1. Given function and regions

We have

Inside the unit square $S=[0,1]\times[0,1]$:

• green region: $G=\{(x,y): y>f(x)\}$
• red region: $R=\{(x,y): y<f(x)\}$

For a horizontal line $L_h: y=h$, we split both regions into parts above and below the line.

2. Useful area formulas

For a fixed $x$:

• green segment is from $y=f(x)$ to $y=1$
• red segment is from $y=0$ to $y=f(x)$

So:

• Green area above $L_h$ at that $x$ is

Equivalently, after integrating over $x$, it is easier to think geometrically.

• Green area below $L_h$ at that $x$ is

• Red area above $L_h$ at that $x$ is

• Red area below $L_h$ at that $x$ is

But for all options, total areas are more useful.

3. First compute total red and green areas

Total red area:

Now

With denominator $36$,

Hence

So total red area and total green area are equal.

4. Range of $f(x)$ on $[0,1]$

Differentiate:

Solve $f'(x)=0$:

Thus critical points are $x=\frac13,\frac53$, and only $x=\frac13$ lies in $[0,1]$.

Now evaluate:

With denominator $324$,

So

Numerically,

Hence for every $x$,

This fact is crucial because the interval

contains the entire range of $f$.

5. Option A: green above = green below

If $h\ge \max f(x)$, then all green region lies above the curve and below the top edge, so:

• green below line = strip between $y=f(x)$ and $y=h$
• green above line = strip between $y=h$ and $y=1$

Since here any $h\in\left[\frac{181}{324},\frac23\right]$ works for this simplification,

Set them equal:

But

If $h<\max f$, equality may still be checked more generally, but note:

Green below + Green above = total green area = $\frac12$. So equality means each must be $\frac14$.

For $h\le \min f=\frac{13}{36}$, green below area is $0$. For $h\in\left[\frac{13}{36},\frac{181}{324}\right]$, green below area increases continuously from $0$ but remains less than the value at $h=\max f$, which is

So green below area never reaches $\frac14$ inside the interval. Therefore equality is impossible.

Hence A is false.

6. Option B: red above = red below

Total red area is $\frac12$. So equality means each part must be $\frac14$.

Define

This is continuous in $h$.

Now evaluate at two heights in the allowed interval:

• At $h=\frac14$, since $\frac14<\min f(x)$ for all $x$,

Thus red below area is $\frac14$, so red above area is

Hence equality holds already for

So B is true.

7. Option C: green above = red below

Let:

• $G_a(h)$ = green area above $L_h$
• $R_b(h)$ = red area below $L_h$

For every fixed $x$, the vertical segment below $L_h$ is split into:

• red below part of length $\min(f(x),h)$,
• green below part of length $\max(h-f(x),0)$.

Similarly, above $L_h$, we have red above and green above.

But there is a simpler identity:

This is because for each fixed $x$:

• below-line red length is $\min(f(x),h)$,
• above-line green length is $1-\max(f(x),h)$, so their sum is

Integrating over $x\in[0,1]$ gives

For equality $G_a(h)=R_b(h)$, each must be half of $1-h$; equivalently we can use a specific value.

Take $h=\frac12$ (which lies in the interval). Since $\frac12$ is within the range of $f$, let us use total areas:

Also,

And since total red = total green = $\frac12$,

Thus

So

Therefore for every function with equal total red and green areas, option C holds at

Since $\frac12\in\left[\frac14,\frac23\right]$, C is true.

8. Option D: red above = green below

Similarly,

Therefore

So at

we get red above = green below.

Since $\frac12\in\left[\frac14,\frac23\right]$, D is true.

9. Final conclusion

• A: False
• B: True
• C: True
• D: True

So the correct options are

10. Comparison with stored answer

Stored correct answer: $B, C, D$

Our derived answer matches exactly.