Let λ * be the largest value of λ for which the function f λ( x )=4 λ x 3-36 λ x 2+36 x +48 is increasing for all x ∈ R. Then fλ *(1)+fλ *(-1) is equal to :

Question

Let λ * be the largest value of λ for which the function f λ( x )=4 λ x 3-36 λ x 2+36 x +48 is increasing for all x ∈ R. Then fλ *(1)+fλ *(-1) is equal to : (A) 36 (B) 48 (C) 64 (D) 72

Tardigrade Admin · Accepted Answer

fλ(x)=4 λ x3-36 λ x2+36 x+48 fλ prime(x)=12 λ x2-72 λ x+36 fλ prime(x)=12(λ x2-6 λ x+3) ≥ 0 ∴ λ>0 D ≤ 0 36 λ2-4 × λ × 3 ≤ 0 9 λ2-3 λ ≤ 0 3 λ(3 λ-1) ≤ 0 λ ∈[0, (1/3)] ∴ λ text largest =(1/3) f ( x )=(4/3) x 3-12 x 2+36 x +48 ∴ f (1)+ f (1)=72

Q. Let λ∗ be the largest value of λ for which the function fλ​(x)=4λx3−36λx2+36x+48 is increasing for all x∈R. Then fλ​∗(1)+fλ​∗(−1) is equal to :

36

48

64

72

fλ​(x)=4λx3−36λx2+36x+48 fλ​′(x)=12λx2−72λx+36 fλ​′(x)=12(λx2−6λx+3)≥0 ∴λ>0&D≤0 36λ2−4×λ×3≤0 9λ2−3λ≤0 3λ(3λ−1)≤0 λ∈[0,31​] ∴λlargest ​=31​ f(x)=34​x3−12x2+36x+48 ∴f(1)+f(1)=72

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