JEE Advanced 2013 · Paper 1 · Mathematics Q46

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MathematicsSingle correct+2Question 46 of 60

A curve passes through the point $\left(1,\frac\pi6\right)$ . Let the slope of the curve at each point $(x,y)$ be $\dfrac yx+\sec\left(\dfrac yx\right)$ , $x>0$ . Then the equation of the curve is

\sin\left(\dfrac yx\right)=\log x+\dfrac12

\operatorname{cosec}\left(\dfrac yx\right)=\log x+2

\sec\left(\dfrac{2y}x\right)=\log x+2

\cos\left(\dfrac{2y}x\right)=\log x+\dfrac12

Answer and worked solution

Official answer: A

Homogeneous equation. With $y=vx$ : $v+x\dfrac{dv}{dx}=v+\sec v$ , so $\cos v\,dv=\dfrac{dx}x$ and $\sin v=\log x+C$ .

Point $(1,\frac\pi6)$ : $C=\frac12$ . So $\sin\left(\dfrac yx\right)=\log x+\dfrac12$ .

(A)

Topics:Homogeneous equations

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JEE Advanced 2013 · Paper 1, Mathematics question 46