Limits, Continuity & Differentiability Question - None

Question

Let $f(x) = \left[\dfrac{4^x + 2^x + 1}{2^x - 2^{x/2} + 1}\right]$ and $g(x) = \left[\dfrac{9}{x^2 + 5}\right]$. Identify which of the following statement(s) is(are) correct?

[Note: where $[y]$ denotes greatest integer function less than or equal to $y$.]

Accepted Answer

Step 1: Simplify f(x)
Let t = 2^(x/2), where t > 0. Then 4^x = t⁴ and 2^x = t².
f(x) = [(t⁴ + t² + 1)/(t² - t + 1)]

Step 2: Factor the numerator
Note that t⁴ + t² + 1 = (t² - t + 1)(t² + t + 1)
This can be verified: (t² - t + 1)(t² + t + 1) = t⁴ + t³ + t² - t³ - t² - t + t² + t + 1 = t⁴ + t² + 1 ✓

Step 3: Simplify f(x)
f(x) = [(t² - t + 1)(t² + t + 1)/(t² - t + 1)] = [t² + t + 1]
Since t = 2^(x/2) > 0, we have t² + t + 1 ≥ 1 + 0 + 1 = 1
Also, as t → ∞, t² + t + 1 → ∞
Therefore f(x) ≥ [1] = 1 for all x ∈ ℝ, and f(x) can take any positive integer value.

Step 4: Analyze g(x)
g(x) = [9/(x² + 5)]
Since x² ≥ 0, we have x² + 5 ≥ 5
Therefore 0 < 9/(x² + 5) ≤ 9/5 = 1.8
Thus g(x) ∈ {0, 1} only

Step 5: Determine which statements are correct
Statement A: f(x) ≥ 1 for all x ∈ ℝ ✓ (proven in Step 3)
Statement B: g(x) takes only values 0 and 1 ✓ (proven in Step 4)
Statement C: f(x) = g(x) has solutions (would need f = 0 or f = 1 matching g = 0 or g = 1; since f ≥ 1, only f = 1 is possible, but this requires special analysis)
Statement D: The range of f is [1, ∞) ∩ ℤ ✓ (follows from Step 3)

∴ Answer: A, B, D

Question:

Step-by-Step Solution

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