Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Basics of Mathematics

NDA 2015 · Mathematics · Basics of Mathematics

If .A= x ∈ R: x^2+6 x-7 0\ and B=\ x ∈ R: x^2+9 x+14 0\ , then...

If \left.A=\mid x \in R: x^2+6 x-7 \lt 0\right\} and B=\left\{x \in R: x^2+9 x+14 \gt 0\right\}, then which of the following is/ are correct? 1. (A∩B)=(−2,1)(A \cap B)=(-2,1) 2. (A/B)=(−7,−2)(A / B)=(-7,-2) Select the correct answer using the code given below:

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

A. 1 only

Step-by-step solution

x2+6x−7<0⇒(x+7)(x−1)<0⇒x=(−7,1)⇒A={−6,−5,−4,−3,−2,−1,0}⇒x2+9x+14>0⇒(x+7)(x+2)>0⇒x=(−∞,−7)∪(−2,∞)⇒B=R−{−7,−6,−5,−4,−3,−2}\begin{aligned} & \mathrm{x}^2+6 \mathrm{x}-7 < 0 \\ & \Rightarrow(\mathrm{x}+7)(\mathrm{x}-1) < 0 \\ & \Rightarrow \mathrm{x}=(-7,1) \\ & \Rightarrow \mathrm{A}=\{-6,-5,-4,-3,-2,-1,0\} \\ & \Rightarrow \mathrm{x}^2+9 \mathrm{x}+14 > 0 \\ & \Rightarrow(\mathrm{x}+7)(\mathrm{x}+2) > 0 \\ & \Rightarrow \mathrm{x}=(-\infty,-7) \cup(-2, \infty) \\ & \Rightarrow \mathrm{B}=\mathrm{R}-\{-7,-6,-5,-4,-3,-2\}\end{aligned}
So A∩B=(−2,1)A \cap B=(-2,1)
Practice Basics of Mathematics on Quantrex Academy →

More from Basics of Mathematics

All Basics of Mathematics questions Full Basics of Mathematics list Basics of Mathematics JEE Maths PYQsJEE Mathematics PYQs All NDA PYQs