Start free

Mathematics · Functions

JEE Main 2026 — 4 April, Evening Shift — Question 25

For the function f:[1,∞)→[1,∞)f:[1,\infty)\to [1,\infty) defined by f(x)=(x−1)4+1f(x) = (x - 1)^4 + 1 , among the two statements :

(I) The set S={x∈[1,∞):f(x)=f−1(x)}S = \{x\in [1,\infty):f(x) = f^{-1}(x)\} contains exactly two elements, and

(II) The set S={x∈[1,∞):f(x)=f−1(x+1)}S = \{x\in [1,\infty):f(x) = f^{-1}(x + 1)\} is an empty set,

  1. Option A:

    only (I) is TRUE

    Correct
  2. Option B:

    only (II) is TRUE

  3. Option C:

    both (I) and (II) are TRUE

  4. Option D:

    neither (I) nor (II) is TRUE

Answer: A

Step-by-step solution

f(x)=(x−1)4+1f(x)=(x-1)^{4}+1 f′(x)=4(x−1)3;f′(x)≥0\mathrm{f}^{\prime}(\mathrm{x})=4(\mathrm{x}-1)^{3} ; \mathrm{f}^{\prime}(\mathrm{x}) \geq 0 ∴f(x)\therefore \mathrm{f}(\mathrm{x}) is increasing ∴(x−1)4+1=x\therefore(\mathrm{x}-1)^{4}+1=\mathrm{x} (x−1)[(x−1)3−1]=0(\mathrm{x}-1)\left[(\mathrm{x}-1)^{3}-1\right]=0 x=1,x=2\mathrm{x}=1, \mathrm{x}=2 are two solutions Now f−1(x)=(x−1)1/4+1\mathrm{f}^{-1}(\mathrm{x})=(\mathrm{x}-1)^{1 / 4}+1 f−1(x+1)=x1/4+1\mathrm{f}^{-1}(\mathrm{x}+1)=\mathrm{x}^{1 / 4}+1 f−1(x+1)=f(x)\mathrm{f}^{-1}(\mathrm{x}+1)=\mathrm{f}(\mathrm{x}) x1/4+1=(x−1)4+1\mathrm{x}^{1 / 4}+1=(\mathrm{x}-1)^{4}+1 x=(x−1)16\mathrm{x}=(\mathrm{x}-1)^{16} Above equation has one solution using graph.

Solution figure

Answer key and solution verified before publishing.

Practise Functions

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2026
Paper
4 April, Evening Shift
Subject
Mathematics
Chapter
Functions
Topic
Inverse of a Function

More Functions questions from this paper

For the function f:[1,∞)to [1,∞) defined by f(x) = (x - 1) 4 + 1 … | JEE Main 2026 PYQ with Solution · DhiX AI