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MHT CET202611 April 2026Morning ShiftMathematicsApplication of DerivativesActual

The equation of the tangent to the curve y = 3x^3 - 3x^2 + x at x = 1 is

Options

  1. A4x - y + 3 = 0
  2. B4x + y - 3 = 0
  3. C4x - y - 3 = 0
  4. D4x + y + 3 = 0

Correct answer

C. 4x - y - 3 = 0

Step-by-step solution

Given curve is y = 3x^3 - 3x^2 + x Substituting x = 1 , we get y = 3(1)^3 - 3(1)^2 + 1 = 1 The point of tangency is (1, 1) Differentiating the given equation with respect to x , we get dy dx = 9x^2 - 6x + 1 The slope of the tangent at x = 1 is m = ( dy dx )_ x=1 = 9(1)^2 - 6(1) + 1 = 4 The equation of the tangent at (1, 1) is y - 1 = 4(x - 1) y - 1 = 4x - 4 4x - y - 3 = 0 Answer: 4x - y - 3 = 0

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