Let 5f(x) + 4f(1/x) = 1/x + 3, x 0. Then 18 ∫₁² f(x) dx is equal to (1) 5 loge 2 + 3 (2) 10 loge 2

Q: Wait, why do we replace $x$ with $1/x$? Is that a standard trick?

It is a classic technique for functional equations involving reciprocal arguments. By swapping $x$ and $1/x$, you create a second equation that allows you to treat $f(x)$ and $f(1/x)$ as variables in a system of linear equations. It turns a functional problem into a simple algebraic one!

Q: Can I just integrate the original equation directly?

If you integrate the original equation, you will have an integral of $f(1/x)$ left over, which you cannot evaluate without knowing $f(x)$. You must isolate $f(x)$ first to make the integration straightforward.

Q: Why did we multiply by 5 and 4 in the elimination step?

We want to eliminate $f(1/x)$. Since the coefficients of $f(1/x)$ in the two equations are 4 and 5, the least common multiple is 20. Multiplying the first equation by 5 and the second by 4 makes the coefficients of $f(1/x)$ both equal to 20, allowing for clean subtraction.

Q: Is there any restriction on $x$ that I should worry about?

The problem states $x > 0$, which is crucial because the function $f(1/x)$ and the term $1/x$ are undefined at $x=0$. Since our integration interval is $[1, 2]$, we are safely within the domain where the function is well-behaved.

Q: I got $10 \log_e 2 - 4$ for the upper limit. Did I miss something?

That is the value at the upper limit $x=2$. Remember, the definite integral is $F(2) - F(1)$. You must also evaluate the expression at $x=1$ and subtract that result from your upper limit value. Don't forget the lower limit!

Question

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

Options:

(A) $5 \log_e 2 + 3$

(B) $10 \log_e 2 + 6$

(C) $10 \log_e 2 - 6$

(D) $5 \log_e 2 - 3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of functions.

We are given the equation $5f(x) + 4f\left(\frac{1}{x}\right) = \frac{1}{x} + 3$. Your goal is to find the value of $18 \int_{1}^{2} f(x) \, dx$.

### The Symmetry Strategy

When you see a functional equation where $x$ and $\frac{1}{x}$ appear together, do not panic. This is a beautiful invitation to use symmetry.

If you replace $x$ with $\frac{1}{x}$ in the original equation, the term $f(x)$ transforms into $f\left(\frac{1}{x}\right)$, and $f\left(\frac{1}{x}\right)$ transforms back into $f(x)$.

Replacing $x$ with $\frac{1}{x}$ in our original equation, we get:

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

### The Art of Elimination

We now have a system of two equations with two unknowns. Let us label them:

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

(2) $4f(x) + 5f\left(\frac{1}{x}\right) = x + 3$

To isolate $f(x)$, we eliminate $f\left(\frac{1}{x}\right)$ by multiplying equation (1) by 5 and equation (2) by 4:

$$25f(x) + 20f\left(\frac{1}{x}\right) = \frac{5}{x} + 15$$

$$16f(x) + 20f\left(\frac{1}{x}\right) = 4x + 12$$

Subtracting these two equations causes the $f\left(\frac{1}{x}\right)$ terms to vanish:

$$(25 - 16)f(x) = \left(\frac{5}{x} + 15\right) - (4x + 12)$$

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

### Scaling to Victory

The question asks for $18 \int_{1}^{2} f(x) \, dx$. Since we have an expression for $9f(x)$, we multiply the entire equation by 2:

$$18f(x) = 2 \left( \frac{5}{x} - 4x + 3 \right) = \frac{10}{x} - 8x + 6$$

We are no longer dealing with an abstract functional equation; we are dealing with a standard calculus problem.

### The Final Integration

Now, we apply the integral from 1 to 2:

$$18 \int_{1}^{2} f(x) \, dx = \int_{1}^{2} \left( \frac{10}{x} - 8x + 6 \right) \, dx$$

Integrating term by term, we obtain:

$$\left[ 10 \ln|x| - 4x^2 + 6x \right]_{1}^{2}$$

Evaluating at the upper limit $x=2$:

$$10 \ln(2) - 4(2)^2 + 6(2) = 10 \ln(2) - 16 + 12 = 10 \ln(2) - 4$$

Evaluating at the lower limit $x=1$:

$$10 \ln(1) - 4(1)^2 + 6(1) = 0 - 4 + 6 = 2$$

Subtracting the lower limit from the upper limit, we arrive at our final result:

$$(10 \ln(2) - 4) - 2 = \mathbf{10 \ln(2) - 6}$$

Animated Solution for Mathematics - Definite Integration: Let $5 f (x) + 4 f (\frac{1}{x}) = \frac{1}{x} + 3, x > 0$ . Then $18 \int_{1}^{2} f (x) d x$ is equal to

Select Answer:

Visualized Solution

Analyze the Functional Equation

Apply the Reciprocal Substitution

Formulate the Second Equation

Strategy for Elimination

Isolate the Function $f (x)$

Scale to the Required Expression

Set up the Definite Integral

Perform the Integration

Evaluate at Upper Limit $x = 2$

Evaluate at Lower Limit $x = 1$

Final Calculation and Result

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Symmetry Strategy

The Art of Elimination

Scaling to Victory

The Final Integration

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let 5f(x)+4f(x1​)=x1​+3,x>0. Then 18∫12​f(x)dx is equal to