Question 14

NUMERICALHARD

For a real number α\alpha, let [α][\alpha] denote the greatest integer less than or equal to α\alpha. For a finite set SS, let ∣S∣|S| denote the number of elements in the set SS.

Consider the functions f:(βˆ’3,3)β†’(βˆ’βˆž,∞)f : (-3, 3) \to (-\infty, \infty) and g:(βˆ’3,3)β†’(βˆ’βˆž,∞)g : (-3, 3) \to (-\infty, \infty) defined by f(x)=[x3]log⁑e(1+sin⁑2(Ο€(xβˆ’[x])))f(x) = [x^3]\log_e(1 + \sin^2(\pi(x - [x]))) and g(x)=x3sin⁑2(Ο€log⁑e(1+xβˆ’[x])).g(x) = x^3 \sin^2(\pi \log_e(1 + x - [x])).

Let A={x∈(βˆ’3,3):fΒ isΒ discontinuousΒ atΒ x}A = \{x \in (-3, 3) : f \text{ is discontinuous at } x\} and B={x∈(βˆ’3,3):gΒ isΒ discontinuousΒ atΒ x}.B = \{x \in (-3, 3) : g \text{ is discontinuous at } x\}.

Then the value of ∣A∣+2∣Bβˆ£βˆ’βˆ£A∩B∣|A| + 2|B| - |A \cap B| is ___________.

Correct Answer: 56

Detailed Solution

  1. Analyze Continuity of f(x)f(x): f(x)=[x3]β‹…h(x)f(x) = [x^3] \cdot h(x), where h(x)=log⁑e(1+sin⁑2(Ο€{x}))h(x) = \log_e(1 + \sin^2(\pi\{x\})).

    • The function h(x)h(x) is continuous for all x∈Rx \in \mathbb{R}. Note that h(x)=0h(x) = 0 if and only if sin⁑2(Ο€{x})=0\sin^2(\pi\{x\}) = 0, which occurs when {x}=0\{x\} = 0, i.e., x∈Zx \in \mathbb{Z}.
    • [x3][x^3] is discontinuous at points where x3=k∈Zx^3 = k \in \mathbb{Z}. In the domain (βˆ’3,3)(-3, 3), x3∈(βˆ’27,27)x^3 \in (-27, 27). The integer values for kk are {βˆ’26,βˆ’25,…,0,…,26}\{-26, -25, \dots, 0, \dots, 26\}, totaling 53 values.
    • If x∈Zx \in \mathbb{Z}, then x3∈Zx^3 \in \mathbb{Z}. At these points, h(x)=0h(x) = 0, making f(x)f(x) continuous regardless of the jump in [x3][x^3]. The integers in (βˆ’3,3)(-3, 3) are {βˆ’2,βˆ’1,0,1,2}\{-2, -1, 0, 1, 2\}.
    • If x3=kx^3 = k but xβˆ‰Zx \notin \mathbb{Z}, then h(x)β‰ 0h(x) \neq 0, so the jump in [x3][x^3] makes f(x)f(x) discontinuous.
    • Thus, A={x=k3:k∈Z,βˆ’27<k<27}βˆ–{βˆ’2,βˆ’1,0,1,2}A = \{x = \sqrt[3]{k} : k \in \mathbb{Z}, -27 < k < 27\} \setminus \{-2, -1, 0, 1, 2\}.
    • ∣A∣=53βˆ’5=48|A| = 53 - 5 = 48.
  2. Analyze Continuity of g(x)g(x): g(x)=x3sin⁑2(Ο€log⁑e(1+{x}))g(x) = x^3 \sin^2(\pi \log_e(1 + \{x\})).

    • g(x)g(x) is continuous on every interval (n,n+1)(n, n+1) for n∈Zn \in \mathbb{Z}. We only need to check integers x∈{βˆ’2,βˆ’1,0,1,2}x \in \{-2, -1, 0, 1, 2\}.
    • At x=0x = 0, lim⁑xβ†’0g(x)=0=g(0)\lim_{x \to 0} g(x) = 0 = g(0), so gg is continuous at x=0x = 0.
    • At x=k∈{βˆ’2,βˆ’1,1,2}x = k \in \{-2, -1, 1, 2\}, the right-hand limit lim⁑xβ†’k+g(x)=k3sin⁑2(Ο€ln⁑1)=0\lim_{x \to k^+} g(x) = k^3 \sin^2(\pi \ln 1) = 0. The left-hand limit lim⁑xβ†’kβˆ’g(x)=k3sin⁑2(Ο€ln⁑2)β‰ 0\lim_{x \to k^-} g(x) = k^3 \sin^2(\pi \ln 2) \neq 0 (since ln⁑2β‰ˆ0.693\ln 2 \approx 0.693, Ο€ln⁑2\pi \ln 2 is not a multiple of Ο€\pi).
    • Thus, gg is discontinuous at x∈{βˆ’2,βˆ’1,1,2}x \in \{-2, -1, 1, 2\}.
    • ∣B∣=4|B| = 4.
  3. Calculate Intersection and Final Value:

    • Since AA contains no integers and BB contains only integers, ∣A∩B∣=0|A \cap B| = 0.
    • The value is ∣A∣+2∣Bβˆ£βˆ’βˆ£A∩B∣=48+2(4)βˆ’0=48+8=56|A| + 2|B| - |A \cap B| = 48 + 2(4) - 0 = 48 + 8 = 56.
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