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

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q1 – JEE-MAIN 2026

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Home Maths Relations and FunctionsQuestion #363

For the function

f : [1, \infty) \to [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,

Clue 1 of 3: Inverse Functions and Monotonicity

For a strictly increasing function $f (x)$ , the equation $f (x) = f^{- 1} (x)$ is equivalent to $f (x) = x$ .

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Step 1: Find the inverse function and analyze statement (I)✦ Active

The function is $f (x) = (x - 1)^{4} + 1$ for $x \in [1, \infty)$ . To find $f^{- 1} (x)$ , let $y = (x - 1)^{4} + 1$ . This implies $y - 1 = (x - 1)^{4}$ . Since $x \geq 1$ , $x - 1 \geq 0$ , so we take the positive fourth root: $x - 1 = \sqrt[4]{y - 1}$ . Thus, $x = 1 + \sqrt[4]{y - 1}$ . So, $f^{- 1} (x) = 1 + \sqrt[4]{x - 1}$ .

For statement (I), we solve $f (x) = f^{- 1} (x)$ . Since $f (x)$ is strictly increasing on $[1, \infty)$ (as $f^{'} (x) = 4 (x - 1)^{3} \geq 0$ for $x \geq 1$ ), this is equivalent to solving $f (x) = x$ .

(x - 1)^{4} + 1 = x

Let $u = x - 1$ . Then $x = u + 1$ . Substituting this into the equation gives:

u^{4} + 1 = u + 1 ⟹ u^{4} - u = 0 ⟹ u (u^{3} - 1) = 0

This yields $u = 0$ or $u^{3} = 1 ⟹ u = 1$ . Substituting back $u = x - 1$ :

If $u = 0$ , then $x - 1 = 0 ⟹ x = 1$ . If $u = 1$ , then $x - 1 = 1 ⟹ x = 2$ . Both $x = 1$ and $x = 2$ are in the domain $[1, \infty)$ . Thus, the set $S$ for statement (I) contains exactly two elements ${1, 2}$ . Statement (I) is TRUE.

Step 2: Analyze statement (II)○ Expand

For statement (II), we need to solve $f (x) = f^{- 1} (x + 1)$ for $x \in [1, \infty)$ . We have $f (x) = (x - 1)^{4} + 1$ . Using the inverse function found in Step 1, $f^{- 1} (x + 1) = 1 + \sqrt[4]{(x + 1) - 1} = 1 + \sqrt[4]{x}$ . So, the equation becomes:

(x - 1)^{4} + 1 = 1 + \sqrt[4]{x} ⟹ (x - 1)^{4} = \sqrt[4]{x}

Let $g (x) = (x - 1)^{4} - \sqrt[4]{x}$ . We need to check if $g (x) = 0$ has any solutions in $[1, \infty)$ . Let's evaluate $g (x)$ at a few points:

$g (1) = (1 - 1)^{4} - \sqrt[4]{1} = 0 - 1 = - 1$ . $g (2) = (2 - 1)^{4} - \sqrt[4]{2} = 1 - \sqrt[4]{2} \approx 1 - 1.189 = - 0.189$ . $g (3) = (3 - 1)^{4} - \sqrt[4]{3} = 2^{4} - \sqrt[4]{3} = 16 - 1.316 \approx 14.684$ .

Since $g (x)$ is continuous on $[1, \infty)$ , and $g (2) < 0$ and $g (3) > 0$ , by the Intermediate Value Theorem, there must be at least one root in the interval $(2, 3)$ . Therefore, the set $S$ for statement (II) is not an empty set. Statement (II) is FALSE.

Step 3: Conclude based on the truth values of statements (I) and (II)○ Expand

Statement (I) is TRUE. Statement (II) is FALSE. Therefore, only statement (I) is TRUE.

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Maths Question 1 – JEE-MAIN 2026

Clue 1 of 3: Inverse Functions and Monotonicity

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