therefore$ $f(t) = {1 \over 3}\left[ {{{{e^{{t^3}}}} \over {{t^6} + 2{t^3} + 2}} - {1 \over 2}} \right]$

So, $f(1) = {1 \over 3}\left[ {{e \over {1 + 2 + 2}} - {1 \over 2}} \right]$

$= {1 \over 3}\left[ {{e \over 5} - {1 \over 2}} \right]$

Given, $f(1) + f'(1) = \alpha e - {1 \over 6}$

$\Rightarrow {1 \over 3}\left[ {{e \over 5} - {1 \over 2}} \right] + {e \over {25}} = \alpha e - {1 \over 6}$

$\Rightarrow {e \over {15}} - {1 \over 6} + {e \over {25}} = \alpha e - {1 \over 6}$

$\Rightarrow {e \over {15}} + {e \over {25}} = \alpha e$

$\Rightarrow {{10e + 6e} \over {150}} = \alpha e$

$\Rightarrow {{16e} \over {150}} = \alpha e$

$\Rightarrow \alpha = {{16} \over {150}}$

$\therefore$ $150\alpha = 150 \times {{16} \over {150}} = 16$