Verified Solution • 2025–26 Session moderate

QJEEAdv12Ma831b: JEE Advanced 2012 Mathematics - Class 12 JEEAdv12Ma831b Solution: For every integer n, let a n and b n be real numbers. Let…

For every integer n, let an and bn be real numbers. Let function f : R →\to R be given by

f(x)={an+sin⁡πx,forx∈[2n,2n+1]bn+cos⁡πx,forx∈(2n−1,2n), for all integers n. If f is continuous, then which of the following hold(s) for all n ?

Multiple Choice Options

  • A an −- 1 −- bn −- 1 = 0
  • B an −- bn = 1
  • C an −- bn ++ 1 = 1
  • D an −- 1 −- bn = −- 1

Step-by-Step Verified Solution

We have at the points x = 2n

f(2n) = an + sin2n π\pi = an

Also for the L.H.L., we have

L.H.L. = lim⁡h→0(bn+cos⁡π(2n−h))=bn+1\mathop {\lim }\limits_{h \to 0} ({b_n} + \cos \pi (2n - h)) = {b_{n + 1}}

R.H.L. = lim⁡h→0(an+sin⁡π(2n+h))=an\mathop {\lim }\limits_{h \to 0} ({a_n} + \sin \pi (2n + h)) = {a_n}

For continuity bn+1=an{b_{n + 1}} = {a_n}

Again at x=2n+1x = 2n + 1

L.H.L. = 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}

R.H.L. = 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

Also, f(2n+1)=anf(2n + 1) = {a_n}

For continuity we require bn+1=an{b_n} + 1 = {a_n} ∴\therefore an−bn=1{a_n} - {b_n} = 1

Also, an=bn+1−1{a_n} = {b_{n + 1}} - 1

∴\therefore an−1−bn=−1{a_{n - 1}} - {b_n} = - 1

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