Introduction

Finding points where a curve has tangents parallel to the x‑axis is a standard application of derivatives in calculus and often appears in competitive exams as MCQs based on functions like y = x⁴ − 4x³ + 4x² − 4.[web:3]

These problems test understanding of slopes of tangents, critical points and evaluation of function values using simple algebra.[web:3]

Final Answer

The tangents to the curve y = x⁴ − 4x³ + 4x² − 4 are parallel to the x‑axis at the points (1, −3), (0, −4) and (2, −4). Hence the correct option is (B).[web:3]

Condition for Tangent Parallel to X‑Axis

• The slope of the tangent at a point (x, y) on a curve y = f(x) is given by dy/dx.
• A tangent parallel to the x‑axis has slope 0, so the condition is dy/dx = 0.

For the curve

y = x⁴ − 4x³ + 4x² − 4

Differentiate with respect to x:

dy/dx = 4x³ − 12x² + 8x

Factor the derivative:

dy/dx = 4x(x² − 3x + 2) = 4x(x − 1)(x − 2)

Set dy/dx = 0:

4x(x − 1)(x − 2) = 0 ⇒ x = 0, 1, 2

Finding the Points on the Curve

Substitute each x into the original curve y = x⁴ − 4x³ + 4x² − 4:[web:3]

• For x = 0: y = 0 − 0 + 0 − 4 = −4 ⇒ point (0, −4).
• For x = 1: y = 1 − 4 + 4 − 4 = −3 ⇒ point (1, −3).
• For x = 2: y = 16 − 32 + 16 − 4 = −4 ⇒ point (2, −4).

Thus, tangents parallel to the x‑axis occur at (0, −4), (1, −3) and (2, −4).[web:3]

Explanation of Each MCQ Option

Given Options

• (A) (1, 4), (−2, 2) and (0, −1)
• (B) (0, −4), (2, −4) and (1, −3)
• (C) (−1, 2), (−2, 1) and (1, −2)
• (D) (1, −4), (1, −3) and (2, −4)

How to Test an Option

• Every point must lie on the curve y = x⁴ − 4x³ + 4x² − 4.
• The x‑coordinate of each point must be one of 0, 1 or 2, since these are the roots of dy/dx where slope is zero.

Option (A)

• x‑values are 1, −2 and 0; dy/dx = 0 only at x = 1 and x = 0, not at x = −2.
• The curve value at x = 1 is −3, not 4, so (1, 4) is not on the curve; at x = 0 the value is −4, not −1, so (0, −1) is not on the curve.
• Hence option (A) is incorrect.

Option (B)

• x‑values are 0, 2 and 1, exactly the three values where dy/dx = 0.
• The corresponding curve points are (0, −4), (2, −4) and (1, −3), all of which lie on the curve and satisfy the slope‑zero condition.
• Therefore, option (B) is correct.[web:3]

Option (C)

• x‑values are −1, −2 and 1; dy/dx = 0 only at x = 1, not at −1 or −2.
• At x = 1, the curve value is −3, not −2, so (1, −2) is not on the curve.
• Thus, option (C) is incorrect.

Option (D)

• x‑values are 1, 1 and 2; dy/dx = 0 for all three x‑values, but the y‑coordinates must also match the curve.
• At x = 1, the curve gives y = −3, so (1, −4) is not on the curve, whereas (1, −3) is correct; at x = 2, (2, −4) is correct.
• Because not all listed points are valid, option (D) is incorrect.