Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Limits Continuity and Differentiability question

2012 · Shift 2 · Q38
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /Limits Continuity and Differentiability
  5. /2012 · Shift 2 · Q38

Limits Continuity and Differentiability question

2012 · Shift 2 · Q38

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityMultiple correct+4 / −1
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,for x∈[2n,2n+1]bn+cos⁡πx,for x∈(2n−1,2n)f(x) = \left\{ {\begin{matrix} {{a_n} + \sin \pi x,} & {for\,x \in [2n,2n + 1]} \\ {{b_n} + \cos \pi x,} & {for\,x \in (2n - 1,2n)} \\ \end{matrix} } \right.f(x)={an​+sinπx,bn​+cosπx,​forx∈[2n,2n+1]forx∈(2n−1,2n)​, for all integers n. If f is continuous, then which of the following hold(s) for all n ?
  1. A
    an −-− 1 −-− bn −-− 1 = 0
  2. B
    an −-− bn = 1
  3. C
    an −-− bn +++ 1 = 1
  4. D
    an −-− 1 −-− bn =−-− 1
View written solutionFree

Correct answer: B, D

  1. Given piecewise function

For each integer nnn,

f(x)={an+sin⁡πx,x∈[2n,2n+1]bn+cos⁡πx,x∈(2n−1,2n)f(x)= \begin{cases} a_n+\sin \pi x, & x\in[2n,2n+1]\\ b_n+\cos \pi x, & x\in(2n-1,2n) \end{cases}f(x)={an​+sinπx,bn​+cosπx,​x∈[2n,2n+1]x∈(2n−1,2n)​

We are told that fff is continuous on R\mathbb RR.

So continuity must hold at the junction points of these intervals.

  1. Find the points where continuity gives conditions

The possible junction points are:

  • x=2nx=2nx=2n,
  • x=2n−1x=2n-1x=2n−1.

We check continuity at both.

  1. Continuity at x=2nx=2nx=2n

At x=2nx=2nx=2n, the value comes from the first branch:

f(2n)=an+sin⁡(2nπ)=an.f(2n)=a_n+\sin(2n\pi)=a_n.f(2n)=an​+sin(2nπ)=an​.

The left-hand limit comes from the interval (2n−1,2n)(2n-1,2n)(2n−1,2n), i.e. the second branch with the same index nnn:

lim⁡x→2n−f(x)=bn+cos⁡(2nπ)=bn+1.\lim_{x\to 2n^-} f(x)=b_n+\cos(2n\pi)=b_n+1.x→2n−lim​f(x)=bn​+cos(2nπ)=bn​+1.

For continuity,

an=bn+1.a_n=b_n+1.an​=bn​+1.

So,

an−bn=1.a_n-b_n=1.an​−bn​=1.

This is exactly Option B.

  1. Continuity at x=2n−1x=2n-1x=2n−1

Now x=2n−1x=2n-1x=2n−1 belongs to the interval [2(n−1),2(n−1)+1]=[2n−2,2n−1][2(n-1),2(n-1)+1]=[2n-2,2n-1][2(n−1),2(n−1)+1]=[2n−2,2n−1], so the function value comes from the first branch with index n−1n-1n−1:

f(2n−1)=an−1+sin⁡((2n−1)π)=an−1+0=an−1,f(2n-1)=a_{n-1}+\sin\big((2n-1)\pi\big)=a_{n-1}+0=a_{n-1},f(2n−1)=an−1​+sin((2n−1)π)=an−1​+0=an−1​,

since sin⁡(kπ)=0\sin(k\pi)=0sin(kπ)=0 for any integer kkk.

The right-hand limit comes from the second branch on (2n−1,2n)(2n-1,2n)(2n−1,2n) with index nnn:

lim⁡x→(2n−1)+f(x)=bn+cos⁡((2n−1)π)=bn−1.\lim_{x\to (2n-1)^+} f(x)=b_n+\cos\big((2n-1)\pi\big)=b_n-1.x→(2n−1)+lim​f(x)=bn​+cos((2n−1)π)=bn​−1.

For continuity,

an−1=bn−1.a_{n-1}=b_n-1.an−1​=bn​−1.

Rearranging,

an−1−bn=−1.a_{n-1}-b_n=-1.an−1​−bn​=−1.

This is exactly Option D.

  1. Check the remaining options

Option A: an−1−bn−1=0a_{n-1}-b_{n-1}=0an−1​−bn−1​=0

From continuity at x=2(n−1)x=2(n-1)x=2(n−1), we get

an−1−bn−1=1,a_{n-1}-b_{n-1}=1,an−1​−bn−1​=1,

not 000. So A is false.

Option C: an−bn+1=1a_n-b_{n+1}=1an​−bn+1​=1

From continuity at x=2n+1x=2n+1x=2n+1 (equivalently using the condition at odd points), we get

an=bn+1−1,a_n=b_{n+1}-1,an​=bn+1​−1,

so

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

not 111. So C is false.

  1. Final answer

The correct options are:

B, D\boxed{\text{B, D}}B, D​
  1. Comparison with stored answer

Stored correct answer: B, D

Our derived answer matches the stored answer.

PreviousNext

More from Limits Continuity and Differentiability

  • Let f : R → R be a function such that f(x+y)=f(x)+f(y),∀x,y∈R. If f(x) is differentiable at x = 0, then2011 · Multiple correct
  • If x→0lim​[1+xln(1+b2)]1/x=2bsin2θ, b>0 and θ∈(−π,π], then the value of θ is2011 · MCQ
  • If f(x)={−x−2π​,−cosx​x≤−2π​−2π​1​, then2011 · Multiple correct
  • Let L=x→0lim​x4a−a2−x2​−4x2​​,a>0. If L is finite, then2009 · Multiple correct
  • Let f(x) be a non-constant twice differentiable function defined on (−∞,∞) such that f(x)=f(1−x) and f′(41​)=0. Then,2008 · Multiple correct
  • Let g(x)=logcosm(x−1)(x−1)n​;00, and let p be the left hand derivative of ∣x−1∣ at x=1. If x→1+lim​g(x)=p, then2008 · MCQ
  • Consider the function f:(−∞,∞)→(−∞,∞) defined by f(x)=x2+ax+1x2−ax+1​,0<a<2.Which of the following is true?2008 · MCQ
  • Let the function g:(−∞,∞)→(−2π​,2π​) be given by g(u)=2tan−1(eu)−2π​. Then, g is2008 · MCQ