Let $$R$$ = the set of real numbers, $$Z$$ = the set of integers, $$N$$ = the set of
natural numbers. If $$S$$ be the solution set of the equation $${\left( x \right)^2} + {\left[ x \right]^2} = {\left( {x - 1} \right)^2} + {\left[ {x + 1} \right]^2},$$ where $$(x)$$ = the least integer greater than or equal to $$x$$ and $$[x]$$ = the greatest integer less than or equal to $$x,$$ then
A.
$$S = R$$
B.
$$S = R - Z$$
C.
$$S = R - N$$
D.
none of these
Answer :
$$S = R - Z$$
Solution :
$$\eqalign{
& {\left\{ {\left( {x - 1} \right)} \right\}^2} = {\left\{ {\left( x \right) - 1} \right\}^2} = {\left( x \right)^2} - 2\left( x \right) + 1 \cr
& {\left\{ {\left[ {x + 1} \right]} \right\}^2} = {\left\{ {\left[ x \right] + 1} \right\}^2} = {\left[ x \right]^2} + 2\left[ x \right] + 1 \cr
& \therefore \,\,{\text{equation gives }}\left[ x \right] - \left( x \right) + 1 = 0. \cr
& {\text{If }}x = n \in Z,n - n + 1 = 0\left( {{\text{absurd}}} \right). \cr
& {\text{If }}x = n + k,n \in Z,0 < k < 1\,\,{\text{then }}n - \left\{ {n + 1} \right\} + 1 = 0. \cr} $$
This is true for all $$n$$ and $$k.$$
Releted MCQ Question on Algebra >> Quadratic Equation
Releted Question 1
If $$\ell ,m,n$$ are real, $$\ell \ne m,$$ then the roots by the equation: $$\left( {\ell - m} \right){x^2} - 5\left( {\ell + m} \right)x - 2\left( {\ell - m} \right) = 0$$ are