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Integral Calculus: JEE Main Mathematics Question with Solution

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

Because the relation holds for all . Replacing by produces another valid identity on the same domain.

Let

5\mathrm{f}(\mathrm{x})+4\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{1}{\mathrm{x}}+3, \mathrm{x}>0

. Then

18\int_{1}^{2}\mathrm{f}(\mathrm{x})\mathrm{dx}

is equal to:

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

10\log_{\mathrm{e}}2-6

10\log_{\mathrm{e}}2+6

5\log_{\mathrm{e}}2+3

5\log_{\mathrm{e}}2-3

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

10\log_{\mathrm{e}}2-6

None. 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 ·

10\log_{\mathrm{e}}2+6

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

C ·

5\log_{\mathrm{e}}2+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 $\int_1^2 f(x)\,dx$ .

D ·

5\log_{\mathrm{e}}2-3

Evaluated $9\int_{1}^{2} f(x)\,dx$ instead of $18\int_{1}^{2} f(x)\,dx$ , forgetting the coefficient 18 from the stem. Check the factor in front of the integral in the question stem: it asks for $18\int_1^2 f(x)\,dx$ , not $9\int_1^2 f(x)\,dx$ .

Reviewed route

Solution

StepWorking

01given

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

02goal

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

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: $5f\left(\frac{1}{x}\right) + 4f(x) = x + 3 \quad \text{--- (2)}$ Given equation is: $5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3 \quad \text{--- (1)}$ Eliminate $f(1/x)$ by computing $5 \times (1) - 4 \times (2)$ : $25f(x) - 16f(x) = 5\left(\frac{1}{x} + 3\right) - 4(x + 3)$ $9f(x) = \frac{5}{x} - 4x + 3 \implies f(x) = \frac{1}{9}\left(\frac{5}{x} - 4x + 3\right)$

05execute

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

✓verify

For $x=1$ , $5f(1)+4f(1)=4 \implies 9f(1)=4 \implies f(1)=4/9$ . 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 $5f(x) + 4f(1/x) = \frac{1}{x} + 3$ and $4f(x) + 5f(1/x) = x + 3$ for $f(x)$ yields:
$9f(x) = \frac{5}{x} - 4x + 3$
Now compute $18\int_{1}^{2} f(x)\,dx = 2\int_{1}^{2} (9f(x))\,dx = 2\int_{1}^{2}\left(\frac{5}{x} - 4x + 3\right)dx$ . What is the result?
$10\log_{\mathrm{e}}2 - 6$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $f(x) = \frac{5}{9x} - \frac{4x}{9} + \frac{1}{3}$ , which gives $18\int_{1}^{2} f(x)\,dx = 10\log_{\mathrm{e}}2 - 6$ .

Why each option works or fails

A · correct: $10\log_{\mathrm{e}}2-6$ - None. 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: $10\log_{\mathrm{e}}2+6$ - Sign error during linear combination when subtracting $4 \times (\text{second equation})$ from $5 \times (\text{first equation})$ , leading to $+6$ instead of $-6$ . Carefully track negative signs when evaluating $5(3) - 4(3) = 3$ and the integral of the linear term $-\frac{4x}{9}$ .
C: $5\log_{\mathrm{e}}2+3$ - 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 $\int_1^2 f(x)\,dx$ .
D: $5\log_{\mathrm{e}}2-3$ - Evaluated $9\int_{1}^{2} f(x)\,dx$ instead of $18\int_{1}^{2} f(x)\,dx$ , forgetting the coefficient 18 from the stem. Check the factor in front of the integral in the question stem: it asks for $18\int_1^2 f(x)\,dx$ , not $9\int_1^2 f(x)\,dx$ .

Step-by-step solution

given: $5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3 \quad \text{for } x > 0$
goal: Find the value of $I = 18 \int_{1}^{2} f(x) \, dx$ .
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: $5f\left(\frac{1}{x}\right) + 4f(x) = x + 3 \quad \text{--- (2)}$ Given equation is: $5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3 \quad \text{--- (1)}$ Eliminate $f(1/x)$ by computing $5 \times (1) - 4 \times (2)$ : $25f(x) - 16f(x) = 5\left(\frac{1}{x} + 3\right) - 4(x + 3)$ $9f(x) = \frac{5}{x} - 4x + 3 \implies f(x) = \frac{1}{9}\left(\frac{5}{x} - 4x + 3\right)$
execute: Now compute $18 \int_{1}^{2} f(x) \, dx$ : $18 \int_{1}^{2} \frac{1}{9}\left(\frac{5}{x} - 4x + 3\right) dx = 2 \int_{1}^{2} \left(\frac{5}{x} - 4x + 3\right) dx$ $= 2 \left[ 5\ln x - 2x^2 + 3x \right]_{1}^{2}$ $= 2 \left( (5\ln 2 - 8 + 6) - (0 - 2 + 3) \right) = 2(5\ln 2 - 2 - 1) = 2(5\ln 2 - 3) = 10\ln 2 - 6$
verify: For $x=1$ , $5f(1)+4f(1)=4 \implies 9f(1)=4 \implies f(1)=4/9$ . Using our formula $f(1)=\frac{1}{9}(5-4+3)=\frac{4}{9}$ , perfectly consistent.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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