The equation of the tangent to the curve y=3 x³-3 x²+x at x=1 is

Correct Answer is : 4x−y−3=04 x-y-3=04x−y−3=0

Correct Answer is : 4x−y−3=04 x-y-3=04x−y−3=0

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MHTCET 11 Apr 2026 Shift 1 Paper

Question : 127 of 150

Marks: +1, -0

y=3 x^{3}-3 x^{2}+x

x=1

Solution:

Concept:

The tangent to a curve at a point has slope equal to the derivative

\frac{dy}{dx}

at that point.

Explanation:

Given curve:

y = 3x^{3} - 3x^{2} + x

x = 1

y = 3(1)^3 - 3(1)^2 + 1 = 3 - 3 + 1 = 1

So the point of tangency is

(1, 1)

Differentiate the curve:

\frac{dy}{dx} = 9x^{2} - 6x + 1

Slope at

x = 1

m = 9(1)^2 - 6(1) + 1 = 9 - 6 + 1 = 4

Equation of tangent:

y - y_1 = m(x - x_1)

Substitute

m = 4

(x_1, y_1) = (1, 1)

y - 1 = 4(x - 1)

y - 1 = 4x - 4

y = 4x - 3

Rearrange:

4x - y - 3 = 0

Answer:

The correct option is Option C:

4x - y - 3 = 0

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