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(d) Nine

| 2x + 6 | < \(\frac{19}{2}\)

⇒ 2x + 6 < \(\frac{19}{2}\) or - (2x + 6) < \(\frac{19}{2}\)

⇒ - \(\frac{19}{2}\) < 2x + 6 < \(\frac{19}{2}\)

⇒ - \(\frac{19}{2}\) - 6 < 2x < \(\frac{19}{2}\) - 6 ⇒ \(-\frac{31}{2}\) < 2x < \(\frac72\)

⇒ \(-\frac{31}{2}\) < x < \(\frac72\) ⇒ - 7.75 < x > 1.75

∴ The integers between –7.75 and 1.75 are –7, –6, –5, –4, –3, –2, –1, 0, 1, i.e., nine in number.

(b) –1 < x ≤ 3

 –1 < 5x + 4 ≤ 19 

⇒ –1 < 5x + 4 and 5x + 4 ≤ 19 

⇒ –1 – 4 < 5x and 5x ≤ 19 – 4 

⇒ –5 < 5x and 5x ≤ 15 

⇒ –1 < x and x ≤ 3 

⇒ –1 < x ≤ 3

(a) (– 5, – 1) ∪ (1, 5)

Let |x| = p, where p > 0            …(i) 

So |p – 3| < 2 and |p – 2| < 3            …(ii) 

⇒ 1 < p < 5 and – 1 < p < 5             …(iii) 

(∴ |x – a| < r ⇒ a – r < x < a + r) 

Therefore, the conditions (i), (ii) and (iii) are satisfied by 1 < p < 5, i.e. 1 < | x | < 5, i.e., | x | > 1 and | x | < 5 

i.e., x < – 1 or x > 1 and – 5 < x < 5 

⇒ \(x\)∈ (– 5, – 1) ∪ (1, 5).

2 ≤ 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4
⇒ 6 ≤ 3x ≤ 9
⇒ 2 ≤ x ≤ 3

Thus, all the real numbers, x, which are greater than or equal to 2 but less than or equal to 3, are the solutions of the given inequality. The solution set for the given inequality is [2, 3].

(c)  n(2c)^\(\frac1n\) 

For positive real numbers, AM > GM

⇒ \(\frac{(a_1+a_2+a_3+.......+2a_n)}{n}\) > (a₁.a₂.a₃.......2a_n)^\(\frac1n\)

⇒ (a₁ + a₂ + a₃ + ......+ 2a_n) > n(c.2)^\(\frac1n\) =  n(2c)^\(\frac1n\)

Hence the least value is  n(2c)^\(\frac1n\) 

We have, 

5x < 24

\(\frac{5x}{4}\) < \(\frac{24}{4}\)

or  x < \(\frac{24}{5}\)

This is true when

x =1,2,3,4 (x being a natural number)

(c) {3, 4, 5}

Given, 9 < 4x – 1 ≤ 19 

⇒ 9 < 4x – 1 and 4x – 1 ≤ 19 

⇒ 9 + 1 < 4x and 4x ≤ 19 + 1 

⇒  x > \(\frac52\) and x ≤ 5 

∴  x < \(\frac52\) ≤ 5

∵ x ∈ Z

 ∴ x = {3, 4, 5}.

(c) x > – 1

x³ + 1 > x² + \(x\) ⇒ x³ + 1 – x² – \(x\) > 0 

⇒ x³ – x² + 1 – \(x\) > 0 ⇒ x² (x – 1) – (x – 1) > 0 

⇒ (x² – 1) (x – 1) > 0 ⇒ (x + 1) (x – 1)² > 0 

As (x – 1)² is +ve so the given expression is > 0, when x + 1 > 0 ⇒ x > – 1.

(d) Both (a) and (b)

(|x – 1| – 3) (|x + 2| – 5) < 0 

The product of the two factors is < 0, i.e., negative when one of the factors is positive and one negative. 

Case I : (|x – 1| – 3) > 0 and (|x + 2| – 5) < 0 

⇒ |x – 1| > 3 and |x + 2| < 5 

⇒ – (x – 1) > 3, (x – 1) > 3 and –(x + 2) < 5, x + 2 < 5 

⇒ – x > 2, x > 4 and – x < 7, x < 3 

⇒ \(x\) < – 2, x > 4 and x > –7, x < 3 

Combining we get – 7 < x < – 2 

Case II : (|x – 1| – 3) < 0 and (|x + 2| – 5) > 0 

⇒ |x – 1| < 3 and |x + 2| > 5 

⇒ – (x – 1) < 3, (x – 1) < 3 and (x + 2) > 5, – (x + 2) > 5 

⇒ – \(x\) < 2, \(x\) < 4 and \(x\) > 3, – \(x\) > 7 

⇒ \(x\) > – 2, \(x\) < 4 and \(x\) > 3, \(x\) < – 7 

⇒ 3 < \(x\) < 4 

∴ –7 < \(x\) < – 2 and 3 < \(x\) < 4 is the solution.

(a) x∈ (– 1, 4)

– x² + 3x + 4 > 0 

⇒ x² – 3x – 4 < 0 ⇒ (x – 4) (x + 1) < 0 

⇒ (x – 4) < 0, (x + 1) > 0 or (x – 4) > 0, (x + 1) < 0 

⇒ x < 4, x > – 1 or x > 4, x < – 1 

⇒ – 1 < x < 4.

(a) – 8 < x < 1

\(x^{\frac23}+x^{\frac13}\) – 2 < 0 

⇒ y² + y – 2 < 0, where y = \(x^{\frac13}\) 

⇒ (y – 1) (y + 2) < 0 

⇒ (y – 1) < 0, (y + 2) > 0 or (y – 1) > 0, (y + 2) < 0 

⇒ y < 1, y > – 2 or y > 1, y < – 2 

⇒ – 2 < y < 1 

⇒ – 2 < \(x^{\frac13}\) < 1 

⇒ – 8 < x < 1.

(a) 1 < x < 2

x² – 3\(x\) + 2 > 0 ⇒ (\(x\) – 2) (\(x\) – 1) > 0 

⇒ (\(x\) – 2) > 0, (\(x\) – 1) < 0 or (\(x\) – 2) < 0, (\(x\) – 1) > 0 

⇒ \(x\) > 2, \(x\) < 1 or \(x\) < 2, \(x\) > 1 ⇒ \(x\) < 1 or \(x\) > 2 

∴ No value of x which lies between these extremes, i.e., 1 and 2 satisfies the inequality, i.e., 1 < \(x\) < 2 is the solution set not satisfying the inequality x² – 3\(x\) + 2 > 0 at all.

(D) is the correct choice, since |x +3| ≥ 10 ,

⇒ x + 3 ≤ – 10 or x + 3 ≥ 10

⇒ x ≤ – 13 or x ≥ 7

⇒ x ∈ (– ∞, – 13] ∪ [7, ∞)

(C) is the correct choice.

If x cm is the breadth, then

2 (3x + x) ≥ 160 ⇒ x ≥ 20

We have 3x – 5 < x + 7

⇒ 3x < x + 12  (Adding 5 to both sides)

⇒ 2x < 12 (Subtracting x from both sides)

⇒ x < 6 (Dividing by 2 on both sides)

(i) Solution set is {1, 2, 3, 4, 5}

(ii) Solution set is {0, 1, 2, 3, 4, 5}

(iii) Solution set is {....– 3, – 2, –1, 0, 1, 2, 3, 4, 5}

(iv) Solution set is {x : x ∈ R and x < 6}, i.e., any real number less than 6.

Given: 4x + 6 - 10 < 6x - 12 

⇒ 12 - 4 < 6x - 4x 

⇒ 8 < 2x

⇒ x > 8/2 = 4

∴ Solution set = (4, ∞)

An inequality is a statement involving quantities (variables or constants) connected to each other with signs of inequality, i.e.,<, <, > , >. 

Thus, x < y, x < y, x > y and x > y represent inequalities.

Given: 4x + 3 < 6x + 7 

⇒ - 7 + 3 < 6x – 4x

⇒ -4 < 2x 

⇒ x > - 4/2 = - 2

∴ Solution set = (-2, ∞).

Given: 4x + 3 < 5x + 7 

⇒ -7 + 3 < 5x – 4x 

⇒ -4 < x 

∴ Solution set = (-4, ∞).

Given: 3x – 7 > 5x – 1 

⇒1 – 7 > 5x – 3x 

⇒ -6 > 2x

⇒ x > - 6/2 = - 3

∴ Solution set = (-3, ∞)

Given: 3(x – 1) < 2(x – 3) 

⇒ 3x - 3 < 2x - 6 

⇒ 3x – 2x < - 6 + 3 

⇒ x < -3 

∴ Solution set = (-∞, -3].

3(2 – x) ≥ 2(1 – x)
⇒ 6 – 3x ≥ 2 – 2x
⇒ 6 – 3x + 2x ≥ 2 – 2x + 2x
⇒ 6 – x ≥ 2
⇒ 6 – x – 6 ≥ 2 – 6
⇒ –x ≥ –4
⇒ x ≤ 4

Thus, all real numbers x, which are less than or equal to 4, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (–∞, 4].

3(x – 1) ≤ 2(x – 3)
⇒ 3x – 3 ≤ 2x – 6
⇒ 3x – 3 + 3 ≤ 2x – 6 + 3
⇒ 3x ≤ 2x – 3
⇒ 3x – 2x ≤ 2x – 3 – 2x
⇒ x ≤ – 3

Thus, all real numbers x, which are less than or equal to –3, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (–∞, –3].

(i) False 

(ii) False 

(iii) True

(iv) False

(v) True 

(vi) False 

(vii) True 

(viii) False

(ix) True 

(x) False

(xi) True 

(xii) False

(xiii)False 

(xiv)True 

(xv) True

(i) ≤

(ii) ≥ 

(iii) > 

(iv) >

(v) > 

(vi) > 

(vii) < , >

(viii) ≤ 

(i) (≥), because same number can be added to both sides of inequality without changing the sign of inequality

(ii) (≥), after multiplying both sides by – 2, the sign of inequality is reversed.

(iii) (≤, ≤ ), x −1 ≤ 2 ⇒ – 2 ≤ x – 1 ≤ 2 ⇒ –1 ≤ x ≤ 3.

(iv) (<), as p is positive and q is negative, therefore, p + q is always smaller than p.

(b) –1 < x < 11

\(\bigg|\frac{5-x}{3}\bigg|<2\)  ⇒ |5-x| < 6

⇒ – 6 < 5 – x < 6 ⇒ –11 < – x < 1 ⇒ 11 > x > – 11 

(∵ a < b < c ⇒ ma > mb > mc, when m is –ve. Here m = –1) 

⇒ –1 < x < 11.

(b) (–∞, –10] ∪ [4, ∞)

|x + 3| > 7 

⇒ \(x\) + 3 < – 7 or x + 3 > 7 

⇒ \(x\) < – 10 or \(x\) > 4 ⇒ \(x\) ∈ (– ∞, – 10] or \(x\)∈ [4, ∞) 

⇒ \(x\)∈ (–∞, –10] ∪ [4, ∞).

(a)  all real numbers except \(-\frac12\).

4x² + 4x + 1 > 0 ⇒ (2x + 1)² > 0 

⇒ (2x + 1) < 0 or (2x + 1) > 0 

⇒ x > \(-\frac12\) or x < \(-\frac12\)

∴ The inequality holds for all real numbers except x = \(-\frac12\).

On LHS of the given inequality, we have two terms both containing modulus. By equating the expression within the modulus to zero, we get x = -1,0 as critical points. These critical points divide the real line in three parts as (−∞, −1),[−1,0),[0. ∞).

Case - I :

When −∞ < x < −1

|x + 1| + |x| > 3 

⇒ −x− 1 − x > 3 

⇒ x < −2.

Case - II 

When −1 ≤ x < 0

x + 1| + |x| > 3 

⇒ x + 1 − x > 3 ⇒ 1 > 3

Case - III 

When 0 ≤ < ∞,

x + 1| + |x| > 3 

⇒ x + 1 + x > 3 

⇒ x > 1.

Combining the results of cases (I) (II) and (III), we get,

x ∈ (−∞, −2),∪ [1, ∞)

(c)  –5 < \(x\) < 13

|x – 4| < 9 ⇒ – 9 < x – 4 < 9 

⇒ – 9 + 4 < \(x\) < 9 + 4 ⇒ –5 < \(x\) < 13.

(d) ac > bc

a > b ⇒ ac > bc when c > 0, but 

a > b ⇒ ac < bc when c < 0. 

∴ ac > bc is not always correct.

We have,

(a + b + c)(bc + ca + ab) > 9abc

= abc+ a²c + a²b + b²c+ abc + b²a + c²a + c²b + abc −9abc

= a (b² + c² ) + b(c² + a² ) + c(a² + b² ) − 6abc

= a (b² + c² − 2bc) + b(c² + a² − 2ca) + c(a² + b² − 2ab)

= a (b − c)² + b(c − a)² + c(a − b)²

Which is positive because each term of RHS is positive.

Thus,

(a + b + c)(bc + ca + ab) − 9abc > 0

(a + b + c)(bc + ca + ab) > 9abc.

We have, 

3 – 2x < 9,

Transposing 3 to RHS,

− 2x < 9 – 3

Or  − 2x < 6

Or \(\frac{-2x}{-2}\) > \(\frac{6}{-2}\)

Or x >−3

∴ When x is real, the solution is (- 3, ∞)

(b) 1

8m + 35 > 75 and 5m + 18 < 53 

⇒ 8m > 40 ⇒ 5m < 35 

⇒ m > 5 ⇒ m < 7 

∴ There is only one integral value of m, i.e., m = 6 satisfying the given inequalities.

(d) (2, 5)

x² – 3x – 10 < 0 

⇒ (x + 2) (x – 5) < 0 ⇒ (x – (–2)) (x – 5) < 0 

⇒ x∈ (–2, 5) 

(If a < b, then (x – a) (x – b) < 0 ⇒ a < x < b) 

10x – x² – 16 > 0 ⇒ x² – 10x + 16 < 0 ⇒ (x – 2) (x – 8) < 0 

⇒ x∈ (2, 8) 

(If a < 0, then (x – a) (x – b) < 0 ⇒ a < x < b) 

∴ x∈ (–2, 5) ∩ (2, 8) 

⇒ x∈ (2, 5).

(b) > \(\frac{2s}{abc}\)

a > 0, b > 0, c > 0 ⇒ ab > 0, bc > 0, ca > 0 

∴ \(\frac{(ab)^2+(bc)^2}{2}\) > \(\big((ab)^2(bc)^2\big)^{\frac12}\)

⇒ a²b² + b²c² > 2 (a²b⁴c²)^\(\frac12\) 

⇒ a²b² + b²c² > 2 (abc).b             …(i) 

Similarly, b²c² + c²a² > 2 (abc).c           …(ii) 

c²a² + a²b² > 2(abc)a           …(iii) 

(i) + (ii) + (iii) 

⇒ 2(a²b² + b²c² + c²a²) > 2 abc (a + b + c) 

⇒ a²b² + b²c² + c²a² > abc (a + b + c)

Dividing both sides by a²b²c², we have

⇒ \(\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{b^2}\) > \(\frac{a+b+c}{abc}\)

⇒ \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) > \(\frac{2s}{abc}\)

⇒ \(\frac{2s}{abc}\) <  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\).

(c) (1 + abc)³

(1 + a³) (1 + b³) (1 + c³) = 1 + a³ + b³ + c³ + a³b³ + b³c³ + c³a³ + a³b³c³     …(i) 

Now for distinct positive reals, a, b, c, AM > GM

\(\frac{a^3+b^3+c^3}{3}>(a^3b^3c^3)^{\frac13}\)

⇒ a³ + b³ + c³ > 3abc                …(ii)

(∵ a, b, c > 0 ⇒ a³, b³, c³ > 0)

Also \(\frac{a^3b^3+b^3c^3+c^3a^3}{3}>(a^3b^3.b^3c^3.c^3a^3)^{\frac13}\)

⇒ a³b³ + b³c³ + c³a³ > 3a²b²c²           …(iii) 

∴ Putting the values from (ii) and (iii) on the RHS of (i), we have

(1 + a³) (1 + b³) (1 + c³) > 1 + 3abc + 3a²b²c² + a³b³c³  = (1 + abc)³ 

∴ (1 + a³) (1 + b³) (1 + c³) > (1 + abc)³.

(a) negative

Let x = b + c – a, y = c + a – b, z = a + b – c 

Since a, b, c are the sides of a non-equilateral triangle, by the triangle inequality a + b > c, b + c > a, c + a > b 

⇒ a + b – c > 0, b + c – a > 0, c + a – b > 0 

⇒ z > 0, x > 0, y > 0 

Also, x, y, z are distinct. 

∴ AM > GM

\(c=\frac{x+y}{2}>\sqrt{xy}\)

\(b=\frac{x+z}{2}>\sqrt{xz}\)

\(a=\frac{y+z}{2}>\sqrt{yz}\)

∴ abc > \(\sqrt{xy}.\sqrt{xz}.\sqrt{yz}\)

⇒ abc > xyz ⇒ abc > (b + c – a) (c + a – b) (a + b – c) 

⇒ (b + c – a) (c + a – b) (a + b – c) – abc < 0, i.e., negative.

(c) \(\frac9p\)

The AM of mth powers of n positive numbers is greater than the mth power of their AM if m < 0 or m > 1. So for m = – 1,

\(\frac{a^{-1}+b^{-1}+c^{-1}}{3}\) > \(\big(\frac{a+b+c}{3}\big)^{-1}\)

⇒ \(\frac13\)\(\big(\frac1a+\frac1b+\frac1c\big)\) > \(\big(\frac{p}{3}\big)^{-1}\)                    (Given: a + b + c ≡ p)

⇒\(\big(\frac1a+\frac1b+\frac1c\big)\) > \(\frac9p\)

Let x be the smaller of the two consecutive even positive integers, then the other even integer is x + 2. 

Given x > 8 and x + (x + 2) < 25. 

⇒ x > 8, and 2x + 2 < 25. 

⇒ x > 8, 2x < 23 ⇒ x > 8, x < 23/2

⇒ 8 < x < 23/2 ⇒ x = 10, 

∴ the required parity even integers is (10, 12)

Let x be the score obtained by Mohan in the last examination.

Then,

80 ≤ \(\frac{94+73+72+84+x}{5}\) < 90

⟹ 80 ≤ \(\frac{323+x}{5}\) < 90

⟹ 400 ≤ 323 + x < 450 

[Multiplying both sides by 5]

⟹ 400 − 323 ≤ 323 + x − 323 < 450 − 323

[Subtracting 3 from both sides]

⟹ 77 ≤ x > 127

Since, the upper limit is 100, 

therefore the required range is 77 ≤ x ≥ 100.

Let x be the marks in the fifth test obtained by Sunita (87 + 92 + 94 + 95 + x)/5 ≥ 90

⇒ 368 + x > 450 ⇒ x > 82 

So, Sunita must obtain a minimum of 82 marks to get grade ‘A’.

Let x be the marks obtained in the third test.

(70 + 75 + x)/3 ≥ 60

⇒ 145+ x≥180

⇒ x> 180-145 = 35.

Thus, Ravi must obtain a minimum of 35 marks to get an average of at least 60 marks.

Given: 24x < 100

x < 100/24 = 4.16

(i) when x ∈ N, solution set ={ 1, 2, 3, 4} 

(ii) when x ∈ Z, 

∴ Solution set = {… -3, -2, -1, 0,1, 2, 3, 4}

Given: 5x – 3 < 7 

⇒ 5x < 7 + 3 = 10

∴ x < 10/5 = 2

(i) If x is an integer, solution set = {…, -3, -2, -1, 0, 1}.

(ii) If x is a real number, solution set = (-∞, 2). 

4x + 3 < 5x + 7
⇒ 4x + 3 – 7 < 5x + 7 – 7
⇒ 4x – 4 < 5x
⇒ 4x – 4 – 4x < 5x – 4x
⇒ –4 < x

Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality.
Hence, the solution set of the given inequality is (–4, ∞).

(b)

| 3 – p | –4 < 1 ⇒ | 3 – p | < 5 

⇒ –5 < 3 – p < 5 

⇒ –8 < –p < 2 

⇒ –2 < p < 8 

(Note the step: Dividing by minus sign reverses the direction of the inequality)

D. x ∈ (5, ∞)

The above graph represents all values of x greater than 5 excluding 5, hence

x > 5 i.e. x ∈ (5, ∞)