Verified Solution • 2025–26 Session moderate

QJEEM23Mafbe4: JEE Main 2023 Mathematics - Class 11 JEEM23Mafbe4 Solution: Let where [t] denotes greatest integer function. Then,

Let A={x∈R:[x+3]+[x+4]≤3},A = \{ x \in R:[x + 3] + [x + 4] \le 3\} ,

B = \left\{ {x \in R:{3^x}{{\left( {\sum\limits_{r = 1}^\infty {{3 \over {{{10}^r}}}} } \right)}^{x - 3}} < {3^{ - 3x}}} \right\}, where [t] denotes greatest integer function. Then,

Multiple Choice Options

  • A B⊂C,A≠BB \subset C,A \ne B
  • B A⊂B,A≠BA \subset B,A \ne B
  • C A=BA = B
  • D A∩B=ϕA \cap B = \phi

Step-by-Step Verified Solution

We have,

amp;A={x∈R:[x+3]+[x+4]≤3}amp; Here, [x+3]+[x+4]≤3amp;⇒[x]+3+[x]+4≤3amp;(∵[x+n]=[x]+n,n∈I)amp;⇒2[x]+4≤0⇒[x]≤−2amp;⇒x∈(−∞,−1)amp;A≡(−∞,−1)...........(i)\begin{aligned}& A=\{x \in R:[x+3]+[x+4] \leq 3\} \\\\& \text { Here, }[x+3]+[x+4] \leq 3 \\\\& \Rightarrow [x]+3+[x]+4 \leq 3 \\\\& (\because[x+n]=[x]+n, n \in I) \\\\& \Rightarrow 2[x]+4 \leq 0 \Rightarrow[x] \leq-2 \\\\& \Rightarrow x \in(-\infty,-1) \\\\& A \equiv(-\infty,-1) ...........(i)\end{aligned}

Also,

B=\left\{x \in R: 3^x\left(\sum\limits_{r=1}^{\infty} \frac{3}{10^r}\right)^{x-3}<3^{-3 x}\right\}

Here,

amp;3x(310+3102+3103+…)x−3lt;3−3x⇒amp;3x(3101−110)x−3lt;3−3x⇒amp;3x(13)x−3lt;3−3x\begin{aligned}& 3^x\left(\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\ldots\right)^{x-3}<3^{-3 x} \\\\\Rightarrow & 3^x\left(\frac{\frac{3}{10}}{1-\frac{1}{10}}\right)^{x-3}<3^{-3 x} \\\\\Rightarrow & 3^x\left(\frac{1}{3}\right)^{x-3}<3^{-3 x}\end{aligned}

⇒amp;3x−x+3lt;3−3x⇒33lt;3−3x⇒amp;−3xgt;3⇒xlt;−1⇒amp;x∈(−∞,−1)⇒amp;B≡(−∞,−1)...........(ii)\begin{array}{ll}\Rightarrow & 3^{x-x+3}<3^{-3 x} \Rightarrow 3^3<3^{-3 x} \\\\\Rightarrow & -3 x>3 \Rightarrow x<-1 \\\\\Rightarrow & x \in(-\infty,-1) \\\\\Rightarrow & B \equiv(-\infty,-1) ...........(ii)\end{array}

From equations (i) and (ii), we get

A=BA=B

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