JEE MAIN - Mathematics (2022 - 24th June Evening Shift - No. 15)
Let $$\lambda$$$$^ * $$ be the largest value of $$\lambda$$ for which the function $${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36x + 48$$ is increasing for all x $$\in$$ R. Then $${f_{{\lambda ^ * }}}(1) + {f_{{\lambda ^ * }}}( - 1)$$ is equal to :
36
48
64
72
Explanation
$$\because$$ $${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36\lambda + 48$$
$$\therefore$$ $$f{'_\lambda }(x) = 12(\lambda {x^2} - 6\lambda x + 3)$$
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, $$f_\lambda ^*(x) = {4 \over 3}{x^3} - 12{x^2} + 36x + 48$$
$$\therefore$$ $$f_\lambda ^*(1) + f_\lambda ^*( - 1) = 73{1 \over 2} - 1{1 \over 2} = 72$$
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