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Functions — JEE Maths practice question

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

Question
The range of the function y=[x2]−[x]2y=[x^{2}]-[x]^{2}, x∈[0,2]x\in[0,2] (where [⋅][\cdot] denotes the integral part) is:
A{0}\{0\}
B{0,1}\{0,1\}
C{1,2}\{1,2\}
D{0,1,2}\{0,1,2\}correct
Solution
Step 1: Evaluate y=[x2]−[x]2y=[x^{2}]-[x]^{2} on subintervals of [0,2][0,2]: For 0≤x<10\le x<1: x2∈[0,1)x^{2}\in[0,1), so [x2]=0[x^{2}]=0; [x]=0[x]=0. Thus y=0−0=0y=0-0=0. For 1≤x<21\le x<\sqrt{2}: x2∈[1,2)x^{2}\in[1,2), so [x2]=1[x^{2}]=1; [x]=1[x]=1. Thus y=1−1=0y=1-1=0. For 2≤x<3\sqrt{2}\le x<\sqrt{3}: x2∈[2,3)x^{2}\in[2,3), so [x2]=2[x^{2}]=2; [x]=1[x]=1. Thus y=2−1=1y=2-1=1. For 3≤x<2\sqrt{3}\le x<2: x2∈[3,4)x^{2}\in[3,4), so [x2]=3[x^{2}]=3; [x]=1[x]=1. Thus y=3−1=2y=3-1=2. At x=2x=2: [x2]=[4]=4[x^{2}]=[4]=4; [x]=2[x]=2. Thus y=4−4=0y=4-4=0. Step 2: Values taken: 0,1,20,1,2. Range ={0,1,2}=\{0,1,2\}. Answer: (4) {0,1,2}\{0,1,2\}
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