Animated Solution for Mathematics - Definite Integration: If for nonzero $x$, $af(x)+bf(1/x)=\frac{1}{x}-5$ where $a\ne b$, then ${\int }_1^{2}f(x)dx=\dots \dots \dots$

Unlocking Functional Equations

Welcome, future engineers. Today, we are going to dissect a problem that, at first glance, might look like a nightmare of unknown functions. You see $f(x)$ and $f(1/x)$ tangled together, and your instinct might be to panic.

But I want you to take a deep breath. In the world of JEE Advanced, when you see a functional equation involving $x$ and its reciprocal $1/x$, you are not looking at a problem to be solved by brute force. You are looking at a puzzle designed to be solved by symmetry.

The First Integration

We are given the equation $af(x)+bf(1/x)=1/x-5$. Our goal is to find the definite integral $I={\int }_1^{2}f(x)dx$.

The most natural, intuitive step is to integrate the entire equation over the interval $[1,2]$. We want to transform the function $f(x)$ into the integral $I$.

When we integrate both sides, we get:

Let us pause here. We have two unknowns. Let $I_1={\int }_1^{2}f(x)dx$ and $I_2={\int }_1^{2}f(1/x)dx$.

The right-hand side is a simple calculus exercise. The integral of $1/x$ is $\ln x$, and the integral of $5$ is $5x$. Evaluating this from $1$ to $2$ gives us $(\ln 2-10)-(0-5)=\ln 2-5$.

So, our first equation is:

The Symmetry Trick

Now, we have one equation with two unknowns ($I_1$ and $I_2$). We are stuck. But this is where the JEE magic happens. We need a second equation.

We use the symmetry of the functional equation. We replace $x$ with $1/x$ in the original equation.

When we substitute $x\to 1/x$, the term $f(x)$ becomes $f(1/x)$, and $f(1/x)$ becomes $f(x)$. The term $1/x$ becomes $x$. The equation transforms into:

Rearranging this, we get:

This is beautiful. We have created a new equation that involves the same functions but with different coefficients. Now, we integrate this new equation from $1$ to $2$:

Solving the System

Let us evaluate the right-hand side of this second equation. The integral of $x$ is $x^2/2$, and the integral of $5$ is $5x$. Evaluating from $1$ to $2$:

So, our second equation is:

Now, look at what we have achieved. We have a system of two linear equations:

1) $a{I}_1+b{I}_2=\ln 2-5$

2) $b{I}_1+a{I}_2=-\frac{7}{2}$

Our goal is to isolate $I_1$. To do this, we eliminate $I_2$. We multiply the first equation by $a$ and the second by $b$:

Subtracting the second from the first, the $I_2$ terms vanish into thin air! We are left with:

Finally, dividing by $(a^2-b^2)$, we arrive at our solution:

Do you see the elegance? We never needed to know what $f(x)$ was. We didn't need to solve for the function itself.

We treated the integral as a variable, used the symmetry of the input to generate a second equation, and solved the resulting linear system. This is the essence of JEE Advanced mathematics: it is not about memorizing formulas; it is about recognizing patterns and manipulating them with confidence.

Keep practicing this mindset, and you will find that even the most intimidating problems become solvable.