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Given, a + b + c = 1, so that a + b = 1 – c, b + c = 1 – a, c + a = 1 – b 

Now, AM > GM ⇒ (a + b) > 2\(\sqrt{ab}\) ⇒ (1 – c) > 2\(\sqrt{ab}\)       …(i) 

Similarly, (b + c) > 2\(\sqrt{bc}\)  ⇒ (1 - a) > 2\(\sqrt{ab}\)            …(ii) 

(c + a) > 2\(\sqrt{ca}\)  ⇒ (1 - b) > 2\(\sqrt{ca}\)                            …(iii) 

Multiplying (i), (ii) and (iii), we get 

⇒ (1 – a) (1 – b) (1 – c) > 8 \(\sqrt{ab}\) \(\sqrt{bc}\) \(\sqrt{ca}\)

⇒ (1 – a) (1 – b) (1 – c) > 8 abc ⇒ K = 8.

Let the length of shortest side be x cm.

According to the given information,

Longest side = 2 x Shortest side = 2x cm

And third side = 2 + Shortest side = (2 + x) cm

Perimeter of triangle = x + 2x + (x + 2) = 4x + 2

But it is given that,

Perimeter > 166 cm

=> 4x + 2 > 166 => 4x> 166-2 => 4x>164

x>164/4 =41 cm

Hence, the minimum length of shortest side is 41 cm.

Let x, 3x, 3x – 2 be the sides of a triangle

Given: x + 3x + 3x – 2 ≥ 61

7x ≥ 63

x ≥ 9.

∴ Minimum length of shortest side = 9 cm.

∴ Other sides are 27 and 25.

a² + b² + c² > ab + bc + ca 

To prove a² + b² + c² – ab – bc – ca > 0 

Let S = a² + b² + c² – ab – bc – ca 

Then S = \(\frac12\)(2a² + 2b² + 2c² – 2ab – 2bc – 2ca) 

= \(\frac12\)(a² + b² - 2ab + b² + c² - 2bc + c² + a² – 2ca) 

= \(\frac12\)[(a - b)² + (b - c)² + (c - a)²] > 0

 As the RHS is the sum of squares which are positive ⇒ S > 0. Also the equality, i.e. = 0 holds when a = b = c.

We have a > 0, b > 0, c > 0, so 

(a – b)² + (b – c)² + (c – a)² > 0 

⇒ a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca > 0 

⇒ 2a² + 2b² + 2c² > 2ab + 2bc + 2ca ⇒ a² + b² + c² > ab + bc + ca 

⇒ 1 > \(\frac{ab+bc+ca}{a^2+b^2+c^2}\) ⇒ \(\frac{ab+bc+ca}{a^2+b^2+c^2}\) < 1               …(i) 

Now for the other part let us using the triangle inequality, 

i.e., the sum of two sides of a triangle is greater than the third side, we have 

∴ a + b > c,  b + c > a,  c + a > b.  Also, a > 0, b > 0, c > 0 

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

⇒ a (b + c – a) + b (c + a – b) + c (a + b – c) > 0 

⇒ ab + ac – a² + bc + ba – b² + ca + cb – c² > 0 

⇒ 2 (ab + bc + ac) > a² + b² + c² ⇒ \(\frac{ab+bc+ca}{a^2+b^2+c^2}≥\frac12\)                   ....(ii) 

∴ From (i) and (ii) 

\(\frac12<\frac{ab+bc+ca}{a^2+b^2+c^2}≥1.\)

Let the length of shortest side = ‘x’ cm

According to the question,

The longest side of a triangle is twice the shortest side

⇒ Length of largest side = 2x

Also, the third side is 2 cm longer than the shortest side

⇒ Length of third side = (x + 2) cm

Perimeter of triangle = sum of three sides

= x + 2x + x + 2

= 4x + 2 cm

Now, we know that,

Perimeter is more than 166 cm

⇒ 4x + 2 ≥ 166

⇒ 4x ≥ 164

⇒ x ≥ 41

Hence, minimum length of the shortest side should be = 41 cm.

(b) < √2c

Since a, b, c are the lengths of the sides of a right triangle having hypotenuse of length c, therefore a² + b² = c² 

(Pythagoras, Theoram) 

Now, a > 0, b > 0 ⇒ (a – b)² > 0 

⇒ a² + b² > 2ab 

⇒ 2(a² + b²) > a² + b² + 2ab 

⇒ 2 (a² + b²) > (a + b)² 

⇒ 2c² > (a + b)² ⇒ (a + b)² < 2c² ⇒ (a + b) < √2c.

(d) 27

By triangle inequality, we have sum of lengths of two sides of a Δ > length of third side 

⇒ a + b > c, b + c > a, c + a > b 

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

All being positive quantities, we apply AM > GM

⇒ \(\frac{(a+b-c)+(b+c-a)+(c+a-b)}{3}\) ≥ [(a+b-c)(b+c-a)(c+a-b)]^\(\frac13\)

⇒ \(\big(\frac{a+b+c}{3}\big)^3\) ≥ [(a+b-c)(b+c-a)(c+a-b)]

(a+b+c)3 ≥ 27(a+b-c)(b+c-a)(c+a-b)

⇒ K = 27.

(a) 2x + y + ≤ 4

The line shown on the graph is 2x + y = 4 

Substituting (0, 0) in the inequality 2x + y ≤ 4. 

2 × 0 + 0 ≤ 4 ⇒ 0 ≤ 4, which hold true

⇒ (0, 0) lies in the required region.

∴ (a) is the correct option.

C.-x > -5

x < 5

Multiplication or dividing by negative number inverts the inequality sign.

⇒ -x > -5

A. x ∈ (10, ∞)

-3x + 17 < -13

⇒ -3x < -30 [Subtracting 17 both side]

⇒ x > 10 [Division by negative number inverts the inequality sign]

⇒ x ∈ (10, ∞)

B. -3 < x < 3

|x| < 3

Hence, there are two cases,

x < 3 [1]

and

-x < 3

⇒ x > -3 [2]

From [1] and [2], we get

⇒ -3 < x < 3

D. x ∈ (-∞, -4) ∪ (6, ∞)

Given,

|x – 1| > 5

There are two cases,

(x – 1) > 5

⇒ x > 6

⇒ x ∈ (6, ∞) [1]

And

-(x – 1) > 5

⇒ -x + 1 > 5

⇒ -x > 4

⇒ x < -4 [Multiplication by negative inverts the inequality sign]

⇒ x ∈ (-∞, -4) [2]

From [1] and [2], we get

⇒ x ∈ (-∞, -4) ∪ (6, ∞)

B. x ∈ [-11, 7]

Given,

|x + 2| ≤ 9

There are two cases,

(x + 2) ≤ 9

⇒ x ≤ 7

⇒ x ∈ (-∞,7] [1]

And

-(x + 2) ≤ 9

⇒ -x – 2 ≤ 9

⇒ -x ≤ 11

⇒ x ≥ -11 [Multiplication by negative inverts the inequality sign]

⇒ x ∈ [-11, ∞) [2]

From [1] and [2], we get

⇒ x ∈ [-11, 7]

D. x ∈ (-∞, -b) ∪ (b, ∞)

|x| > b

Hence, there are two cases,

x > b

⇒ x ∈ (b, ∞) [1]

and

-x > b

⇒ x < -b

⇒ x ∈ (-∞, -b) [2]

From [1] and [2], we get

⇒ x ∈ (-∞, -b) ∪ (b, ∞)

We have, 

\(\frac{1}{x-2}\) < 0

⇒ − 2 < 0 

[∵ \(\frac{a}{b}\) < 0 and > 0 ⇒ < 0]

⇒ x − 2 

⇒ x ∈ (−∞, 2)

x/4 < (5x - 2)/3 - (7x - 3)/5

⇒ 15x < 16x - 4

⇒ 4 < 16x - 15x

⇒ x > 4

∴ Solution set = (4, ∞).  

We have, 

0 < \(\frac{-x}{3}\)< 1,

Multiplying inequality by 3, we get

0 < −x < 3

or 0 > x > 3

or −3 < x > 0 

⇒ x ∈ (−3,0)

We have, 

−3 ≤ −3x + 2 <4,

⇒ −3 − 2 ≤ −3x + 2 − 2 < 4 − 2

(Adding -2 in inequality)

⇒ −5 ≤ −3x < 2

⇒ \(\frac{-5}{-3}\) ≥ \(\frac{-3x}{-3}\) >\(\frac{2}{-3}\)  (Dividing by −3)

Or ⇒ \(-\frac{2}{3}\) > x ≤ \(-\frac{5}{3}\)

Or x ∈\(-\frac{2}{3}\),\(-\frac{5}{3}\)

We have, 

\(\frac{2x−3}{4}\)−3 < \(\frac{x−4}{3}\)−2,

⟹ \(\frac{2x−9}{4}\) < \(\frac{x−10}{3}\)

⟹ \(\frac{x}{2}\) − \(\frac{x}{3}\) < \(\frac{−10}{3}\) + \(\frac{9}{4}\)

[Transposing \(\frac{x}{2}\)to LHS and \(\frac{9}{4}\) to RHS]

⟹ \(\frac{x}{6}\) < \(\frac{13}{12}\)

⟹ x < \(\frac{13}{2}\)

∴ x ∈ (−∞, \(\frac{13}{2}\))

According to the question,

4x + 3 ≥ 2x + 17

⇒ 4x – 2x ≥ 17 – 3

⇒ 2x ≥ 14

⇒ x ≥ 7 …(i)

Also,

3x – 5 < – 2

⇒ 3x < 3

⇒ x < 1 …(2)

Since, equations [i] and [ii] cannot be possible simultaneously,

We conclude that x has no solution.

For positive real numbers, x, y applying AM > GM, we have

\(\frac{x+y}{2}\) > \(\sqrt{xy}\)

⇒ \(\frac12\)  > \(\sqrt{xy}\)         (∵ x + y = 1)

⇒ 1 ≥ 2\(\sqrt{xy}\)         ⇒ 1 > 4xy ⇒ 2 > 8xy ⇒ 1 + 1 > 8xy

⇒ 1 + \(x\) + y > 8xy ⇒ 1 + \(x\) + y + xy > 9xy ⇒ (1 + \(x\)) (1 + y) > 9xy

⇒ \(\frac{(1+x)(1+y)}{xy}\) > 9  ⇒   \(\big(\frac{x+1}{x}\big)\big(\frac{y+1}{y}\big)\) > 9 ⇒  \(\big(1+\frac1x\big)\big(1+\frac1y\big)\) > 9

(c) 3.

a > 0, b > 0, c > 0 ⇒ loga, logb, logc are all defined.

Also a^{logb – logc}, b^logc–loga, c^loga–logb are all positive quantities. 

∴ Applying AM > GM, we have 

\(\frac13\)[ a^{logb – logc} + b^logc–loga + c^loga–logb ] > [a^{logb – logc}. b^logc–loga. c^loga–logb]^\(\frac13\)      ...(i) 

Let x = a^{log b – log c} . b^{log c – log a}. c^{log a – log b} 

⇒ log x = (log b – log c) log a + (log c – log a) log b + (log a – log b) log c 

Now, log_e x = 0 ⇒ x = e⁰ = 1. 

∴ (i) ⇒ \(\frac13\)[ a^{logb – logc} + b^logc–loga + c^loga–logb ] > 1

⇒ a^{log b – log c} + b^{log c – log a} + c^{log a – log b} > 3 

⇒ The least value of a^{log b – log c} + b^{log c – log a} + c^{log a – log b} is 3.

(a) abcd > 81(s – a) (s – b) (s – c) (s – d)

3s = a + b + c + d ⇒ 3s – b – c – d = a 

⇒ a = (s – b) + (s – c) + (s – d) 

For distinct positive real numbers AM > GM

⇒ \(\frac13\)[(s – b) + (s – c) + (s – d)]> {(s – b) + (s – c) + (s – d)}^\(\frac13\)

⇒ (s – b) + (s – c) + (s – d) > 3 {(s – b) + (s – c) + (s – d)}^\(\frac13\)

⇒ a > 3 {(s – b) + (s – c) + (s – d)}^\(\frac13\)               ....(i)

Similarly,  b > 3 {(s – a) + (s – c) + (s – d)}^\(\frac13\)               ....(ii)

 c > 3 {(s – a) + (s – b) + (s – d)}^\(\frac13\)               ....(iii)

 a > 3 {(s – a) + (s – b) + (s – c)}^\(\frac13\)               ....(iv)

∴ (i) × (ii) × (iii) × (iv)

⇒ abcd > 81{(s - a)³ (s – b)³ (s – c)³ (s – d)³}^\(\frac13\)

⇒ abcd > 81(s – a) (s – b) (s – c) (s – d).

– 3 < x – 2 < 9 – 2x 

⇒ – 3 < x – 2 and x – 2 < 9 – 2x ⇒ – 3 + 2 < x and x + 2x < 9 + 2 

⇒ – 1 < x and 3x < 11 or x < \(\frac{11}{3}\) ⇒ − 1 < \(x\) ≤ \(\frac{11}{3}\)

Since \(x\) ∈ Z, so the solution set = {0, 1, 2, 3}.

(b) {2,4,6}

x + 2 < 9 ⇒ x < 9 – 2 ⇒ x < 7 

∴ The positive even integers less than 7 are 2, 4, 6.

(c)  x < \(-\frac45\)

 –5x > 4  ⇒  5x < – 4  ⇒  x < \(-\frac45\)

(b) 2x -6 <10

2x – 6 < 10 ⇒ 2x < 16 ⇒ x < 8

(d) x ≥ -20

–3.5x – 12 ≤ 58 ⇒ –3.5 x ≤ 70 

⇒ x ≥ \(-\frac{70}{3.5}\) ⇒ x ≥ -20

Answer is (B) x ∈ [– 11, 7]

Answer is (A) | x| < 5 

Given: 3x + 8 > 2 

⇒ 3x > 2 – 8 = – 6

∴ x > - 6/3 = - 2

(i) When x ∈ Z, solution set = {-1, 0, 1, 2, 3,…} 

(ii) When x∈M, solution set = (-2, ∞). 

Given: 5x – 3 < 3x + 1 

⇒ 5x – 3x <1 + 3

⇒ 2x<4 

⇒ x < 4/2 = 2

(i) When x ∈ Z, solution set ={…, -3, -2, -1, 0,1} 

(ii) When x ∈ M, solution set = (- ∞, 2).

Given: -12x > 30

⇒ x < 30/-12 = - 2.5

(i) When x ∈ N, solution set = φ 

(ii) When x ∈ Z, solution set = {…, -5, -4, -3}

The set of all those values of x which satisfy the given inequation is called the solution set of inequation.

Rules for solving inequations: 

• Rule 1: Adding the same number or expression to each side of an inequation docs not change the inequality. 
• Rule 2: Subtracting the same number or expression from each side of an inequation does not change the inequality. 
• Rule 3: Multiplying (or dividing) each side of an inequation by the same positive number does not change the inequality. 
• Rule 4: Multiplying (or dividing) each side of an inequation by same negative number reverses the inequality.

Cost function: C(x) = 26000 + 3Ox Revenue function: R(x) = 43x For profit, R(x) > C(x)

⟹ 26000 + 30x < 43x

⟹ 43x – 30x > 26000

⟹ 13x > 26000

⟹ x > 2000

Hence, more than 2000 cassettes must be produced to get profit.

Given: 30x < 200

x < 200/30 = 6.6

(i) x is a natural number: 

Solution set = { 1, 2, 3,4, 5, 6}

(ii) x is a real number:

∴ Solution set = {… -3, -2, -1,0, 1,2, 3,4, 5, 6}

Let x be the smaller of the two consecutive odd natural number, so that the other one is x + 2. 

Given x > 10 and x + (x + 2) < 40 

⇒ 2x < 38 

⇒ x < 19 

∴10 < x < 19 

Since x is an odd number, x can take the values 11,13, 15 and 17. 

∴ Required possible pairs are (11, 13), (13, 15), (15, 17), (17,19).

Two real numbers or two algebraic expressions related by the symbol ‘<’, V, ‘<’ or ‘>’ form an inequality.

For example: 3 < 5, 7 > 5 are numerical inequalities x < 5, y > 2, x > 3 are literal inequalities. ax + b < 0, ax + b > 0 are strick inequalities in one variable. 

ax + b < 0, ax + b >0 are slack inequalities in one variable. 

ax + by < c, ax + by > c are strick inequalities in two variables. 

ax + by < c, ax + by > c are slack inequalities in two variables. 

ax²  + bx + c < 0, ax²  + bx + c > 0 are quadratic inequalities in one variable.

Answer is (C) – x > – 5

Answer is (A) x ∈ (10, ∞) 

Answer is (D) x ∈ (– ∞, – b) ∪ (b, ∞)

Answer is (B) – 3 < x < 3

We have, profit = Revenue – Cost

= (60x + 2000) – (20x + 4000)

= 40x – 2000

To earn some profit, 40x – 2000 > 0

⇒ x > 50

Hence, the manufacturer must sell more than 50 items to realise some profit.

(c) {1}

3x – 4 < 8  ⇒  3x < 12  ⇒  x < \(\frac{12}3\) ⇒ x < 4 

The solution set = {1} as 1 is the only square number less than 4.

(d) 

Temperature was less than 10° C, it never touched 10° C

(a) – 1 < x < 5 

log_e (x² – 16) < log_e (4x – 11) 

(x² – 16) < 4x – 11 

(∵ a > 1, log_a f (x) > log_a(gx) ⇒ f (x) > g(x) > 0) 

⇒ x² – 4x – 5 < 0 ⇒ (x – 5) (x + 1) < 0 

⇒ – 1 < x < 5 

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

(b) x ≤ –1

– 2x + (3³ – 5² ) ≥ 4 

⇒ – 2x + (27 – 25) ≥ 4 

⇒ – 2x + 2 ≥ 4  ⇒ –2x ≥ 2  ⇒  x ≤ –1

(d) (−∞ , -√2) ∪ ( √2, ∞)

Case I: When \(x\) + 2 > 0, then \(x\) > – 2 

⇒ |\(x\) + 2| = \(x\) + 2 

∴ x² – |\(x\) + 2| + \(x\) > 0 = x² – (\(x\) + 2) + \(x\) > 0 

⇒ x² – 2 > 0 ⇒ x² > 2 

⇒ \(x\) < -√2 or 2 \(x\) > √2

⇒ x∈ (– 2, − √2 ) ∪ ( 2 , ∞)                [∵ \(x\) ≥ − 2]          …(i) 

Case II: When \(x\) + 2 < 0, then \(x\) < – 2 

⇒ |\(x\) + 2| = – (\(x\) + 2) 

∴ x² – |\(x\) + 2| + \(x\) > 0 ⇒ x² + (\(x\) + 2) + \(x\) > 0 

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

⇒ (x + 1)² + 1 > 0, which is true for all value of x. 

∴ \(x\) < – 2 ⇒ \(x\)∈ (– ∞, – 2)                …(ii) 

From (i) and (ii) 

\(x\)∈ (– 2, −√2) ∪ ( √2, ∞ ) ∪ (– ∞, – 2) 

⇒ \(x\)∈ (−∞ , −√2) ∪ ( √2, ∞)

(d) x < 2 or x > 8

| 10 – 2x | > 6  ⇒  10 – 2x > 6 or – (10 – 2x) > 6 

⇒ –2x > 6 – 10 or 2x > 6 + 10 

⇒ –2x > – 4 or 2x > 16 

⇒ x < 2 or x > 8