Why substitute x → π/2 - x specifically?

Because the functional equation involves f(x) and f(π/2 - x). Applying the transformation x → π/2 - x creates a system of two linear equations in two unknowns, f(x) and f(π/2 - x).

JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Functional Equation Resolution and Area Between Curves

Let f: ℝ → ℝ be a function such that f(x) + 3f(π/2 - x) = sin x, x ∈ ℝ. Let maximum value of f on ℝ be α. If the area of the region bounded by the curves g(x) = x^2 and h(x) = β x^3, β > 0, is α^2, then 30β^3 is equal to......

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Correct answer

The maximum value of the function is $α = \frac{1}{8}$ , and equating the enclosed area $\frac{1}{12 β^{2}}$ to $α^{2} = \frac{1}{64}$ gives $30 β^{3} = 40 \sqrt{3} \approx 69.28$ or solving for $β$ : $30 β^{3} = 40 \sqrt{3}$ or let's recheck the exact math.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Given functional relation: $f (x) + 3 f (\frac{π}{2} - x) = \sin x$ , and curves $g (x) = x^{2}$ , $h (x) = β x^{3}$ (where $β > 0$ ) bounding an area equal to $α^{2}$ , where $α = \max f (x)$ .

02Goal

Find the value of $30 β^{3}$ .

03Approach

First, substitute $x \to \frac{π}{2} - x$ to form a system of two linear equations and solve for $f (x)$ . Next, determine $α = \max f (x)$ using the maximum of $A \cos x + B \sin x$ . Finally, find the intersection points of $g (x)$ and $h (x)$ , compute the bounded area in terms of $β$ , equate it to $α^{2}$ , and solve for $30 β^{3}$ .

04Execute

Replacing $x$ with $\frac{π}{2} - x$ in $f (x) + 3 f (\frac{π}{2} - x) = \sin x$ : $f (\frac{π}{2} - x) + 3 f (x) = \cos x$ . Multiplying this equation by $3$ gives: $3 f (\frac{π}{2} - x) + 9 f (x) = 3 \cos x$ . Subtracting the original equation: $8 f (x) = 3 \cos x - \sin x ⟹ f (x) = \frac{3 \cos x - \sin x}{8}$ .

05Execute

The maximum value of $a \cos x + b \sin x$ is $\sqrt{a^{2} + b^{2}}$ . Thus: $α = \max f (x) = \frac{\sqrt{3^{2} + (- 1)^{2}}}{8} = \frac{\sqrt{10}}{8}$ . Then $α^{2} = \frac{10}{64} = \frac{5}{32}$ .

06Execute

Find the region bounded by $y = x^{2}$ and $y = β x^{3}$ for $β > 0$ : Intersection points: $x^{2} = β x^{3} ⟹ x^{2} (1 - β x) = 0 ⟹ x = 0, x = \frac{1}{β}$ . The bounded area is: $Area = \int_{0}^{1 / β} (x^{2} - β x^{3}) d x = {[\frac{x^{3}}{3} - \frac{β x^{4}}{4}]}_{0}^{1 / β} = \frac{1}{3 β^{3}} - \frac{1}{4 β^{3}} = \frac{1}{12 β^{3}}$ .

07Execute

Equating the area to $α^{2}$ : $\frac{1}{12 β^{3}} = \frac{5}{32} ⟹ β^{3} = \frac{32}{12 \times 5} = \frac{8}{15}$ . Therefore, $30 β^{3} = 30 \times \frac{8}{15} = 16$ .

✓Verify

Check: $1 / (12 β^{3}) = 1 / (12 \times (8 / 15)) = 15 / 96 = 5 / 32 = α^{2}$ . Everything is consistent.

Hints that build this answer step by step

Given $f (x) + 3 f (\frac{π}{2} - x) = \sin x$ , what is the appropriate substitution to determine $f (x)$ explicitly?
Replace $x$ with $\frac{π}{2} - x$ to form a second equation in $f (x)$ and $f (\frac{π}{2} - x)$ .
Eliminating $f (\frac{π}{2} - x)$ from the system yields what expression and maximum value $α$ for $f (x)$ ?
$f (x) = \frac{3 \cos x - \sin x}{8}$ , with maximum $α = \frac{\sqrt{10}}{8}$
What is the area of the region bounded by $g (x) = x^{2}$ and $h (x) = β x^{3}$ for $β > 0$ in terms of $β$ ?
$\int_{0}^{1 / β} (x^{2} - β x^{3}) d x = \frac{1}{12 β^{3}}$
Equating the area $\frac{1}{12 β^{3}}$ to $α^{2} = {(\frac{\sqrt{10}}{8})}^{2} = \frac{10}{64} = \frac{5}{32}$ , what is the value of $30 β^{3}$ ?
$16$

✓ 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 substitute

x \to π / 2 - x

specifically?

Because the functional equation involves $f (x)$ and $f (π / 2 - x)$ . Applying the transformation $x \to π / 2 - x$ creates a system of two linear equations in two unknowns, $f (x)$ and $f (π / 2 - x)$ .

Answer

Accepted answer: 16

The maximum value of the function is $α = \frac{1}{8}$ , and equating the enclosed area $\frac{1}{12 β^{2}}$ to $α^{2} = \frac{1}{64}$ gives $30 β^{3} = 40 \sqrt{3} \approx 69.28$ or solving for $β$ : $30 β^{3} = 40 \sqrt{3}$ or let's recheck the exact math.

Why each option works or fails

Step-by-step solution

Given: Given functional relation: $f (x) + 3 f (\frac{π}{2} - x) = \sin x$ , and curves $g (x) = x^{2}$ , $h (x) = β x^{3}$ (where $β > 0$ ) bounding an area equal to $α^{2}$ , where $α = \max f (x)$ .
Goal: Find the value of $30 β^{3}$ .
Approach: First, substitute $x \to \frac{π}{2} - x$ to form a system of two linear equations and solve for $f (x)$ . Next, determine $α = \frac{1}{8}$ using the maximum of $A \cos x + B \sin x$ . Finally, find the intersection points of $g (x)$ and $h (x)$ , compute the bounded area in terms of $β$ , equate it to $α^{2}$ , and solve for $30 β^{3}$ .
Execute: Replacing $x$ with $\frac{π}{2} - x$ in $f (x) + 3 f (\frac{π}{2} - x) = \sin x$ : $f (\frac{π}{2} - x) + 3 f (x) = \cos x$ . Multiplying this equation by $3$ gives: $3 f (\frac{π}{2} - x) + 9 f (x) = 3 \cos x$ . Subtracting the original equation: $8 f (x) = 3 \cos x - \sin x ⟹ f (x) = \frac{3 \cos x - \sin x}{8}$ .
Execute: The maximum value of $a \cos x + b \sin x$ is $\sqrt{a^{2} + b^{2}}$ . Thus: $α = \max f (x) = \frac{\sqrt{3^{2} + (- 1)^{2}}}{8} = \frac{\sqrt{10}}{8}$ . Then $α^{2} = \frac{10}{64} = \frac{5}{32}$ .
Execute: Find the region bounded by $y = x^{2}$ and $y = β x^{3}$ for $β > 0$ : Intersection points: $x^{2} = β x^{3} ⟹ x^{2} (1 - β x) = 0 ⟹ x = 0, x = \frac{1}{β}$ . The bounded area is: $Area = \int_{0}^{1 / β} (x^{2} - β x^{3}) d x = {[\frac{x^{3}}{3} - \frac{β x^{4}}{4}]}_{0}^{1 / β} = \frac{1}{3 β^{3}} - \frac{1}{4 β^{3}} = \frac{1}{12 β^{3}}$ .
Execute: Equating the area to $α^{2}$ : $\frac{1}{12 β^{3}} = \frac{5}{32} ⟹ β^{3} = \frac{32}{12 \times 5} = \frac{8}{15}$ . Therefore, $30 β^{3} = 30 \times \frac{8}{15} = 16$ .
Verify: Check: $1 / (12 β^{3}) = 1 / (12 \times (8 / 15)) = 15 / 96 = 5 / 32 = α^{2}$ . Everything is consistent.

Functional Equation Resolution and Area Between Curves

What feels right?

The maximum value of the function is $α = \frac{1}{8}$ , and equating the enclosed area $\frac{1}{12 β^{2}}$ to $α^{2} = \frac{1}{64}$ gives $30 β^{3} = 40 \sqrt{3} \approx 69.28$ or solving for $β$ : $30 β^{3} = 40 \sqrt{3}$ or let's recheck the exact math.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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