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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.

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.

Grade:11

3 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
5 years ago
576-1329_1.PNG
Parshu Jauhari
25 Points
5 years ago
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+1
y(x^2+x+1)=x^2+x+2
x^2(y-1)+x(y-1)+y-2
Since x belongs to real numbers.
then D>=0
(y-1)^2-4(y-1)(y-2)
-3y^2+10y-7>=0
3y^2-10y+7
(3y-7)(y-1)
so 1
Now verify then end point 
You will see the that original equation will never have f(x)=1
Verification
1=x^2+x+2/x^2+x+1
x^2+x+1 != x^2+x+2
So our Range is 1
Hope it helps :-)
Parshu Jauhari
25 Points
5 years ago
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+1
y(x^2+x+1)=x^2+x+2
x^2(y-1)+x(y-1)+y-2
Since x belongs to real numbers.
then D>=0
(y-1)^2-4(y-1)(y-2)
-3y^2+10y-7>=0
3y^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)=1
Verification
1=x^2+x+2/x^2+x+1
x^2+x+1 != x^2+x+2
So our Range is (1,7/3]
Hope it helps :-)

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