JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Functional Equations and Definite Integration

Q: Why substitute x → 1/x?

Because the equation involves f(x) and f(1/x). Replacing x with 1/x produces a second linear equation in terms of the two unknowns f(x) and f(1/x).

Let f(x) + 2f(1/x) = x^2 + 5 and 2g(x) - 3g(1/2) = x, x > 0. If α = integral from 1 to 2 of f(x) dx, and β = integral from 1 to 2 of g(x) dx, then the value of 9α + β is:

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A1B0C10D11

Step-by-step solutionView

Correct answer

By solving the functional equations for $f (x)$ and $g (x)$ , we find $f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} - \frac{5}{3}$ and $g (x) = \frac{x}{2} - \frac{3}{2}$ , yielding $9 α + β = 11$ .

Option analysis

Why each option works or fails

A · 1

Computing $9 α$ or evaluating the integration limits with arithmetic errors, leading to an incorrect net sum. Carefully integrate each term $\int_{1}^{2} \frac{1}{x^{2}} d x = \frac{1}{2}$ and $\int_{1}^{2} x^{2} d x = \frac{7}{3}$ before multiplying by 9.

B · 0

Confusing the values of the integrals or dropping constant terms during the substitution for $g (1 / 2)$ . Substitute $x = 1 / 2$ into $2 g (x) - 3 g (1 / 2) = x$ to get $- g (1 / 2) = 1 / 2 ⟹ g (1 / 2) = - 1 / 2$ , then integrate $g (x) = \frac{x - 3 / 2}{2}$ .

C · 10

Making a sign or factor slip when eliminating $f (1 / x)$ from the system of functional equations. Multiply $f (1 / x) + 2 f (x) = 1 / x^{2} + 5$ by 2 and subtract the original equation: $3 f (x) = \frac{2}{x^{2}} - x^{2} + 5$ is incorrect, the constant is $2 (5) - 5 = 5$ .

D · 11

The value 11 is obtained correctly through systematic solving of both functional equations and definite integration.

Reviewed route

Solution

StepWorking

01Given

$f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ for $x > 0$ , with $α = \int_{1}^{2} f (x) d x$ and $β = \int_{1}^{2} g (x) d x$ .

02Approach

Solve the system of functional equations for $f (x)$ by replacing $x \to \frac{1}{x}$ . For $g (x)$ , substitute $x = \frac{1}{2}$ to find the constant value $g (1 / 2)$ , then find $g (x)$ . Finally, integrate both functions from 1 to 2 to compute $9 α + β$ .

03Execute

Given (1): $f (x) + 2 f (1 / x) = x^{2} + 5$ . Replacing $x$ with $1 / x$ gives (2): $f (1 / x) + 2 f (x) = \frac{1}{x^{2}} + 5$ . Multiplying (2) by 2 and subtracting (1) yields: $3 f (x) = \frac{2}{x^{2}} - x^{2} + 5 ⟹ f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}$

04Execute

Integrate $f (x)$ from 1 to 2: $α = \int_{1}^{2} (\frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}) d x = {[- \frac{2}{3 x} - \frac{x^{3}}{9} + \frac{5 x}{3}]}_{1}^{2}$ $α = (- \frac{1}{3} - \frac{8}{9} + \frac{10}{3}) - (- \frac{2}{3} - \frac{1}{9} + \frac{5}{3}) = \frac{19}{9} - \frac{8}{9} = \frac{11}{9}$

05Execute

For $g (x)$ : $2 g (x) - 3 g (1 / 2) = x$ . Substitute $x = 1 / 2$ : $2 g (1 / 2) - 3 g (1 / 2) = 1 / 2 ⟹ - g (1 / 2) = 1 / 2 ⟹ g (1 / 2) = - 1 / 2$ Then, $2 g (x) - 3 (- 1 / 2) = x ⟹ 2 g (x) = x - 3 / 2 ⟹ g (x) = \frac{x}{2} - \frac{3}{4}$ .

06Execute

Integrate $g (x)$ from 1 to 2: $β = \int_{1}^{2} (\frac{x}{2} - \frac{3}{4}) d x = {[\frac{x^{2}}{4} - \frac{3 x}{4}]}_{1}^{2} = (1 - \frac{3}{2}) - (\frac{1}{4} - \frac{3}{4}) = - \frac{1}{2} - (- \frac{1}{2}) = 0$ Thus, $9 α + β = 9 (\frac{11}{9}) + 0 = 11$ .

✓Verify

Check at $x = 1$ : $f (1) + 2 f (1) = 6 ⟹ 3 f (1) = 6 ⟹ f (1) = 2$ . From our formula, $f (1) = 2 / 3 - 1 / 3 + 5 / 3 = 6 / 3 = 2$ , which matches.

Hints that build this answer step by step

Given $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ , what is the explicit expression for $f (x)$ ?
$f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}$
What is the value of $α = \int_{1}^{2} f (x) d x$ ?
$α = \frac{13}{9}$
Given $2 g (x) - 3 g (\frac{1}{2}) = x$ , what is $g (x)$ and its integral $β = \int_{1}^{2} g (x) d x$ ?
$g (x) = \frac{2 x - 3}{4}$ , so $β = - 2$
What is the final value of $9 α + β$ ?
$9 (\frac{13}{9}) + (- 2) = 11$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why substitute

x \to 1 / x

Because the equation involves $f (x)$ and $f (1 / x)$ . Replacing $x$ with $1 / x$ produces a second linear equation in terms of the two unknowns $f (x)$ and $f (1 / x)$ .

Answer

By solving the functional equations for $f (x)$ and $g (x)$ , we find $f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} - \frac{5}{3}$ and $g (x) = \frac{x}{2} - \frac{3}{2}$ , yielding $9 α + β = 11$ .

Why each option works or fails

A: 1 - Computing $9 α$ or evaluating the integration limits with arithmetic errors, leading to an incorrect net sum. Carefully integrate each term $\int_{1}^{2} \frac{1}{x^{2}} d x = \frac{1}{2}$ and $\int_{1}^{2} x^{2} d x = \frac{7}{3}$ before multiplying by 9.
B: 0 - Confusing the values of the integrals or dropping constant terms during the substitution for $g (1 / 2)$ . Substitute $x = 1 / 2$ into $2 g (x) - 3 g (1 / 2) = x$ to get $- g (1 / 2) = 1 / 2 ⟹ g (1 / 2) = - 1 / 2$ , then integrate $g (x) = \frac{x - 3 / 2}{2}$ .
C: 10 - Making a sign or factor slip when eliminating $f (1 / x)$ from the system of functional equations. Multiply $f (1 / x) + 2 f (x) = 1 / x^{2} + 5$ by 2 and subtract the original equation: $3 f (x) = \frac{2}{x^{2}} - x^{2} + 5$ is incorrect, the constant is $2 (5) - 5 = 5$ .
D · correct: 11 - The value 11 is obtained correctly through systematic solving of both functional equations and definite integration.

Step-by-step solution

Given: $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ for $x > 0$ , with $α = \int_{1}^{2} f (x) d x$ and $β = \int_{1}^{2} g (x) d x$ .
Approach: Solve the system of functional equations for $f (x)$ by replacing $x \to \frac{1}{x}$ . For $g (x)$ , substitute $x = \frac{1}{2}$ to find the constant value $g (1 / 2)$ , then find $g (x)$ . Finally, integrate both functions from 1 to 2 to compute $9 α + β$ .
Execute: Given (1): $f (x) + 2 f (1 / x) = x^{2} + 5$ . Replacing $x$ with $1 / x$ gives (2): $f (1 / x) + 2 f (x) = \frac{1}{x^{2}} + 5$ . Multiplying (2) by 2 and subtracting (1) yields: $3 f (x) = \frac{2}{x^{2}} - x^{2} + 5 ⟹ f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}$
Execute: Integrate $f (x)$ from 1 to 2: $α = \int_{1}^{2} (\frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}) d x = {[- \frac{2}{3 x} - \frac{x^{3}}{9} + \frac{5 x}{3}]}_{1}^{2}$ $α = (- \frac{1}{3} - \frac{8}{9} + \frac{10}{3}) - (- \frac{2}{3} - \frac{1}{9} + \frac{5}{3}) = \frac{19}{9} - \frac{8}{9} = \frac{11}{9}$
Execute: For $g (x)$ : $2 g (x) - 3 g (1 / 2) = x$ . Substitute $x = 1 / 2$ : $2 g (1 / 2) - 3 g (1 / 2) = 1 / 2 ⟹ - g (1 / 2) = 1 / 2 ⟹ g (1 / 2) = - 1 / 2$ Then, $2 g (x) - 3 (- 1 / 2) = x ⟹ 2 g (x) = x - 3 / 2 ⟹ g (x) = \frac{x}{2} - \frac{3}{4}$ .
Execute: Integrate $g (x)$ from 1 to 2: $β = \int_{1}^{2} (\frac{x}{2} - \frac{3}{4}) d x = {[\frac{x^{2}}{4} - \frac{3 x}{4}]}_{1}^{2} = (1 - \frac{3}{2}) - (\frac{1}{4} - \frac{3}{4}) = - \frac{1}{2} - (- \frac{1}{2}) = 0$ Thus, $9 α + β = 9 (\frac{11}{9}) + 0 = 11$ .
Verify: Check at $x = 1$ : $f (1) + 2 f (1) = 6 ⟹ 3 f (1) = 6 ⟹ f (1) = 2$ . From our formula, $f (1) = 2 / 3 - 1 / 3 + 5 / 3 = 6 / 3 = 2$ , which matches.

Functional Equations and Definite Integration

What feels right?

By solving the functional equations for $f (x)$ and $g (x)$ , we find $f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} - \frac{5}{3}$ and $g (x) = \frac{x}{2} - \frac{3}{2}$ , yielding $9 α + β = 11$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

06Execute

✓Verify

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