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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Advanced 2012 · Paper 2 · Mathematics · Continuity and Differentiability

For every integer n , let a_ n and b_ n be real numbers. Let...

For every integer nn, let ana_{n} and bnb_{n} be real numbers. Let function f:IR→IRf: I R \rightarrow I R be given by f(x)=\left\{an+sin⁡πx, for x∈[2n,2n+1]bn+cos⁡πx, for x∈(2n−1,2n)\begin{array}{lll} 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{array}\right. for all integers nn. If ff is continuous, then which of the following hold (s)(s) for all nn ?

Options

  1. Aan−1−bn−1=0a_{n-1}-b_{n-1}=0
  2. Ban−bn=1a_{n}-b_{n}=1
  3. Can−bn+1=1a_{n}-b_{n+1}=1
  4. Dan−1−bn=−1a_{n-1}-b_{n}=-1

Correct answer

B. an−bn=1a_{n}-b_{n}=1

Step-by-step solution

Given : f(x)=\left\{an+sin⁡πx,x∈[2n,2n+1]bn+cos⁡πx,x∈(2n−1,2n)\begin{array}{ll}a_{n}+\sin \pi x, & x \in[2 n, 2 n+1] \\ b_{n}+\cos \pi x, & x \in(2 n-1,2 n)\end{array}\right. ∵f\because f is continuous for all nn ∴\therefore At x=2n,x=2 n,\mathrm{LHL}=\mathrm{RHL}=f(2n)=f(2 n) ⇒bn+cos⁡2πn=an+sin⁡2πn=an+sin⁡2πn\Rightarrow b_{n}+\cos 2 \pi n=a_{n}+\sin 2 \pi n=a_{n}+\sin 2 \pi n ⇒bn+1=an⇒an−bn=1,∴\Rightarrow b_{n}+1=a_{n} \Rightarrow a_{n}-b_{n}=1, \therefore option (b) is correct. Also at x=2n+1,x=2 n+1,\mathrm{LHL}=\mathrm{RHL}=f(2n+1)=f(2 n+1) ⇒lim⁡h→0an+sin⁡π(2n+1−h)\Rightarrow \lim _{h \rightarrow 0} a_{n}+\sin \pi(2 n+1-h) =lim⁡h→0bn+1+cos⁡π(2n+1−h)=an+sin⁡(2n+1)π=\lim _{h \rightarrow 0} b_{n+1}+\cos \pi(2 n+1-h)=a_{n}+\sin (2 n+1) \pi ⇒an=bn+1−1=an⇒an−bn+1=−1\Rightarrow a_{n}=b_{n+1}-1=a_{n} \Rightarrow a_{n}-b_{n+1}=-1 ∴\therefore option (c) is incorrect. ⇒an−1−bn=−1,∴\Rightarrow a_{n-1}-b_{n}=-1, \therefore option (d) is correct.
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