Question

A point on the curve \( y = x^3 - 3x \), where the tangent is parallel to the chord joining \( (1, -2) \) and \( (2, 2) \), is:

• A. \( (1, -1) \)
• B. \( (1, -1.53) \)
• C. \( (1.53, -1) \)
• D. \( (1.53, -1.53) \)

Hint:

To find a point where the tangent is parallel to a chord, first find the slope of the chord, then differentiate the curve and set the derivative equal to the slope of the chord.

Answer 1

The Correct Option is B

We are given the curve \( y = x^3 - 3x \) and need to find a point where the tangent is parallel to the chord joining \( (1, -2) \) and \( (2, 2) \). 1. Find the slope of the chord: The slope of the chord joining \( (1, -2) \) and \( (2, 2) \) is given by the formula: \[ \text{Slope of chord} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-2)}{2 - 1} = \frac{4}{1} = 4 \] 2. Find the derivative of the curve: The derivative of the curve \( y = x^3 - 3x \) will give us the slope of the tangent at any point on the curve. Differentiate the given equation: \[ \frac{dy}{dx} = 3x^2 - 3 \] 3. Set the derivative equal to the slope of the chord: Since the tangent is parallel to the chord, the slope of the tangent at the point of interest must be equal to the slope of the chord, which is 4: \[ 3x^2 - 3 = 4 \] Solving for \( x \): \[ 3x^2 = 7 \quad \Rightarrow \quad x^2 = \frac{7}{3} \quad \Rightarrow \quad x = \pm \sqrt{\frac{7}{3}} \approx \pm 1.53 \] 4. Find the corresponding \( y \)-coordinates: Now, substitute \( x = 1.53 \) into the original equation \( y = x^3 - 3x \): \[ y = (1.53)^3 - 3(1.53) \approx 3.576 - 4.59 \approx -1.53 \] Thus, the point on the curve is approximately \( (1.53, -1.53) \), and the correct answer is (b) \( (1, -1.53) \).