Let f(x) be a function satisfying f(x) + f( - x) = ^2, for all x R. Then 0^ f(x) x dx is equal to:

Given: f(x) + f( - x) = ^2 for all x R. Find: I = 0^ f(x) x dx Let I = 0^ f(x) x dx Using the symmetry transformation, I = 0^ f( - x) ( - x) dx = 0^ f( - x) x dx Adding the two expressions, 2I = 0^ (f(x) + f( - x)) x dx Now substitute f(x) + f( - x) = ^2: 2I = 0^ ^2 x dx So, 2I = ^2 0^ x dx = ^2 2 Hence, I = ^2 Therefore, the correct option is B, and the value of the integral is ^2. Given: f(x) + f( - x) = ^2. Find: 0^ f(x) x dx Because ( - x) = x, the weight function remains unchanged under the transformation x - x. So averaging the integral with its transformed form gives I = 12 0^ (f(x) + f( - x)) x dx Now use the given relation: I = 12 0^ ^2 x dx = ^22 2 = ^2 Therefore, the correct option is B.

Let f(x) be a function satisfying f(x) + f( - x) = ^2, for all x… — Solution | JEE Mathematics

MCQMediumJEE 2023Properties of Definite Integrals

Let $f(x)$ be a function satisfying $f(x) + f(\pi - x) = \pi^2$ , for all $x \in \mathbb{R}$ . Then $\int_0^{\pi} f(x)\sin x \, dx$ is equal to:

A
$\frac{\pi^2}{2}$
B
$\pi^2$
C
$2\pi^2$
D
$\frac{\pi^2}{4}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $f(x) + f(\pi - x) = \pi^2$ for all $x \in \mathbb{R}$ .

Find: $I = \int_0^{\pi} f(x)\sin x \, dx$

Let

I = \int_0^{\pi} f(x)\sin x \, dx

Using the symmetry transformation,

I = \int_0^{\pi} f(\pi - x)\sin(\pi - x) \, dx = \int_0^{\pi} f(\pi - x)\sin x \, dx

Adding the two expressions,

2I = \int_0^{\pi} \left(f(x) + f(\pi - x)\right)\sin x \, dx

Now substitute $f(x) + f(\pi - x) = \pi^2$ :

2I = \int_0^{\pi} \pi^2 \sin x \, dx

So,

2I = \pi^2 \int_0^{\pi} \sin x \, dx = \pi^2 \times 2

Hence,

I = \pi^2

Therefore, the correct option is B, and the value of the integral is $\pi^2$ .

Symmetry Trick

Given: $f(x) + f(\pi - x) = \pi^2$ .

Find: $\int_0^{\pi} f(x)\sin x \, dx$

Because $\sin(\pi - x) = \sin x$ , the weight function remains unchanged under the transformation $x \mapsto \pi - x$ . So averaging the integral with its transformed form gives

I = \frac{1}{2}\int_0^{\pi} \left(f(x) + f(\pi - x)\right)\sin x \, dx

Now use the given relation:

I = \frac{1}{2}\int_0^{\pi} \pi^2 \sin x \, dx = \frac{\pi^2}{2} \cdot 2 = \pi^2

Therefore, the correct option is B.

Common mistakes

Using $\sin(\pi - x) = -\sin x$ is incorrect. The correct identity is $\sin(\pi - x) = \sin x$ . Always verify the trigonometric transformation before substituting.
Substituting $f(\pi - x) = \pi^2$ is wrong because the given relation is $f(x) + f(\pi - x) = \pi^2$ , not either term individually. Add the two integral forms first, then apply the condition.
Forgetting the factor of $2$ after adding the two equal expressions for $I$ leads to a wrong final answer. Once the two forms are added, the left side becomes $2I$ , not $I$ .

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Answer

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