For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [-10, 10] by $$f(x)=\begin{cases} x-[x] & \text{if }[x]\text{ is odd,} \\ 1+[x]-x & \text{if }[x]\text{ is even} \end{cases}$$ Then the value of $\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x \, dx$ is

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Definite Integrals

For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real valued function defined on the interval [-10, 10] by

f (x) = {x - [x] 1 + [x] - x if [x] is odd, if [x] is even

Then the value of

\frac{π ^{2}}{10} \int_{- 10}^{10} f (x) cos π x d x

Answer: 4

Solution

The function

f (x)

is periodic with period 2 and symmetric. For

[x]

odd,

f (x) = x - [x]

is the fractional part (sawtooth). For

[x]

even,

f (x) = 1 + [x] - x

is the inverted sawtooth.

The function

f (x)

can be written as a triangular wave. Over one period

[n, n + 1]

If $n$ is odd: $f (x) = x - n$ (ramp up from 0 to 1)
If $n$ is even: $f (x) = n + 1 - x$ (ramp down from 1 to 0)

This creates a symmetric triangular wave. Using Fourier series or direct integration:

\int_{- 10}^{10} f (x) cos (π x) d x = 2 \int_{0}^{10} f (x) cos (π x) d x

By the periodicity and orthogonality of cosine with the triangular wave, the integral evaluates to

\frac{40}{π ^{2}}

.

Thus

\frac{π ^{2}}{10} \cdot \frac{40}{π ^{2}} = 4

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Definite Integrals

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