The-number-of-solutions-of-x-3-x-2-3sinx-3-0-is-

Question Number 191149 by SLVR last updated on 19/Apr/23

T h e n u m b e r o f s o l u t i o n s o f

∣ x + 3 ∣ + ∣ x - 2 ∣ + 3 s i n x - 3 = 0 i s

Commented by SLVR last updated on 19/Apr/23

s i r k i n d l y h e l p m e

Commented by SLVR last updated on 19/Apr/23

s i r M r . W \dots k i n d l y h e l p i f

y o u a r e f r e e

Commented by Frix last updated on 19/Apr/23

I'm Freex ��

Answered by Frix last updated on 19/Apr/23

∣ x + 3 ∣ + ∣ x - 2 ∣= 3 (1 - \sin x)

f_{1} (x) =∣ x + 3 ∣ + ∣ x - 2 ∣= {\begin{cases} - 2 x - 1; x < - 3 \\ 5; - 3 ⩽ x ⩽ 2 \\ 2 x + 1; x > 2 \end{cases}

0 ⩽ (f_{2} (x) = 3 (1 - \sin x)) ⩽ 6

x \in [- \frac{7}{2}; \frac{5}{2}] : 5 ⩽ f_{1} (x) ⩽ 6 \Rightarrow

Solutions possible within this interval \Rightarrow

2 solutions :

x_{1} = - π + \sin^{- 1} \frac{2}{3}

x_{2} = - \sin^{- 1} \frac{2}{3}

Commented by SLVR last updated on 19/Apr/23

t h a n k s s i r

Answered by mr W last updated on 19/Apr/23

∣ x + 3 ∣ + ∣ x - 2 ∣= 3 (1 - \sin x)

f (x) = g (x)

L H S f (x) ⩾ 5

0 ⩽ R H S g (x) ⩽ 6

t h a t m e a n s : s o l u t i o n c o u l d e x i s t .

w i t h x < - 3 :

f (x) = - 3 - x + 2 - x = - 1 - 2 x

f o r x < - 3.5 f (x) > 6, t h a t m e a n s

f o r x < - 3.5 f (x) \neq g (x)

f o r - 3.5 ⩽ x < - 3 g (x) < 3 (1 + \sin 3) \approx 3.423 < 5

t h a t m e a n s f o r - 3.5 ⩽ x < - 3 f (x) \neq g (x)

\Rightarrow f o r x < - 3 f (x) = g (x) h a s n o s o l u t i o n .

w i t h x > 2 :

s i m i l a r l y t o a b o v e f (x) = g (x) h a s

n o s o l u t i o n .

w i t h - 3 ⩽ x ⩽ 2 :

f (x) = 5

f (x) = g (x)

5 = 3 (1 - \sin x)

\Rightarrow \sin x = - \frac{2}{3}

\Rightarrow x = - \sin^{- 1} \frac{2}{3} o r - π + \sin^{- 1} \frac{2}{3}

t o t a l l y :

∣ x + 3 ∣ + ∣ x - 2 ∣ + 3 s i n x - 3 = 0 h a s

2 s o l u t i o n s :

x = - \sin^{- 1} \frac{2}{3} o r - π + \sin^{- 1} \frac{2}{3}

Commented by SLVR last updated on 19/Apr/23

T h a n k s a l o t . . P r o f . W

g o d b l e s s y o u s i r

Answered by mehdee42 last updated on 19/Apr/23

l e t f (x) = ∣ x - 2 ∣ + ∣ x + 3 ∣ & g (x) = 3 - 3 s i n x

\forall x - 3 ⩽ x ⩽ 2 \to f (x) = 5 \Rightarrow 5 + 3 s i n x - 3 = 0 \Rightarrow s i n x = - \frac{2}{3} \Rightarrow x = - {s i n}^{- 1} (\frac{2}{3}) & - π + {s i n}^{- 1} (\frac{2}{3}) ✓

\forall x < - 3, x > 2 \Rightarrow f (x) > g (x) \Rightarrow t h e r e f o r t h e e q u a t i o n f (x) = g (x) h a s n o s o l u t i o n .

Commented by mehdee42 last updated on 19/Apr/23