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Question: For every integer \[n\]let \[{a_n}\]and \[{b_n}\]be real numbers. Let function \[f:IR \to IR\]be giv...

For every integer nnlet an{a_n}and bn{b_n}be real numbers. Let function f:IR→IRf:IR \to IRbe given by

{{a_n} + \sin \pi x}&{{\text{for }}x\varepsilon \left[ {2n,2n + 1} \right]} \\\ {{b_n} + \cos \pi x}&{{\text{for }}x\varepsilon \left( {2n - 1,2n} \right)} \end{array}$$, for all integers $$n$$. If $$f$$ is continuous, then which of the following holds (s) for all $$n$$? This question has multiple correct options A. $${a_{n - 1}} - {b_{n - 1}} = 0$$ B. $${a_n} - {b_n} = 1$$ C. $${a_n} - {b_{n + 1}} = 1$$ D. $${a_{n - 1}} - {b_n} = - 1$$
Explanation

Solution

In calculus, a function is continuous at x = a if and only if all three of the following conditions are met:
i) The function is defined at x = a; that is, f(a) equals a real number
ii) The limit of the function as x approaches a exists
iii) The limit of the function as x approaches a is equal to the function value at x = a

Complete step by step answer:
First check the continuity at x=2nx = 2n
Left Hand Limit =lim⁡x→2n−[f(x)] = \mathop {\lim }\limits_{x \to 2{n^ - }} [f(x)]
=lim⁡h→0[f(2n−h)]= \mathop {\lim }\limits_{h \to 0} [f(2n - h)]
=lim⁡h→0[bn+cos⁡π(2n−h)]= \mathop {\lim }\limits_{h \to 0} [{b_n} + \cos \pi (2n - h)]
=lim⁡h→0[bn+cos⁡πh]= \mathop {\lim }\limits_{h \to 0} [{b_n} + \cos \pi h]
== bn+1{b_n} + 1
Right Hand Limit =lim⁡x→2n+[f(x)]=lim⁡h→0[f(2n+h)] = \mathop {\lim }\limits_{x \to 2{n^ + }} [f(x)] = \mathop {\lim }\limits_{h \to 0} [f(2n + h)]
=lim⁡h→0[an+sin⁡π(2n+h)]=an= \mathop {\lim }\limits_{h \to 0} [{a_n} + \sin \pi (2n + h)] = {a_n}
f(2n)=an+sin⁡2πn=anf(2n) = {a_n} + \sin 2\pi n = {a_n}
For continuity, lim⁡x→2n−f(x)=lim⁡x→2n+f(x)=f(2n)\mathop {\lim }\limits_{x \to 2{n^ - }} f(x) = \mathop {\lim }\limits_{x \to 2{n^ + }} f(x) = f(2n)
So, an=bn+1⇒an−bn=1{a_n} = {b_n} + 1 \Rightarrow {a_n} - {b_n} = 1
Now, check the continuity at x=2n+1x = 2n + 1
Left Hand Limit =lim⁡h→0[an+sin⁡π(2n+1−h)]=an = \mathop {\lim }\limits_{h \to 0} [{a_n} + \sin \pi (2n + 1 - h)] = {a_n}
Right Hand Limit =lim⁡h→0[bn+1+cos⁡(π(2n+1−h))]=bn+1−1 = \mathop {\lim }\limits_{h \to 0} [{b_{n + 1}} + \cos (\pi (2n + 1 - h))] = {b_{n + 1}} - 1
f(2n+1)=anf(2n + 1) = {a_n}
For continuity, an=bn+1−1⇒an−1=bn−1{a_n} = {b_{n + 1}} - 1 \Rightarrow {a_{n - 1}} = {b_n} - 1

Hence both option B and D are the right answer.

Note: A function is a process or a relation that associates each element x of a set X, the domain of the function, to a single element y of another set Y, the codomain of the function.
An integer is known colloquially as a number that is writable without a fractional part. It is a whole number that can be positive, negative, or zero.