JEE Advanced 2022 Paper 2 · Mathematics · Differential Equations · hard

A curve y=f(x), x>0, passes through (1,1) and has the property that at every point P(x,y) on it the tangent cuts the y-axis at a point whose ordinate…

Numerical answer

A curve

y=f(x)

x>0

, passes through

(1,1)

and has the property that at every point

P(x,y)

on it the tangent cuts the

y

-axis at a point whose ordinate equals

2x^{2}

. The curve meets the positive

x

-axis again at one point, and the region enclosed between the curve and the

x

-axis has area

\mathcal A

. The value of

8\mathcal A

Show answer and solution

Answer: 9

The tangent at

(x,y)

has

y

-intercept

y-xy'

. The condition gives

y-xy'=2x^2

, i.e.

y'-\dfrac yx=-2x

. This is linear with integrating factor

1/x

, so

\dfrac{d}{dx}\!\left(\dfrac yx\right)=-2

, hence

y=-2x^{2}+cx

. The point

(1,1)

gives

c=3

, so

y=3x-2x^{2}=x(3-2x)

. The curve meets the

x

-axis again at

x=\tfrac32

. The enclosed area is

\mathcal A=\int_0^{3/2}(3x-2x^2)\,dx=\dfrac{3}{2}\cdot\dfrac94-\dfrac23\cdot\dfrac{27}{8}=\dfrac{27}{8}-\dfrac94=\dfrac98

. Therefore

8\mathcal A=9

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