Question

The function \( y = 2x^3 - 8x^2 + 10x - 4 \) is defined on \([1,2]\). If the tangent drawn at a point \( (a,b) \) on the graph of this function is parallel to the X-axis and \( a \in (1,2) \), then \( a = \) ? 

• A. \( 0 \)
• B. \( 5 \)
• C. \( 1 \)
• D. \( \frac{5}{3} \)

Hint:

To find where a function's tangent is parallel to the X-axis, set the first derivative equal to zero and solve for \( x \).

Answer 1

The Correct Option is D

Step 1: Understanding the Condition for a Tangent Parallel to X-axis The equation of the given function is: \[ y = 2x^3 - 8x^2 + 10x - 4 \] For the tangent to be parallel to the X-axis, its slope must be zero. The slope of the tangent is given by the first derivative of \( y \), i.e., \[ \frac{dy}{dx} = 0 \]

Step 2: Compute the First Derivative \[ \frac{dy}{dx} = \frac{d}{dx} (2x^3 - 8x^2 + 10x - 4) \] \[ = 6x^2 - 16x + 10 \] Setting \( \frac{dy}{dx} = 0 \) for the required condition: \[ 6x^2 - 16x + 10 = 0 \]

Step 3: Solve for \( x \) Solving the quadratic equation: \[ 6x^2 - 16x + 10 = 0 \] Using the quadratic formula: \[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(6)(10)}}{2(6)} \] \[ x = \frac{16 \pm \sqrt{256 - 240}}{12} \] \[ x = \frac{16 \pm \sqrt{16}}{12} \] \[ x = \frac{16 \pm 4}{12} \] \[ x = \frac{16 + 4}{12} = \frac{20}{12} = \frac{5}{3}, \quad x = \frac{16 - 4}{12} = \frac{12}{12} = 1 \]

Step 4: Selecting the Correct Value of \( a \) Since \( a \in (1,2) \), the valid solution is: \[ a = \frac{5}{3} \]