First slide

Theory of expressions

Question

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

Easy

A

2,1
B

5, $\frac{1}{5}$
C

7, $\frac{1}{7}$
D

None of these

Solution

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

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