JEE Main 2026 Apr 4 Shift 2, Mathematics Q1: Invertible And Composite Functions

JEE Main2026Apr 4, Shift 2Mathematics

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,

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

Solution

$f^{'} (x) = 4 (x - 1)^{3} \geq 0$ on $[1, \infty)$ , so $f$ is increasing. For an increasing function, $f (x) = f^{- 1} (x)$ reduces to $f (x) = x$ .

Statement I: Solve $(x - 1)^{4} + 1 = x \Rightarrow (x - 1) [(x - 1)^{3} - 1] = 0$ . The factors give $x = 1$ and $x = 2$ , so $S$ has exactly two elements. True.

Statement II: $f^{- 1} (y) = (y - 1)^{1/4} + 1$ , so $f^{- 1} (x + 1) = x^{1/4} + 1$ . Set $f (x) = f^{- 1} (x + 1)$ :

$(x - 1)^{4} + 1 = x^{1/4} + 1 \Rightarrow (x - 1)^{4} = x^{1/4} \Rightarrow x = (x - 1)^{16}$ .

Graphically the LHS line $y = x$ and the RHS curve $y = (x - 1)^{16}$ intersect at exactly one point in $[1, \infty)$ , so $S$ is non-empty. False.

$∴$ Only (I) is TRUE.

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JEE Main 2026 Apr 4 Shift 2, Mathematics Q1: Invertible And Composite Functions

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