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`        Range of the function f(x)=x²+x+2/x²+x+1 where x is a real number is Please solve fully I can't understand this chapter.`
one year ago

Saurabh Koranglekar
3287 Points
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one year ago
Parshu Jauhari
25 Points
```							There is a better approch to solve such problems.convert the above equation into a quadratic equation in which cofficient of the equation are in term of y i.e f(x).y=x^2+x+2/x^2+x+1y(x^2+x+1)=x^2+x+2x^2(y-1)+x(y-1)+y-2Since x belongs to real numbers.then D>=0(y-1)^2-4(y-1)(y-2)-3y^2+10y-7>=03y^2-10y+7(3y-7)(y-1)so 1Now verify then end point You will see the that original equation will never have f(x)=1Verification1=x^2+x+2/x^2+x+1x^2+x+1 != x^2+x+2So our Range is 1Hope it helps :-)
```
9 months ago
Parshu Jauhari
25 Points
```							There is a better approch to solve such problems.convert the above equation into a quadratic equation in which cofficient of the equation are in term of y i.e f(x).y=x^2+x+2/x^2+x+1y(x^2+x+1)=x^2+x+2x^2(y-1)+x(y-1)+y-2Since x belongs to real numbers.then D>=0(y-1)^2-4(y-1)(y-2)-3y^2+10y-7>=03y^2-10y+7(3y-7)(y-1)so [1,7/3]Now verify then end point You will see the that original equation will never have f(x)=1Verification1=x^2+x+2/x^2+x+1x^2+x+1 != x^2+x+2So our Range is (1,7/3]Hope it helps :-)
```
9 months ago
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### Course Features

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