Because the equation involves and . Replacing with produces a second linear equation in terms of the two unknowns and .

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Integral Calculus: JEE Main Mathematics Question with Solution

Let

f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5

and

2g(x) - 3g\left(\frac{1}{2}\right) = x

x > 0

. If

\alpha = \int_{1}^{2} f(x) \, dx

, and

\beta = \int_{1}^{2} g(x) \, dx

, then the value of

9\alpha + \beta

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 $f(x) = \frac{2}{3x^2} - \frac{x^2}{3} - \frac{5}{3}$ and $g(x) = \frac{x}{2} - \frac{3}{2}$ , yielding $9\alpha + \beta = 11$ .

Option analysis

Why each option works or fails

A · 1

Computing $9\alpha$ or evaluating the integration limits with arithmetic errors, leading to an incorrect net sum. Carefully integrate each term $\int_1^2 \frac{1}{x^2}\,dx = \frac{1}{2}$ and $\int_1^2 x^2\,dx = \frac{7}{3}$ 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 $2g(x) - 3g(1/2) = x$ to get $-g(1/2) = 1/2 \implies g(1/2) = -1/2$ , 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 $f(1/x) + 2f(x) = 1/x^2 + 5$ by 2 and subtract the original equation: $3f(x) = \frac{2}{x^2} - x^2 + 5$ is incorrect, the constant is $2(5) - 5 = 5$ .

D · 11

None. The value 11 is obtained correctly through systematic solving of both functional equations and definite integration. None.

Reviewed route

Solution

StepWorking

01given

$f(x) + 2f\left(\frac{1}{x}\right) = x^2 + 5$ and $2g(x) - 3g\left(\frac{1}{2}\right) = x$ for $x > 0$ , with $\alpha = \int_{1}^{2} f(x)\,dx$ and $\beta = \int_{1}^{2} g(x)\,dx$ .

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 $9\alpha + \beta$ .

03execute

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

04execute

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

05execute

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

06execute

Integrate $g(x)$ from 1 to 2: $\beta = \int_{1}^{2} \left( \frac{x}{2} - \frac{3}{4} \right) dx = \left[ \frac{x^2}{4} - \frac{3x}{4} \right]_{1}^{2} = \left( 1 - \frac{3}{2} \right) - \left( \frac{1}{4} - \frac{3}{4} \right) = -\frac{1}{2} - \left(-\frac{1}{2}\right) = 0$ Thus, $9\alpha + \beta = 9\left(\frac{11}{9}\right) + 0 = 11$ .

✓verify

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

✓ 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: 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 $f(x) = \frac{2}{3x^2} - \frac{x^2}{3} - \frac{5}{3}$ and $g(x) = \frac{x}{2} - \frac{3}{2}$ , yielding $9\alpha + \beta = 11$ .

Why each option works or fails

A: 1 - Computing $\alpha = \int_{1}^{2} f(x) \, dx$ or evaluating the integration limits with arithmetic errors, leading to an incorrect net sum. Carefully integrate each term $\int_1^2 \frac{1}{x^2}\,dx = \frac{1}{2}$ and $\int_1^2 x^2\,dx = \frac{7}{3}$ 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 $2g(x) - 3g(1/2) = x$ to get $-g(1/2) = 1/2 \implies g(1/2) = -1/2$ , 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 $f(1/x) + 2f(x) = 1/x^2 + 5$ by 2 and subtract the original equation: $3f(x) = \frac{2}{x^2} - x^2 + 5$ is incorrect, the constant is $2(5) - 5 = 5$ .
D · correct: 11 - None. The value 11 is obtained correctly through systematic solving of both functional equations and definite integration. None.

Step-by-step solution

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

Integral Calculus: JEE Main Mathematics Question with Solution

By solving the functional equations for f(x)f(x)f(x) and g(x)g(x)g(x), we find f(x)=23x2−x23−53f(x) = \frac{2}{3x^2} - \frac{x^2}{3} - \frac{5}{3}f(x)=3x22​−3x2​−35​ and g(x)=x2−32g(x) = \frac{x}{2} - \frac{3}{2}g(x)=2x​−23​, yielding 9α+β=119\alpha + \beta = 119α+β=11.

01given

02approach

03execute

04execute

05execute

06execute

✓verify

Students also ask

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