JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Sets Relations and Functions

On $[1,\infty )$, $f(x)=(x-1{)}^4+1$ has $f^{\prime }(x)=4(x-1{)}^3\ge 0$, so $f$ is increasing and invertible onto $[1,\infty )$. For an increasing function, $f(x)=f^{-1}(x)$ holds only where $f(x)=x$; here $(x-1{)}^4+1=x$ means $(x-1{)}^4=x-1$, so with $t=x-1\ge 0$, $t^4=t$ gives $t=0$ or $t=1$, i.e. $x=1$ and $x=2$ — exactly two elements, so statement (I) is true. For (II), $f^{-1}(x)=(x-1{)}^{1/4}+1$, so $f^{-1}(x+1)=x^{1/4}+1$; the equation $f(x)=f^{-1}(x+1)$ becomes $(x-1{)}^4=x^{1/4}$, and raising to the fourth power, $(x-1{)}^{16}=x$. With $t=x-1\ge 0$ the continuous function $t^{16}-(t+1)$ is negative at $t=1$ ($1-2<0$) and positive at $t=1.1$, so it has a root; the solution set is non-empty, making statement (II) false. Hence only (I) is true.