Let-f-x-x-1-x-2-x-3-then-find-the-value-of-k-for-which-f-x-k-has-1-no-solution-2-only-one-solution-3-two-solutions-of-same-sign-4-two-solutions-of-opposite-sign-

Question Number 21168 by Tinkutara last updated on 15/Sep/17

Let f (x) = ∣ x - 1 ∣ + ∣ x - 2 ∣ + ∣ x - 3 ∣,

then find the value of k for which f (x)

= k has

1 . no solution

2 . only one solution

3 . two solutions of same sign

4 . two solutions of opposite sign

Answered by alex041103 last updated on 15/Sep/17

f (x) = {\begin{cases} 3 x - 6, x \in (3, \infty), f (x) \in (3, \infty) \\ 3, x = 3 \\ x, x \in (2, 3), f (x) \in (2, 3) \\ 2, x = 2 \\ 4 - x, x \in (1, 2), f (x) \in (2, 3) \\ 3, x = 1 \\ 6 - 3 x, x \in (- \infty, 1), f (x) \in (3, \infty) \end{cases}

\Rightarrow f (x) \in [2, \infty)

\Rightarrow 1 . k \in (- \infty, 2)

\Rightarrow 2 . k = 2

\Rightarrow 3 . k = 3

\Rightarrow 4 . k \in [6, \infty)

Commented by Tinkutara last updated on 15/Sep/17

Answer of 3^{rd} part is wrong .

Commented by alex041103 last updated on 15/Sep/17

f (x) = {\begin{cases} 3 x - 6, x \in (3, \infty), f (x) \in (3, \infty) \\ 3, x = 3 \\ x, x \in (2, 3), f (x) \in (2, 3) \\ 2, x = 2 \\ 4 - x, x \in (1, 2), f (x) \in (2, 3) \\ 3, x = 1 \\ 6 - 3 x, x \in (- \infty, 1), f (x) \in (3, \infty) \end{cases}

Y o u c a n s e e t h a t f o r x = 1, 3 f (x) = k = 3

s i n c e s g n (1) = s g n (3) a m d b e c a u s e

t h e i n t e r v a l s i n b l u e a r e o p e n t h e r e {a r e n}^{'} t

a n y m o r e x f o r w h i c h f (x) = 3

A m I w r o n g ?