Find set of k value so that ∣x∣ + ∣x−1∣ + ∣x−4∣ = k a. has one solution b. has two solutions c. has many solutions d. has no solution


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Question Number 153257 by naka3546 last updated on 06/Sep/21

$F i n d s e t o f k v a l u e s o t h a t$ $∣ x ∣ + ∣ x - 1 ∣ + ∣ x - 4 ∣ = k$ $a . h a s o n e s o l u t i o n$ $b . h a s t w o s o l u t i o n s$ $c . h a s m a n y s o l u t i o n s$ $d . h a s n o s o l u t i o n$
	Answered by ajfour last updated on 06/Sep/21

	$l e t x - 2 = t, t h e n$ $∣ x ∣ + ∣ x - 1 ∣ + ∣ x - 3 ∣ + ∣ x - 4 ∣$ $= k + ∣ x - 3 ∣$ $\Rightarrow ∣ t + 2 ∣ + ∣ t + 1 ∣ + ∣ t - 1 ∣ + ∣ t - 2 ∣$ $= k + ∣ t - 1 ∣$ $n o w w e b e t t e r u s e g r a p h :$ $b l u e g r a p h = r e d g r a p h$ $s o f o r n o s o l u t i o n k < 4$ $o n e s o l u t i o n k = 4$ $f o r t h r e e s o l u t i o n s, (t w o e q u a l)$ $k = 6$ $t w o s o l u t i o n s k > 4$ $F o r m o r e t h a n t w o (d i s t i n c t)$ $n o r e a l v a l u e o f k .$
	Commented by ajfour last updated on 06/Sep/21

	Answered by MJS_new last updated on 06/Sep/21

	$f (x) =∣ x ∣ + ∣ x - 1 ∣ + ∣ x - 4 ∣$ $f (x) = {\begin{cases} - 3 x + 5; x ⩽ 0 \land 5 ⩽ f (x) < + \infty \\ - x + 5; 0 < x ⩽ 1 \land 4 ⩽ f (x) < 5 \\ x + 3; 1 < x ⩽ 4 \land 4 < f (x) ⩽ 7 \\ 3 x - 5; 4 < x \land 7 < f (x) < + \infty \end{cases}$ $\Rightarrow$ $k < 4 no solution$ $k = 4 one solution$ $k > 4 two solutions$
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