Question

The figure shows a portion of the graph $y=2x–4x^3.$ The line $y=c$ is such that the areas of the regions marked
I and II are equal. If $a,b$ are the $x-$coordinates of $A,B$ respectively, then $a+b$ equals

• A. $\frac{2}{\sqrt{7}}$
• B. $\frac{3}{\sqrt{7}}$
• C. $\frac{4}{\sqrt{7}}$
• D. $\frac{5}{\sqrt{7}}$

Answer 1

The correct option is A $\frac{2}{\sqrt{7}}$


Given that Area-I and Area-II are equally separted by line $y=c.$
Area-II $=(b-a)c$
Therefore, ${\int }_a^b(2x-4x^3)dx=2(b-a)c$
$[x^2-x^4{]}_a^b=2(b-a)c(a+b)(1-(a^2+b^2\left)\right)=2c(a+b)(1-(a+b{)}^2+2ab)=2c.........(I)$
Again $2x-4x^3=c$
$\Rightarrow 4x^3-2x+c=0\text{having root}a,b,\alpha \left(\text{let}\right)$
So,$a+b+\alpha =0\Rightarrow a+b=-\alpha .....\left(2\right)$
$ab+a\alpha +b\alpha =-\frac{1}{2}\Rightarrow ab={\alpha }^2-\frac{1}{2}......\left(3\right)$
$ab\alpha =-\frac{c}{4}\Rightarrow c=-4\alpha ({\alpha }^2-\frac{1}{2}).....\left(4\right)$
Put the value of $(a+b),ab,c$ from equation $\left(2\right),\left(3\right),\left(4\right)$ in equation $\left(1\right)$ we get,
$\Rightarrow -\alpha \{1-{\alpha }^2+2{\alpha }^2-1\}=-8\alpha ({\alpha }^2-\frac{1}{2})$
$\Rightarrow {\alpha }^2=8{\alpha }^2-4\Rightarrow {\alpha }^2=\frac{4}{7}\Rightarrow \alpha =\pm \sqrt{\frac{4}{7}}$
$\therefore a+b=\frac{2}{\sqrt{7}}$