QJEEM22Ma2c77: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2c77 Solution: Let be the largest value of for which the function is…

$\because$ ${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36\lambda + 48$

$\therefore$ $f{'_\lambda }(x) = 12(\lambda {x^2} - 6\lambda x + 3)$

For ${f_\lambda }(x)$ increasing : ${(6\lambda )^2} - 12\lambda \le 0$

$\therefore$ $\lambda \in \left[ {0,\,{1 \over 3}} \right]$

$\therefore$ $\lambda^ * = {1 \over 3}$

Now, $f_\lambda ^*(x) = {4 \over 3}{x^3} - 12{x^2} + 36x + 48$

$\therefore$ $f_\lambda ^*(1) + f_\lambda ^*( - 1) = 73{1 \over 2} - 1{1 \over 2} = 72$