For every integer n, let an and bn be real numbers. Let function f : R → R be given by f (x) = {an+sinπx,for x∈ [2n, 2n+1]bn + cosπx,for x∈ (2n−1, 2n) , for all integers n. If f is continuous, then which of the following hold(s) for all n?

For every integer n, let an and bn be real numbers. Let function f : R \rightarrow R be given by f (x) = \left{\right. a_{n} + sinπx , for \textrm{ }\textrm{ } x \in \textrm{ } \left[\right. 2 n , \textrm{ } 2 n + 1 \left]\right. \\ b_{n} \textrm{ }\textrm{ } + \textrm{ }\textrm{ } cosπx , for \textrm{ }\textrm{ } x \in \textrm{ }\textrm{ } \left(\right. 2 n - 1 , \textrm{ }\textrm{ } 2 n \left.\right) , for all integers n. If f is continuous, then which of the following hold(s) for all n?

Engineering

Mathematics

Basic Definition of Continuity

Question

For every integer n, let an and bn be real numbers. Let function f : R $\to$ R be given by
f (x) = ${\begin{cases} a_{n} + sinπx, for x \in [2 n, 2 n + 1] \\ b_{n} + cosπx, for x \in (2 n - 1, 2 n) \end{cases}$ , for all integers n.

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

a_n – b_n = 1

a_{n – 1} – b_{n – 1} = 0

a_n – b_{n + 1}= 1

a_{n – 1} – b_n = – 1

Since, f(x) is continuous for all x at x = 2n, n $\in$ I

LHL = $\underset{h \to 0}{Lim} (b_{n} + cosπ (2 n - h))$ = b_n + 1

RHL = $\underset{h \to 0}{Lim} (a_{n} + sinπ (2 n + h))$ = a_n

Also, f(2n) = a_n.

$∴$ a_n = b_n + 1 $\Rightarrow$ a_n – b_n = 1.

Understanding the Problem

The function f(x) is defined piecewise on intervals of length 2, with different expressions on intervals [2n, 2n+1] and (2n-1, 2n) for every integer n. For f to be continuous, the left-hand limit, right-hand limit, and function value must be equal at every point where the definition changes, i.e., at x = 2n and x = 2n+1 for all integers n.

Step 1: Identify the Points of Discontinuity

The function changes its definition at x = 2n and x = 2n+1. We need to ensure continuity at these points.

Step 2: Ensure Continuity at x = 2n

For x = 2n:

Left-hand limit (x → 2n⁻): x is in (2n-1, 2n), so f(x) = b_n + cos(πx). Therefore, $\lim_{x \to 2^{n} ⁻} f (x) = b_{n} + \cos (2 n π) = b_{n} + 1$
Right-hand limit (x → 2n⁺): x is in [2n, 2n+1], so f(x) = a_n + sin(πx). Therefore, $\lim_{x \to 2^{n} ⁺} f (x) = a_{n} + \sin (2 n π) = a_{n} + 0 = a_{n}$
Function value at x = 2n: f(2n) = a_n + sin(2nπ) = a_n

For continuity at x = 2n, left-hand limit = right-hand limit = function value:

b_{n} + 1 = a_{n}

Which implies:

a_{n} - b_{n} = 1

This corresponds to option (a).

Step 3: Ensure Continuity at x = 2n+1

For x = 2n+1:

Left-hand limit (x → (2n+1)⁻): x is in [2n, 2n+1], so f(x) = a_n + sin(πx). Therefore, $\lim_{x \to (^{2 n + 1)} ⁻} f (x) = a_{n} + \sin (π (2 n + 1)) = a_{n} + \sin (2 n π + π) = a_{n} + 0 = a_{n}$
Right-hand limit (x → (2n+1)⁺): x is in (2n+1, 2n+2), which corresponds to the interval for n+1, i.e., (2(n+1)-1, 2(n+1)) = (2n+1, 2n+2). So f(x) = b_n+1 + cos(πx). Therefore, $\lim_{x \to (^{2 n + 1)} ⁺} f (x) = b_{n + 1} + \cos (π (2 n + 1)) = b_{n + 1} + \cos (2 n π + π) = b_{n + 1} + (- 1) = b_{n + 1} - 1$
Function value at x = 2n+1: f(2n+1) = a_n + sin(π(2n+1)) = a_n + 0 = a_n

For continuity at x = 2n+1, left-hand limit = right-hand limit = function value:

a_{n} = b_{n + 1} - 1

Which implies:

a_{n} - b_{n + 1} = - 1

This is not directly in the options, but let's see its implications.

Step 4: Analyze the Options

From Step 2, we have $a_{n} - b_{n} = 1$ for all n. This is option (a).

From Step 3, we have $a_{n} - b_{n + 1} = - 1$ . This is not listed, but if we shift the index by replacing n with n-1, we get $a_{n - 1} - b_{n} = - 1$ , which is option (d).

Now, check option (b): a_n-1 - b_n-1 = 0. But from our result in Step 2, for any integer, a_k - b_k = 1, not 0. So this is false.

Check option (c): a_n - b_n+1 = 1. But from Step 3, we have a_n - b_n+1 = -1, not 1. So this is false.

Final Answer

For the function f to be continuous, the following must hold for all integers n:

a_n - b_n = 1 (Option a)
a_n-1 - b_n = -1 (Option d)

Options (b) and (c) do not hold.

Key Formulae and Theory

Definition of Continuity at a Point: A function f is continuous at x = c if:

\lim_{x \to c^{⁻}} f (x) = \lim_{x \to c^{⁺}} f (x) = f (c)</mo

Detailed Solution

Understanding the Problem

Step 1: Identify the Points of Discontinuity

Step 2: Ensure Continuity at x = 2n

Step 3: Ensure Continuity at x = 2n+1

Step 4: Analyze the Options

Final Answer

Related Topics

Key Formulae and Theory

Detailed Solution

Understanding the Problem

Step 1: Identify the Points of Discontinuity

Step 2: Ensure Continuity at x = 2n

Step 3: Ensure Continuity at x = 2n+1

Step 4: Analyze the Options

Final Answer

Related Topics

Key Formulae and Theory

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