Question

For values of $x$ less than $1$ but greater than $-4$ , the expression
$\frac{x^2-2x+2}{2x-2}$ has

• A. a maximum value of $-1$
• B. a minimum value of $+1$
• C. a maximum value of $+1$
• D. a minimum value of $-1$

Answer 1

Answer: (D)

a minimum value of $-1$

$y=\frac{1}{2}.\frac{x^2-2x+2}{x-1}=\frac{1}{2}[x-1+\frac{1}{x-1}]$
the sum of a number and its reciprocal is numerically least when the number is $\pm 1$. For $x-1=1,x=2$ which is excluded
$\therefore$ $x-1=-1$ and $y=-1$. All other values of $x$ in the interval given yield values of $y$ less than $-1$; or
$\frac{dy}{dx}=\frac{(x-2)(2x-2)-(x^2-2x+2)}{{(2x-2)}^2}=0;x=0,x=2$ (excluded)
By testing with values close to $x=0$ to the right and to the left, we find that $y=-1$ which is the minimum value