Question

Let f :[0,1] → [0,1] be the function defined by \(\frac{x^3}{3}-x^2+\frac{5}{9}x+\frac{17}{36}.\) Consider the square region S = [0,1] x [0,1]. Let G = {(x,y) ∈ S: y > f(s)} be called the green region and R = {(x,y) ∈ S: y < f(s)} be called the red region. Let L_h = {(x,h) ∈ S: x ∈ [0,1] be the horizontal line drawn at a height h ∈ [0,1]. Then which of the following statements is(are) true?

• A. There exists an h ∈ [\(\frac{1}{4},\frac{2}{3}\)] such that the area of the green region above the line L_h equals the area of the green region below the line L_h
• B. There exists an h ∈ [\(\frac{1}{4},\frac{2}{3}\)] such that the area of the red region above the line L_h equals the area of the red region below the line L_h

• C. There exists an h ∈ [\(\frac{1}{4},\frac{2}{3}\)] such that the area of the green region above the line L_h equals the area of the red region below the line L_h
• D. There exists an h ∈ [\(\frac{1}{4},\frac{2}{3}\)] such that the area of the red region above the line L_h equals the area of the green region below the line L_h

Answer 1

The Correct Option is B, C, D

\(f(x)=\frac{x^3}{3}-x^2+\frac{5}{9}x+\frac{17}{36},f'(x)=x^2-2x+\frac{5}{9}\)
For maxima or minima, \(f'(x)=0\Rightarrow x=\frac{1}{3}\)
\(A_R = \int_{0}^{1}f(x)dx=\frac{1}{2}\Rightarrow A_G=\frac{1}{2}\)
Now, checking each options :
(A) 1 - h = h - \(\frac{1}{2}\)
⇒ \(h=\frac{3}{4},\frac{3}{4}\gt\frac{2}{3}\)    So, the option (A) is incorrect
(B) h = \(\frac{1}{2}-h\)
⇒ h =  \(\frac{1}{4}\)      So, the option (B) is correct
(C) \(\int^1_0f(x)dx=\frac{1}{2},\int_0^1\frac{1}{2}dx=\frac{1}{2}\)
⇒ \(\int_0^1(f(x)-\frac{1}{2})dx=0\)
⇒ \(h=\frac{1}{2}\)    So, the option (B) is correct.
(D) As the option (C) is correct, the option (D) is also correct.
So, the correct options are (B), (C) and (D).