If the line x=α divides the area of region R= (x, y) ∈ R 2: x3 ≤ y ≤ x, 0 ≤ x ≤ 1 into two equal parts, then

Question

If the line x=α divides the area of region R= (x, y) ∈ R 2: x3 ≤ y ≤ x, 0 ≤ x ≤ 1 into two equal parts, then (A) 0 < α le (1/2) (B)  (1/2) < α < 1 (C)  2α4-4α2 + 1 = 0 (D) α4+4α2 - 1 = 0

Tardigrade Admin · Accepted Answer

(1/2)(∫ limits01( x - x 3) dx )=∫ limits0α( x - x 3) dx (1/8)=|( x 2/2)-( x 4/4)|0α (1/2)=2 α2-α4 2 α4-4 α2+1=0 f (α)=2 α4-4 α2+1 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/acd8459982e0d4c4de0290bbb3f8ab1f-.png" />

Q. If the line $x = α$ divides the area of region $R = {(x, y) \in R^{2} : x^{3} \leq y \leq x, 0 \leq x \leq 1}$ into two equal parts, then

$0 < α \leq \frac{1}{2}$

$\frac{1}{2} < α < 1$

$2 α^{4} - 4 α^{2} + 1 = 0$

$α^{4} + 4 α^{2} - 1 = 0$

$\frac{1}{2} (0 \int 1 (x - x^{3}) d x) = 0 \int α (x - x^{3}) d x$
$\frac{1}{8} = ∣ ∣ \frac{x ^{2}}{2} - \frac{x ^{4}}{4} ∣ ∣_{0}^{α}$
$\frac{1}{2} = 2 α^{2} - α^{4}$
$2 α^{4} - 4 α^{2} + 1 = 0$
$f (α) = 2 α^{4} - 4 α^{2} + 1$

Q. If the line x=α divides the area of region R={(x,y)∈R2:x3≤y≤x,0≤x≤1} into two equal parts, then

0<α≤21​

21​<α<1

2α4−4α2+1=0

α4+4α2−1=0