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: Understanding the Concept:
We are looking for values of $x \geq 0$ where the exponential function $2^x$ equals the linear function $x + 1$.
Step 2: Key Formula or Approach:
1. Check small integer values by substitution.
2. Use properties of convexity (second derivative) to determine the maximum number of possible roots.
Step 3: Detailed Explanation:
Let $f(x) = 2^x - x - 1$.
1. For $x = 0$: $f(0) = 2^0 - 0 - 1 = 1 - 1 = 0$. (Root 1)
2. For $x = 1$: $f(1) = 2^1 - 1 - 1 = 2 - 2 = 0$. (Root 2)
Now we analyze the derivative:
$f'(x) = 2^x \ln 2 - 1$.
$f''(x) = 2^x (\ln 2)^2$.
Since $f''(x)>0$ for all real $x$, $f(x)$ is a strictly convex function.
A strictly convex function can have at most two real roots. Since we have found $x = 0$ and $x = 1$, no other roots exist in the real number system.
Step 4: Final Answer:
The number of non-negative real roots is 2.