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let \mathrm{\lambda^{*}}  be the largest value of \mathrm{\lambda} for which the function \mathrm{f_{\lambda}(x)=4 \lambda x^{3}-36 \lambda x^{2}+36 x+48} is increasing for all \mathrm{x \in \mathbb{R}}. Then ion: \mathrm{f_{\lambda^{*}}(1)+f_{\lambda_{*} *}(-1)} is equal to :

Option: 1

36


Option: 2

48


Option: 3

64


Option: 4

72


Answers (1)

best_answer

\mathrm{\begin{aligned} &f_{\lambda}(x)=4 \lambda x^{3}-36 \lambda x^{2}+36 x+48\\ &f_{\lambda}^{\prime}(x)=12 \lambda x^{2}-72 \lambda x+36 \geq 0\\ &D \leq 0\\ \end{aligned}}

\mathrm{\begin{aligned} &(72 \lambda)^{2}-4(12 \lambda)(36) \leq 0\\ &144 \lambda^{2}-48 \lambda \leq 0\\ &\lambda(3 \lambda-1) \leq 0\\ &\lambda \in\left[0, \frac{1}{3}\right]\\ \end{aligned}}

\mathrm{\begin{aligned} &\lambda^{*}=\frac{1}{3}\\ &\therefore f_{\left(\frac{1}{3}\right)} 1+f_{\left(\frac{1}{3}\right)}{ }^{-1}=-36\left(\frac{1}{3}\right)(1+1)+48(1+1)\\ &=-24+96=72 \end{aligned}}

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