Question

The inverse function of $f(x) = ax + 1$, $x \in \mathbb{R}$ is itself. Then the value of $a$ is

• A. 1
• B. -1
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$
• E. 2

Hint:

Logic Tip: Graphically, a function that is its own inverse is symmetric about the line $y = x$. For a linear function $y = ax + b$ to be symmetric about $y = x$ (and not be the line $y=x$ itself), its slope $a$ must be $-1$.

Answer 1

The Correct Option is B

Concept:
If a function is its own inverse, it means that $f^{-1}(x) = f(x)$. By definition of inverse functions, substituting $f(x)$ into itself must return $x$. Therefore: $$f(f(x)) = x$$
Step 1: Formulate the composite function $f(f(x))$.
Given $f(x) = ax + 1$, we substitute $f(x)$ in place of $x$: $$f(f(x)) = a(ax + 1) + 1$$ $$f(f(x)) = a^2x + a + 1$$
Step 2: Apply the self-inverse condition.
Since $f(x)$ is its own inverse, we equate the composite function to $x$: $$a^2x + a + 1 = x$$
Step 3: Compare coefficients to solve for a.
For the equation $a^2x + (a + 1) = 1x + 0$ to hold true for all real values of $x$, the coefficients of corresponding powers of $x$ must be equal. Equating the $x$ coefficients: $$a^2 = 1 \implies a = 1 \text{ or } a = -1$$ Equating the constant terms: $$a + 1 = 0 \implies a = -1$$ The only value that satisfies both conditions is $a = -1$.