Question

The number of non-negative real roots of \(2^x - x - 1 = 0\) is:

• A. 1
• B. 2
• C. 3
• D. 0
• E. 4

Hint:

For equations of the form $a^x = x+1$, check the behavior at $x=0$ and $x=1$ and the derivative to understand the shape.

Answer 1

The Correct Option is B

Step 1:
Consider the function \(f(x) = 2^x - x - 1\).

Step 2:
Find \(f(0) = 2^0 - 0 - 1 = 1 - 0 - 1 = 0\). So \(x = 0\) is a root.

Step 3:
Find \(f(1) = 2^1 - 1 - 1 = 2 - 1 - 1 = 0\). So \(x = 1\) is a root.

Step 4:
For \(x > 1\), \(2^x\) grows faster than \(x + 1\), so \(f(x) > 0\).

Step 5:
For \(0 < x < 1\), check \(f(0.5) = \sqrt{2} - 0.5 - 1 \approx 1.414 - 1.5 = -0.086 < 0\). Since \(f(0) = 0\) and \(f(1) = 0\), and \(f(0.5) < 0\), the function dips below zero between 0 and 1 but returns to zero at 1. Hence, no additional roots exist in this interval.

Step 6:
Final Answer: 2