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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 Rthis imlplies  a^2 – 4{1 – 2(a)^2} is less than 0I 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.  `
one year ago

Deepak Kumar Shringi
4398 Points
```							wait you have misunderstood the concept here x real number mean when you are putting any real value of x the given inequality is positivenow a quadratic will be positive when its garph is above x axis means D<0 and a>0
```
one year ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions