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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Basics of Mathematics

JEE Main · Mathematics · Basics of Mathematics

The set \ x ∈ ℝ: 16(2^X)>16^ (-1)/(x) \ =

The set \left\{x \in \mathbb{R}: 16\left(2^X\right)>16^{\frac{-1}{x}}\right\}=

Options

  1. A{x∈R:x>0}\{\mathrm{x} \in \mathbb{R}: \mathrm{x}>0\}
  2. B{x∈R:x<0}\{\mathrm{x} \in \mathbb{R}: \mathrm{x} < 0\}
  3. CR\mathbb{R}
  4. D{x∈R:x>2}\{x \in \mathbb{R}: x>2\}

Correct answer

A. {x∈R:x>0}\{\mathrm{x} \in \mathbb{R}: \mathrm{x}>0\}

Step-by-step solution

16(2x)>16−1x⇒2x+4>2−4/x⇒x+4>−4/x⇒x2+4x+4>0⇒(x+2)2>0\begin{aligned} & 16\left(2^x\right)>16^{-\frac{1}{x}} \Rightarrow 2^{x+4}>2^{-4 / x} \\ & \Rightarrow x+4>-4 / x \Rightarrow x^2+4 x+4>0 \\ & \Rightarrow(x+2)^2>0 \end{aligned} which is true for all x∈Rx \in R
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