If | x - 1 | + | x - 3 | 8 , then the values of x lie in the interval

Question

If $| x - 1 |$ $+$ $| x - 3 |$ $\leq 8$ , then the values of $x$ lie in the interval

Smart Achievers · Accepted Answer

(2)

Here two cases arises,

Case 1: When $- \infty < x \leq 1$

Then, $\leq 8$

$\Rightarrow$ $| x - 1 |$ $= - (x - 1)$ and $| x - 3 |$ $= - (x - 3)$

Now, $| x - 1 |$ $+$ $| x - 3 |$ $\leq 8$ $\Rightarrow$ $- (x - 1) - (x - 3) \leq 8$

$\Rightarrow$ $- 2 x + 4 \leq 8 \Rightarrow x \geq - 2$ $\dots (i)$

Case 2: When $3 \leq x < \infty$

$∴$ $| x - 1 |$ $= (x - 1)$ and $| x - 3 |$ $= (x - 3)$

Now, $| x - 1 |$ $+$ $| x - 3 |$ $\leq 8$

$\Rightarrow$ $x - 1 + x - 3 \leq 8 \Rightarrow 2 x - 4 \leq 8 \Rightarrow x \leq 6$ $\dots (i i)$

From (i) and (ii), we get $x \in [- 2, 6]$

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