JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q10

← JEE Main 2022 · 27 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 10 of 90

If the solution curve of the differential equation $\left(\left(\tan^{-1}y\right)-x\right)dy=\left(1+y^2\right)dx$ passes through the point $(1,0)$ , then the abscissa of the point on the curve whose ordinate is $\tan(1)$ , is

2e

\dfrac2e

\dfrac1e

Answer and worked solution

Official answer: B

Linear in x. $\dfrac{dx}{dy}+\dfrac x{1+y^2}=\dfrac{\tan^{-1}y}{1+y^2}$ ; with $t=\tan^{-1}y$ , $xe^t=(t-1)e^t+C$ .

Through (1, 0). $1=-1+C$ , $C=2$ .

At $y=\tan1$ ( $t=1$ ): $xe=2$ , $x=\dfrac2e$ .

(B)

Topics:Linear first-order equations

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JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 10