JEE Main 2026 — Functions Question with Solution

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

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

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

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

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

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

Let $t=x-1\ge 0$.

$t^4-t=0\Rightarrow t(t-1)(t^2+t+1)=0$.

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

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

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

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

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

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

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

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

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

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

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

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

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

Therefore, only (I) is TRUE.

Answer: only (I) is TRUE