First slide

Introduction to linear inequalities

Question

$Number of integers satisfying \sqrt{8 + 2 x - x^{2}} > 6 - 3 x is$

Moderate

A

0
B

2
C

3
D

4

Solution

$\begin{array}{l} \sqrt{8 + 2 x - x^{2}} > 6 - 3 x \\ We must have 8 + 2 x - x^{2} \geq 0 \end{array}$
$\Rightarrow x \in [- 2, 4] - - - - (1)$
$For 6 - 3 x < 0 or x > 2, inequality is satisfied.$
$Let 6 - 3 x \geq 0$
$\Rightarrow x \leq 2 - - - - (2)$
$\begin{array}{l} \Rightarrow 8 + 2 x - x^{2} > 36 + 9 x^{2} - 36 x \\ \Rightarrow 5 x^{2} - 19 x + 14 < 0 \\ \Rightarrow (5 x - 14) (x - 1) < 0 \end{array}$
$\Rightarrow x \in (1, \frac{14}{5}) - - - (3)$
$\begin{array}{l} From (ii) and (iii), x \in (1, 2] \\ ∴ finally x \in (1, 4] \end{array}$

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