Let f: [-1, 2] → [0, ∞) be a continuous function such that f(x) = f(1-x) for all x ∈ [-1, 2]. Let R₁

Q: Wait, why is the axis of symmetry $x = 1/2$ and not something else?

Great question! The axis of symmetry for a function satisfying $f(x) = f(c - x)$ is always $x = c/2$. Here, $f(x) = f(1 - x)$, so $c = 1$. Thus, the axis is $x = 1/2$. You can also think of it as the midpoint of the interval $[-1, 2]$, which is $(-1 + 2) / 2 = 1/2$.

Q: Can I solve this by using integration by parts on $R_1$?

Technically, you could try, but it would be a trap! Integration by parts would introduce $f'(x)$, and we don't know anything about the derivative of $f(x)$. The King's Property is the elegant, intended path that bypasses the need to know the specific form of $f(x)$.

Q: Why does $f(1-x) = f(x)$ allow us to replace $f(1-x)$ with $f(x)$ in the integral?

Because the problem explicitly states that $f(x) = f(1-x)$ for all $x$ in the domain. When we perform the substitution $x \to 1-x$ inside the integral using the King's Property, the term $f(1-x)$ appears naturally. Since the function is symmetric, that term is identical to $f(x)$, allowing us to simplify the expression significantly.

Question

Let $f: [-1, 2] 	o [0, \infty)$ be a continuous function such that $f(x) = f(1-x)$ for all $x \in [-1, 2]$. Let $R_1 = \int_{-1}^{2} x f(x) dx$, and $R_2$ be the area of the region bounded by $y = f(x)$, $x = -1$, $x = 2$, and the x-axis. Then

Options:

(A) $R_1 = 2R_2$

(B) $R_1 = 3R_2$

(C) $2R_1 = R_2$

(D) $3R_1 = R_2$

Sigma Percentile · Accepted Answer

### Analyzing the Symmetry

Imagine you are standing before a complex, unknown function $f(x)$ defined on the interval $[-1, 2]$. At first glance, it looks like a daunting, abstract curve.

But the problem gives us a secret key: $f(x) = f(1-x)$. This isn't just an equation; it is a geometric revelation.

It tells us that the function is a mirror image of itself, perfectly balanced about the vertical line $x = 1/2$. In the world of JEE Advanced, symmetry is not just a property; it is a shortcut to brilliance.

### Defining Our Players

We are tasked with finding the relationship between two quantities. First, $R_2 = \int_{-1}^{2} f(x) dx$, which is the area under the curve.

Second, $R_1 = \int_{-1}^{2} x f(x) dx$, which represents the first moment of that area. At first, $R_1$ looks intimidating because of that extra $x$ factor.

It feels like we need to know the exact function to solve it. But we don't; we only need the symmetry.

### The King's Property: Our Secret Weapon

To bridge the gap between $R_1$ and $R_2$, we invoke the King's Property of definite integrals:

$$ \int_{a}^{b} g(x) dx = \int_{a}^{b} g(a + b - x) dx $$

Here, our limits are $a = -1$ and $b = 2$, so $a + b = 1$. This means we can replace every $x$ in our integral with $(1 - x)$ without changing the value of the integral.

Let's apply this to $R_1$:

$$ R_1 = \int_{-1}^{2} x f(x) dx $$

By the property, we substitute $x$ with $(1-x)$:

$$ R_1 = \int_{-1}^{2} (1 - x) f(1 - x) dx $$

### The Elegant Cancellation

Now, we use our symmetry condition. Since $f(1 - x) = f(x)$, we substitute this into our new integral:

$$ R_1 = \int_{-1}^{2} (1 - x) f(x) dx $$

Now, distribute $f(x)$ inside the parentheses:

$$ R_1 = \int_{-1}^{2} (f(x) - x f(x)) dx $$

Using the linearity of integrals, we split this into two:

$$ R_1 = \int_{-1}^{2} f(x) dx - \int_{-1}^{2} x f(x) dx $$

Look closely at what we have created. The first term is exactly $R_2$, and the second term is our original $R_1$. So, we have the beautiful, simple equation:

$$ R_1 = R_2 - R_1 $$

### The Final Victory

With a simple algebraic step, we add $R_1$ to both sides:

$$ 2R_1 = R_2 $$

We have arrived at the solution without ever knowing the explicit form of $f(x)$. This is the power of mathematical insight.

We didn't fight the function; we understood its nature, and the answer revealed itself. Keep this mindset for your JEE journey: look for the symmetry, apply the properties, and let the math do the heavy lifting for you.

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Visualized Solution

Understanding the Symmetry of $f (x)$

Defining the Area $R_{2}$

Defining the Integral $R_{1}$

The King's Property of Definite Integrals

Applying the Property to $R_{1}$

Utilizing the Symmetry Condition

Expanding the Integral

Splitting the Integral

Substituting $R_{1}$ and $R_{2}$

Solving for the Final Relationship

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Symmetry

Defining Our Players

The King's Property

The Elegant Cancellation

The Final Victory

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Visualized Solution

Understanding the Symmetry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Defining the Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Defining the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​