JEE 2023 — Mathematics PYQ

To find the area $A$, we first examine the functions defining the boundaries within the interval $0\le x\le 1$.

1. Simplify the Functions

For $x\in [0,1]$:

• $y_1=|2x-1|$

  • $y_1=1-2x$ for 0 \le x < \frac{1}{2}

  • $y_1=2x-1$ for $\frac{1}{2}\le x\le 1$

• $y_2=|x^2-x|=|x(x-1)|$

  • Since $x(x-1)$ is negative or zero on $[0,1]$, $y_2=-(x^2-x)=x-x^2$.

2. Check the Condition $|2x-1|\le y\le x-x^2$

For a valid area to exist, we must have $y_1\le y_2$. Let's check if $x-x^2\ge |2x-1|$ ever occurs in the interval $[0,1]$.

Case 1: 0 \le x < \frac{1}{2}

$1-2x\le x-x^2$

$x^2-3x+1\le 0$

The roots of $x^2-3x+1=0$ are $x=\frac{3\pm \sqrt{5}}{2}$.

$\frac{3-\sqrt{5}}{2}\approx \frac{3-2.236}{2}\approx 0.382$

So, the inequality holds for $x\in [\frac{3-\sqrt{5}}{2},\frac{1}{2})$.

Case 2: $\frac{1}{2}\le x\le 1$

$2x-1\le x-x^2$

$x^2+x-1\le 0$

The roots of $x^2+x-1=0$ are $x=\frac{-1\pm \sqrt{5}}{2}$.

$\frac{-1+\sqrt{5}}{2}\approx \frac{-1+2.236}{2}\approx 0.618$

So, the inequality holds for $x\in [\frac{1}{2},\frac{\sqrt{5}-1}{2}]$.

Thus, the region exists for $x\in [\frac{3-\sqrt{5}}{2},\frac{\sqrt{5}-1}{2}]$.

3. Calculate the Area $A$

The area is given by:

Due to the symmetry of both functions about $x=\frac{1}{2}$:

Integrating:

Let $\alpha =\frac{\sqrt{5}-1}{2}$. Note that ${\alpha }^2+\alpha -1=0$, so ${\alpha }^2=1-\alpha$.

After substituting the limits and simplifying (which involves radical arithmetic):

4. Find the Final Value

We need to find $(6A+11{)}^2$:

Final Answer:

The value is 125.