Verified Solution • 2025–26 Session moderate

QJEEM21Ma90e1: JEE Main 2021 Mathematics - Class 12 JEEM21Ma90e1 Solution: Let us consider a curve, y = f(x) passing through the point (…

Let us consider a curve, y = f(x) passing through the point ( −- 2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :

Multiple Choice Options

  • A x2+2xf(x)−12=0{x^2} + 2xf(x) - 12 = 0
  • B x3+xf(x)+12=0{x^3} + xf(x) + 12 = 0
  • C x3−3xf(x)−4=0{x^3} - 3xf(x) - 4 = 0
  • D x2+2xf(x)+4=0{x^2} + 2xf(x) + 4 = 0

Step-by-Step Verified Solution

y+xdydx=x2y + {{xdy} \over {dx}} = {x^2} (given)

⇒dydx+yx=x\Rightarrow {{dy} \over {dx}} + {y \over x} = x

If e∫1xdx=x{e^{\int {{1 \over x}dx} }} = x

Solution of DE

⇒y . x=∫x . x dx\Rightarrow y\,.\,x = \int {x\,.\,x\,dx}

⇒xy=x33+c3\Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}

Passes through ( −- 2, 2), so

−- 12 = −- 8 + c ⇒\Rightarrow c = −- 4

∴\therefore 3xy = x3 −- 4

i.e. 3x . f(x) = x3 −- 4

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