Let ^ * be the largest value of for which the function f _ ( x )=… - Vidyadip

MCQ

Let $\lambda^{*}$ be the largest value of $\lambda$ for which the function $f _{\lambda}( x )=4 \lambda x ^{3}-36 \lambda x ^{2}+36 x +48$ is increasing for all $x \in R$. Then $f _{\lambda} *(1)+ f _{\lambda} *(-1)$ is equal to

A
$36$
B
$48$
C
$64$
✓
$72$

✓

Answer

Correct option: D.

$72$

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

$f_{\lambda}^{\prime}(x)=12 \lambda x^{2}-72 \lambda x+36$

$f_{\lambda}^{\prime}(x)=12\left(\lambda x^{2}-6 \lambda x+3\right) \geq 0$

$\therefore \lambda>0 \ and \,D \leq 0$

$36 \lambda^{2}-4 \times \lambda \times 3 \leq 0$

$9 \lambda^{2}-3 \lambda \leq 0$

$3 \lambda(3 \lambda-1) \leq 0$

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

$\therefore \lambda_{\text {largest }}=\frac{1}{3}$

$f ( x )=\frac{4}{3} x ^{3}-12 x ^{2}+36 x +48$

$\therefore f (1)+ f (1)=72$

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