Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • What is the value of m so that both roots of the equation x2 +mx +1 are less than unity?Option: 1 -2 ≥ m<div class='qna-o

What is the value of m so that both roots of the equation x2 +mx +1 are less than unity?

Option: 1

-2 ≥ m


Option: 2

2 ≥ m


Option: 3

m ≥ -2


Option: 4

m ≥ 2


Answers (1)

best_answer

 

 

Location of roots (1) -

let f(x) = ax2 + bx + c where a,b,c is from real number and ‘a’ is non-zero number. Let ? and ? be the solution of the function. And let k is number from real number.  Then:

If both roots of f(x) are less than k then

 

   


 

i) D ≥ 0 (as roots may be equal)

ii) af(k) > 0. As if a < 0 then f(k) < 0. So multiplying two -ve value will give us a positive value, so af(k) > 0 satisfies

iii) \\\mathrm{k > \frac{-b}{2a}\; \;since, \; \frac{-b}{2a}}   will lies between  ? and ?, and ?, ? are less than k so \\\frac{-b}{2a}  will be less than k.

 

 

-

 

 

since roots are less than unity it implies that af(1) > 0, here a = 1 f(1) = m+2

So m+2 >0 ⇒ m > -2 (i)

2nd for two root to exist \\\mathrm{D \geq 0 \Rightarrow m^2 - 4 \geq 0}

So m ≥ 2 or m ≤ -2 (ii)

Combining (i) and (ii) equation, we get

m ≥ 2 

Correct option is (d)

Posted by

Info Expert 30

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions