JEE Advance - Mathematics (2012 - Paper 2 Offline - No. 18)
For every integer n, let an and bn be real numbers. Let function f : R $$\to$$ R be given by
$$f(x) = \left\{ {\matrix{ {{a_n} + \sin \pi x,} & {for\,x \in [2n,2n + 1]} \cr {{b_n} + \cos \pi x,} & {for\,x \in (2n - 1,2n)} \cr } } \right.$$, for all integers n. If f is continuous, then which of the following hold(s) for all n ?
Explanation
We have at the points x = 2n
f(2n) = an + sin2n$$\pi$$ = an
Also for the L.H.L., we have
L.H.L. = $$\mathop {\lim }\limits_{h \to 0} ({b_n} + \cos \pi (2n - h)) = {b_{n + 1}}$$
R.H.L. = $$\mathop {\lim }\limits_{h \to 0} ({a_n} + \sin \pi (2n + h)) = {a_n}$$
For continuity $${b_{n + 1}} = {a_n}$$
Again at $$x = 2n + 1$$
L.H.L. = $$\mathop {\lim }\limits_{h \to 0} ({a_n} + \sin (\pi (2n + 1 - h))) = {a_n}$$
R.H.L. = $$\mathop {\lim }\limits_{h \to 0} ({b_{n + 1}} + \cos (\pi (2n + 1) - h)) = {b_{n + 1}} - 1$$
Also, $$f(2n + 1) = {a_n}$$
For continuity we require $${b_n} + 1 = {a_n}$$ $$\therefore$$ $${a_n} - {b_n} = 1$$
Also, $${a_n} = {b_{n + 1}} - 1$$
$$\therefore$$ $${a_{n - 1}} - {b_n} = - 1$$
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