Question

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
• D. 72

Answer 1

Answer: (D)

$f_{\lambda}(x)=4 \lambda x^{3}-36 \lambda x^{2}+36 x+48$
$f_{\lambda}{ }^{\prime}(x)=12 \lambda x^{2}-72 \lambda x+36$
$f_{\lambda}{ }^{\prime}(x)=12\left(\lambda x^{2}-6 \lambda x+3\right) \geq 0$
$\therefore \lambda>0 \& D \leq 0$
$36 \lambda^{2}-4 \times \lambda \times 3 \leq 0$
$9 \lambda^{2}-3 \lambda \leq 0$
$3 \lambda(3 \lambda-1) \leq 0$
$\lambda \in\left[0, \frac{1}{3}\right]$
$\therefore \lambda_{\text {largest }}=\frac{1}{3}$
$f ( x )=\frac{4}{3} x ^{3}-12 x ^{2}+36 x +48$
$\therefore f (1)+ f (1)=72$