Understanding the Problem

We need to find a point on the curve defined by the equation $y = x^2 + 1$. A tangent line drawn at this point should create a trapezium with the greatest possible area. This trapezium is formed within a specific figure bounded by the curve $y = x^2 + 1$, the x-axis ($y=0$), the y-axis ($x=0$), and the vertical line $x=1$.

Steps to Find the Maximum Area Trapezium

Step 1: Define the Curve and Boundaries

The curve is given by $f(x) = y = x^2 + 1$. The figure is bounded by $x=0$, $x=1$, $y=0$, and $y=x^2+1$.

Step 2: Equation of the Tangent Line

Let the point of tangency on the curve be $(x_0, y_0)$, where $y_0 = x_0^2 + 1$. The derivative of the curve gives the slope of the tangent line:

$ \frac{dy}{dx} = 2x $

At $x=x_0$, the slope $m = 2x_0$. The equation of the tangent line is:

$ y - y_0 = m(x - x_0) $

$ y - (x_0^2 + 1) = 2x_0(x - x_0) $

$ y = 2x_0 x - 2x_0^2 + x_0^2 + 1 $

$ y = 2x_0 x - x_0^2 + 1 $

Step 3: Determine Trapezium Vertices

The tangent line cuts off a trapezium. The figure is bounded by $x=0$ and $x=1$. The trapezium's vertices are formed by the intersections of the tangent line with the boundary lines $x=0$ and $x=1$, and the base is on the x-axis ($y=0$). The vertices are:

• $(0, 0)$ (Origin)
• $(1, 0)$ (On the x-axis at $x=1$)
• Intersection with $x=0$: Substitute $x=0$ into the tangent equation: $y = 2x_0(0) - x_0^2 + 1 = 1 - x_0^2$. Vertex is $(0, 1 - x_0^2)$.
• Intersection with $x=1$: Substitute $x=1$ into the tangent equation: $y = 2x_0(1) - x_0^2 + 1 = 1 + 2x_0 - x_0^2$. Vertex is $(1, 1 + 2x_0 - x_0^2)$.

This forms a trapezium with parallel vertical sides at $x=0$ and $x=1$. The height of the trapezium is $h = 1 - 0 = 1$.

Step 4: Calculate Trapezium Area

The lengths of the parallel sides ($b_1$ and $b_2$) are the y-coordinates of the vertices at $x=0$ and $x=1$ respectively:

• $b_1 = 1 - x_0^2$
• $b_2 = 1 + 2x_0 - x_0^2$

The area $A$ of the trapezium is given by $A = \frac{1}{2}(b_1 + b_2)h$.

$ A(x_0) = \frac{1}{2}((1 - x_0^2) + (1 + 2x_0 - x_0^2)) \times 1 $

$ A(x_0) = \frac{1}{2}(2 + 2x_0 - 2x_0^2) $

$ A(x_0) = 1 + x_0 - x_0^2 $

Step 5: Maximize the Area

To find the maximum area, we need to find the value of $x_0$ (where $0 \le x_0 \le 1$) that maximizes $A(x_0)$. We take the derivative of $A(x_0)$ with respect to $x_0$ and set it to zero:

$ \frac{dA}{dx_0} = 1 - 2x_0 $

Set the derivative to zero:

$ 1 - 2x_0 = 0 $

$ x_0 = \frac{1}{2} $

To confirm this is a maximum, we check the second derivative:

$ \frac{d^2A}{dx_0^2} = -2 $

Since the second derivative is negative, $x_0 = \frac{1}{2}$ corresponds to a maximum area. This value is within the required range $[0, 1]$.

Step 6: Find the Point on the Curve

The value $x_0 = \frac{1}{2}$ is the x-coordinate of the point of tangency. To find the y-coordinate, substitute $x_0 = \frac{1}{2}$ into the curve's equation $y = x^2 + 1$:

$ y_0 = \left(\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4} $

Therefore, the point at which the tangent should be drawn is $\left(\frac{1}{2}, \frac{5}{4}\right)$.