Question

let {x}and [ X] denotes the fractional and integral part of a real number x respectively 5hen no.of solutions of 4{x}= x+{x}​

Answer 1

The only solution is x=0

4[x]=x+{x} where 0≤{x}<1 and [x} belongs to natural numbers and [x]≥0.

Simplifying, the original problem reduces to :—

3[x]=2{x}

putting the restrictions in place, 3[x]>2{x}, for x>0.

when x=0 3[x]=2{x}.

I hope it helps.

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