JEE Main 2022 · 29 Jul · Shift 2 · Mathematics Q9

MathematicsSingle correct+4 / -1Question 9 of 90

If the solution curve of the differential equation $\dfrac{dy}{dx}=\dfrac{x+y-2}{x-y}$ passes through the points (2, 1) and (k + 1, 2), k > 0, then

2\tan^{-1}\left(\dfrac1k\right)=\log_e(k^2+1)

\tan^{-1}\left(\dfrac1k\right)=\log_e(k^2+1)

2\tan^{-1}\left(\dfrac1{k+1}\right)=\log_e(k^2+2k+2)

2\tan^{-1}\left(\dfrac1k\right)=\log_e\left(\dfrac{k^2+1}{k^2}\right)

Answer and worked solution

Official answer: A

Shift $X=x-1$ , $Y=y-1$ : $\dfrac{dY}{dX}=\dfrac{X+Y}{X-Y}$ , whose solution is $\frac12\ln(X^2+Y^2)=\tan^{-1}\frac YX+C$ .

Points. (2, 1) is $X=1$ , $Y=0$ : $C=0$ . $(k+1,2)$ is $X=k$ , $Y=1$ : $\frac12\ln(k^2+1)=\tan^{-1}\frac1k$ .

(A)

Topics:Homogeneous equations

JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 9