For which of the following values of m,the ar | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equation of the curve is $y = x - x^2$.The points of intersection of the curve $y = x - x^2$ and the line $y = mx$ are found by setting $x - x

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) The equation of the curve is $y = x - x^2$.The points of intersection of the curve $y = x - x^2$ and the line $y = mx$ are found by setting $x - x^2 = mx$,which gives $x(1 - x - m) = 0$. Thus,$x = 0$ or $x = 1 - m$.The area $A$ of the region bounded by the curve and the line is given by the integral:$A = \int_{0}^{1-m} (x - x^2 - mx) dx = \int_{0}^{1-m} ((1 - m)x - x^2) dx$Evaluating the integral:$A = \left[ (1 - m)\frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1-m}$$A = (1 - m)\frac{(1 - m)^2}{2} - \frac{(1 - m)^3}{3} = \frac{(1 - m)^3}{2} - \frac{(1 - m)^3}{3} = \frac{(1 - m)^3}{6}$Given that the area is $\frac{9}{2}$,we have:$\frac{(1 - m)^3}{6} = \frac{9}{2}$$(1 - m)^3 = 27$$1 - m = 3$$m = -2$Thus,the correct value of $m$ is $-2$.

For which of the following values of $m$,the area of the region bounded by the curve $y = x - x^2$ and the line $y = mx$ equals $\frac{9}{2}$?

Similar Questions

The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is $24$ sq. units. Then,$a = $ . . . . . . .

Area of the region bounded by the curve $y = x^3$,$x$-axis and the ordinates $x = -1$ and $x = 2$ is . . . . . . . (in $/4$)

The area of the figure bounded by $y = e^x$,$y = e^{-x}$,and the straight line $x = 1$ is

The area (in sq. units) bounded by the parabola $y=x^2+3$,the tangent to the parabola at $(3,12)$ and the coordinate axes and lying in the first quadrant is

Area enclosed by the curve $y = f(x)$ that is defined parametrically as $x = \frac{1 - t^2}{1 + t^2}, y = \frac{2t}{1 + t^2}$ (where $t \in R$) is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module