First slide

Theory of expressions

Question

If x is real, the expression $\frac{x^{2} + 2 x - 11}{x - 3}$ takes all real values except those which lie between a and b, then a and b are

Moderate

A

-12,-4
B

-12,2
C

4,12
D

-4,4

Solution

$\begin{array}{lc} Let & y = \frac{x^{2} + 2 x - 11}{x - 3} \\ \Rightarrow & xy - 3 y = x^{2} + 2 x - 11 \\ \Rightarrow & x^{2} + (2 - y) x + (3 y - 11) = 0 \\ \Rightarrow & (2 - y)^{2} - 4 (3 y - 11) \geq 0 \\ \Rightarrow & 4 + y^{2} - 4 y - 12 y + 44 \geq 0 \\ \Rightarrow & y^{2} - 16 y + 48 \geq 0 \\ \Rightarrow & y^{2} - 12 y - 4 y + 48 \geq 0 \\ \Rightarrow & (y - 4) (y - 12) \geq 0 \\ \Rightarrow & y \leq 4 or y \geq 12 \end{array}$

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