If [x]2+[x]-2

Question

If [x]2+[x]-2<0 and x =(1/2), then the number of possible real values of x, is [Note: [ x ] and x denote the greatest integer less than or equal to x and fractional part of x respectively.] (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

[x]2+[x]<2 ⇒([x]+2)([x]-1)<0 ⇒-2<[x]<1 ∴[ x ]=-1,0 ⇒ x ∈[-1,1) So, x =(1/2) ⇒ x=(-1/2), (1/2) ]

Q. If $[x]^{2} + [x] - 2 < 0$ and ${x} = \frac{1}{2}$ , then the number of possible real values of $x$ , is
[Note: $[x]$ and ${x}$ denote the greatest integer less than or equal to $x$ and fractional part of $x$ respectively.]

1

2

3

4

$[x]^{2} + [x] < 2 \Rightarrow ([x] + 2) ([x] - 1) < 0 \Rightarrow - 2 < [x] < 1$
$∴ [x] = - 1, 0 \Rightarrow x \in [- 1, 1)$
So, ${x} = \frac{1}{2} \Rightarrow x = \frac{- 1}{2}, \frac{1}{2}$ ]

Q. If [x]2+[x]−2<0 and {x}=21​, then the number of possible real values of x, is [Note: [x] and {x} denote the greatest integer less than or equal to x and fractional part of x respectively.]

1

2

3

4

[x]2+[x]<2⇒([x]+2)([x]−1)<0⇒−2<[x]<1 ∴[x]=−1,0⇒x∈[−1,1) So, {x}=21​⇒x=2−1​,21​ ]

Q. If $[x]^{2} + [x] - 2 < 0$ and ${x} = \frac{1}{2}$ , then the number of possible real values of $x$ , is
[Note: $[x]$ and ${x}$ denote the greatest integer less than or equal to $x$ and fractional part of $x$ respectively.]

$[x]^{2} + [x] < 2 \Rightarrow ([x] + 2) ([x] - 1) < 0 \Rightarrow - 2 < [x] < 1$
$∴ [x] = - 1, 0 \Rightarrow x \in [- 1, 1)$
So, ${x} = \frac{1}{2} \Rightarrow x = \frac{- 1}{2}, \frac{1}{2}$ ]