Question

The value of the greatest integer \(k\) satisfying the inequation \(2^{n+4} + 12 \ge k(n + 4)\) for all \(n \in N\) is

• A. \(7\)
• B. \(8\)
• C. \(9\)
• D. \(10\)

Hint:

When dealing with mixed exponential and polynomial functions evaluated over natural numbers, calculating the first few terms often quickly reveals whether the function is strictly increasing or decreasing.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given an inequality that must hold for all natural numbers \(n \in \mathbb{N}\) (i.e., \(n = 1, 2, 3, \dots\)). We need to find the greatest integer \(k\) that makes this inequality true universally.
Step 2: Key Formula or Approach:
Rewrite the inequality to isolate \(k\): \[ k \le \frac{2^{n+4} + 12}{n+4} \] To find the greatest possible integer \(k\), we must find the minimum value of the sequence \(f(n) = \frac{2^{n+4} + 12}{n+4}\) for \(n \ge 1\).
Step 3: Detailed Explanation:
Let's test the first few natural numbers:
For \(n = 1\): \[ f(1) = \frac{2^{1+4} + 12}{1+4} = \frac{32 + 12}{5} = \frac{44}{5} = 8.8 \] For \(n = 2\): \[ f(2) = \frac{2^{2+4} + 12}{2+4} = \frac{64 + 12}{6} = \frac{76}{6} \approx 12.67 \] For \(n = 3\): \[ f(3) = \frac{2^{3+4} + 12}{3+4} = \frac{128 + 12}{7} = \frac{140}{7} = 20 \] As \(n\) increases, the numerator (\(2^{n+4}\)) grows exponentially while the denominator (\(n+4\)) only grows linearly. Thus, the function \(f(n)\) is strictly increasing for \(n \ge 1\).
The minimum value occurs at \(n = 1\), giving a value of \(8.8\).
For the inequality to hold for all \(n\), we must have: \[ k \le 8.8 \] The greatest integer satisfying this condition is \(k = 8\).
Step 4: Final Answer:
The greatest integer \(k\) is \(8\).