JEE Main 2023 — Mathematics Algebra

Given,

A={x\in \mathbb{R}:[x+3]+[x+4]\leq 3}&B={x\in \mathbb{R}:{3}^{x}{(\sum _{r=1}^{\infty }\frac{3}{{10}^{r}})}^{x-3}<{3}^{-3x}},

Now solving $A$A we get,

$$[x+3]+[x+4]\leq 3$$[x+3]+[x+4]≤3

$$\Rightarrow [x]+3+[x]+4\leq 3$$⇒[x]+3+[x]+4≤3

$$\Rightarrow 2[x]\leq -4$$⇒2[x]≤−4

$$\Rightarrow [x]\leq -2$$⇒[x]≤−2

$$\Rightarrow x<-1\Rightarrow A=(-\infty ,-1)$$⇒x<−1⇒A=(−∞,−1)

Now solving $B$B we get,

$${3}^{x}{(\sum _{r=1}^{\infty }\frac{3}{{10}^{r}})}^{x-3}<{3}^{-3x}$$3x(r=1∑∞​10r3​)x−3<3−3x

Now using infinite $G.P$G.P formula ${S}_{\infty }=\frac{a}{1-r}$S∞​=1−ra​ we get,

$$\Rightarrow {3}^{x}{(\frac{3\cdot \frac{1}{10}}{1-\frac{1}{10}})}^{x-3}<{3}^{-3x}$$⇒3x(1−101​3⋅101​​)x−3<3−3x

$$\Rightarrow {3}^{x}{(\frac{1}{3})}^{x-3}<{3}^{-3x}$$⇒3x(31​)x−3<3−3x

$$\Rightarrow {3}^{x-x+3+3x}<1$$⇒3x−x+3+3x<1

$$\Rightarrow {3}^{3(x+1)}<1$$⇒33(x+1)<1

$$\Rightarrow 3(x+1)<0$$⇒3(x+1)<0

$$\Rightarrow x<-1\Rightarrow B=(-\infty ,-1)$$⇒x<−1⇒B=(−∞,−1)

$$\Rightarrow A=B$$⇒A=B