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If x^2 – ax + 1 – 2(a)^2 is greater than 0 for all x is an element of R , then a)a is an element of ( – 2/3 , 2/3) b)a is an element of [-2/3 , 2/3] c)a is element of ( – 2/3 , 1) d)a is an element of (0 , 2/3) In the book this was solved like this..... x^2 – ax + 1 – 2(a)^2 is greater than 0 for all x is an element of R this imlplies a^2 – 4{1 – 2(a)^2} is less than 0 I understood that here we are taking Discriminant but as x is real so D should be greater than or equal to 0 and not less than 0 then why here D is less than 0?? Please explain and solve this question.

If x^2 – ax + 1 – 2(a)^2 is greater than 0 for all x is an element of R , then 
a)a is an element of ( – 2/3 , 2/3)
b)a is an element of [-2/3 , 2/3]
c)a is element of ( – 2/3 , 1)
d)a is an  element of (0 , 2/3)
In the book this was solved like this.....
x^2 – ax + 1 – 2(a)^2 is greater than 0  for all x is an element of R
this imlplies  a^2 – 4{1 – 2(a)^2} is less than 0
I understood that here we are taking Discriminant but as x is real so D should be greater than or equal to 0 and not less than 0 then why here D is less than 0??
Please explain and solve this question.
 
 

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1 Answers

Deepak Kumar Shringi
askIITians Faculty 4404 Points
5 years ago
wait you have misunderstood the concept here x real number mean when you are putting any real value of x the given inequality is positive
now a quadratic will be positive when its garph is above x axis means D<0 and a>0

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