The solution set of inequality 1 2 x - 1 > 1 1 - 2 x - 1 is

0 , log 2 ( 4 3 ) ∪ ( 1 , ∞ )

Solve: x - 1 - | x | < 0

Given: x < 1 - | x | . . . ( 1 ) First we will define the term under the square root. 1 - | x | ≥ 0 ⇒ | x | ≤ 1 ⇒ - 1 ≤ x ≤ 1 ( If | x | ≤ a ( a ≥ 0 ) then - a ≤ x ≤ a ) x = 1 does not satisfy the above inequality. Now we take 2 cases: ( a ) - 1 ≤ x < 0 x < 1 + x as LHS is negative and RHS is positive, we know that negative is less than positive so, x ∈ [ - 1 , 0 ) . . . ( i ) ( b ) Now consider 0 ≤ x < 1 . . . ( a ) Since x < 1 - x squaring both sides, we get x 2 < 1 - x ⇒ x 2 + x - 1 < 0 ⇒ x + 1 2 2 < 5 4 ⇒ x + 1 2 < 5 2 Use modulus inequality - 5 2 < x + 1 2 < 5 2 ⇒ - 5 - 1 2 < x < 5 - 1 2 . . . ( b ) Taking ( a ) ∩ ( b ) , 0 ≤ x < 5 - 1 2 . . . ( ii ) So, combining ( i ) and ( ii ) , we get: x ∈ [ - 1 , 5 - 1 2 ) .

E M B I B E

Chapterwise/Topicwise Daily Practice Problems (DPP) Algebra JEE Main & Advanced>Chapter 1 - Inequalities and Modulus>DPP>Q 10

HARD

JEE Main

IMPORTANT

Earn 100

The solution set of inequality $\frac{(e^{x} - 1) (2 x - 3) (x^{2} + x + 2)}{(\sin x - 2) (x + 1) x} \leq 0$ is

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Important Questions on Inequalities and Modulus

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Algebra JEE Main & Advanced>Chapter 1 - Inequalities and Modulus>DPP>Q 11

The solution set of inequality

\frac{1}{2^{x} - 1} > \frac{1}{1 - 2^{x - 1}}

is

MEDIUM

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Algebra JEE Main & Advanced>Chapter 1 - Inequalities and Modulus>DPP>Q 12

Let $A = \{x |x^{2} - 4 x + 3 < 0, x \in R\}$ and $B = \{x |2^{1 - x} + P \leq 0 & x^{2} - 2 (P + 7) x + 5 \leq 0\}$ . If $B \subseteq A,$ then $P \in$