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MathHardMarks to AllAdv 2017 • Paper 1

Discontinuity Points of Function Involving Greatest Integer Function

Let [x][x] denote the greatest integer less than or equal to xx. At which of the given point(s) is the real function f(x)=xcos⁡(π(x+[x]))f(x) = x \cos(\pi(x + [x])) discontinuous?

Official Notice: Marks Awarded to All

This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.

Options

A

x=−1x = -1

B

x=0x = 0

C

x=1x = 1

D

x=2x = 2

Topics & Concepts

Continuity and DifferentiabilityContinuity

Step-by-Step Solution

To determine the points of discontinuity of the real function f(x)=xcos⁡(π(x+[x]))f(x) = x \cos(\pi(x + [x])) where [x][x] denotes the greatest integer function (floor function), let us analyze the behavior of f(x)f(x) at an arbitrary integer x=k∈Zx = k \in \mathbb{Z}.

Step 1: Evaluation of f(x)f(x) at x=kx = k

For any integer k∈Zk \in \mathbb{Z}, we have [k]=k[k] = k. Thus, f(k)=kcos⁡(π(k+k))=kcos⁡(2kπ)f(k) = k \cos(\pi(k + k)) = k \cos(2k\pi)

Since cos⁡(2kπ)=1\cos(2k\pi) = 1 for all integers kk, f(k)=k⋅1=kf(k) = k \cdot 1 = k

Step 2: Right-Hand Limit (RHL) as x→k+x \to k^+

When xx approaches kk from the right (x∈[k,k+1)x \in [k, k+1)), the value of [x][x] is equal to kk.

RHL=lim⁡x→k+f(x)=lim⁡x→k+xcos⁡(π(x+k))=kcos⁡(π(k+k))=kcos⁡(2kπ)=k\text{RHL} = \lim_{x \to k^+} f(x) = \lim_{x \to k^+} x \cos(\pi(x + k)) = k \cos(\pi(k + k)) = k \cos(2k\pi) = k

Step 3: Left-Hand Limit (LHL) as x→k−x \to k^-

When xx approaches kk from the left (x∈(k−1,k)x \in (k-1, k)), the value of [x][x] is equal to k−1k - 1.

LHL=lim⁡x→k−f(x)=lim⁡x→k−xcos⁡(π(x+k−1))=kcos⁡(π(k+k−1))=kcos⁡((2k−1)π)\text{LHL} = \lim_{x \to k^-} f(x) = \lim_{x \to k^-} x \cos(\pi(x + k - 1)) = k \cos(\pi(k + k - 1)) = k \cos((2k-1)\pi)

Since cos⁡((2k−1)π)=−1\cos((2k-1)\pi) = -1 for all integers kk, LHL=k(−1)=−k\text{LHL} = k(-1) = -k

Step 4: Condition for Continuity at x=kx = k

For f(x)f(x) to be continuous at an integer x=kx = k, the left-hand limit must equal the right-hand limit: LHL=RHL  ⟹  −k=k  ⟹  2k=0  ⟹  k=0\text{LHL} = \text{RHL} \implies -k = k \implies 2k = 0 \implies k = 0

  • At x=0x = 0: LHL=0\text{LHL} = 0, RHL=0\text{RHL} = 0, and f(0)=0f(0) = 0. Therefore, f(x)f(x) is continuous at x=0x = 0.
  • At x=k≠0x = k \neq 0 (k∈Zk \in \mathbb{Z}): LHL=−k≠k=RHL\text{LHL} = -k \neq k = \text{RHL}. Therefore, f(x)f(x) is discontinuous at all non-zero integers.

Conclusion for Given Options:

  • (A) x=−1x = -1: Discontinuous (LHL=1≠−1=RHL\text{LHL} = 1 \neq -1 = \text{RHL})
  • (B) x=0x = 0: Continuous (LHL=0=RHL\text{LHL} = 0 = \text{RHL})
  • (C) x=1x = 1: Discontinuous (LHL=−1≠1=RHL\text{LHL} = -1 \neq 1 = \text{RHL})
  • (D) x=2x = 2: Discontinuous (LHL=−2≠2=RHL\text{LHL} = -2 \neq 2 = \text{RHL})

Mathematically, the function is discontinuous at (A), (C), and (D).

(Note: Due to a printing inconsistency in the original examination test paper, full marks were awarded to all candidates for this question).

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