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EAMCET Engineering 2005 Solved Paper

Section: Mathematics

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Question : 94 of 160

Marks: +1, -0

If

x

is real, then the minimum value of

\;\frac{x^2-x+1}{x^2+x+1}

$\;\frac{1}{3}$
3
$\;\frac{1}{2}$
1

Solution:

Let

f(x)=\;\frac{x^2-x+1}{x^2+x+1}

. . . (i)

On differentiating w.r.t.

x

, we get

\;f'(x)=\;\frac{(x^2+x+1)(2x-1)-(x^2-x+1)(2x+1)}{(x^2+x+1)^2}

\;\;\text{ for maximum or minimum, put }\; f'(x)=0

\;\Rightarrow (x^2+x+1)(2x-1)-(x^2-x+1)(2x+1)=0

\;\Rightarrow \;\; x^2+x-1-(-x^2+x+1)=0

\;\Rightarrow \;\; 2x^2-2=0 \Rightarrow x=\pm 1

\;\;\text{ Now, }\; f'(c)=\;\frac{2x^2-2}{(x^2+x+1)^2}

Again differentiating, we get

\;(x^2+x+1)^2(4x)-(2x^2-2)

\;f''(x)=\;\frac{2(x^2+x+1)(2x+1)}{(x^2+x+1)^4}

at

x=1, f''(x) > 0

Therefore it is minimum at

x=1

Put

x=1

in equation (i), we get

f(1)=\;\frac{1-1+1}{1+1+1}=\;\frac{1}{3}

\therefore

The minimum value is

\;\frac{1}{3}

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