The local maximum value of x(1-x)2, 0 ≤ x ≤ 2 is-

Question

The local maximum value of x(1-x)2, 0 ≤ x ≤ 2 is- (A) 2 (B) (4/27) (C) 5 (D) 2, (4/27)

Tardigrade Admin · Accepted Answer

f(x)=x(1-x)2 f prime(x)=-2 x(1-x)+(1-x)2=0 (1-x)(-2 x+1-x)=0 ⇒ (1-x)(1-3 x)=0, x=1,1 / 3 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/48ee12af3839dc7393f06835e1be0180-.png" /> Local max. at x=1 / 3 which is f(x)=(1/3)(1-(1/3))2=(4/27)

Q. The local maximum value of $x (1 - x)^{2}, 0 \leq x \leq 2$ is-

2

$\frac{4}{27}$

5

$2, \frac{4}{27}$

$f (x) = x (1 - x)^{2}$
$f^{'} (x) = - 2 x (1 - x) + (1 - x)^{2} = 0$
$(1 - x) (- 2 x + 1 - x) = 0$
$\Rightarrow (1 - x) (1 - 3 x) = 0, x = 1, 1/3$

Local max. at $x = 1/3$
which is $f (x) = \frac{1}{3} (1 - \frac{1}{3})^{2} = \frac{4}{27}$

Q. The local maximum value of x(1−x)2,0≤x≤2 is-

2

274​

5

2,274​

f(x)=x(1−x)2 f′(x)=−2x(1−x)+(1−x)2=0 (1−x)(−2x+1−x)=0 ⇒(1−x)(1−3x)=0,x=1,1/3 Local max. at x=1/3 which is f(x)=31​(1−31​)2=274​

Q. The local maximum value of $x (1 - x)^{2}, 0 \leq x \leq 2$ is-

$\frac{4}{27}$

$2, \frac{4}{27}$

$f (x) = x (1 - x)^{2}$
$f^{'} (x) = - 2 x (1 - x) + (1 - x)^{2} = 0$
$(1 - x) (- 2 x + 1 - x) = 0$
$\Rightarrow (1 - x) (1 - 3 x) = 0, x = 1, 1/3$

Local max. at $x = 1/3$
which is $f (x) = \frac{1}{3} (1 - \frac{1}{3})^{2} = \frac{4}{27}$