Question

The equation of the tangent to the curve \(y = (2x-1)e^{2(1-x)}\) at the point of its maximum, is

• A. \(y - 1 = 0\)
• B. \(x - 1 = 0\)
• C. \(x + y - 1 = 0\)
• D. \(x - y + 1 = 0\)

Hint:

At a maximum or minimum, the tangent is horizontal.

Answer 1

The Correct Option is A

To find the equation of the tangent to the curve \( y = (2x-1)e^{2(1-x)} \) at the point of its maximum, we need to first determine the point where the curve reaches its maximum value. This is done by performing the following steps:

1. Find the derivative of \( y \) with respect to \( x \), i.e., \( \frac{dy}{dx} \).
2. Set \( \frac{dy}{dx} = 0 \) to find the critical points of \( y \).
3. Determine which critical point corresponds to the maximum by using the second derivative test or analyzing the behavior around the critical points.
4. Once the point of maximum is found, use it to find the tangent equation at that point.

Let's proceed with these steps:

Step 1: Finding \( \frac{dy}{dx} \)

Using the product and chain rule, differentiate \( y = (2x-1)e^{2(1-x)} \):

\[\frac{dy}{dx} = \frac{d}{dx}[(2x-1) \cdot e^{2(1-x)}]\]

Applying the product rule:

\[\frac{dy}{dx} = (2x-1) \cdot \frac{d}{dx}[e^{2(1-x)}] + e^{2(1-x)} \cdot \frac{d}{dx}[2x-1]\]

Evaluate each derivative:

• \(\frac{d}{dx}[e^{2(1-x)}] = e^{2(1-x)} \cdot \left( -2 \right)\)(using the chain rule)
• \(\frac{d}{dx}[2x-1] = 2\)

Thus, substituting back gives:

\[\frac{dy}{dx} = (2x-1) \cdot (-2) \cdot e^{2(1-x)} + 2 \cdot e^{2(1-x)}\]

Simplifying:

\[\frac{dy}{dx} = e^{2(1-x)} \cdot [-2(2x-1) + 2] = e^{2(1-x)} \cdot (-4x + 4)\]\[\frac{dy}{dx} = 2e^{2(1-x)} \cdot (1-x)\]

Step 2: Setting \( \frac{dy}{dx} = 0 \)

\[2e^{2(1-x)} \cdot (1-x) = 0\]

Since \( e^{2(1-x)} \) is never zero, we set:

\[1-x = 0 \implies x = 1\]

Step 3: Check if \( x = 1 \) is a maximum

Check the concavity by evaluating the second derivative at \( x = 1 \). For this function, upon testing the behavior around \( x = 1 \), we find \( x = 1 \) is a point of maximum because the function changes slope from positive to negative.

Step 4: Find the tangent at \( (x=1, y=(2(1)-1)e^{2(1-1)}) \)

Calculate the value of \( y \) at \( x = 1 \):

\[y = (2 \cdot 1 - 1) \cdot e^{0} = 1\]

Tangent Equation

The slope of the tangent at maximum is zero (horizontal tangent). Hence, the equation of the tangent is:

\[y - 1 = 0\]

This means the tangent is parallel to the x-axis. Therefore, the correct answer is \( y - 1 = 0 \).