JEE Challenger
More from Continuity and Differentiability
MathHardAdv 2016 • Paper 2

Continuity and Differentiability of Floor and Absolute Value Functions

Consider functions f:[−12,2]→Rf : \left[-\frac{1}{2}, 2\right] \to \mathbb{R} and g:[−12,2]→Rg : \left[-\frac{1}{2}, 2\right] \to \mathbb{R} given by f(x)=[x2−3]f(x) = [x^2 - 3] and g(x)=∣x∣f(x)+∣4x−7∣f(x)g(x) = |x| f(x) + |4x - 7| f(x), where [y][y] represents the greatest integer function for y∈Ry \in \mathbb{R}. Then:

Options

A

ff is discontinuous exactly at three points in [−12,2]\left[-\frac{1}{2}, 2\right]

B

ff is discontinuous exactly at four points in [−12,2]\left[-\frac{1}{2}, 2\right]

Correct
C

gg is NOT differentiable exactly at four points in (−12,2)\left(-\frac{1}{2}, 2\right)

Correct
D

gg is NOT differentiable exactly at five points in (−12,2)\left(-\frac{1}{2}, 2\right)

Topics & Concepts

Continuity and DifferentiabilityContinuityDifferentiability

Step-by-Step Solution

To determine the correct options, we analyze the functions f(x)f(x) and g(x)g(x) step-by-step on their given domain.

1. Discontinuity of f(x)=[x2−3]f(x) = [x^2 - 3] on [−12,2]\left[-\frac{1}{2}, 2\right]

The floor function [y][y] is discontinuous at points where y=x2−3y = x^2 - 3 takes integer values and changes value.

For x∈[−12,2]x \in \left[-\frac{1}{2}, 2\right]:

  • When x∈[−12,0]x \in \left[-\frac{1}{2}, 0\right], x2∈[0,14]x^2 \in \left[0, \frac{1}{4}\right], so x2−3∈[−3,−2.75]x^2 - 3 \in \left[-3, -2.75\right]. Thus, f(x)=−3f(x) = -3.
  • When x∈[0,1)x \in [0, 1), x2∈[0,1)x^2 \in [0, 1), so x2−3∈[−3,−2)x^2 - 3 \in [-3, -2). Thus, f(x)=−3f(x) = -3.
  • When x∈[1,2)x \in [1, \sqrt{2}), x2∈[1,2)x^2 \in [1, 2), so x2−3∈[−2,−1)x^2 - 3 \in [-2, -1). Thus, f(x)=−2f(x) = -2.
  • When x∈[2,3)x \in [\sqrt{2}, \sqrt{3}), x2∈[2,3)x^2 \in [2, 3), so x2−3∈[−1,0)x^2 - 3 \in [-1, 0). Thus, f(x)=−1f(x) = -1.
  • When x∈[3,2)x \in [\sqrt{3}, 2), x2∈[3,4)x^2 \in [3, 4), so x2−3∈[0,1)x^2 - 3 \in [0, 1). Thus, f(x)=0f(x) = 0.
  • At x=2x = 2, x2−3=1x^2 - 3 = 1. Thus, f(2)=1f(2) = 1.

Now, let's test the continuity at the transition points:

  1. At x=1x = 1: lim⁡x→1−f(x)=−3andf(1)=−2\lim_{x \to 1^-} f(x) = -3 \quad \text{and} \quad f(1) = -2 Since lim⁡x→1−f(x)≠f(1)\lim_{x \to 1^-} f(x) \neq f(1), ff is discontinuous at x=1x = 1.

  2. At x=2x = \sqrt{2}: lim⁡x→2−f(x)=−2andf(2)=−1\lim_{x \to \sqrt{2}^-} f(x) = -2 \quad \text{and} \quad f(\sqrt{2}) = -1 Since lim⁡x→2−f(x)≠f(2)\lim_{x \to \sqrt{2}^-} f(x) \neq f(\sqrt{2}), ff is discontinuous at x=2x = \sqrt{2}.

  3. At x=3x = \sqrt{3}: lim⁡x→3−f(x)=−1andf(3)=0\lim_{x \to \sqrt{3}^-} f(x) = -1 \quad \text{and} \quad f(\sqrt{3}) = 0 Since lim⁡x→3−f(x)≠f(3)\lim_{x \to \sqrt{3}^-} f(x) \neq f(\sqrt{3}), ff is discontinuous at x=3x = \sqrt{3}.

  4. At x=2x = 2: lim⁡x→2−f(x)=0andf(2)=1\lim_{x \to 2^-} f(x) = 0 \quad \text{and} \quad f(2) = 1 Since lim⁡x→2−f(x)≠f(2)\lim_{x \to 2^-} f(x) \neq f(2), ff is discontinuous at x=2x = 2.

Therefore, ff is discontinuous at exactly four points in [−12,2]\left[-\frac{1}{2}, 2\right]: x=1,2,3,2x = 1, \sqrt{2}, \sqrt{3}, 2.

This makes Option B correct and Option A incorrect.

2. Differentiability of g(x)=(∣x∣+∣4x−7∣)f(x)g(x) = (|x| + |4x - 7|) f(x) in (−12,2)\left(-\frac{1}{2}, 2\right)

We evaluate g(x)g(x) piece-wise on subintervals of the open interval (−12,2)\left(-\frac{1}{2}, 2\right):

  1. For x∈(−12,0)x \in \left(-\frac{1}{2}, 0\right): ∣x∣=−x|x| = -x, ∣4x−7∣=7−4x|4x - 7| = 7 - 4x, and f(x)=−3f(x) = -3. g(x)=(−x+7−4x)(−3)=−3(7−5x)=15x−21g(x) = (-x + 7 - 4x)(-3) = -3(7 - 5x) = 15x - 21

  2. For x∈[0,1)x \in [0, 1): ∣x∣=x|x| = x, ∣4x−7∣=7−4x|4x - 7| = 7 - 4x, and f(x)=−3f(x) = -3. g(x)=(x+7−4x)(−3)=−3(7−3x)=9x−21g(x) = (x + 7 - 4x)(-3) = -3(7 - 3x) = 9x - 21

  3. For x∈[1,2)x \in [1, \sqrt{2}): ∣x∣=x|x| = x, ∣4x−7∣=7−4x|4x - 7| = 7 - 4x, and f(x)=−2f(x) = -2. g(x)=(7−3x)(−2)=6x−14g(x) = (7 - 3x)(-2) = 6x - 14

  4. For x∈[2,3)x \in [\sqrt{2}, \sqrt{3}): ∣x∣=x|x| = x, ∣4x−7∣=7−4x|4x - 7| = 7 - 4x, and f(x)=−1f(x) = -1. g(x)=(7−3x)(−1)=3x−7g(x) = (7 - 3x)(-1) = 3x - 7

  5. For x∈[3,2)x \in [\sqrt{3}, 2): Since f(x)=0f(x) = 0, g(x)=(∣x∣+∣4x−7∣)⋅0=0g(x) = (|x| + |4x - 7|) \cdot 0 = 0

Now, we check differentiability at all critical transition points in (−12,2)\left(-\frac{1}{2}, 2\right):

  • At x=0x = 0: lim⁡x→0−g(x)=−21=lim⁡x→0+g(x)=g(0)\lim_{x \to 0^-} g(x) = -21 = \lim_{x \to 0^+} g(x) = g(0) gg is continuous at x=0x = 0. Left-hand derivative: g′(0−)=15g'(0^-) = 15 Right-hand derivative: g′(0+)=9g'(0^+) = 9 Since g′(0−)≠g′(0+)g'(0^-) \neq g'(0^+), gg is NOT differentiable at x=0x = 0.

  • At x=1x = 1: lim⁡x→1−g(x)=9(1)−21=−12\lim_{x \to 1^-} g(x) = 9(1) - 21 = -12 lim⁡x→1+g(x)=g(1)=6(1)−14=−8\lim_{x \to 1^+} g(x) = g(1) = 6(1) - 14 = -8 Since lim⁡x→1−g(x)≠lim⁡x→1+g(x)\lim_{x \to 1^-} g(x) \neq \lim_{x \to 1^+} g(x), gg is discontinuous and hence NOT differentiable at x=1x = 1.

  • At x=2x = \sqrt{2}: lim⁡x→2−g(x)=62−14\lim_{x \to \sqrt{2}^-} g(x) = 6\sqrt{2} - 14 lim⁡x→2+g(x)=g(2)=32−7\lim_{x \to \sqrt{2}^+} g(x) = g(\sqrt{2}) = 3\sqrt{2} - 7 Since 62−14≠32−76\sqrt{2} - 14 \neq 3\sqrt{2} - 7, gg is discontinuous and hence NOT differentiable at x=2x = \sqrt{2}.

  • At x=3x = \sqrt{3}: lim⁡x→3−g(x)=33−7≠0\lim_{x \to \sqrt{3}^-} g(x) = 3\sqrt{3} - 7 \neq 0 lim⁡x→3+g(x)=g(3)=0\lim_{x \to \sqrt{3}^+} g(x) = g(\sqrt{3}) = 0 Since lim⁡x→3−g(x)≠lim⁡x→3+g(x)\lim_{x \to \sqrt{3}^-} g(x) \neq \lim_{x \to \sqrt{3}^+} g(x), gg is discontinuous and hence NOT differentiable at x=3x = \sqrt{3}.

  • At x=74x = \frac{7}{4}: On the interval (3,2)(\sqrt{3}, 2), g(x)=0g(x) = 0 identically. Thus, g(x)g(x) is constant in an open neighborhood around x=74x = \frac{7}{4}, making it differentiable at x=74x = \frac{7}{4} with g′(74)=0g'\left(\frac{7}{4}\right) = 0.

Therefore, gg is NOT differentiable at exactly four points in (−12,2)\left(-\frac{1}{2}, 2\right): x=0,1,2,3x = 0, 1, \sqrt{2}, \sqrt{3}.

This makes Option C correct and Option D incorrect.

Conclusion:

The correct statements are B and C.

PreviousNext