Question

If x is real, then the maximum and minimum values of the expression $\frac{x^2-3x+4}{x^2+3x+4}$ will be

• A. 2,1
• B. $5,\frac{1}{5}$
• C. $7,\frac{1}{7}$
• D. None of these

Answer 1

The correct option is C $7,\frac{1}{7}$

Let $y=\frac{x^2-3x+4}{x^2+3x+4}$
$\Rightarrow (y-1)x^2+3(y+1)x+4(y-1)=0$
For x is real $D\ge 0$
$\Rightarrow 9(y+1{)}^2-16(y-1{)}^2\ge 0$
$\Rightarrow -7y^2+50y-7\ge 0\Rightarrow 7y^2-50y+7\le 0$
$\Rightarrow (y-7)(7y-1)\le 0$
Now, the product of two factors is negative if one in -ve and one in +ve.
Case I : $(y-7)\ge 0and(7y-1)\le 0$
$\Rightarrow y\ge 7andy\le \frac{1}{7}.$ But it is impossible
Case II : $(y-7)\le 0and(7y-1)\ge 0$
$\Rightarrow y\le 7andy\ge \frac{1}{7}\Rightarrow \frac{1}{7}\le y\le 7$
Hence maximum value is 7 and minimum value is $\frac{1}{7}$