The figure shows a portions of the graph $y = 2 x - 4 x^{3}$ . 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

\frac{2}{\sqrt{7}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3}{\sqrt{7}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4}{\sqrt{7}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{5}{\sqrt{7}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{2}{\sqrt{7}}$
$\int_{a}^{b} (2 x - 4 x^{3}) d x = 2 (b - a) C$
$(x^{2} - x^{4})_{a}^{b} = 2 (b - a) C$
$(b + a) [1 - (a^{2} + b^{2})] = 2 C$
$(a + b) [1 - (a + b)^{2} + 2 a b] = 2 C$ ......(i)
Again
$2 x - 4 x^{3} = C$
$4 x^{3} - 2 x + C = 0$
Roots, a, b, $α$
$4 x^{3} + 0. x^{2} - 2 x + C = 0$
$a + b + α = 0$ ........(ii)
$a b + (a + b) α = - \frac{1}{2}$ ......(iii)
$7 α^{3} = 4 α \Rightarrow α = \frac{2}{\sqrt{7}}$

Suggest Corrections

Similar questions

The figure shows a portion of the graph $y = 2 x - 4 x^{3}$ . 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

The coordinates of A, B, C are (6, 3), (–3, 5) and (4, – 2) respectively and P is any point (x, y). Show that the ratio of the areas of triangle PBC and ABC is .
| $\frac{x + y - 2}{7}$ | [4 MARKS]