NIMCET 2019 Question Paper with solutions

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Question : 26 of 120

Marks: +1, -0

If x is real, then the minimum value of

{x^2-x+1}/{x^2+x-1}

is

$1/2$
2
3
$1/3$

Solution:

Let

y=\;\;{x^{2}-x+1}/{x^{2}+x+1}

x^{2}y+xy+y=x^{2}-x+1

x^{2}(y-1)+x(y+1)+y-1=0

x ∊ R

∴\;(y+1)^{2}-4(y-1)^{2}≥0

[∵D ≥ 0]

y^{2}+2y+1-4y^{2}+8y-4 ≥ 0

⇒3y^{2}-10y+3 ≤ 0

3y^{2}-9y-y+3 ≤ 0

(3y-1)(y-3) ≤ 0

y∈[\;\;{1}/{3},3]

Hence, minimum value is

1/3

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