Question

The set of all values of $a$ for which $\displaystyle\lim _{x \rightarrow a}([x-5]-[2 x+2])=0$,where $[\propto]$ denotes the greatest integer less than or equal to $\propto$ is equal to

• A. $[-7.5,-6.5)$
• B. $(-7.5,-6.5)$
• C. $(-7.5,-6.5]$
• D. $[-7.5,-6.5]$

Answer 1

The Correct Option is B

The correct answer is (B) : $(-7.5,-6.5)$
$\underset{x\to a}{\lim }([x-5]-[2x+2])=0$
$\underset{x\to a}{\lim }([x]-5-[2x]-2)=0$
$\underset{x\to a}{\lim }([x]-[2x])=7$
$[a]-[2a]=7$
$a\in I,a=-7$
$a\notin I,a=I+f$
Now, $[a]-[2a]=7$
$-I-[2f]=7$
Case-I: $f\in \left(0,\frac{1}{2}\right)$
$2f\in (0,1)$
$-I=7$
$I=-7\Rightarrow a\in (-7,-6.5)$
Case-II: $f\in \left(\frac{1}{2},1\right)$
$2f\in (1,2)$
$-I-1=7$
$I=-8\Rightarrow a\in (-7.5,-7)$
Hence, $a\in (-7.5,-6.5)$

Answer 2

Given: \[ \lim_{{x \to a}} \left( \left\lfloor x - 5 \right\rfloor - \left\lfloor 2x + 2 \right\rfloor \right) = 0. \] 

Step 1: Analyze the limits of \(\left\lfloor x - 5 \right\rfloor\) and \(\left\lfloor 2x + 2 \right\rfloor\) 

The greatest integer function \(\left\lfloor x \right\rfloor\) satisfies:

\[ \left\lfloor x \right\rfloor \leq x < \left\lfloor x \right\rfloor + 1. \]

At \(x = a\), the limit becomes:

\[ \left\lfloor a - 5 \right\rfloor - \left\lfloor 2a + 2 \right\rfloor = 0 \quad \Rightarrow \quad \left\lfloor a - 5 \right\rfloor = \left\lfloor 2a + 2 \right\rfloor. \] 

Step 2: Define cases based on the equality

1. Let \(\left\lfloor a - 5 \right\rfloor = k\), where \(k\) is an integer. Then:

\[ k \leq a - 5 < k + 1 \quad \Rightarrow \quad k + 5 \leq a < k + 6. \]

2. Similarly, \(\left\lfloor 2a + 2 \right\rfloor = k\) gives:

\[ k \leq 2a + 2 < k + 1 \quad \Rightarrow \quad \frac{k - 2}{2} \leq a < \frac{k - 1}{2}. \] 

Step 3: Solve for intersection

For \(k = -7\):

\[ -7 + 5 \leq a < -7 + 6 \quad \Rightarrow \quad -2 \leq a < -1. \]

For \(k = -6\):

\[ a \in \left(-7.5, -6.5\right). \]