2x - 3 ≥ 3x + 5 can be represented as ____.

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The given inequality is 2x - 3 ≥ 3x + 5. Subtracting 3x from both sides of the inequality, - x - 3 ≥ 5 Adding 3 to both sides of the inequality, - x ≥ 8 Multiplying both sides of the inequality by -1, the inequality is reversed. x ≤ - 8

The solution set in R for the inequality (x + 2)5 - (x - 4)5 ≤ 3126 is ____.

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The given inequality is (x + 2)5 - (x - 4)5 ≤ 3126. The average of x + 2 and x - 4 is (x+2) + (x-4)/2 = 2x-2/2 = x – 1 So the given inequality can be written as : ((x - 1) + 3)5 - ((x - 1) - 3)5 ≤ 3126 Putting y = x - 1, it becomes (y + 3)5 - (y - 3)5≤ 3126 Expanding the LHS, (y5 + 5 . y4 . 3 + 10 . y3 . 9 + 10 . y2 . 27 + 5 . y . 81 + 243) - (y5 - 5 . y4 . 3 + 10 . y3 . 9 - 10 . y2. 27 + 5 . y . 81 - 243) ≤ 3126 => 2 ( 5 . y4 . 3 + 10 . y2. 27 + 243) ≤ 3126 => 30y4 + 540y2 + 486 ≤ 3126 On simplification, y4 + 18y2- 88 ≤ 0 => (y2 - 4)(y2 + 22) ≤ 0 But y2 + 22 ≥ 0 for all real values of y. So the given inequality is satisfied whenever y2 – 4 ≤ 0. That is, (y + 2)(y - 2) ≤ 0 This inequality holds when -2 ≤ y ≤ 2 Therefore, -2 ≤ x - 1 ≤ 2 => - 1 ≤ x ≤ 3 So the solution set for the given inequality is : x ∈ [-1, 3]

Solve for x ε R, where 3x + 31 - x – 4 < 0.

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The given inequality is: 3x + 31 - x – 4 < 0 That is , 3x + 3/3x - 4 < 0 Put 3x = y. Then y > 0 ∀x∈R The inequality becomes : y + 3/y - 4 0, multiplying throughout by y, y2 + 3 - 4y (y - 3)(y - 1) < 0 The solution set for this inequality is : 1 < y < 3 That is, 1 < 3x 30 < 3x < 31 => 0 < x < 1

x is an integer; one-third of the next integer is greater than one-fourth of its previous integer by 2. The minimum value of x is____.

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Given, One-third of the next integer = (x+1)/3 One-fourth of the previous integer = (x−1)/4 (x+1)/3 - (x−1)/4 > 2 ⇒ 4x + 4 - 3x + 3 > 24 ⇒ x > 17 Hence, the minimum value of the x is 18.

The solution set for the inequality |x + 1| + |x - 1| + |x - 3| ≥ 9 is ____.

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The definition of |x| is: |x| = x, when x ≥ 0 = -x, when x < 0 So, |x + 1| = x + 1, when x + 1 ≥ 0, i.e., when x ≥ - 1 = 1 - x, when x + 1 < 0, i.e., when x < - 1 Similarly, |x - 1| = x - 1, when x - 1 ≥ 0, i.e., when x ≥ 1 = 1 - x, when x - 1 < 0, i.e., when x < 1 And, |x - 3| = x - 3, when x - 3 ≥ 0, i.e., when x ≥ 3 = 3 - x, when x - 3 < 0, i.e., when x < 3 When x 3 - 3x ≥ 9 => - 3x ≥ 6 Dividing both sides of the inequality by -3, the inequality is reversed: x ≤ -2 So the given inequality holds in the interval (- ∞, -2]. When - 1 ≤ x - x ≥ 4 Multiplying both side of the inequality by -1, the inequality is reversed : x ≤ - 4 This is not possible, as we have considered -1 ≤ x < 1. When 1 ≤ x 3 + x ≥ 9 => x ≥ 6 When x ≥ 3: |x + 1| + |x - 1| + |x - 3| = (x + 1) + (x - 1) + (x - 1) + (x - 3) = 3x – 3 Therefore, |x + 1|+ |x - 1| + |x - 3| ≥ 9 => 3x - 3 ≥ 12 => x ≥ 4 So the given inequality holds in the interval [4, ∞). Hence, the solution set is: ( - ∞, - 2] U [4, ∞)

The solution set of ||x + 1| - 2 | ≤ 4, where x ∈ R is ____.

Correct! Wrong!

The given inequality is ||x + 1| -2 ≤ 4. When x ≥ - 1: x + 1 ≤ 0 => |x + 1 | = x + 1 Therefore, the given inequality becomes: |(x + 1) - 2| ≤ 4 => |x - 1| ≤ 4 When x ≥ 1 : x - 1 ≥ 0 => |x - 1| = x -1 Therefore |x - 1| ≤ 4 => x ≤ 5 Therefore the given inequality is satisfied in the interval 1 ≤ x ≤ 5 When x < 1 : x - 1 | x - 1| = 1 – x Therefore |x - 1| ≤ 4 => 1 - x ≤ 4 => -x ≤ 3 => x ≥ -3 Therefore the given inequality is satisfied in the interval – 3 ≤ x ≤ 1 When x < -1: x + 1 |x + 1| = - (x + 1) Therefore the given inequality becomes : | - (x + 1) - 2 | ≤ 4 => |x + 3| ≤ 4 When x ≥ -3 : x + 3 ≥0 => |x + 3| = x + 3 Therefore, |x + 3| ≤ 4 => x + 3 ≤ 4 => x ≤ 1 Therefore the given inequality is satisfied in the interval -3 ≤ x ≤ 1 When x < -3 : x + 3 |x + 3| = -(x + 3) = -x – 3 Therefore, |x + 3| ≤ 4 => -x - 3 ≤ 4 => -x ≤ 7 => x ≥ - 7 Therefore the given inequality is satisfied in the interval -7 ≤ x < -3 So the solution set is [1, 5] U [-3, 1) U [-3, 1] U [ -7, -3) which is nothing but [-7, 5]. Therefore the solution set for the given inequality is: x ∈ [ -7, 5].

Mohan has Rs. 500 and Sohan has Rs. 600. Mohan wants to buy two similar pens and Sohan wants to buy three pens of the same type. With the remaining amount with them, they wish to buy a chess board. If x represents the price of a pen and y represents the price of a chess board, an expression using inequalities, to ascertain the maximum price possible for a pen and a chess board is ____.

Correct! Wrong!

Mohan has Rs.500. He wants to buy two similar pens. If x represents the price of a pen, 2x ≤ 500 Sohan has Rs.600. He wants to buy three pens of the same type. Therefore, 3x ≤ 600 With the remaining amount with them, they wish to buy a chess board. The amount remaining with them after purchasing pens is 1100 - 5x. Hence if y represents the price of a chess board, then y ≤ 1100 - 5x. That is, 5x + y ≤ 1100

If 4 is added to three times a positive integer, then it is greater than four times that positive integer reduced by 1. The solution set is ____.

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Let the positive integer be x. 4 is added to three times the positive integer = 3x + 4 Four times the positive integer reduced by 1 = 4x – 1 Given condition is 3x + 4 > 4x – 1 ⇒ -x > -5 ⇒ x < 5 The solution set is {1, 2, 3, 4}

If 2 ≤ 4x2-9x+20/x2+2 ≤ 3, x ∈ R, then x is ____.

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the given inequalities are 2 ≤ 4 x2-9x+20/ x2+2 ≤ 3 Multiplying thoughout by x2 + 2, 2 (x2 + 2) ≤ 4 x2 - 9x + 20 ≤ 3(x2+ 2) Consider the first inequality 2(x2 + 2) ≤ 4 x2 - 9x + 20 That is, 2 x2 + 4 ≤ 4 x2 - 9x + 20 Simplifying, 2 x2 - 9x + 16 ≥ 0 The corresponding equation 2 x2 - 9x + 16 = 0 has D = B2 - 4AC = 81 - 4 × 2 × 16 = 47 As D < 0, the value of the expression 2 x2 - 9x + 16 will have the same sign as that of its leading coefficient, A. Here A = 2 >0. Therefore, 2 x2 - 9x + 16 is always positive for all real values of x. So the first inequality holds for all x ε R. Now consider the second inequality 4 x2 - 9x + 20 ≤ 3 (x2 + 2) That is, 4x2 - 9x + 20 ≤ 3x2 + 6 Simplifying, x2 - 9x + 14 ≤ 0 => (x - 7)(x - 2) ≤ 0 Therefore, x = 7 or x = 2 The product (x - 7)(x - 2) is negative when x lies between 2 and 7. Hence the solution set for the second inequality is [2, 7] Therefore, the solution for the given inequality is R ∩[2,7]=[2,7]

Point satisfying the inequation 3x + 4y < 11 is ____.

Correct! Wrong!

Given, 3x + 4y < 11 - - - - (1) Substituting the point (2, 4) in (1), we get 3(2) + 4(4) = 22 < 11, which is false Substituting the point (-1, 0) in (1), we get 3(-1) + 4(0) = -3 < 11, which is true Substituting the point (1, 6) in (1), we get 3(1) + 4(6) = 27 < 11, which is false Substituting the point (1, 3) in (1), we get 3(1) + 4(3) = 15 < 11, which is false

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CBSE Class 11th Math 6 - Linear Inequalities MCQs
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