Find-the-range-of-values-of-x-x-1-2x-x-2-1-

Question Number 52282 by Tawa1 last updated on 05/Jan/19

Find the range of values of x

∣ \frac{x - 1}{2 x} ∣ ⩾ x^{2} + 1

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

f r o m g r a p h i t c l e a r

∣ \frac{x - 1}{2 x} ∣= x^{2} + 1 a t x = - 0.59 a n d x = 0.313

a n d ∣ \frac{x - 1}{2 x} ∣> x^{2} + 1 w h e n x (- 0.59, 0.313)

s o g i v e n ∣ \frac{x - 1}{2 x} ∣⩾ x^{2} + 1 [- 0.59, 0.313]

Commented by Tawa1 last updated on 05/Jan/19

God bless you sir .

But sir . can it be solve without graph sir . if yes .

please help when you are chanced .

Commented by Tawa1 last updated on 06/Jan/19

Please sir, where can i study questions like this, on which textbook

sir

Commented by Abdo msup. last updated on 05/Jan/19

f o r x \neq 0 (e) \Leftrightarrow∣ x - 1 ∣⩾ 2 ∣ x ∣ (x^{2} + 1) \Leftrightarrow

∣ x - 1 ∣ - 2 ∣ x ∣ (x^{2} + 1) ⩾ 0 l e t f (x) =∣ x - 1 ∣ - 2 ∣ x ∣ (x^{2} + 1)

s o (e) \Leftrightarrow f (x) ⩾ 0

x - \infty 0 1 + \infty

∣ x - 1 ∣ - x + 1 - x + 1 0 x - 1

∣ x ∣ - x 0 x x

f (x) 2 x^{3} + x + 1 - 2 x^{3} - 3 x + 1 - 2 x^{3} - x - 1

c a s e 1 x ⩽ 0 (e) \Leftrightarrow 2 x^{3} + x + 1 ⩾ 0

l e t f i n d r o o t s o f p (x) = 2 x^{3} + x + 1

t h e r o o t s a r e x_{1} \sim - 0, 5898 (r e a l)

x_{2} = 0, 2949 + 0, 8723 i (c o m p l e x)

x_{3} = 0, 2949 - 0, 8723 i (c o m p l e x) \Rightarrow

f (x) = 2 (x - x_{1}) (x - x_{2}) (x - x_{3})

= 2 (x - x_{1}) (x - α - i β) (x - α + i β)

= 2 (x - x_{1}) ({(x - α)}^{2} + β^{2}) s o

f (x) ⩾ 0 \Leftrightarrow x ⩾ x_{1} \Rightarrow S_{1} = [x_{1}, 0 [

c a s e 2 0 < x ⩽ 1 (e) \Leftrightarrow - 2 x^{3} - 3 x + 1 ⩾ 0 \Leftrightarrow

2 x^{3} + 3 x - 1 ⩽ 0 \dots b e c o n t i n u e d \dots

Commented by Tawa1 last updated on 06/Jan/19

God bless you sir

Commented by maxmathsup by imad last updated on 06/Jan/19

i h a v n t t h e n a m e b u t y o u c a n f i n d t h i s i n o l d b o o k s \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

1) x \neq 0

2) \frac{x - 1}{2 x} = x^{2} + 1

2 x^{3} + 2 x - x + 1 = 0

2 x^{3} + x + 1 = 0

x = - 0.59

\frac{- (x - 1)}{2 x} = x^{2} + 1

2 x^{3} + 2 x + x - 1 = 0

2 x^{3} + 3 x - 1 = 0

x = 0.313

c r i t i c a l v a l u e o f x a r e - 0.59 a n d 0.313

n o w p u t c o n d i t i o n

i) 1 > x > 0.313

i i) x < - 0.59

i i i) 0.313 > x > - 0.59

∣ \frac{x - 1}{2 x} ∣> x^{2} + 1

w h e n 1 > x > 0.313

r i g h t h a n d s i d e x^{2} + 1 i s + v e w h e n 1 > x > 0.313

b u t \frac{x - 1}{2 x} i s n e g e t i v e s o ∣ \frac{x - 1}{2 x} ∣ ⪈ x^{2} + 1 w h e n 1 > x > 0.313

b u t f o r x = 0.313 \frac{(x - 1)}{2 x} = x^{2} + 1 t r u e

i i) x < - 0.59

∣ \frac{x - 1}{2 x} ∣ i s \frac{x - 1}{2 x} (p u t a n y v a l u e o f x < - 0.59

s a y x = - 1)

L H S = \frac{- 1 - 1}{2 \times - 1} = 1

R H S {(- 1)}^{2} + 1 = 2

L H S ≯ R H S

s o w h e n x < - 0.59 g i v e n c o n d i t i o n

d o e s n o t h o l d

s o x c a n n o t b e l e s s t h a n - 0.59

i i i) n o w w h e n 0.313 > x > - 0.59

[s a y x = 0.2

L H S

∣ \frac{x - 1}{2 x} ∣

=∣ \frac{0.2 - 1}{2 \times 0.2} ∣

= 2

R H S {(0.2)}^{2} + 1 = 1.04

L H S > R H S

s o w h e n 0.313 > x > - 0.59

g i v e n c l n d i t i o n h o l d

s o l u t i o n s e t f o r x [- 0.59, 0.313]

Commented by Tawa1 last updated on 05/Jan/19

God bless you sir . Is that all sir

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19

p l s s e e f i n a l a n s w e r

Commented by Tawa1 last updated on 05/Jan/19

God bless you sir

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