Question

Let {x} and [x] denotes the fractional and integral part of a real number x respectively. Then number of solutions of 4 {x} = x+ {x}. is_​

Answer 1

it is given that 4{x} = x + [x] , where {x} is fractional part of x and [x] is greatest integer.

we have to find the number of solutions.

4{x} = x + [x]

we know, [x] + {x} = x ⇒{x} = x - [x]

4(x - [x]) = x + [x]

⇒4x - 4[x] = x + [x]

⇒3x = 5[x]

⇒x = 5/3 [x]..........(1)

we know, [x] + {x} = x

so, [x] + {x} = 5/3[x]

⇒{x} = 5/3 [x] - [x] = 2/3 [x]

we know, 0 ≤ {x} < 1

0 ≤ 2/3 [x] < 1

⇒0 ≤ [x] < 3/2

when [x] = 0 ⇒0 ≤ x < 1

when [x] = 1 ⇒1 ≤ x < 3/2 for [x] ∈ [0, 3/2)

putting, [x] = 0 and 1 in equation (1) we get,

x = 0, 5/3

therefore number of solutions = 2