The maximum value of the function f(x) = 3x8 - 18x2 + 27x - 40 on the set S = x ∫ R: x2 + 30 le 11 x is :

Question

The maximum value of the function f(x) = 3x8 - 18x2 + 27x - 40 on the set S = x ∫ R: x2 + 30 le 11 x is : (A) 122 (B) -222 (C) -122 (D) 222

Tardigrade Admin · Accepted Answer

S = x ∈ R, x2 +30 -11x le0 = x∈ R , 5 le x le 6 Now f(x) =3x3 -18x2 +27x-40 ⇒ f'(x) =9(x-1)(x-3), which is positive in [5, 6] ⇒ f(x) increasing in [5, 6] Hence maximum value = f(6) = 122

Q. The maximum value of the function $f (x) = 3 x^{8} - 18 x^{2} + 27 x - 40$ on the set $S = {x \int R : x^{2} + 30 \leq 11 x}$ is :

122

-222

-122

222

$S = {x \in R, x^{2} + 30 - 11 x \leq 0}$
$= {x \in R, 5 \leq x \leq 6}$
Now $f (x) = 3 x^{3} - 18 x^{2} + 27 x - 40$
$\Rightarrow f^{'} (x) = 9 (x - 1) (x - 3)$ ,
which is positive in $[5, 6]$
$\Rightarrow f (x)$ increasing in $[5, 6]$
Hence maximum value = $f (6) = 122$

Q. The maximum value of the function f(x)=3x8−18x2+27x−40 on the set S={x∫R:x2+30≤11x} is :

122

-222

-122

222

S={x∈R,x2+30−11x≤0} ={x∈R,5≤x≤6} Now f(x)=3x3−18x2+27x−40 ⇒f′(x)=9(x−1)(x−3), which is positive in [5,6] ⇒f(x) increasing in [5,6] Hence maximum value = f(6)=122

Q. The maximum value of the function $f (x) = 3 x^{8} - 18 x^{2} + 27 x - 40$ on the set $S = {x \int R : x^{2} + 30 \leq 11 x}$ is :

$S = {x \in R, x^{2} + 30 - 11 x \leq 0}$
$= {x \in R, 5 \leq x \leq 6}$
Now $f (x) = 3 x^{3} - 18 x^{2} + 27 x - 40$
$\Rightarrow f^{'} (x) = 9 (x - 1) (x - 3)$ ,
which is positive in $[5, 6]$
$\Rightarrow f (x)$ increasing in $[5, 6]$
Hence maximum value = $f (6) = 122$