Test Index

EAMCET Engineering 2010 Solved Paper

Section: Mathematics

© examsnet.com

Question : 61 of 160

Marks: +1, -0

For

x \in R

, the least value of

\;\frac{x^2-6x+5}{x^2+2x+1}

-1
$-\;\frac{1}{2}$
$-\;\frac{1}{4}$
$-\;\frac{1}{3}$

Solution:

Let

f(x)=\;\frac{x^2-6x+5}{x^2+2x+1}

for

x \in R

Let

y=\;\frac{x^2-6x+5}{x^2+2x+1}

\; yx^2+2yx+y=x^2-6x+5

\; (y-1)x^2+(2y+6)x+(y-5)=0

\; x=\;\frac{-2(y+3)+\sqrt{4(y+3)^2-4(y-1)(y-5)}}{2 \times (y-1)}

Since,

x

is real.

Then, its discriminant should be

\geq 0

.

\therefore \; 4(y+3)^2-4(y-1)(y-5) \geq 0

\Rightarrow (y+3)^2-(y-1)(y-5) \geq 0

\Rightarrow y^2+9+6y-(y^2-6y+5) \geq 0

\Rightarrow y^2+9+6y-y^2+6y-5 \geq 0

\Rightarrow 12y+4 \geq 0

\Rightarrow 4(3y+1) \geq 0

\Rightarrow y \geq -\frac{1}{3}

So, least value of given expression is

\left( \; -\frac{1}{3} \right)

.

© examsnet.com

Go to Question:

Prev Question

Next Question