Question

question of inequality class 11.
it is urgent . *NO SPAM*
ANSWER ONLY IF YOU KNOW

please provide solution also​

Answer 1

AnsWer :

d ) None of these.

QuestioN :

Solve : $\tt \dagger \: \: \: \: \: \dfrac{ |x + 2| - x }{x} < 2.$

• a ) ( - ∞ , 0 ) ∪ ( 1 , ∞ )
• b ) ( ∞ , 0 ) ∪ ( 1 , - ∞ )
• c ) ( - ∞ , 0 ) ∩ ( 1 , ∞ )
• d ) None of these.

SolutioN :

$\tt \dagger \: \: \: \: \: \dfrac{ |x + 2| - x }{x} < 2.$

» Taking Modules part,

☛ Condition 1.

$\tt : \implies |x + 2|$

» Let's take Condition,

• If x + 2 equal to and larger than.

$\tt : \implies x + 2 \geq0.$

• Both sides taking ( - 2 )

$\tt : \implies x + 2 - 2 \geq0- 2$

$\tt : \implies x \geq - 2$

$\rule{100}5$

☛ Condition 2.

• If x + 2 is less than.

$\tt : \implies x + 2 < 0.$

• Both sides taking ( - 2 )

$\tt : \implies x + 2 - 2 < 0 - 2.$

$\tt : \implies x < - 2.$

✐ Let's Taking Condition :

$\tt \dagger \: \: \: \: \: \dfrac{ x + 2 - x }{x} < 2.$

$\tt : \implies x + 2 - x < 2x$

$\tt : \implies \cancel{x} + 2 \cancel{- x} < 2x$

$\tt : \implies 2 < 2x$

$\tt : \implies \dfrac{2}{2} < x$

$\tt : \implies 1 < x$

$\tt : \implies x> 1.$

According to Condition,

• x lies on - 2 and greater than - 2,
• Now, x > 1.

If we combine, So We can write as,

$\tt \dagger \: \: \: \: \: \: \: \: \: \: \: \: x \in [ -2 , 0 ) \cup[ 1, \infty )$

$\rule{100}5$

✐ Condition 2. ( Modules )

$\tt \dagger \: \: \: \: \: \dfrac{ - x - 2 - x }{x} < 2.$

$\tt : \implies \dfrac{ - x - 2 - x }{x} < 2.$

$\tt : \implies \dfrac{ - 2x - 2 }{x} < 2.$

• Both sides multiple by x.

$\tt : \implies \dfrac{ \cancel x (- 2x - 2) }{ \cancel{x}} < 2x.$

$\tt : \implies 0 < 2x + 2x + 2$

$\tt : \implies 0 < 4x + 2$

$\tt : \implies - \dfrac{1}{2} < x$

$\tt : \implies x > - \dfrac{1}{2}$

# According to Condition,

• x lies on less than 1 / 2, x > 1 / 2.
• Now, x < - 2
• If we combine, So We can write as,

$\tt \dagger \: \: \: \: \: \: \: \: \: \: \: \: x \in ( - \infty , - 2 )$

Now, according to observation, condition 1 , 2 We can say that.

$\tt \dagger \: \: \: \: \: \: \: \: \: \: \: \: x \in ( - \infty , - 2 ) \cup[ 1, \infty )$

Therefore, the correct option is d ) none of these.

Answer 2

Answer:

AnsWer :

d ) None of these.

QuestioN :

Solve : \tt \dagger \: \: \: \: \: \dfrac{ |x + 2| - x }{x} < 2.†

x

∣x+2∣−x

<2.

a ) ( - ∞ , 0 ) ∪ ( 1 , ∞ )

b ) ( ∞ , 0 ) ∪ ( 1 , - ∞ )

c ) ( - ∞ , 0 ) ∩ ( 1 , ∞ )

d ) None of these.

SolutioN :

\tt \dagger \: \: \: \: \: \dfrac{ |x + 2| - x }{x} < 2.†

x

∣x+2∣−x

<2.

» Taking Modules part,

☛ Condition 1.

\tt : \implies |x + 2|:⟹∣x+2∣

» Let's take Condition,

If x + 2 equal to and larger than.

\tt : \implies x + 2 \geq0.:⟹x+2≥0.

Both sides taking ( - 2 )

\tt : \implies x + 2 - 2 \geq0- 2:⟹x+2−2≥0−2

\tt : \implies x \geq - 2:⟹x≥−2

\rule{100}5

☛ Condition 2.

If x + 2 is less than.

\tt : \implies x + 2 < 0.:⟹x+2<0.

Both sides taking ( - 2 )

\tt : \implies x + 2 - 2 < 0 - 2.:⟹x+2−2<0−2.

\tt : \implies x < - 2.:⟹x<−2.

✐ Let's Taking Condition :

\tt \dagger \: \: \: \: \: \dfrac{ x + 2 - x }{x} < 2.†

x

x+2−x

<2.

\tt : \implies x + 2 - x < 2x:⟹x+2−x<2x

\tt : \implies \cancel{x} + 2 \cancel{- x} < 2x:⟹

x

+2

−x

<2x

\tt : \implies 2 < 2x:⟹2<2x

\tt : \implies \dfrac{2}{2} < x:⟹

2

2

<x

\tt : \implies 1 < x:⟹1<x

\tt : \implies x > 1.:⟹x>1.

According to Condition,

x lies on - 2 and greater than - 2,

Now, x > 1.

If we combine, So We can write as,

\tt \dagger \: \: \: \: \: \: \: \: \: \: \: \: x \in [ -2 , 0 ) \cup[ 1, \infty )†x∈[−2,0)∪[1,∞)

\rule{100}5

✐ Condition 2. ( Modules )

\tt \dagger \: \: \: \: \: \dfrac{ - x - 2 - x }{x} < 2.†

x

−x−2−x

<2.

\tt : \implies \dfrac{ - x - 2 - x }{x} < 2.:⟹

x

−x−2−x

<2.

\tt : \implies \dfrac{ - 2x - 2 }{x} < 2.:⟹

x

−2x−2

<2.

Both sides multiple by x.

\tt : \implies \dfrac{ \cancel x (- 2x - 2) }{ \cancel{x}} < 2x.:⟹

x

x

(−2x−2)

<2x.

\tt : \implies 0 < 2x + 2x + 2:⟹0<2x+2x+2

\tt : \implies 0 < 4x + 2:⟹0<4x+2

\tt : \implies - \dfrac{1}{2} < x:⟹−

2

1

<x

\tt : \implies x > - \dfrac{1}{2}:⟹x>−

2

1

# According to Condition,

x lies on less than 1 / 2, x > 1 / 2.

Now, x < - 2

If we combine, So We can write as,

\tt \dagger \: \: \: \: \: \: \: \: \: \: \: \: x \in ( - \infty , - 2 )†x∈(−∞,−2)

Now, according to observation, condition 1 , 2 We can say that.

\tt \dagger \: \: \: \: \: \: \: \: \: \: \: \: x \in ( - \infty , - 2 ) \cup[ 1, \infty )†x∈(−∞,−2)∪[1,∞)

Therefore, the correct option is d ) none of these.