Solve: x - 6 - 10 - x ≥ 1

We have, x - 6 - 10 - x ≥ 1 By the definition of square root function, x − 6 ⩾ 0 ⇒ x ⩾ 6 . . . . . 1 and 10 − x ⩾ 0 ⇒ x ≤ 10 . . . . . . . . ( 2 ) ⇒ x ∈ [ 6 , 10 ] … ( 3 ) Now, x - 6 ≥ 1 + 10 - x Squaring both sides, we get x − 6 ≥ ( 1 + 10 − x ) 2 ⇒ x − 6 ≥ 1 + ( 10 − x ) + 2 10 − x ⇒ 2 x − 17 ≥ 2 10 − x Squaring again, we get ( 2 x − 17 ) 2 ≥ 4 ( 10 − x ) ⇒ 4 x 2 + 289 − 68 x ≥ 40 − 4 x ⇒ 4 x 2 − 64 x + 249 ≥ 0 ⇒ x = 64 ± 64 2 − 4 4 ( 249 ) 2 4 = 64 ± 4 16 2 − 249 2 4 = 16 ± 7 2 ⇒ x − 16 − 7 2 x − 16 + 7 2 ⩾ 0 x ∈ ( − ∞ , 16 − 7 2 ] ∪ [ 16 + 7 2 , ∞ ) . . . ( 4 ) Taking intersection with 3 , we get: x ∈ 6 , 16 − 7 2 ∪ 16 + 7 2 , 10

Solve: 1 - 21 - 4 x - x 2 x + 1 ≥ 0

We have 1 - 21 - 4 x - x 2 x + 1 ≥ 0 Expression inside square root cannot be negative, therefore, 21 - 4 x - x 2 ≥ 0 ⇒ x 2 + 4 x - 21 ≤ 0 ⇒ x 2 + 7 x - 3 x - 21 ≤ 0 ⇒ ( x + 7 ) ( x - 3 ) ≤ 0 ⇒ x ∈ [ - 7 , 3 ] . . . . ( 1 ) Also, x + 1 ≠ 0 ⇒ x ≠ - 1 Now, 1 - 21 - 4 x - x 2 x + 1 ≥ 0 Case I : When 1 − 21 − 4 x − x 2 ⩾ 0 and x + 1 > 0 So, 21 − 4 x − x 2 ≤ 1 Squaring both sides, we get: ⇒ 21 − 4 x − x 2 ≤ 1 ⇒ x 2 + 4 x − 20 ⩾ 0 ⇒ x - ( - 2 + 2 6 ) ) ( x - ( - 2 - 2 6 ) ≥ 0 x ∈ ( - ∞ , - 2 - 2 6 ] ∪ [ - 2 + 2 6 , ∞ ) . . . ( 2 ) And also, x + 1 > 0 x > − 1 ⇒ x ∈ [ − 2 + 2 6 , ∞ ) . . . ( 4 ) Case II : When x + 1 < 0 x < - 1 . . . ( 5 ) and 1 - 21 - 4 x - 2 - x 2 ≤ 0 So, 21 - 4 x - 2 - x 2 ≥ 1 Squaring both sides, we get: ⇒ 21 - 4 x - x 2 ≥ 1 ⇒ x 2 + 4 x - 20 ≤ 0 ⇒ x ∈ [ - 2 - 2 6 , - 1 ) . . . ( 6 ) From 5 and 6 , we get: x ∈ [ - 2 - 2 6 , - 1 ) . . . ( 7 ) Taking union of 4 and 7 , we get: x ∈ [ − 2 − 2 6 , − 1 ) ∪ [ − 2 + 2 6 , ∞ ) … ( 8 ) Taking intersection of 8 with 1 , we get: x ∈ [ − 2 − 2 6 , − 1 ) ∪ [ − 2 + 2 6 , 3 ]

The complete set of values of x for which x 3 ( x - 1 ) 2 ( x + 4 ) ( x + 1 ) ( x - 3 ) ≥ 0 , is

( - ∞ , - 4 ] ∪ ( - 1 , 0 ] ∪ ( 3 , ∞ ) ∪ { 1 }

( - ∞ , - 4 ] ∪ ( - 1 , 0 ] ∪ ( 3 , ∞ ) ∪ { - 1 }

[ - 4 , - 1 ) ∪ [ 0 , 1 ) ∪ ( 3 , ∞ )

[ - 4 , - 1 ) ∪ [ 0 , 1 ] ∪ ( 3 , ∞ )

The set of all values of x for which ( x + 1 ) ( x - 3 ) 2 ( x - 5 ) ( x - 4 ) 3 ( x - 2 ) x < 0 is

( - ∞ , - 1 ) ∪ ( 0 , 2 ) ∪ ( 4 , 5 )

( - 1 , 0 ) ∪ ( 2 , 4 ) ∪ ( 5 , ∞ )

( - 1 , 0 ) ∪ ( 2 , 3 ) ∪ ( 4 , 5 )

( - ∞ , - 1 ) ∪ ( 0 , 2 ) ∪ [ 3 , 5 )

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

0 , log 2 ( 4 3 ) ∪ ( 1 , ∞ )

E M B I B E

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

G Tewani Solutions for Chapter: Inequalities and Modulus, Exercise 1: DPP

Author:G Tewani

G Tewani Mathematics Solutions for Exercise - G Tewani Solutions for Chapter: Inequalities and Modulus, Exercise 1: DPP

Attempt the free practice questions on Chapter 1: Inequalities and Modulus, Exercise 1: DPP with hints and solutions to strengthen your understanding. Chapterwise/Topicwise Daily Practice Problems (DPP) Algebra JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.

Questions from G Tewani Solutions for Chapter: Inequalities and Modulus, Exercise 1: DPP with Hints & Solutions

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Algebra JEE Main & Advanced>Chapter 1 - Inequalities and Modulus>DPP>Q 5

Solve: $\sqrt{x - 6} - \sqrt{10 - x} \geq 1$

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Algebra JEE Main & Advanced>Chapter 1 - Inequalities and Modulus>DPP>Q 6

Solve: $\frac{1 - \sqrt{21 - 4 x - x^{2}}}{x + 1} \geq 0$

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Algebra JEE Main & Advanced>Chapter 1 - Inequalities and Modulus>DPP>Q 8

The complete set of values of $x$ for which $\Rightarrow 4 x^{2} - 64 x + 249 \geq 0$ , is

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Algebra JEE Main & Advanced>Chapter 1 - Inequalities and Modulus>DPP>Q 9

The set of all values of $x$ for which $= \frac{16 \pm \sqrt{7}}{2}$ is

HARD

JEE Main

IMPORTANT

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

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

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 $P \in$