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MISCELLANEOUS EXERCISE 6 (II) · Q191

Q.Solve the following for xx, where ∣x∣|x| is the modulus function, [x][x] is the greatest integer function, {x}\{x\} is the fractional part function: [x−2]+[x+2]+{x}=0[x-2]+[x+2]+\{x\}=0.

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Using [x+n]=[x]+n[x+n]=[x]+n for integer nn: [x−2]=[x]−2[x-2]=[x]-2 and [x+2]=[x]+2[x+2]=[x]+2, so [x−2]+[x+2]=([x]−2)+([x]+2)=2[x][x-2]+[x+2]=([x]-2)+([x]+2)=2[x].

The equation becomes 2[x]+{x}=02[x]+\{x\}=0.

Write x=[x]+{x}x=[x]+\{x\} with 0≤{x}<10\le\{x\}<1. From 2[x]+{x}=02[x]+\{x\}=0: {x}=−2[x]\{x\}=-2[x]. …

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