JEE Challenger
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MathMedAdv 2012 • Paper 2

Continuity of Piecewise Trigonometric Function for All Integers

For every integer nn, let ana_n and bnb_n be real numbers. Let function f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be given by

f(x)={an+sin⁡πx,for x∈[2n,2n+1]bn+cos⁡πx,for x∈(2n−1,2n),for all integers n.f(x) = \begin{cases} a_n + \sin \pi x, & \text{for } x \in [2n, 2n + 1] \\ b_n + \cos \pi x, & \text{for } x \in (2n - 1, 2n) \end{cases}, \quad \text{for all integers } n.

If ff is continuous, then which of the following hold(s) for all nn?

Options

A

an−1−bn−1=0a_{n-1} - b_{n-1} = 0

B

an−bn=1a_n - b_n = 1

Correct
C

an−bn+1=1a_n - b_{n+1} = 1

D

an−1−bn=−1a_{n-1} - b_n = -1

Correct

Topics & Concepts

Continuity and DifferentiabilityContinuity

Step-by-Step Solution

To determine which relations between ana_n and bnb_n hold for all integers nn, we analyze the continuity of f(x)f(x) at the transition points between the piecewise definitions, specifically at x=2nx = 2n and x=2n−1x = 2n - 1.

The function is given by:

f(x)={an+sin⁡πx,for x∈[2n,2n+1]bn+cos⁡πx,for x∈(2n−1,2n),∀n∈Zf(x) = \begin{cases} a_n + \sin \pi x, & \text{for } x \in [2n, 2n + 1] \\ b_n + \cos \pi x, & \text{for } x \in (2n - 1, 2n) \end{cases}, \quad \forall n \in \mathbb{Z}

1. Continuity at x=2nx = 2n:

At x=2nx = 2n, the function values and one-sided limits are:

  • For x∈(2n−1,2n)x \in (2n - 1, 2n), we have f(x)=bn+cos⁡πxf(x) = b_n + \cos \pi x. Thus, the left-hand limit is: lim⁡x→2n−f(x)=lim⁡x→2n−(bn+cos⁡πx)=bn+cos⁡(2nπ)=bn+1\lim_{x \to 2n^-} f(x) = \lim_{x \to 2n^-} (b_n + \cos \pi x) = b_n + \cos(2n\pi) = b_n + 1

  • For x∈[2n,2n+1]x \in [2n, 2n + 1], we have f(x)=an+sin⁡πxf(x) = a_n + \sin \pi x. Thus, the right-hand limit and the value of the function are: f(2n)=lim⁡x→2n+f(x)=an+sin⁡(2nπ)=an+0=anf(2n) = \lim_{x \to 2n^+} f(x) = a_n + \sin(2n\pi) = a_n + 0 = a_n

Since f(x)f(x) is continuous everywhere on R\mathbb{R}, it must be continuous at x=2nx = 2n: lim⁡x→2n−f(x)=f(2n)\lim_{x \to 2n^-} f(x) = f(2n) bn+1=an  ⟹  an−bn=1b_n + 1 = a_n \implies a_n - b_n = 1

This verifies that Option (B) holds for all integers nn.

Consequently, for n−1n - 1, we have: an−1−bn−1=1≠0a_{n-1} - b_{n-1} = 1 \neq 0 Hence, Option (A) is incorrect.

2. Continuity at x=2n−1x = 2n - 1:

At x=2n−1x = 2n - 1, the function values and one-sided limits are:

  • For x∈[2(n−1),2(n−1)+1]=[2n−2,2n−1]x \in [2(n-1), 2(n-1) + 1] = [2n - 2, 2n - 1], we have f(x)=an−1+sin⁡πxf(x) = a_{n-1} + \sin \pi x. Thus, the left-hand limit and the value of the function are: f(2n−1)=lim⁡x→(2n−1)−f(x)=an−1+sin⁡((2n−1)π)=an−1+0=an−1f(2n - 1) = \lim_{x \to (2n - 1)^-} f(x) = a_{n-1} + \sin((2n - 1)\pi) = a_{n-1} + 0 = a_{n-1}

  • For x∈(2n−1,2n)x \in (2n - 1, 2n), we have f(x)=bn+cos⁡πxf(x) = b_n + \cos \pi x. Thus, the right-hand limit is: lim⁡x→(2n−1)+f(x)=lim⁡x→(2n−1)+(bn+cos⁡πx)=bn+cos⁡((2n−1)π)=bn−1\lim_{x \to (2n - 1)^+} f(x) = \lim_{x \to (2n - 1)^+} (b_n + \cos \pi x) = b_n + \cos((2n - 1)\pi) = b_n - 1

Since f(x)f(x) is continuous at x=2n−1x = 2n - 1: lim⁡x→(2n−1)−f(x)=lim⁡x→(2n−1)+f(x)\lim_{x \to (2n - 1)^-} f(x) = \lim_{x \to (2n - 1)^+} f(x) an−1=bn−1  ⟹  an−1−bn=−1a_{n-1} = b_n - 1 \implies a_{n-1} - b_n = -1

This verifies that Option (D) holds for all integers nn.

Replacing nn with n+1n + 1 in the relation above gives: an−bn+1=−1a_n - b_{n+1} = -1 Hence, an−bn+1=1a_n - b_{n+1} = 1 is false, so Option (C) is incorrect.

Conclusion:

The correct statements are (B) and (D).

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