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JEE Advanced 2025 Paper 2 · Mathematics · Differential Equations · medium

The curve y=y(x), x>0, y>0, passes through (1,√3) and satisfies dy/dx=(x^2+y^2/(2xy).

One or more options correct

The curve y=y(x)y=y(x), x>0x>0, y>0y>0, passes through (1,3)(1,\sqrt3) and satisfies dydx=x2+y22xy\dfrac{dy}{dx}=\dfrac{x^2+y^2}{2xy}. Which of the following statements is(are) correct?
  1. A. y(2)=22y(2)=2\sqrt2
  2. B. yy is an increasing function of xx on (0,∞)(0,\infty)
  3. C. The curve is a part of a rectangular hyperbola of eccentricity 2\sqrt2
  4. D. The slope of the tangent at (1,3)(1,\sqrt3) is 13\dfrac{1}{\sqrt3}
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Answer: A, B, C

Put v=y2v=y^2: then dvdx=2ydydx=x2+vx=x+vx\frac{dv}{dx}=2y\frac{dy}{dx}=\frac{x^2+v}{x}=x+\frac vx, a linear equation with solution v=x2+cxv=x^2+cx. Using v(1)=3v(1)=3 gives c=2c=2, so y2=x2+2xy^2=x^2+2x, i.e. (x+1)2−y2=1(x+1)^2-y^2=1. (A) True: y(2)=8=22y(2)=\sqrt8=2\sqrt2. (B) True: y′=x2+y22xy>0y'=\frac{x^2+y^2}{2xy}>0 for x,y>0x,y>0. (C) True: (x+1)2−y2=1(x+1)^2-y^2=1 has a=b=1a=b=1, so e=2e=\sqrt2. (D) False: y′(1)=1+323=23y'(1)=\frac{1+3}{2\sqrt3}=\frac{2}{\sqrt3}. Correct: A, B, C.
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Original question written in the JEE Advanced 2025 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake