For the function f:[1,∞) → [1,∞) defined by f(x)=(x-1)^4+1, among the two statements: (I) The set S=x ∈ [1,∞): f(x)=f^-1(x) contains exactly two…

Given $f(x) = (x-1)^4 + 1$f(x)=(x−1)4+1 for $x \ge 1$x≥1.

To find the inverse function $f^{-1}(x)$f−1(x), let $y = (x-1)^4 + 1$y=(x−1)4+1.

$\Rightarrow (x-1)^4 = y-1 \Rightarrow x-1 = (y-1)^{1/4} \Rightarrow x = (y-1)^{1/4} + 1$⇒(x−1)4=y−1⇒x−1=(y−1)1/4⇒x=(y−1)1/4+1.

Thus, $f^{-1}(x) = (x-1)^{1/4} + 1$f−1(x)=(x−1)1/4+1.

Evaluating the first statement, since $f'(x) = 4(x-1)^3 \ge 0$f′(x)=4(x−1)3≥0 for $x \ge 1$x≥1, $f(x)$f(x) is strictly increasing. For a strictly increasing function, the solutions to $f(x) = f^{-1}(x)$f(x)=f−1(x) are the same as the solutions to $f(x) = x$f(x)=x.

$(x-1)^4 + 1 = x \Rightarrow (x-1)^4 - (x-1) = 0$(x−1)4+1=x⇒(x−1)4−(x−1)=0.

Let $t = x-1 \ge 0$t=x−1≥0.

$t^4 - t = 0 \Rightarrow t(t-1)(t^2+t+1) = 0$t4−t=0⇒t(t−1)(t2+t+1)=0.

Since $t \ge 0$t≥0, the real roots are $t = 0$t=0 and $t = 1$t=1.

$$\Rightarrow x-1 = 0 \Rightarrow x = 1$$⇒x−1=0⇒x=1

$$\Rightarrow x-1 = 1 \Rightarrow x = 2$$⇒x−1=1⇒x=2

Both $x=1$x=1 and $x=2$x=2 lie in $[1, \infty)$[1,∞). The set contains exactly two elements, making Statement (I) TRUE.

Evaluating the second statement, the equation is $f(x) = f^{-1}(x+1)$f(x)=f−1(x+1).

$f^{-1}(x+1) = (x+1-1)^{1/4} + 1 = x^{1/4} + 1$f−1(x+1)=(x+1−1)1/4+1=x1/4+1.

Equating $f(x)$f(x) and $f^{-1}(x+1)$f−1(x+1):

$(x-1)^4 + 1 = x^{1/4} + 1 \Rightarrow (x-1)^4 - x^{1/4} = 0$(x−1)4+1=x1/4+1⇒(x−1)4−x1/4=0.

Let $g(x) = (x-1)^4 - x^{1/4}$g(x)=(x−1)4−x1/4.

$g(2) = (2-1)^4 - 2^{1/4} = 1 - 2^{1/4} < 0$g(2)=(2−1)4−21/4=1−21/4<0.

$g(3) = (3-1)^4 - 3^{1/4} = 16 - 3^{1/4} > 0$g(3)=(3−1)4−31/4=16−31/4>0.

Since $g(x)$g(x) is a continuous function on $[1, \infty)$[1,∞) and changes sign between $x=2$x=2 and $x=3$x=3, by the Intermediate Value Theorem, there exists at least one real root in the interval $(2, 3)$(2,3).

Thus, the set is not empty, making Statement (II) FALSE.

Therefore, only (I) is TRUE.

Answer: only (I) is TRUE