Let f: R → R and g: R → R be continuous functions. Then the value of the integral ∫_(-π/2)^(π/2) [f(

Q: Wait, why do we reject the negative root here? Or rather, why do we assume the integral is zero just because it's an odd function?

That is a fantastic question! It is not an assumption; it is a geometric necessity. For any odd function, the graph is symmetric about the origin. This means the area accumulated from $-a$ to $0$ is exactly the negative of the area from $0$ to $a$. When you sum them up, they cancel out perfectly. It is like walking five steps forward and then five steps backward—you end up exactly where you started.

Q: Can I just expand the brackets and integrate term by term? It seems safer.

You *could*, but you would be walking into a trap set by the paper setters. Since $f(x)$ and $g(x)$ are generic continuous functions, you cannot integrate them explicitly. The problem is designed to test your ability to recognize properties of definite integrals. Brute force is the enemy of elegance here!

Q: What if the function was neither even nor odd? How would I solve it then?

If the function is neither even nor odd, the property $\int_{-a}^{a} h(x) dx = 0$ does not apply. In such a case, you would likely need to use other properties, such as the King's Property: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. Always check for symmetry first, but if it fails, look for properties that transform the limits or the integrand.

Question

Let $f: R 	o R$ and $g: R 	o R$ be continuous functions. Then the value of the integral $\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)] [g(x) - g(-x)] dx$ is

Options:

(A) $\pi$

(B) 1

(C) -1

(D) 0

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the Integral

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering a hidden truth about the nature of functions.

When you first look at the integral

$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x) + f(-x)] [g(x) - g(-x)] dx$$

your instinct might be to panic. You see generic functions $f(x)$ and $g(x)$, and you think, 'How can I integrate functions I don't even know?'

But here is the secret of JEE Advanced: whenever you see an integral with limits from $-a$ to $a$, stop. Breathe. You are not meant to integrate; you are meant to observe.

### Phase 1: The Symmetric Interval

The limits $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ are a massive clue. They are symmetric about the origin.

In the world of calculus, this is a giant neon sign pointing toward the properties of even and odd functions. If we can prove that the integrand—the entire expression inside the integral—is an odd function, the answer collapses to zero instantly.

Let us define our integrand as a single function, $h(x)$:

$$h(x) = [f(x) + f(-x)] [g(x) - g(-x)]$$

Our mission is simple: determine the parity of $h(x)$. Is it even, where $h(-x) = h(x)$? Or is it odd, where $h(-x) = -h(x)$?

### Phase 2: The Algebraic Dance

To test for parity, we perform the standard substitution: replace every $x$ with $-x$. Let us watch the algebra unfold:

$$h(-x) = [f(-x) + f(-(-x))] [g(-x) - g(-(-x))]$$

Simplifying the double negatives, we get:

$$h(-x) = [f(-x) + f(x)] [g(-x) - g(x)]$$

Now, look closely at the first bracket: $[f(-x) + f(x)]$. Because addition is commutative, this is identical to $[f(x) + f(-x)]$. The first bracket is invariant!

But look at the second bracket: $[g(-x) - g(x)]$. This is the negative of our original second bracket, $[g(x) - g(-x)]$. If we factor out a $-1$, we get:

$$h(-x) = [f(x) + f(-x)] 	imes (-1) [g(x) - g(-x)]$$

When we pull that negative sign to the front, the transformation is complete:

$$h(-x) = - [f(x) + f(-x)] [g(x) - g(-x)] = -h(x)$$

### Phase 3: The Geometric Revelation

We have proven that $h(x)$ is an odd function. Geometrically, this means the graph of $h(x)$ has rotational symmetry about the origin.

Imagine the area under the curve from $0$ to $\frac{\pi}{2}$. Let us call this area $A$. Because the function is odd, the area from $-\frac{\pi}{2}$ to $0$ is exactly $-A$.

When we integrate from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, we are summing these two regions:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} h(x) dx = \int_{-\frac{\pi}{2}}^{0} h(x) dx + \int_{0}^{\frac{\pi}{2}} h(x) dx$$

Substituting our values:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} h(x) dx = -A + A = 0$$

### Conclusion

There is no complex integration, no substitution, and no headache. Just pure, elegant logic.

The answer is **$0$**.

This problem teaches us that in physics and mathematics, observation is often more powerful than calculation. Keep this perspective, and you will conquer the most difficult problems the JEE can throw at you.

Animated Solution for Mathematics - Definite Integration: Let $f : R \to R$ and $g : R \to R$ be continuous functions. Then the value of the integral $\int_{- π /2}^{π /2} [f (x) + f (- x)] [g (x) - g (- x)] d x$ is

Select Answer:

Visualized Solution

The Symmetric Integral

Defining the Integrand

Checking Parity: Substitute $- x$

Simplifying the First Bracket

Simplifying the Second Bracket

The Odd Function Revelation

Geometry of Odd Functions

Final Integral Value

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Conclusion

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