Limits, Continuity & Differentiability Question - None

Question

Let $f: R \to R$ and $g: (-2, 2) \to R$ be two functions defined by $f(x) = \max(|1-|x||, x^3+1)$ and $g(x) = [f(x)]$. Identify which of the following statement(s) is(are) correct?
[Note: $[y]$ denotes greatest integer function less than or equal to $y$.]

Accepted Answer

Step 1: Analyze |1-|x||

For |x| ≤ 1: |1-|x|| = 1-|x| (decreasing from 1 to 0)

For |x| > 1: |1-|x|| = |x|-1 (increasing)

Step 2: Find intersection of 1-|x| and x³+1

For x ≥ 0: At x=0, 1-0=1 and 0+1=1 ✓ (intersection at origin)

For x ∈ [0,1]: 1-x = x³+1 gives x³+x=0, so x=0 only

For x > 1: x-1 = x³+1 gives x³-x+2=0. Testing x=-1: -1+1+2=2≠0. For large x, x³ dominates so intersection exists around x≈1.5

Step 3: Determine f(x) structure

• At x=0: f(0)=max(1,1)=1

• For x ∈ (0,1): f(x)=1-x (since 1-x > x³+1 in this interval)

• For x ≥ 1: f(x)=x³+1 or |x|-1 depending on which is larger

• By symmetry: f is even function

Step 4: Construct g(x)=[f(x)]

For x ∈ (-1,0): f(x) ∈ (0,1), so g(x)=0

For x=0: f(0)=1, so g(0)=1

For x ∈ (0,1): f(x)=1-x ∈ (0,1), so g(x)=0

For x ∈ [1,2): f(x) ∈ [2,9), so g(x) ∈ {2,3,...,8}

Step 5: Verify continuity

• f(x) is continuous everywhere (max of continuous functions)

• g(x) is discontinuous at points where f(x) is an integer (x=0, and points in [1,2) where x³+1 equals integers)

• g(x) is continuous on open intervals where f(x) avoids integer values

∴ Answer: ACD

Question:

Step-by-Step Solution

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