Let lambda^{*} be the largest value of lambda | vedclass

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(D) Given $f_{\lambda}(x) = 4\lambda x^{3} - 36\lambda x^{2} + 36x + 48$.For $f_{\lambda}(x)$ to be increasing for all $x \in \mathbb{R}$,we must have

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$72$

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(D) Given $f_{\lambda}(x) = 4\lambda x^{3} - 36\lambda x^{2} + 36x + 48$.For $f_{\lambda}(x)$ to be increasing for all $x \in \mathbb{R}$,we must have $f_{\lambda}^{\prime}(x) \geq 0$ for all $x \in \mathbb{R}$.$f_{\lambda}^{\prime}(x) = 12\lambda x^{2} - 72\lambda x + 36$.Setting $f_{\lambda}^{\prime}(x) \geq 0$,we get $12(\lambda x^{2} - 6\lambda x + 3) \geq 0$,which implies $\lambda x^{2} - 6\lambda x + 3 \geq 0$.For this quadratic to be non-negative for all $x$,we must have $\lambda > 0$ and the discriminant $D \leq 0$.$D = (-6\lambda)^{2} - 4(\lambda)(3) = 36\lambda^{2} - 12\lambda \leq 0$.$12\lambda(3\lambda - 1) \leq 0$,which gives $\lambda \in [0, 1/3]$.Since $\lambda > 0$,the largest value is $\lambda^{*} = 1/3$.Now,$f_{\lambda^{*}}(x) = 4(1/3)x^{3} - 36(1/3)x^{2} + 36x + 48 = \frac{4}{3}x^{3} - 12x^{2} + 36x + 48$.$f_{\lambda^{*}}(1) = \frac{4}{3} - 12 + 36 + 48 = \frac{4}{3} + 72 = \frac{220}{3}$.$f_{\lambda^{*}}(-1) = \frac{4}{3}(-1) - 12(1) + 36(-1) + 48 = -\frac{4}{3} - 12 - 36 + 48 = -\frac{4}{3}$.$f_{\lambda^{*}}(1) + f_{\lambda^{*}}(-1) = \frac{220}{3} - \frac{4}{3} = \frac{216}{3} = 72$.

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 $x \in \mathbb{R}$. Then $f_{\lambda^{}}(1) + f_{\lambda^{*}}(-1)$ is equal to:

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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 $x \in \mathbb{R}$. Then $f_{\lambda^{*}}(1) + f_{\lambda^{*}}(-1)$ is equal to: