A student is searching for a solution to the equation $f(x) = 0$ - AQA - A-Level Maths Mechanics - Question 2 - 2020 - Paper 1

To determine which function leads to an incorrect conclusion about the existence of a root between -1 and 1, we need to analyze each function:

1. For $f(x) = \frac{1}{x}$:

   • At $f(-1) = -1$ and $f(1) = 1$
   • This function is undefined at $x = 0$ and indeed has no root in the interval (-1, 1).

2. For $f(x) = x$:

   • At $f(-1) = -1$ and $f(1) = 1$; it changes sign and has a root at $x = 0$.

3. For $f(x) = x^3$:

   • At $f(-1) = -1$ and $f(1) = 1$; it changes sign and has a root at $x = 0$.

4. For $f(x) = \frac{2x + 1}{x + 2}$:

   • This function also has $f(-1) = 0$ and $f(1) = \frac{3}{3} = 1$, indicating a continuous change in sign.

Since only $f(x) = \frac{1}{x}$ does not guarantee a root in the interval, the correct answer to circle is: $f(x) = \frac{1}{x}$.