For $x \in R$, the least value of $\frac{x^2-6 x+5}{x^2+2 x+1}$ is
For $x \in R$, the least value of $\frac{x^2-6 x+5}{x^2+2 x+1}$ is
- -1
- $-\frac{1}{2}$
- $-\frac{1}{4}$
- $-\frac{1}{3}$
Solution
Let $f(x)=\frac{x^2-6 x+5}{x^2+2 x+1}$ for $x \in R$
Let $\quad y=\frac{x^2-6 x+5}{x^2+2 x+1}$
$y x^2+2 y x+y=x^2-6 x+5$
$(y-1) x^2+(2 y+6) x+(y-5)=0$
$x=\frac{-2(y+3)+\sqrt{4(y+3)^2-4(y-1)(y-5)}}{2 \times(y-1)}$
Since, $x$ is real.
Then, its discriminant should be $\geq 0$.
$\therefore \quad 4(y+3)^2-4(y-1)(y-5) \geq 0$
$\Rightarrow \quad(y+3)^2-(y-1)(y-5) \geq 0$
$\Rightarrow y^2+9+6 y-\left(y^2-6 y+5\right) \geq 0$
$\Rightarrow \quad y^2+9+6 y-y^2+6 y-5 \geq 0$
$\Rightarrow \quad 12 y+4 \geq 0$
$\Rightarrow \quad 4(3 y+1) \geq 0$
$\Rightarrow \quad y \geq-1 / 3$
So, least value of given expression is $\left(\frac{-1}{3}\right)$.
Asked in: AP EAMCET 2010
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