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For every integer n, let \mathrm{a_n\, \, and\, \, b_n } be real numbers. Let function \mathrm{ f: R \rightarrow R } be given by

\mathrm{ f(x)= \begin{cases}a_n+\sin \pi x, & \text { for } x \in[2 n, 2 n+1] \\ b_n+\cos \pi x, & \text { 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 ?
 

Option: 1

a_{n-1}-b_n=-1
 


Option: 2

a_n-b_n=1
 


Option: 3

 Both (a) & (b)
 


Option: 4

None of these


Answers (1)

best_answer

We have at the points x=2 n
\mathrm{f(2 n)=a_n+\sin 2 n \pi=a_n }
Also,
\mathrm{\begin{aligned} & \text { L.H.L. }=\lim _{h \rightarrow 0}\left(b_n+\cos \pi(2 n-h)\right)=b_n+\cos 2 n \pi=b_n+1 \\ & \text { R.H.L. }=\lim _{h \rightarrow 0}\left(a_n+\sin \pi(2 n+h)\right)=a_n+\sin 2 n \pi=a_n \end{aligned} }
For continuity,\mathrm{ b_n+1=a_n }
Again, at x=2 n+1
\mathrm{\begin{aligned} & \text { L.H.L. }=\lim _{h \rightarrow 0}\left(a_n+\sin (\pi(2 n+1-h))\right) \\ & =a_n+\sin (2 n+1) \pi=a_n \\ & \text { R.H.L. }=\lim _{h \rightarrow 0}\left(b_{n+1}+\cos (\pi(2 n+1)-h)\right)=b_{n+1}-1 \end{aligned} }
Also \mathrm{f(2 n+1)=a_n \Rightarrow a_n=b_{n+1}-1 }
For continuity we require \mathrm{b_n+1=a_n \Rightarrow a_n-b_n=1 }
Also,\mathrm{ a_n=b_{n+1}-1 \therefore a_{n-1}-b_n=-1 }

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manish painkra

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