Solution

The correct answer is D, C, B, A

Key Points

To determine the least integer ( n ) such that f(x) is \(O(x^n)\) for each function, we need to analyze the growth of each function as \(x \to \infty\).

Let's examine each function:

A. \(f(x) = 3x^3 + (\log x)^4\)

• The term \(3x^3\) dominates as \(x \to \infty\) since \((\log x)^4 \) grows much slower than \(3x^3\). Therefore, f(x) is dominated by \(3x^3\) and f(x) is \(O(x^3)\).
• n = 3

B. \(f(x) = \frac{x^5 + x^2 + 1}{x^3 + 1}\)

• To find the dominant term, we note that as \(x \to \infty\):
• x⁵ in the numerator dominates x² + 1.
• x³ in the denominator dominates 1.
• Thus, \(f(x) \sim \frac{x^5}{x^3} = x^2\)
• So, f(x) is O(x²).
• n = 2

C. \(f(x) = \frac{x^3 + x^2 + 1}{x^2 + 1}\)

• Similarly, as \( x \to \infty\):
• x³ in the numerator dominates x² + 1.
• x² in the denominator dominates 1.
• Thus, \( f(x) \sim \frac{x^3}{x^2} = x\)
• So, f(x) is O(x).
• n = 1

D. \(f(x) = \frac{x^2 + 5 \log x}{x^2 + 1}\)

• As \(x \to \infty\):
• x² in the numerator dominates \(5 \log x\).
• x² in the denominator dominates 1.
• Thus, \(f(x) \sim \frac{x^2}{x^2} = 1\)
• So, f(x) is O(1).
• n = 0

Now, we arrange the functions in increasing order of ( n ):

• ( D: n = 0 )
• ( C: n = 1 )
• ( B: n = 2 )
• ( A: n = 3 )

Thus, the correct answer is: D, C, B, A
The integer ( n ) ordering is correct for the first option according to the dominance factors determined.