First slide

Introduction to linear inequalities

Question

$Number of integers satisfying \sqrt{x - 6} - \sqrt{10 - x} \geq 1 is$

Difficult

A

4
B

5
C

3
D

1

Solution

$\begin{aligned} We have \sqrt{x - 6} - \sqrt{10 - x} \geq 1 \\ We must have x \geq 6 and x \leq 10 \end{aligned}$
$\begin{aligned} Then \sqrt{x - 6} \geq 1 + \sqrt{10 - x} \\ Squaring given inequality, \end{aligned}$
$\begin{array}{l} x - 6 \geq 1 + 10 - x + 2 \sqrt{10 - x} \\ \Rightarrow 2 x - 17 \geq 2 \sqrt{10 - x} \end{array}$
Clearly LHS must be positive.
$∴ x \geq \frac{17}{2}$
so, we have
$\begin{array}{l} \begin{aligned} 4 x^{2} + 289 - 68 x \geq 4 (10 - x) \\ \Rightarrow & 4 x^{2} - 64 x + 249 \geq 0 \\ \Rightarrow & x^{2} - 16 x + \frac{249}{4} \geq 0 \\ \Rightarrow & (x - 8)^{2} - {(\frac{\sqrt{7}}{2})}^{2} \geq 0 \\ \Rightarrow & (x - 8 - \frac{\sqrt{7}}{2}) (x - 8 + \frac{\sqrt{7}}{2}) \geq 0 \end{aligned} \\ From the sign scheme, x \in [(16 + \sqrt{7}) / 2, 10] \end{array}$

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