JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Solving a functional equation by reciprocal substitution and integrating

Q: Why can we replace x with 1/x?

Because the relation holds for all x > 0. Replacing x by 1/x produces another valid identity on the same domain.

Let 5f(x)+4f(1/x)=1/x+3, x>0. Then 18integral from 1 to 2 of f(x)dx is equal to:

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A10log_e2-6B10log_e2+6C5log_e2+3D5log_e2-3

Step-by-step solutionView

Correct answer

By substituting $1 / x$ into the given equation and eliminating $f (1 / x)$ , we find $f (x) = \frac{5}{9 x} - \frac{4 x}{9} + \frac{1}{3}$ , which gives $18 \int_{1}^{2} f (x) d x = 10 \log_{e} 2 - 6$ .

Option analysis

Why each option works or fails

A · 10log_e2-6

This is the correct evaluation. Correctly eliminate $f (1 / x)$ to solve for $f (x)$ and integrate term-by-term from $1$ to $2$ multiplied by $18$ .

B · 10log_e2+6

Sign error during linear combination when subtracting $4 \times (second equation)$ from $5 \times (first equation)$ , leading to $+ 6$ instead of $- 6$ . Carefully track negative signs when evaluating $5 (3) - 4 (3) = 3$ and the integral of the linear term $- \frac{4 x}{9}$ .

C · 5log_e2+3

Student forgot to multiply the integrated result by $2$ after factoring out $9$ (or confused the multiplier $18$ with $9$ ) and made a sign slip. Ensure the overall factor $18$ multiplies the entire integral $\int_{1}^{2} f (x) d x$ .

D · 5log_e2-3

Evaluated $9 \int_{1}^{2} f (x) d x$ instead of $18 \int_{1}^{2} f (x) d x$ , forgetting the coefficient 18 from the stem. Check the factor in front of the integral in the question stem: it asks for $18 \int_{1}^{2} f (x) d x$ , not $9 \int_{1}^{2} f (x) d x$ .

Reviewed route

Solution

StepWorking

01Given

$5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3 for x > 0$

02Goal

Find the value of $I = 18 \int_{1}^{2} f (x) d x$ .

03Approach

Replace $x$ with $1 / x$ to obtain a second linear equation in $f (x)$ and $f (1 / x)$ , then eliminate $f (1 / x)$ to solve for $f (x)$ explicitly before integrating.

04Execute

Replace $x \to \frac{1}{x}$ in the given equation: $5 f (\frac{1}{x}) + 4 f (x) = x + 3 — (2)$ Given equation is: $5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3 — (1)$ Eliminate $f (1 / x)$ by computing $5 \times (1) - 4 \times (2)$ : $25 f (x) - 16 f (x) = 5 (\frac{1}{x} + 3) - 4 (x + 3)$ $9 f (x) = \frac{5}{x} - 4 x + 3 ⟹ f (x) = \frac{1}{9} (\frac{5}{x} - 4 x + 3)$

05Execute

Now compute $18 \int_{1}^{2} f (x) d x$ : $18 \int_{1}^{2} \frac{1}{9} (\frac{5}{x} - 4 x + 3) d x = 2 \int_{1}^{2} (\frac{5}{x} - 4 x + 3) d x$ $= 2 {[5 \ln x - 2 x^{2} + 3 x]}_{1}^{2}$ $= 2 ((5 \ln 2 - 8 + 6) - (0 - 2 + 3)) = 2 (5 \ln 2 - 2 - 1) = 2 (5 \ln 2 - 3) = 10 \ln 2 - 6$

✓Verify

For $x = 1$ , $5 f (1) + 4 f (1) = 4 ⟹ 9 f (1) = 4 ⟹ f (1) = 4 / 9$ . Using our formula $f (1) = \frac{1}{9} (5 - 4 + 3) = \frac{4}{9}$ , perfectly consistent.

Hints that build this answer step by step

What substitution allows us to form a second equation and solve for $f (x)$ ?
Replace $x$ with $\frac{1}{x}$
Solving the system $5 f (x) + 4 f (1 / x) = \frac{1}{x} + 3$ and $4 f (x) + 5 f (1 / x) = x + 3$ for $f (x)$ yields:
$9 f (x) = \frac{5}{x} - 4 x + 3$
Now compute $18 \int_{1}^{2} f (x) d x = 2 \int_{1}^{2} (9 f (x)) d x = 2 \int_{1}^{2} (\frac{5}{x} - 4 x + 3) d x$ . What is the result?
$10 \log_{e} 2 - 6$

✓ 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: 8 September 2026

Quick checks

Students also ask

Why can we replace x with 1/x?

Because the relation holds for all $x > 0$ . Replacing $x$ by $1 / x$ produces another valid identity on the same domain.

Answer

By substituting $1 / x$ into the given equation and eliminating $f (1 / x)$ , we find $f (x) = \frac{5}{9 x} - \frac{4 x}{9} + \frac{1}{3}$ , which gives $18 \int_{1}^{2} f (x) d x = 10 \log_{e} 2 - 6$ .

Why each option works or fails

A · correct: 10log_e2-6 - This is the correct evaluation. Correctly eliminate $f (1 / x)$ to solve for $f (x)$ and integrate term-by-term from $1$ to $2$ multiplied by $18$ .
B: 10log_e2+6 - Sign error during linear combination when subtracting $4 \times (second equation)$ from $5 \times (first equation)$ , leading to $+ 6$ instead of $- 6$ . Carefully track negative signs when evaluating $5 (3) - 4 (3) = 3$ and the integral of the linear term $- \frac{4 x}{9}$ .
C: 5log_e2+3 - Student forgot to multiply the integrated result by $2$ after factoring out $9$ (or confused the multiplier $18$ with $9$ ) and made a sign slip. Ensure the overall factor $18$ multiplies the entire integral $\int_{1}^{2} f (x) d x$ .
D: 5log_e2-3 - Evaluated $9 \int_{1}^{2} f (x) d x$ instead of $18 \int_{1}^{2} f (x) d x$ , forgetting the coefficient 18 from the stem. Check the factor in front of the integral in the question stem: it asks for $18 \int_{1}^{2} f (x) d x$ , not $9 \int_{1}^{2} f (x) d x$ .

Step-by-step solution

Given: $5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3 for x > 0$
Goal: Find the value of $I = 18 \int_{1}^{2} f (x) d x$ .
Approach: Replace $x$ with $1 / x$ to obtain a second linear equation in $f (x)$ and $f (1 / x)$ , then eliminate $f (1 / x)$ to solve for $f (x)$ explicitly before integrating.
Execute: Replace $x \to \frac{1}{x}$ in the given equation: $5 f (\frac{1}{x}) + 4 f (x) = x + 3 — (2)$ Given equation is: $5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3 — (1)$ Eliminate $f (1 / x)$ by computing $5 \times (1) - 4 \times (2)$ : $25 f (x) - 16 f (x) = 5 (\frac{1}{x} + 3) - 4 (x + 3)$ $9 f (x) = \frac{5}{x} - 4 x + 3 ⟹ f (x) = \frac{1}{9} (\frac{5}{x} - 4 x + 3)$
Execute: Now compute $18 \int_{1}^{2} f (x) d x$ : $18 \int_{1}^{2} \frac{1}{9} (\frac{5}{x} - 4 x + 3) d x = 2 \int_{1}^{2} (\frac{5}{x} - 4 x + 3) d x$ $= 2 {[5 \ln x - 2 x^{2} + 3 x]}_{1}^{2}$ $= 2 ((5 \ln 2 - 8 + 6) - (0 - 2 + 3)) = 2 (5 \ln 2 - 2 - 1) = 2 (5 \ln 2 - 3) = 10 \ln 2 - 6$
Verify: For $x = 1$ , $5 f (1) + 4 f (1) = 4 ⟹ 9 f (1) = 4 ⟹ f (1) = 4 / 9$ . Using our formula $f (1) = \frac{1}{9} (5 - 4 + 3) = \frac{4}{9}$ , perfectly consistent.

Solving a functional equation by reciprocal substitution and integrating

What feels right?

By substituting $1 / x$ into the given equation and eliminating $f (1 / x)$ , we find $f (x) = \frac{5}{9 x} - \frac{4 x}{9} + \frac{1}{3}$ , which gives $18 \int_{1}^{2} f (x) d x = 10 \log_{e} 2 - 6$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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