mathematics-class-11-unit-07-chapter-02-linear-inequalities-l-2-6-h7ccoorfync

### <Success>Solution</Success>
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<p>$\text { from } (ii) x \in[-4,6] $</p>
<p>$|x| \geqslant 2$</p>
<p>$|x| \geqslant a\Leftrightarrow x \leqslant a$ $\text{or}$ $x \geqslant a $</p>
<p>$\Rightarrow x\leqslant - 2$   or   $x\geqslant 2$</p>
<p>$\Rightarrow x \in(-\infty,-2] \cup[2, \infty)$</p>
<p>from (i) $|x-1| \leqslant 5 \Rightarrow x \in[-4,6] $</p>
<p>from (ii) $|x| \geqslant 2 \Rightarrow x \in(-\infty,-2] \cup[2, \infty)$</p>
<p>Combining solution (i) & (ii), we get</p>
<p>Solution set =$[-4,-2] \cup[2,6]$</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-02-linear-inequalities-l-2-6-h7ccoorfync-12a.jpg"></p>
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### <Success>Example 5</Success>
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<p>Solve $ \frac{|x|-1}{|x|-2} \geqslant 0, \quad x \in \R, \ \ x \neq \pm 2$</p>
<p><strong>Solution</strong>
Given : $ \frac{|x|-1}{|x|-2} \geqslant 0$ Let $|x| = z$</p>
<p>$\Rightarrow \frac{z-1}{z-2} \geqslant 0 $</p>
<p>$ \Rightarrow z \leqslant 1 \text { or } z>2$</p>
<p>(i) $ \frac{x-a}{x-b} \geqslant 0, a<b $</p>
<p>$ \Rightarrow x \leqslant a \text { a } x \geqslant b $</p>
<p>(ii) $ \frac{x-a}{x-b} \leqslant 0, a<b $</p>
<p>$ \Rightarrow a \leqslant x \leqslant b$</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-02-linear-inequalities-l-2-6-h7ccoorfync-15a.jpg"></p>

### <Success>Solution</Success>
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<p>$ \Rightarrow|x| \leqslant 1 \text { or }|x|>2$</p>
<p>$ \Rightarrow(-1 \leqslant x \leqslant 1) \text { or }(x \leqslant-2 \text { or } x>2) $</p>
<p>$ \Rightarrow x \in[-1,1] \text { or } x \in(-\infty,-2) \cup(2, \infty)$</p>
<p>On the basis of solution represented on number line</p>
<p>Solution set $=[-1,1] \cup(-\infty,-2) \cup(2, \infty)$</p>
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<p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-02-linear-inequalities-l-2-6-h7ccoorfync-16.jpg"></p>
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### <Success>Solution</Success>
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<p>We will discuss in three intervals</p>
<p>i.e. $(-\infty, 1),[1,2),[2, \infty)$</p>
<p><ins>Case I:</ins> When $x \in(-\infty, 1)$</p>
<p>$|x-1|=-(x-1) $</p>
<p>$|x-2|=-(x-2) $</p>
<p>$\therefore|x-1|+|x-2| \geqslant 4 $</p>
<p>$\Rightarrow-(x-1)-(x-2) \geqslant 4 $</p>
<p>$|x| =x, x \geqslant 0$</p>
<p>$=-x, x<0$</p>
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### <Success>Solution</Success>
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<p>$ \Rightarrow \quad-2 x+3 \geqslant 4 $</p>
<p>$ \Rightarrow \quad-2 x \geqslant 4-3 $</p>
<p>$ \Rightarrow \quad-2 x \geqslant 1 $</p>
<p>$ \Rightarrow \quad 2 x \leqslant-1 $</p>
<p>$ \Rightarrow \quad x \leqslant-\frac{1}{2} \Rightarrow x \in\left(-\infty,-\frac{1}{2}\right] $</p>
<p>$ \Rightarrow \quad \text { but } x \in(-\infty, 1)$</p>
<p>$\Rightarrow x \in\left(-\infty,-\frac{1}{2}\right]$ -(i)</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-02-linear-inequalities-l-2-6-h7ccoorfync-23a.jpg"></p>
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### <Success>Solution</Success>
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<p>$ \Rightarrow \quad x \in\left[\frac{7}{2}, \infty\right)$</p>
<p>Case I: $\quad x \in\left(-\infty,-\frac{1}{2}\right]$</p>
<p>Case-II: No Solution-</p>
<p>Case-III: $\quad x \in\left[\frac{7}{2}, \infty\right)$</p>
<p>Combining cases I, II & III we get</p>
<p>$ x \in\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{7}{2}, \infty\right) $</p>
<p>$ \therefore \text { Solution set }=\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{7}{2}, \infty\right) $</p>
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<p>======</p>
<h3 id="successthank-yousuccess"><Success>Thank you</Success></h3>
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