JEE Main 12 Jan 2019 Evening Question 14
Question: The number of integral values of m for which the quadratic expression, $ (1+2m)x^{2}-2 $ $ (1+3m)x+4(1+m),x\in R, $ is always positive is
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 8
B) 7
C) 3
D) 6
Show Answer
Answer:
Correct Answer: B
Solution:
- The given quadratic expression $ (1+2m)x^{2}-2(1+3m)x+4(1+m), $ $ x\in R $ is always positive
if $ 1+2m>0\Rightarrow m>\frac{-1}{2} $ …(i) And, D < 0
$ \Rightarrow $ $ 4{{(1+3m)}^{2}}-4(1+2m)\times 4(1+m)<0 $
$ \Rightarrow $ $ (1+9m^{2}+6m)-4(1+3m+2m^{2})<0 $
$ \Rightarrow $ $ m^{2}-6m-3<0 $
$ \Rightarrow $ $ 3-\sqrt{12}<m<3+\sqrt{12} $ -(ii)
From (i) and (ii), common interval is $ 3-\sqrt{12}<m<3A+\sqrt{12} $ So, integral values of m are 0, 1, 2, 3,4, 5, 6.
