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f(x)=(x-1)^4+1 Inverse Statements: Only (I) True | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
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, consider 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.
Aonly (I) is TRUEcorrect
Bonly (II) is TRUE
Cboth (I) and (II) are TRUE
Dneither (I) nor (II) is TRUE
Solution
Step 1: f′(x)=4(x−1)3≥0f'(x)=4(x-1)^3\ge0 on [1,∞)[1,\infty), so ff is increasing. For an increasing function, f(x)=f−1(x)f(x)=f^{-1}(x) is equivalent to f(x)=xf(x)=x. Step 2: Solve f(x)=xf(x)=x: (x−1)4+1=x⇒(x−1)4−(x−1)=0⇒(x−1)[(x−1)3−1]=0(x-1)^4+1=x\Rightarrow (x-1)^4-(x-1)=0\Rightarrow (x-1)\big[(x-1)^3-1\big]=0, giving x=1x=1 and x=2x=2. So statement (I) has exactly two elements — (I) is TRUE. Step 3: Here f−1(x)=(x−1)1/4+1f^{-1}(x)=(x-1)^{1/4}+1, so f−1(x+1)=x1/4+1f^{-1}(x+1)=x^{1/4}+1. Set f−1(x+1)=f(x)f^{-1}(x+1)=f(x):
x1/4+1=(x−1)4+1 ⇒ x1/4=(x−1)4 ⇒ x=(x−1)16.x^{1/4}+1=(x-1)^4+1\ \Rightarrow\ x^{1/4}=(x-1)^4\ \Rightarrow\ x=(x-1)^{16}.
 From the graph this equation has one solution, so the set is NOT empty — (II) is FALSE. Step 4: Therefore only (I) is TRUE.
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