JEE MainMathematics

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Area Between Curves Defined by Absolute Value Functions

Q: Why replace x^2 and x^3 with linear terms in x?

Since x_1 satisfies x^2 - 3x + 1 = 0, reducing higher powers of x_1 to linear polynomials prevents tedious cubic binomial expansions and avoids arithmetic mistakes.

Let the area of the region {(x, y):|2 x-1| ≤ y ≤|x^2-x|, 0 ≤ x ≤ 1} be A. Then (6 A+11)^2 is equal to

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Correct answer

The value of $(6 A + 11)^{2}$ is 125.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

The region is defined by $∣ 2 x - 1 ∣ \leq y \leq ∣ x^{2} - x ∣$ for $0 \leq x \leq 1$ .

02Goal

Find the area $A$ of the region and evaluate $(6 A + 11)^{2}$ .

03Approach

For $x \in [0, 1]$ , $x^{2} - x \leq 0$ , so $∣ x^{2} - x ∣ = x - x^{2}$ . Both functions $y = ∣ 2 x - 1 ∣$ and $y = x - x^{2}$ are symmetric about the line $x = 1 / 2$ . Thus, we can find the intersection points for $0 \leq x \leq 1 / 2$ where $∣ 2 x - 1 ∣ = 1 - 2 x$ , set up the integral over this half, double it by symmetry, and evaluate the antiderivative using polynomial reduction.

04Execute

Solve for the intersection point in $0 \leq x \leq 1 / 2$ : $x - x^{2} = 1 - 2 x ⟹ x^{2} - 3 x + 1 = 0$ $x = \frac{3 \pm \sqrt{5}}{2}$ Since $x \in [0, 1 / 2]$ , we choose $x_{1} = \frac{3 - \sqrt{5}}{2}$ . The upper curve is $y = x - x^{2}$ and the lower curve is $y = 1 - 2 x$ on $[x_{1}, 1 / 2]$ .

05Execute

By symmetry about $x = 1 / 2$ , the total area is: $A = 2 \int_{x_{1}}^{1 / 2} [(x - x^{2}) - (1 - 2 x)] d x = 2 \int_{x_{1}}^{1 / 2} (- x^{2} + 3 x - 1) d x$ Let $F (x) = \frac{3 x^{2}}{2} - \frac{x^{3}}{3} - x$ . Then $A = 2 [F (1 / 2) - F (x_{1})]$ . At $x = 1 / 2$ : $F (1 / 2) = \frac{3}{8} - \frac{1}{24} - \frac{1}{2} = \frac{9 - 1 - 12}{24} = - \frac{4}{24} = - \frac{1}{6}$ Using $x_{1}^{2} - 3 x_{1} + 1 = 0 ⟹ x_{1}^{2} = 3 x_{1} - 1$ and $x_{1}^{3} = 8 x_{1} - 3$ : $F (x_{1}) = \frac{3}{2} (3 x_{1} - 1) - \frac{1}{3} (8 x_{1} - 3) - x_{1} = \frac{5 x_{1} - 3}{6}$ Substitute $x_{1} = \frac{3 - \sqrt{5}}{2}$ : $F (x_{1}) = \frac{5 (\frac{3 - \sqrt{5}}{2}) - 3}{6} = \frac{9 - 5 \sqrt{5}}{12}$ Thus, $A = 2 [- \frac{1}{6} - \frac{9 - 5 \sqrt{5}}{12}] = 2 [\frac{- 2 - 9 + 5 \sqrt{5}}{12}] = \frac{5 \sqrt{5} - 11}{6}$

06Execute

Calculate $(6 A + 11)^{2}$ : $6 A + 11 = 5 \sqrt{5}$ $(6 A + 11)^{2} = (5 \sqrt{5})^{2} = 125$

✓Verify

Verify that $x - x^{2} \geq 1 - 2 x$ on $[x_{1}, 1 / 2]$ : at $x = 0.4$ , $x - x^{2} = 0.24$ and $1 - 2 x = 0.20$ , so the parabola is indeed above the line. Also, $(1 / 2 - x_{1}) = \frac{\sqrt{5} - 2}{2} \approx 0.118$ , height at midpoint $x = 0.44$ is $\approx 0.2464 - 0.12 = 0.1264$ , area of each wing $\approx \frac{2}{3} \times 0.118 \times 0.19 \approx 0.015$ , total area $2 \times 0.015 = 0.030$ , matching $\frac{5 \sqrt{5} - 11}{6} \approx 0.03005$ .

Hints that build this answer step by step

For $x \in [0, 1]$ , what are the simplified forms of the upper and lower bounding curves without absolute value signs?
The upper curve is $y = x - x^{2}$ , while the lower curve is $y = 1 - 2 x$ on $[0, 1 / 2]$ and $y = 2 x - 1$ on $[1 / 2, 1]$ .
What is the lower intersection point $x_{1} \in [0, 1 / 2]$ where the lower boundary meets the upper boundary?
$x_{1} = \frac{3 - \sqrt{5}}{2}$
Using the symmetry of the region about $x = 1 / 2$ , what is the total area $A$ ?
$A = \frac{5 \sqrt{5} - 11}{6}$
Given $A = \frac{5 \sqrt{5} - 11}{6}$ , what is the value of $(6 A + 11)^{2}$ ?
125

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why replace

x^{2}

and

x^{3}

with linear terms in

x

Since $x_{1}$ satisfies $x^{2} - 3 x + 1 = 0$ , reducing higher powers of $x_{1}$ to linear polynomials prevents tedious cubic binomial expansions and avoids arithmetic mistakes.

Answer

Accepted answer: 125

The value of $(6 A + 11)^{2}$ is 125.

Why each option works or fails

Step-by-step solution

Given: The region is defined by $∣ 2 x - 1 ∣ \leq y \leq ∣ x^{2} - x ∣$ for $0 \leq x \leq 1$ .
Goal: Find the area $A$ of the region and evaluate $(6 A + 11)^{2}$ .
Approach: For $x \in [0, 1]$ , $x^{2} - x \leq 0$ , so $∣ x^{2} - x ∣ = x - x^{2}$ . Both functions $y = ∣ 2 x - 1 ∣$ and $y = x - x^{2}$ are symmetric about the line $x = 1 / 2$ . Thus, we can find the intersection points for $0 \leq x \leq 1 / 2$ where $∣ 2 x - 1 ∣ = 1 - 2 x$ , set up the integral over this half, double it by symmetry, and evaluate the antiderivative using polynomial reduction.
Execute: Solve for the intersection point in $0 \leq x \leq 1 / 2$ : $x - x^{2} = 1 - 2 x ⟹ x^{2} - 3 x + 1 = 0$ $x = \frac{3 \pm \sqrt{5}}{2}$ Since $x \in [0, 1 / 2]$ , we choose $x_{1} = \frac{3 - \sqrt{5}}{2}$ . The upper curve is $y = x - x^{2}$ and the lower curve is $y = 1 - 2 x$ on $[x_{1}, 1 / 2]$ .
Execute: By symmetry about $x = 1 / 2$ , the total area is: $A = 2 \int_{x_{1}}^{1 / 2} [(x - x^{2}) - (1 - 2 x)] d x = 2 \int_{x_{1}}^{1 / 2} (- x^{2} + 3 x - 1) d x$ Let $F (x) = \frac{3 x^{2}}{2} - \frac{x^{3}}{3} - x$ . Then $A = 2 [F (1 / 2) - F (x_{1})]$ . At $x = 1 / 2$ : $F (1 / 2) = \frac{3}{8} - \frac{1}{24} - \frac{1}{2} = \frac{9 - 1 - 12}{24} = - \frac{4}{24} = - \frac{1}{6}$ Using $x_{1}^{2} - 3 x_{1} + 1 = 0 ⟹ x_{1}^{2} = 3 x_{1} - 1$ and $x_{1}^{3} = 8 x_{1} - 3$ : $F (x_{1}) = \frac{3}{2} (3 x_{1} - 1) - \frac{1}{3} (8 x_{1} - 3) - x_{1} = \frac{5 x_{1} - 3}{6}$ Substitute $x_{1} = \frac{3 - \sqrt{5}}{2}$ : $F (x_{1}) = \frac{5 (\frac{3 - \sqrt{5}}{2}) - 3}{6} = \frac{9 - 5 \sqrt{5}}{12}$ Thus, $A = 2 [- \frac{1}{6} - \frac{9 - 5 \sqrt{5}}{12}] = 2 [\frac{- 2 - 9 + 5 \sqrt{5}}{12}] = \frac{5 \sqrt{5} - 11}{6}$
Execute: Calculate $(6 A + 11)^{2}$ : $6 A + 11 = 5 \sqrt{5}$ $(6 A + 11)^{2} = (5 \sqrt{5})^{2} = 125$
Verify: Verify that $x - x^{2} \geq 1 - 2 x$ on $[x_{1}, 1 / 2]$ : at $x = 0.4$ , $x - x^{2} = 0.24$ and $1 - 2 x = 0.20$ , so the parabola is indeed above the line. Also, $(1 / 2 - x_{1}) = \frac{\sqrt{5} - 2}{2} \approx 0.118$ , height at midpoint $x = 0.44$ is $\approx 0.2464 - 0.12 = 0.1264$ , area of each wing $\approx \frac{2}{3} \times 0.118 \times 0.19 \approx 0.015$ , total area $2 \times 0.015 = 0.030$ , matching $\frac{5 \sqrt{5} - 11}{6} \approx 0.03005$ .

Area Between Curves Defined by Absolute Value Functions

What feels right?

The value of $(6 A + 11)^{2}$ is 125.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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