JEE Main 2023 — Mathematics Calculus

We have,

$$\underset{x\rightarrow a}{\mathrm{lim}}([x-5]-[2x+2])=0$$x→alim​([x−5]−[2x+2])=0

$$\Rightarrow \underset{x\rightarrow a}{\mathrm{lim}}([x]-5-[2x]-2)=0$$⇒x→alim​([x]−5−[2x]−2)=0

$$\Rightarrow \underset{x\rightarrow a}{\mathrm{lim}}([x]-[2x]-7)=0$$⇒x→alim​([x]−[2x]−7)=0

$$\Rightarrow \underset{x\rightarrow a}{\mathrm{lim}}([x]-[2x])=7$$⇒x→alim​([x]−[2x])=7

Now when $a\in I$a∈I (integer) then $a-2a=7\Rightarrow a=-7$a−2a=7⇒a=−7

Now when $a\in I+f$a∈I+f where $f$f is fraction,

So, $[I+f]-[2I+2f]=7$[I+f]−[2I+2f]=7

$$\Rightarrow I+[f]-2I-[2f]=7$$⇒I+[f]−2I−[2f]=7

$$\Rightarrow I+0-2I-[2f]=7$$⇒I+0−2I−[2f]=7

$$\Rightarrow -I-[2f]=7$$⇒−I−[2f]=7

Now taking case $(1)$(1) when $f\in (0,\frac{1}{2})$f∈(0,21​)

Then $\Rightarrow -I-0=7\Rightarrow I=-7\Rightarrow a\in (-7,-6.5)$⇒−I−0=7⇒I=−7⇒a∈(−7,−6.5)

Now taking case $(2)$(2) when $f\in (\frac{1}{2},1)$f∈(21​,1)

So, $-I-1=7\Rightarrow I=-8\Rightarrow a\in (-7.5,7)$−I−1=7⇒I=−8⇒a∈(−7.5,7)

Hence, from both the cases we can say that $a\in (-7.5,-6.5)$a∈(−7.5,−6.5).