Let λ ^* be the largest value of λ for which the function f_λ (x)=4λ x^3-36λ x^2+36x+48 is increasing for all x∈ mathbbR. Then f_λ ^*(1)+f_λ ,*(-1)…

Given,

$${f}_{\lambda }(x)=4\lambda {x}^{3}-36\lambda {x}^{2}+36x+48$$fλ​(x)=4λx3−36λx2+36x+48

Differentiating w.r.t $x$x both side we get,

$${f}_{\lambda }^{'}(x)=12\lambda {x}^{2}-72\lambda x+36$$fλ′​(x)=12λx2−72λx+36

$${f}_{\lambda }(x)'=12(\lambda {x}^{2}-6\lambda x+3)\geq 0$$fλ​(x)′=12(λx2−6λx+3)≥0

$\therefore \lambda >0$∴λ>0 and $D\leq 0$D≤0

$$36{\lambda }^{2}-4\times \lambda \times 3\leq 0$$36λ2−4×λ×3≤0

$$9{\lambda }^{2}-3\lambda \leq 0$$9λ2−3λ≤0

$$3\lambda (3\lambda -1)\leq 0$$3λ(3λ−1)≤0

$$\lambda \in [0,\frac{1}{3}]$$λ∈[0,31​]

So, ${\lambda }_{\mathrm{largest}}=\frac{1}{3}$λlargest​=31​

$$f(x)=\frac{4}{3}{x}^{3}-12{x}^{2}+36x+48$$f(x)=34​x3−12x2+36x+48

So,$f(1)+f(-1)=72$f(1)+f(−1)=72