Plz expain how to solve dis...

​Q1. The range of  f (x) =  x 2 -   x   + 1 x 2   +   x   +   1  is 

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y=x2-x+1x2+x+1yx2+yx+y=x2-x+1yx2+yx+y-x2+x-1=0x2y-1+xy+1+y-1=0Now this equation is quadratic in y if y1, hence if xR, thenD0y+12-4y-1y-10y+12-2y-120y+12-2y-220y+1+2y-2y+1-2y+203y-1-y+30Multiply by -13y-1y-30Using wavy curve method-+++++13----3++++, we gety13,3Now we need to check y=1 separately. Put y=1 in given question1=x2-x+1x2+x+1x2+x+1=x2-x+1x=-xx=0 which is real, hence y=1 will be part of range, henceRange is y13,3

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Let f(x) =  (x2 - x + 1) / (x2 + x + 1)  = y

=> ​x2 - x + 1 = y (​x2 + x + 1)

=> (1 - y)x2 - (1 + y)x (1 - y) = 0

This essentially is a quadratic expression in x where x is real. So for x to be real,  D = b2 - 4ac ​≥ 0

Thus, (1 + y)2 - 4(1- y)(1- y) ≥ 0
     => 1 + y2 + 2y - 4(y2 + 1 - 2y) ≥ 0
     => -3y2 + 10y  - 3 ≥ 0
     =>  3y2 - 10y + 3 ≤  0
      =>  (3y - 1)(y - 3) ≤  0
      =>   1/3 ≤ ≤  3            (solving inequality using wavy curve method)

Since, y = f(x) thus,   

f(x) = (x2 - x + 1) / (x2 + x + 1)​  ​∈ [ 1/3 , 3]

Hope that helps!!
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