Solve ( x + 3 ) 5 - x - 1 5 ≥ 244 .

Given, ( x + 3 ) 5 - ( x - 1 ) 5 ≥ 244 . . . ( i ) ( x + 1 + 2 ) 5 - ( x + 1 - 2 ) 5 ≥ 244 Let λ = x + 1 . . . ( i i ) ( λ + 2 ) 5 - ( λ - 2 ) 5 ≥ 244 Using the binomial theorem we get, 2 { c 1 5 . λ 4 . 2 1 + c 3 5 . λ 2 . 2 3 + c 5 5 . λ 0 . 2 5 } ≥ 244 2 { 10 . λ 4 + 80 . λ 2 + 32 } ≥ 244 5 λ 4 + 40 λ 2 + 16 ≥ 61 5 λ 4 + 40 λ 2 - 45 ≥ 0 λ 4 + 8 λ 2 - 9 ≥ 0 Now, let λ 2 = t ... i i i t 2 + 8 t - 9 ≥ 0 ( t + 9 ) ( t - 1 ) ≥ 0 t ≤ - 9 or t ≥ 1 Using equation i i i we get, λ 2 ≤ - 9 (This is not a possible solution.) Hence, λ 2 ≥ 1 λ 2 - 1 ≥ 0 ( λ + 1 ) ( λ - 1 ) ≥ 0 Using equation (ii) we get, ( x + 1 + 1 ) ( x + 1 - 1 ) ≥ 0 ( x + 2 ) ( x ) ≥ 0 x ∈ ( - ∞ , - 2 ] ∪ [ 0 , ∞ ) .

Solve x - 2 - 1 ≥ 3 .

Given, x - 2 - 1 ≥ 3 When x ≥ a , x ∈ ( - ∞ , - a ] ∪ [ a , ∞ ) : a > 0 Using the above concept, we can write, x - 2 - 1 ∈ ( - ∞ , - 3 ] ∪ [ 3 , ∞ ) Adding one on both sides of the equation, we get, x - 2 ∈ ( - ∞ , - 2 ] ∪ [ 4 , ∞ ) . x cannot be negative, so neglecting the negative portion we can write, x - 2 ≥ 4 Using the concept x ≥ a , x ∈ ( - ∞ , - a ] ∪ [ a , ∞ ) : a > 0 . Now, we can write, x - 2 ∈ ( - ∞ , - 4 ] ∪ [ 4 , ∞ ) Adding two on both sides of the equation, we get, x ∈ ( - ∞ , - 2 ] ∪ [ 6 , ∞ ) This is the required solution set for the given equation.

Solve 1 ≤ 3 x 2 - 7 x + 8 x 2 + 1 ≤ 2 .

Given, 1 ≤ 3 x 2 - 7 x + 8 x 2 + 1 ≤ 2 The basic concept of inequality tells us that if the denominator is positive for all x , then we can directly cross multiply it. x 2 + 1 ≤ 3 x 2 - 7 x + 8 ≤ 2 ( x 2 + 1 ) Now we can break it into two cases: Case I: x 2 + 1 ≤ 3 x 2 - 7 x + 8 3 x 2 - 7 x + 8 - x 2 - 1 ≥ 0 2 x 2 - 7 x + 7 ≥ 0 Let y = a x 2 + b x + c . If a > 0 and b 2 - 4 a c < 0 , then y > 0 . Using the above concept, we can say that 2 x 2 - 7 x + 7 is always positive. Hence, x ∈ R . Case II: 3 x 2 - 7 x + 8 ≤ 2 ( x 2 + 1 ) 3 x 2 - 7 x + 8 - 2 ( x 2 + 1 ) ≤ 0 3 x 2 - 7 x + 8 - 2 x 2 - 2 ≤ 0 x 2 - 7 x + 6 ≤ 0 ( x - 1 ) ( x - 6 ) ≤ 0 After solving the above inequality, we get, x ∈ [ 1 , 6 ] Our final solution set will be the intersection of Case I and Case II. Hence, our final solution set is x ∈ [ 1 , 6 ] .

Let f x = 2 x 2 x 2 + 5 x + 2 and g ( x ) = 1 x + 1 . Find the set of real values of x for which f x > g x .

Given, f ( x ) = 2 x 2 x 2 + 5 x + 2 . . . ( i ) g ( x ) = 1 x + 1 . . . ( ii ) We have to solve f ( x ) > g ( x ) using equations ( i ) and ( ii ) . 2 x 2 x 2 + 5 x + 2 > 1 x + 1 1 x + 1 - 2 x 2 x 2 + 5 x + 2 < 0 ( 2 x 2 + 5 x + 2 ) - 2 x ( x + 1 ) ( x + 1 ) ( 2 x 2 + 5 x + 2 ) < 0 2 x 2 + 5 x + 2 - 2 x 2 - 2 x ( x + 1 ) ( x + 2 ) ( 2 x + 1 ) < 0 3 x + 2 ( x + 1 ) ( x + 2 ) ( 2 x + 1 ) < 0 ( x + 2 3 ) ( x + 1 ) ( x + 2 ) ( x + 1 2 ) < 0 Now plot all zeros on a number line. x ∈ ( - 2 , - 1 ) ∪ ( - 2 3 , - 1 2 )

For x ∈ R , x is defined as follows: x = x + 1 : 0 ≤ x < 2 x - 4 : 2 ≤ x . Then, solve the equation x 2 + x = x + x 2 .

Given, x = x + 1 ; 0 ≤ x < 2 x - 4 ; x ≥ 2 . . . ( 1 ) We have to solve x 2 + x = x + x 2 . . . ( 2 ) Case I: When x ∈ [ 0 , 2 ) , Using equations ( 1 ) and ( 2 ) we get, ( x + 1 ) 2 + x = ( x + 1 ) + x 2 x 2 + 2 x + 1 = x 2 + 1 x = 0 In this case, we consider x ∈ [ 0 , 2 ) . Hence, x = 0 is a solution for this case. Case II : When x ∈ [ 2 , ∞ ) Using equation 1 and equation 2 we can write, ( x - 4 ) 2 + x = x - 4 + x 2 ( x - 4 ) 2 + x = x - 4 + x 2 x 2 - 8 x + 16 + x - x - 4 - x 2 = 0 16 - 7 x - x - 4 = 0 Now, if x ∈ [ 2 , 4 ) . . . ( 3 ) , 16 - 7 x - { - ( x - 4 ) } = 0 16 - 7 x + x - 4 = 0 12 = 6 x x = 2 . . . ( 4 ) Using equations 3 and 4 , we can say, x = 2 . . . ( 5 ) is a solution for this case. Now, if x ∈ [ 4 , ∞ ) . . . ( 6 ) , 16 - 7 x - ( x - 4 ) = 0 16 - 7 x - x + 4 = 0 20 = 8 x x = 5 2 but x ≥ 4 . Hence, there is no solution for this case. Hence, the equation x 2 + x = x + x 2 has only two solutions, x ∈ { 0 , 2 } .

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 1 - Essential Mathematical Tools>Proficiency in 'Essential Mathematical Tools' (Exercise)

Amit M Agarwal Solutions for Chapter: Essential Mathematical Tools, Exercise 3: Proficiency in 'Essential Mathematical Tools' (Exercise)

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Essential Mathematical Tools, Exercise 3: Proficiency in 'Essential Mathematical Tools' (Exercise)

Attempt the practice questions on Chapter 1: Essential Mathematical Tools, Exercise 3: Proficiency in 'Essential Mathematical Tools' (Exercise) with hints and solutions to strengthen your understanding. Skills in Mathematics Differential Calculus for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.

Questions from Amit M Agarwal Solutions for Chapter: Essential Mathematical Tools, Exercise 3: Proficiency in 'Essential Mathematical Tools' (Exercise) with Hints & Solutions

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 1 - Essential Mathematical Tools>Proficiency in 'Essential Mathematical Tools' (Exercise)>Q 4

Solve $2^{| x + 1 |} - 2^{x} = |2^{x} - 1| + 1$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 1 - Essential Mathematical Tools>Proficiency in 'Essential Mathematical Tools' (Exercise)>Q 5

Find the set of all real ‘ $a$ ’ such that $5 a^{2} - 3 a - 2, a^{2} + a - 2$ and $2 a^{2} + a - 1$ are the lengths of the sides of a triangle.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 1 - Essential Mathematical Tools>Proficiency in 'Essential Mathematical Tools' (Exercise)>Q 6

Solve ${(x + 3)}^{5} - {(x - 1)}^{5} \geq 244$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 1 - Essential Mathematical Tools>Proficiency in 'Essential Mathematical Tools' (Exercise)>Q 7

Solve $2^{| x + 1 |} - 2^{x} = |2^{x} - 1| + 1$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 1 - Essential Mathematical Tools>Proficiency in 'Essential Mathematical Tools' (Exercise)>Q 8

Solve $1 \leq \frac{3 x^{2} - 7 x + 8}{x^{2} + 1} \leq 2$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 1 - Essential Mathematical Tools>Proficiency in 'Essential Mathematical Tools' (Exercise)>Q 9

Let $f (x) = \frac{2 x}{2 x^{2} + 5 x + 2}$ and $g (x) = \frac{1}{x + 1}$ . Find the set of real values of $x$ for which $f (x) > g (x)$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 1 - Essential Mathematical Tools>Proficiency in 'Essential Mathematical Tools' (Exercise)>Q 10

For $x \in R$ , $||x||$ is defined as follows: $||x|| = \{\begin{matrix} x + 1 : 0 \leq x < 2 \\ |x - 4| : 2 \leq x \end{matrix}$ . Then, solve the equation ${||x||}^{2} + x = ||x|| + x^{2}$ .