First slide

Theory of expressions

Question

If x is rea, then the maximum and minimum values of expression $\frac{x^{2} + 14 x + 9}{x^{2} + 2 x + 3}$ will be

Moderate

A

$4, - 5$
B

$5, - 4$
C

$- 4, 5$
D

$- 4, - 5$

Solution

Let $y = \frac{x^{2} + 14 x + 9}{x^{2} + 2 x + 3}$
$\Rightarrow y (x^{2} + 2 x + 3) - x^{2} - 14 x - 9 = 0$
$\Rightarrow (y - 1) x^{2} + (2 y - 14) x + 3 y - 9 = 0$
For real x, its discriminant $\geq 0$
i.e. $4 {(y - 7)}^{2} - 4 (y - 1) 3 (y - 3) \geq 0$
$\Rightarrow y^{2} + y - 20 \leq 0 or (y - 4) (y + 5) \leq 0$
Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:
Case I: $y - 4 \geq 0 or y \geq 4 and y + 5 \leq 0 or y \leq - 5$
This is not possible.
Case II: $y - 4 \leq 0 or y \leq 4 and y + 5 \geq 0 or y \geq - 5$
Both of these are satisfied if $- 5 \leq y \leq 4$
Hence maximum value of y is 4 and minimum value is -5.

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