Question

A figure is bounded by the curves \(y=x^{2}+1\), \(y=0\), \(x=0\) and \(x=1\). The point at which a tangent should be drawn to the curve \(y=x^{2}+1\) for it to cut off a trapezium of the greatest area from the figure is:

• A. $(1,2)$
• B. $(-1,2)$
• C. $\left(\frac{1}{2},\frac{5}{4}\right)$
• D. $(0,1)$

Hint:

The area function $A(t) = 1 + t - t^2$ forms an downward-opening parabola. Its absolute vertex turning point occurs exactly at the midpoint of the horizontal range interval, which lets you identify $t = 1/2$ by inspection!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem asks to maximize the area of a trapezium formed by a tangent line under the curve \( y = x^2 + 1 \) between \( x=0 \) and \( x=1 \). The tangent at a point \( (h, h^2+1) \) will intersect the vertical lines \( x=0 \) and \( x=1 \) to form the parallel sides of the trapezium.
Step 2: Key Formula or Approach:
1. Tangent at \( (h, h^2+1) \): \( y - (h^2+1) = 2h(x - h) \).
2. Height of trapezium \( = 1 \).
3. Parallel sides are the y-values at \( x=0 \) and \( x=1 \).
4. Area \( = \frac{1}{2} (y_0 + y_1) \times 1 \).
Step 3: Detailed Explanation:
Let the point of tangency be \( P(h, h^2+1) \), where \( 0 \le h \le 1 \).
Derivative \( y' = 2x \), so slope \( m = 2h \).
Equation of tangent: \( y = 2h(x - h) + h^2 + 1 = 2hx - h^2 + 1 \).
For the trapezium between \( x=0 \) and \( x=1 \):
At \( x=0, y_0 = 1 - h^2 \).
At \( x=1, y_1 = 2h - h^2 + 1 \).
Area \( A(h) = \frac{1}{2} (y_0 + y_1) \cdot (1 - 0) \)
\[ A(h) = \frac{1}{2} [(1 - h^2) + (2h - h^2 + 1)] = \frac{1}{2} [2 + 2h - 2h^2] = 1 + h - h^2 \]
To maximize \( A(h) \), find \( A'(h) \):
\[ A'(h) = 1 - 2h = 0 \implies h = \frac{1}{2} \]
Verify with second derivative: \( A''(h) = -2<0 \), so it's a maximum.
At \( h = 1/2, y = (1/2)^2 + 1 = 5/4 \).
The point is \( (1/2, 5/4) \).
Step 4: Final Answer:
The point is \( (1/2, 5/4) \), which matches option (C).