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Question: The number of solutions of the equation \({x^3} - \left[ x \right] = 3\) is (where \(\left[ . \right...

The number of solutions of the equation x3−[x]=3{x^3} - \left[ x \right] = 3 is (where [.]\left[ . \right] represents the greatest integer function)
A) 0
B) 1
C) 2
D) 3

Explanation

Solution

First assume x=[x]+fx = \left[ x \right] + f and substitute the value of [x]\left[ x \right] in the equation. Then write the equation (x3−x)\left( {{x^3} - x} \right) in the form of less than and greater than the value of the equation. Then find the value of (x3−x)\left( {{x^3} - x} \right) by substituting different values of xx and check whether it satisfies the condition. It will give the value of [x]\left[ x \right]. Substitute it in the original equation and find the number of solutions.

Complete step-by-step answer:
Given:- x3−[x]=3{x^3} - \left[ x \right] = 3 ….. (1)
Let x=[x]+fx = \left[ x \right] + f, where [x]\left[ x \right] is the greatest integer function and f is a fraction.
Then,
[x]=x−f\left[ x \right] = x - f
Substitute the value of [x]\left[ x \right] in equation (1),
x3−(x−f)=3{x^3} - \left( {x - f} \right) = 3
Open the brackets and change the sign accordingly,
x3−x+f=3{x^3} - x + f = 3
Move f to the other side of the equation,
x3−x=3−f{x^3} - x = 3 - f
It can also be written as,
2<(x3−x)<32 < \left( {{x^3} - x} \right) < 3 ….. (2)
Substitute the different value of xx to check for what value of xx, the value of the function lies between 2 and 3.
Let x=−2x = - 2,
x3−x=(−2)3−(−2){x^3} - x = {\left( { - 2} \right)^3} - \left( { - 2} \right)
Open the brackets and change the sign accordingly,
x3−x=−8+2{x^3} - x = - 8 + 2
Add the like terms,
x3−x=−6{x^3} - x = - 6
Let x=−1x = - 1,
x3−x=(−1)3−(−1){x^3} - x = {\left( { - 1} \right)^3} - \left( { - 1} \right)
Open the brackets and change the sign accordingly,
x3−x=−1+1{x^3} - x = - 1 + 1
Add the like terms,
x3−x=0{x^3} - x = 0
Let x=0x = 0,
x3−x=(0)3−(0){x^3} - x = {\left( 0 \right)^3} - \left( 0 \right)
Add the like terms,
x3−x=0{x^3} - x = 0
Let x=1x = 1,
x3−x=(1)3−(1){x^3} - x = {\left( 1 \right)^3} - \left( 1 \right)
Open the brackets and change the sign accordingly,
x3−x=1−1{x^3} - x = 1 - 1
Add the like terms,
x3−x=0{x^3} - x = 0
Let x=2x = 2,
x3−x=(2)3−(2){x^3} - x = {\left( 2 \right)^3} - \left( 2 \right)
Open the brackets and change the sign accordingly,
x3−x=8−2{x^3} - x = 8 - 2
Subtract 2 from 6,
x3−x=6{x^3} - x = 6
The condition in equation (2) is satisfied if
1<x<21 < x < 2.
So,
[x]=1\left[ x \right] = 1
Substitute the value of [x]\left[ x \right] in equation (1),
x3−1=3{x^3} - 1 = 3
Move 1 on the right side and add,
x3=4{x^3} = 4
Take cube root on both sides,
x=43x = \sqrt[3]{4}
Thus, there is only one solution.

Hence, option (b) is the correct answer.

Note: The greatest integer function is a piecewise defined function. If the number is an integer, use that integer. If the number is not an integer, use the next smaller integer.
It is also known as the floor of an integer.
If n⩽X<n+1n \leqslant X < n + 1. Then, [X]=n\left[ X \right] = n. It means if X lies in [n,n+1)\left[ {n,n + 1} \right) then the Greatest Integer Function of X will be n.