First slide

Introduction to linear inequalities

Question

The number of integral values of x satisfying $\sqrt{- x^{2} + 10 x - 16} < x - 2 is$

Moderate

A

0
B

1
C

2
D

3

Solution

$\begin{array}{l} \sqrt{- x^{2} + 10 x - 16} < x - 2 \\ We must have - x^{2} + 10 x - 16 \geq 0 \end{array}$ $(\sin c e \sqrt{a} < b \Rightarrow a \geq 0 a n d a < b^{2})$
$\Rightarrow 2 \leq x \leq 8 - - - - (1)$
$\begin{array}{l} Also - x^{2} + 10 x - 16 < x^{2} - 4 x + 4 \\ \Rightarrow 2 x^{2} - 14 x + 20 > 0 \\ \Rightarrow x^{2} - 7 x + 10 > 0 \end{array}$
$\Rightarrow x > 5 or x < 2 - - - (2)$
$From (1) and (2), 5 < x \leq 8 \Rightarrow x = 6, 7, 8$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App