MathematicsEasy38×since 2002Q3535Let 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 :Ax2+2xf(x)−12=0{x^2} + 2xf(x) - 12 = 0x2+2xf(x)−12=0Bx3+xf(x)+12=0{x^3} + xf(x) + 12 = 0x3+xf(x)+12=0Cx3−3xf(x)−4=0{x^3} - 3xf(x) - 4 = 0x3−3xf(x)−4=0Dx2+2xf(x)+4=0{x^2} + 2xf(x) + 4 = 0x2+2xf(x)+4=0Check answerSkip