Question

For the function \(f: [1, \infty) \rightarrow [1, \infty)\) defined by \(f(x) = (x - 1)^4 + 1\), among the two statements:
(I) The set \(S = \{x \in [1, \infty) : f(x) = f^{-1}(x)\}\) contains exactly two elements, and
(II) The set \(S = \{x \in [1, \infty) : f(x) = f^{-1}(x + 1)\}\) is an empty set,
Options:

• A. only (I) is TRUE
• B. only (II) is TRUE
• C. both (I) and (II) are TRUE
• D. neither (I) nor (II) is TRUE

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a strictly increasing function \(f(x)\), the points of intersection of the function and its inverse, \(f(x) = f^{-1}(x)\), always lie on the line \(y = x\). For statement (II), we evaluate the equation by manipulating it to \(f(f(x)) = x+1\) to verify real roots.
Step 2: Key Formula or Approach:
For strictly increasing functions: \[ f(x) = f^{-1}(x) \iff f(x) = x \] For the second statement, apply the inverse function definition: \[ f(x) = f^{-1}(x+1) \implies f(f(x)) = x+1 \] Step 3: Detailed Explanation:
Let's analyze statement (I):
Given \(f(x) = (x - 1)^4 + 1\) for \(x \ge 1\).
Since \(f'(x) = 4(x - 1)^3 \ge 0\) for \(x \ge 1\), the function \(f(x)\) is strictly increasing.
Thus, the equation \(f(x) = f^{-1}(x)\) is equivalent to \(f(x) = x\).
\[ (x - 1)^4 + 1 = x \implies (x - 1)^4 - (x - 1) = 0 \] Let \(t = x - 1 \ge 0\).
\[ t^4 - t = 0 \implies t(t^3 - 1) = 0 \] This gives real roots \(t = 0\) or \(t = 1\).
Substituting back \(x - 1 = t\):
If \(t = 0 \implies x = 1\).
If \(t = 1 \implies x = 2\).
Both values belong to \([1, \infty)\). Thus, set \(S\) has exactly two elements \(\{1, 2\}\). Statement (I) is TRUE.
Let's analyze statement (II):
We are given \(f(x) = f^{-1}(x+1)\).
Let \(y = f(x)\). Then \(y = f^{-1}(x+1) \implies f(y) = x+1\).
\[ (y - 1)^4 + 1 = x + 1 \implies (y - 1)^4 = x \] Substitute \(y = (x-1)^4 + 1\):
\[ \left( (x-1)^4 + 1 - 1 \right)^4 = x \implies (x-1)^{16} = x \] Let \(x - 1 = t \ge 0 \implies x = t + 1\).
\[ t^{16} = t + 1 \implies t^{16} - t - 1 = 0 \] Let \(g(t) = t^{16} - t - 1\).
We evaluate \(g(t)\): \(g(1) = -1<0\) and \(g(2) = 2^{16} - 3>0\).
By the Intermediate Value Theorem, there exists at least one real root in \((1, 2)\).
Therefore, the set \(S\) is not empty. Statement (II) is FALSE.
Step 4: Final Answer:
Only (I) is TRUE.