Question

The curve y = f(x) is such that the area of the trapezium formed by the coordinate axis ordinate of an arbitrary point and the tangent at this point equals half the square of its abscissa. The curve is

• A. y = cx² ± x
• B. y = cx² ± 1
• C. y = cx ± x²
• D. y = cx² ± x ± 1

Answer 1

Answer: (A)

y = cx² ± x

Answer 2

length of intercept on y-axis by any tangent at

P(x, y) = OT = y – x $\frac{dy}{dx}$

Area of trapezium OLPTO = $\frac{1}{2}$ (PL + OT) OL

= $\frac{1}{2}\left( y + y - x\frac{dy}{dx} \right)$x

= $\frac{1}{2}\left( 2y - x\frac{dy}{dx} \right)$x

Now Area of trapezium OLPTO = $\frac{1}{2}$x²

$\frac{1}{2}\left( 2y - x\frac{dy}{dx} \right)$= ± $\frac{1}{2}$x²

2y – x $\frac{dy}{dx}$ = ± x

$\frac{dy}{dx}–\frac{2y}{x}$ = ± 1

Solve diff. equation

y = ± x + cx²