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, follow these steps:

1. First, find the derivative of the function to locate the points of maxima and minima. The given function is \( y = (2x-1)e^{2(1-x)} \).
2. Apply the product rule to differentiate:
   \(\frac{dy}{dx} = \frac{d}{dx} [(2x-1) e^{2(1-x)}]\)
   Use the product rule: \((uv)' = u'v + uv'\). Here, \(u = (2x-1)\) and \(v = e^{2(1-x)}\).
3. Differentiate each part:
   • For \(u = 2x-1\), \(u' = 2\).
   • For \(v = e^{2(1-x)}\), use chain rule:
     \(\frac{d}{dx} e^{2(1-x)} = e^{2(1-x)} \cdot (-2)\)
4. Combine these using the product rule:
   \(\frac{dy}{dx} = 2e^{2(1-x)} + (2x-1)(-2e^{2(1-x)})\)
   Simplify:
   \(\frac{dy}{dx} = 2e^{2(1-x)} - 2(2x-1)e^{2(1-x)}\)
   \(\frac{dy}{dx} = [2 - 2(2x-1)]e^{2(1-x)}\)
   \(\frac{dy}{dx} = (2 - 4x + 2)e^{2(1-x)}\)
   \(\frac{dy}{dx} = (-4x + 4)e^{2(1-x)}\)
   \(\frac{dy}{dx} = -4(x-1)e^{2(1-x)}\)
5. Set \(\frac{dy}{dx} = 0\) to find critical points:
   -4(x - 1)e^{2(1-x)} = 0\)
   (x-1) = 0 \Rightarrow x=1\)
6. Substitute \(x = 1\) back into the original function to find the corresponding \(y\) coordinate:
   y = (2 \times 1 - 1)e^{2(1-1)} = 1 \times e^{0} = 1
   So, the point of maximum is \((1, 1)\).
7. To find the equation of the tangent at this point, use the formula for a line \(y - y_1 = m(x - x_1)\), where \(m\) is the slope at the tangent point: The slope \(m = \frac{dy}{dx} \) at \(x = 1\). Substitute \(x = 1\) to get \(m = 0\).
8. The tangent line is horizontal since \(m = 0\). Therefore, the equation of the tangent is:
   y - 1 = 0\)

Thus, the equation of the tangent to the curve at the point of maximum is \(\boxed{y - 1 = 0}\).