Logo
Jeeify
MCQMediumJEE 2023Definite Integrals

JEE Mathematics 2023 Question with Solution

Let 5f(x)+4f(1x)=1x+3,  x>05f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3, \; x>0. Then 18∫12f(x) dx18\int_{1}^{2} f(x)\,dx is equal to:

  • A

    10log⁡e2−610\log_e 2 - 6

  • B

    10log⁡e2+610\log_e 2 + 6

  • C

    5log⁡e2−35\log_e 2 - 3

  • D

    5log⁡e2+35\log_e 2 + 3

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: 5f(x)+4f(1x)=1x+35f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3 for x>0x>0.

Find: 18∫12f(x) dx18\int_{1}^{2} f(x)\,dx.

Replace xx by 1x\frac{1}{x} in the given equation:

5f(1x)+4f(x)=x+35f\left(\frac{1}{x}\right) + 4f(x) = x + 3

Now solve the two linear equations:

5f(x)+4f(1x)=1x+34f(x)+5f(1x)=x+3\begin{aligned} 5f(x) + 4f\left(\frac{1}{x}\right) &= \frac{1}{x} + 3 \\ 4f(x) + 5f\left(\frac{1}{x}\right) &= x + 3 \end{aligned}

Multiply the first equation by 55 and the second equation by 44, then subtract:

25f(x)+20f(1x)=5x+1516f(x)+20f(1x)=4x+12\begin{aligned} 25f(x) + 20f\left(\frac{1}{x}\right) &= \frac{5}{x} + 15 \\ 16f(x) + 20f\left(\frac{1}{x}\right) &= 4x + 12 \end{aligned}

So,

9f(x)=5x−4x+39f(x) = \frac{5}{x} - 4x + 3

Hence,

f(x)=59x−4x9+13f(x) = \frac{5}{9x} - \frac{4x}{9} + \frac{1}{3}

Integrate from 11 to 22:

∫12f(x) dx=∫12(59x−4x9+13)dx\int_{1}^{2} f(x)\,dx = \int_{1}^{2} \left(\frac{5}{9x} - \frac{4x}{9} + \frac{1}{3}\right) dx

Evaluating,

∫12f(x) dx=59∫121x dx−49∫12x dx+13∫12dx=59log⁡e2−49⋅22−122+13(2−1)=59log⁡e2−49⋅32+13=59log⁡e2−23+13=59log⁡e2−13\begin{aligned} \int_{1}^{2} f(x)\,dx &= \frac{5}{9}\int_{1}^{2} \frac{1}{x}\,dx - \frac{4}{9}\int_{1}^{2} x\,dx + \frac{1}{3}\int_{1}^{2} dx \\ &= \frac{5}{9}\log_e 2 - \frac{4}{9}\cdot \frac{2^2-1^2}{2} + \frac{1}{3}(2-1) \\ &= \frac{5}{9}\log_e 2 - \frac{4}{9}\cdot \frac{3}{2} + \frac{1}{3} \\ &= \frac{5}{9}\log_e 2 - \frac{2}{3} + \frac{1}{3} \\ &= \frac{5}{9}\log_e 2 - \frac{1}{3} \end{aligned}

Therefore,

18∫12f(x) dx=18(59log⁡e2−13)=10log⁡e2−618\int_{1}^{2} f(x)\,dx = 18\left(\frac{5}{9}\log_e 2 - \frac{1}{3}\right) = 10\log_e 2 - 6

So the correct option is A.

Direct Elimination

Given: 5f(x)+4f(1x)=1x+35f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3.

Find: 18∫12f(x) dx18\int_{1}^{2} f(x)\,dx.

Write the companion equation by replacing xx with 1x\frac{1}{x}:

4f(x)+5f(1x)=x+34f(x) + 5f\left(\frac{1}{x}\right) = x + 3

The coefficients are symmetric, so elimination is immediate.

Subtracting after matching the coefficient of f(1x)f\left(\frac{1}{x}\right) gives

9f(x)=5x−4x+39f(x) = \frac{5}{x} - 4x + 3

Hence

f(x)=59x−4x9+13f(x) = \frac{5}{9x} - \frac{4x}{9} + \frac{1}{3}

Now integrate and multiply by 1818:

18∫12f(x) dx=18∫12(59x−4x9+13)dx=10log⁡e2−618\int_{1}^{2} f(x)\,dx = 18\int_{1}^{2} \left(\frac{5}{9x} - \frac{4x}{9} + \frac{1}{3}\right) dx = 10\log_e 2 - 6

Therefore, the correct option is A.

Common mistakes

  • Replacing f(1x)f\left(\frac{1}{x}\right) incorrectly when substituting x→1xx \to \frac{1}{x}. The correct transformed equation is 5f(1x)+4f(x)=x+35f\left(\frac{1}{x}\right) + 4f(x) = x + 3, because 1(1/x)=x\frac{1}{(1/x)} = x.

  • Solving the simultaneous equations with wrong elimination. Keep the coefficients aligned carefully before subtracting; otherwise the expression for f(x)f(x) becomes incorrect.

  • Making an error in ∫12x dx\int_{1}^{2} x\,dx. It is [x22]12=4−12=32\left[\frac{x^2}{2}\right]_{1}^{2} = \frac{4-1}{2} = \frac{3}{2}, not 11 or 22.

  • Forgetting the final factor of 1818 after evaluating ∫12f(x) dx\int_{1}^{2} f(x)\,dx. The question asks for 18∫12f(x) dx18\int_{1}^{2} f(x)\,dx, not only the integral.

Practice more Definite Integrals questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Definite Integrals

Related questions