JEE Mathematics 2023 Question with Solution

Given:

$$\lim_{x \to a} \left([x-5]-[2x+2]\right)=0$$

Find: the set of all values of $a$.

Using the property of greatest integer function shown in the extracted solution,

$$[x-5] = [x]-5, \qquad [2x+2] = [2x]+2$$

So the condition becomes

$$\lim_{x \to a} \left([x]-[2x]\right)=7$$

Hence,

$$[a]-[2a]=7$$

Let

$$a=I+f$$

where $I \in \mathbb{Z}$ and $f \in [0,1)$. Then

$$[a]=I, \qquad [2a]=[2I+2f]=2I+[2f]$$

Therefore,

$$I-(2I+[2f])=7$$ $$-I-[2f]=7$$

Now consider cases for $f$.

Case I: $f \in (0, \tfrac{1}{2})$, so $[2f]=0$. Then

$$-I=7$$ $$I=-7$$

Thus,

$$a \in (-7,-6.5)$$

Case II: $f \in (\tfrac{1}{2},1)$, so $[2f]=1$. Then

$$-I-1=7$$ $$I=-8$$

Thus,

$$a \in (-7.5,-7)$$

Combining both cases,

$$a \in (-7.5,-6.5)$$

The extracted solution concludes this interval, which corresponds to option B in the given options list. There is a discrepancy because the option texts place $(-7.5,-6.5)$ as A, but the solution explicitly states The Correct Option is B. Therefore, following the solution as authority, the correct option is B.](streamdown:incomplete-link)

Given:

$$\lim_{x \to a} \left([x-5]-[2x+2]\right)=0$$

Find: all possible values of $a$.

Rewrite the expression:

$$[x-5]-[2x+2]=([x]-5)-([2x]+2)=[x]-[2x]-7$$

So for the limit to be $0$,

$$\lim_{x \to a}([x]-[2x])=7$$

Hence the extracted working uses

$$[a]-[2a]=7$$

Write

$$a=I+f, \qquad I \in \mathbb{Z}, \quad 0 \le f < 1$$

Then

$$[a]=I$$

and

$$[2a]=[2I+2f]=2I+[2f]$$

Thus,

$$[a]-[2a]=I-(2I+[2f])=-I-[2f]$$

Given this equals $7$,

$$-I-[2f]=7$$

Now $[2f]$ can only be $0$ or $1$.

If

$$0$$