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Let r be the radius of the sphere and ∆r be the error in measuring the radius then r =7m and ∆r = 0.02 m

We have;

V = \(\frac{4}{3}\) πr³
dV = \(\frac{dV}{dr}\) ∆r = \(\frac{4}{3}\) π3r² × ∆r
= 4π(7)² × 0.02 = 3.92 π m³.

Answer is : C. -39

Given: f(x) = 3x⁴ - 8x³ - 48x+25.

F’(x) = 12x³ - 24x² - 48 = 0

F’(x) = 12(x³ - 2x² - 4) = 0

Differentiating again, we get,

F’’(x) = 3x² – 4x = 0

x(3x – 4) = 0

x = 0 or x = 4/3

Putting the value in equation, we get,

f(x) = -39

Hence, C is the correct answer

f (x) = 3x⁴ – 8x³ + 12x² – 48 x + 25 

f’(x) = 12x³ – 24x² + 24x – 48

f‘(x) = 0 ⇒ 12 (x³ – 2x² + 2x - 4) = 0 

⇒ 12 (x² (x – 2) + 2 (x – 2)) 

⇒ 12 ( (x – 2) (x² + 2) ) = 0 (x – 2)(x² + 2) = 0 

⇒ x = 2 but x² + 2 ≠ 0 

The points are f(0), f(2), f(3) 

f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25 

f(0) = 25 

f(2) = 3(16) – 8(8) + 12(4) – 48(2) + 25 = 48 – 64 + 48 – 96 + 25 = -39 

f(3) = 3(34) – 8 (33) + 12(32) – 48(3) + 25 = 243 – 216 + 108 – 144 + 25 

376 – 360 = 16 

∴ maximum of f (x) at x = 0 is 25 

minimum of f (x) at x = 2 is – 39.

Condition (1):

Since, f(x) = sinx - sin2x is a trigonometric function and we know every trigonometric function is continuous.

⇒ f(x) = sinx - sin2x is continuous on [0,2π].

Condition (2):

Here, f’(x) = cosx - 2cos2x which exist in [0,2π].

So, f(x) = sinx - sin2x is differentiable on (0,2π)

Condition (3):

Here, f(0) = sin0 - sin0 = 0

And f(2π) = sin(2π) - sin(4π) = 0

i.e. f(0) = f(2π)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one c ϵ (0,2π) such that f’(c) = 0

i.e. cosx - 2cos2x = 0

i.e. cosx - 4cos²x+2 = 0

i.e. 4cos²x - cosx - 2 = 0

i.e. cos x = \(\frac{1\pm\sqrt{33}}{8}\)

i.e. c = 32° 32’ or c = 126°23’

Value of c = 32°32’ ϵ (0,2π)

Thus, Rolle’s theorem is satisfied.

max. value is 139 at x = −2 and min. value is 89 at x = 3

F’(x) = 6x² - 24 = 0

6(x² - 4) = 0

6(x² - 2²) = 0

6(x -2)(x+2) = 0

X = 2,-2

Now, we shall evaluate the value of f at these points and the end points

F(2) = 2(2)³ - 24(2)+107 = 75

F(-2) = 2(-2)³ - 24(-2)+107 = 139

F(-3) = 2(-3)³ - 24(-3)+107 = 125

F(3) = 2(3)³ - 24(3)+107 = 89

local max. value is 0 at x = 3

F’(x) = 4(x - 3)³ = 0

⇒ x = 3

local max. value is 0.

local min. value is 0 at x = 0

F’(x) = 2x = 0

x = 0

local min.value is

max. value is 257 at x = 4 and min. value is −63 at x = 2

F^|(x) = 12x³ - 24x²+24x - 48 = 0

12(x³ - 2x²+2x - 4) = 0

Since for x = 2, x³ - 2x²+2x - 4 = 0, x - 2 is a factor

On dividing x³ - 2x²+2x - 4 by x - 2, we get,

12(x - 2)(x²+2) = 0

X = 2,4

Now, we shall evaluate the value of f at these points and the end points

F(1) = 3(1)⁴ - 8(1)³+12(1)² - 48(1)+1 = -40

F(2) = 3(2)⁴ - 8(2)³+12(2)² - 48(2)+1 = -63

F(4) = 3(4)⁴ - 8(4)³+12(4)² - 48(4)+1 = 257

Condition (1):

Since, f(x) = x(x+2)e^x is a combination of exponential and polynomial function which is continuous for all x ϵ R.

⇒ f(x) = x(x+2)e^x is continuous on [-2,0].

Condition (2):

Here, f’(x) = (x²+4x+2)e^x which exist in [-2, 0].

So, f(x) = x(x+2)e^x is differentiable on (-2,0).

Condition (3):

Here, f(-2) = (-2)(-2+2)e^-2 = 0

And f(0) = 0(0+2)e⁰ = 0

i.e. f(-2) = f(0)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one c ϵ (-2,0) such that f’(c) = 0

i.e. (c²+4c+2)e^c = 0

i.e. (c+√2)² = 0

i.e. c = -√2

Value of c = -√2 ϵ (-2,0)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x) = x(x - 5)² is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = x(x - 5)² is continuous on [0,5].

Condition (2):

Here, f’(x) = (x-5)²+ 2x(x - 5) which exist in [0,5].

So, f(x) = x(x - 5)² is differentiable on (0,5).

Condition (3):

Here, f(0) = 0(0 - 5)² = 0

And f(5) = 5(5 - 5)² = 0

i.e. f(0) = f(5)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,5) such that f’(c) = 0

i.e. (c - 5)²+ 2c(c - 5) = 0

i.e.(c - 5)(3c - 5) = 0

i.e. c = 5/3 or c = 5

Value of c = 5/3 ∈ (0, 5)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x) = (x - 1)(2x - 3) is a polynomial and we know every polynomial function is continuous for all x ϵ R.

⇒ f(x) = (x -1)(2x - 3) is continuous on [1,3].

Condition (2):

Here, f’(x) = (2x - 3)+ 2(x - 1) which exist in [1,3].

So, f(x) = (x -1)(2x - 3) is differentiable on (1,3).

Condition (3):

Here, f(1) = [1 - 1][2(1) - 3] = 0

And f(5) = [3 -1][2(3) - 3] = 6

i.e. f(1) ≠ f(3)

Condition (3) of Rolle’s theorem is not satisfied.

So, Rolle’s theorem is not applicable.

Condition (1):

Since, f(x) = [x] which is discontinuous at x = 0

⇒ f(x) = [x] is not continuous on [-1,1].

Condition (1) of Rolle’s theorem is not satisfied.

So, Rolle’s theorem is not applicable.

Let the two numbers be x and y 

Given x + y = 14 & S = x² + y²

where y = 14 – x

∴ S = x² + (14 – x)² = x² + 14² + x² – 28x = 2x² – 28x + 14² 

\(\frac{ds}{dx}\) = 4x – 28 → (1) \(\frac{ds}{dx}\) = 0 

⇒ 4x – 28 = 0 ⇒ x = 7 

\(\frac{d^2s}{dx^2}\) = 4 > 0, sum is minimum. 

∴ y = 14 – x = 14 – 7 = 7 

∴ the two positive number are 7 & 7.

Since the tape is stretchable, its will show the same measurements for different lenths. Therefore we cannot use it as measuring tape. While measuring a distance, we need to tell someone how much tape has been stretched which is difficult to measure. It leads to incorrect measurements.

(d) Any of the three scales.

Solution:
(1) Handle bar or seat : The handle of bicycle will always move in rectilinear path because it cannot execute circular or rotatory motion. 
(2) Pedal : The pedals of bicycle will always move circularly around its chain fixing system because they cannot move in forward direction without the whole chain system. 
(3) Wheel : Both rectilinear and circular The wheels of bicycle will move in rectilinear as well as in circular motion because the wheel as a whole will move forward and its point or particles around the rim will execute circular motion.

(b) longer distance with a higher speed.

Let the original speed be x kmph and let the time taken to complete the journey be y hours. 

∴ Length of the whole journey = (xy) km 

Case I: When the speed is (x + 5) kmph and the time taken is (y – 3) hrs: 

Total journey = (x + 5) (y – 3) km 

⇒ (x + 5) (y – 3) = xy 

⇒ xy + 5y – 3x – 15 = xy 

⇒ 5y – 3x = 15 ………(i) 

Case II: When the speed is (x – 4) kmph and the time taken is (y + 3) hrs: 

Total journey = (x – 4) (y + 3) km 

⇒ (x – 4) (y + 3) = xy 

⇒ xy – 4y + 3x – 12 = xy 

⇒ 3x – 4y = 12 ………(ii) 

On adding (i) and (ii), we get: 

y = 27 On substituting y = 27 in (i), we get: 

5 × 27 – 3x = 15 

⇒135 – 3x = 15 

⇒3x = 120 

⇒x = 40 

∴ Length of the journey = (xy) km = (40 × 27) km = 1080 km 

Let the speed of the cars be x km/h and y km/h.

Relative speed in same direction = (x - y) km/h

Relative speed in opposite direction = (x + y) km/h

Now, \(\frac{70}{x+y} = 1\)

=> x + y = 70 .....(i)

and \(\frac{70}{x-y} = 7\)

=> x - y = 10

Solving Eqs. (i) and (ii),

x = 40 km/h and y = 30 km/h

Let’s consider the car starting from point A as X and its speed as x km/hr. 

And, the car starting from point B as Y and its speed as y km/hr. 

It’s seen that there are two cases in the question: 

# Case 1: Car X and Y are moving in the same direction 

# Case 2: Car X and Y are moving in the opposite direction 

Let’s assume that the meeting point in case 1 as P and in case 2 as Q. 

Now, solving for case 1: 

The distance travelled by car X = AP 

And, the distance travelled by car Y = BP

As the time taken for both the cars to meet is 7 hours, 

The distance travelled by car X in 7 hours = 7x km [∵ distance = speed x time] 

⇒ AP = 7x 

Similarly, 

The distance travelled by car Y in 7 hours = 7y km 

⇒ BP = 7Y 

As the cars are moving in the same direction (i.e. away from each other), we can write 

AP – BP = AB 

So, 7x – 7y = 70 

x – y = 10 ………………………. (i) [after taking 7 common out] 

Now, solving for case 2: 

In this case as it’s clearly seen that, 

The distance travelled by car X = AQ

And, 

The distance travelled by car Y = BQ 

As the time taken for both the cars to meet is 1 hour, 

The distance travelled by car x in 1 hour = 1 x km 

⇒ AQ = 1x 

Similarly, 

The distance travelled by car y in 1 hour = 1y km 

⇒ BQ = 1y 

Now, since the cars are moving in the opposite direction (i.e. towards each other), we can write 

AQ + BQ = AB 

⇒ x + y = 70 …………… (ii) 

Hence, by solving (i) and (ii) we get the required solution 

From (i), we have x = 10 + y……. (iii) 

Substituting this value of x in (ii). 

⇒ (10 + y) + y = 70 

⇒ y = 30 

Now, using y = 30 in (iii), we get 

⇒ x = 40 

Therefore, 

– Speed of car X = 40km/hr. 

– Speed of car Y = 30 km/hr.

 20x³ – 40x² + 80x

= 20x (x² – 2x +4)

If a pair of linear equations in two variables is consistent, then the lines represented by two equations are Intersecting or coincident.

Conditions for pair of linear equations to be consistent are:

a₁/a₂ ≠ b₁/b_2. [unique solution]

a₁/a₂ = b₁/b₂ = c₁/c₂ [coincident or infinitely many solutions]

(i) No.

The given pair of linear equations

– 3x – 4y – 12 = 0 and 4y + 3x – 12 = 0

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = – 3, b₁ = – 4, c₁ = – 12;

a₂ = 3, b₂ = 4, c₂ = – 12;

a₁ /a₂ = – 3/3 = – 1

b₁ /b₂ = – 4/4 = – 1

c₁ /c₂ = – 12/ – 12 = 1

Here, a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Hence, the pair of linear equations has no solution, i.e., inconsistent.

(ii) Yes.

The given pair of linear equations

(3/5)x – y = ½

(1/5)x – 3y= 1/6

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = 3/5, b₁ = – 1, c₁ = – ½;

a₂ = 1/5, b₂ = 3, c₂ = – 1/6;

a₁ /a₂ = 3

b₁ /b₂ = – 1/ – 3 = 1/3

c₁ /c₂ = 3

Here, a₁/a₂ ≠ b₁/b_2.

Hence, the given pair of linear equations has unique solution, i.e., consistent.

(iii) Yes.

The given pair of linear equations –

2ax + by –a = 0 and 4ax + 2by – 2a = 0

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = 2a, b₁ = b, c₁ = – a;

a₂ = 4a, b₂ = 2b, c₂ = – 2a;

a₁ /a₂ = ½

b₁ /b₂ = ½

c₁ /c₂ = ½

Here, a₁/a₂ = b₁/b₂ = c₁/c₂

Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent

(iv) No.

The given pair of linear equations

x + 3y = 11 and 2x + 6y = 11

Comparing the above equations with ax + by + c = 0;

We get,

a₁ = 1, b₁ = 3, c₁ = 11

a₂ = 2, b₂ = 6, c₂ = 11

a₁ /a₂ = ½

b₁ /b₂ = ½

c₁ /c₂ = 1

Here, a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Hence, the given pair of linear equations has no solution.

(C) intersecting or coincident

Explanation:

Condition for a pair of linear equations to be consistent are:

Intersecting lines having unique solution,

a₁/a₂ ≠ b₁/b₂

Coincident or dependent

a₁/a₂ = b₁/b₂ = c₁/c₂

“Dependent pair of linear equations in two variables is always consistent”. This is true because dependent pair of linear equations represent coincident lines which have infinite number of solutions. Hence they are consistent.

Correct option is D) All of the above

No solution will be when

a₁/b₁ = b₁/b₂ ≠ C₁/C₂

kx + 3y = 3

12x + ky = 6

k/12 = 3/k

⇒ k = ± 6

Since P, Q, R and S divides AB, BC, CD and DA in half.

∴ AP = PB = BQ = QC = CR = RD = DS = SA = 6 cm.

Given, side of a square BC = 12 cm

Area of the square = 12 × 12 = 144 cm²

Area of the shaded region = Area of the square - (Area of the four quadrants)

Area of one quadrant = (π/4)×(Radius)² = (3.14/4)× 36 = 113.04/4 cm²

Area of four quadrants = 113.04 cm²

Area of the shaded region = 144-113.04 = 30.96 cm²

Let the number of runs required by the cricket player to score in the third match be x. 

Number of runs scored by the player in first match = 180 

Number of runs scored in second match = 257 

∴ Total runs scored by the player = 180 + 257 + x = 437 + x 

Average of runs in the three matches = (437 + x)/3

Since, the average of runs should be 230.

(437 + x)/3 = 230

∴ 437 + x = 230 × 3 

∴ 437 + x = 690 

∴ x = 690 – 437 

∴ x = 253

∴ The cricket player should score 253 runs in the third match.

Let the son’s present age be x years. 

∴ Mother’s present age = (x + 25) years

After 8 years, 

Son’s age = (x + 8) years 

Mother’s age = (x + 25 + 8) = (x + 33) years 

Since, the ratio of the son’s age to mother’s age after 8 years is 4/9

∴ (x + 8/x + 33) = 4/9

∴ 9 (x + 8) = 4 (x + 33) 

∴ 9x + 72 = 4x + 132

∴ 9x – 4x = 132 – 72 

∴ 5x = 60

∴ x = 12

∴ Son’s present age is 12 years.

a) Social stratification is a feature of society: 

Social stratification is a feature of society. It is not merely the difference between individuals. It is something that permeates (covers) the entire society. Stratification implies that the societal resources are inequitably distributed among the different classes of people. In backward societies, production was very little and therefore stratification was also only nominal, or very minimal. Societies that lived by hunting and collecting foods are examples for this. But where societies achieved technological progress, stratification was seen in its full color. In such societies, people produced surplus, more than what they required. These resources were unequally distributed among the people of different classes. The innate abilities of people were not considered here. 

b) Social stratification has been in existence for generations:

Social stratification is related to the family and also to the societal resources that one generation receives from the previous one as hereditary property. The social status of a person is imposed on him. A person gets the social status of his parents. For example, a Dalit is supposed to do only traditional jobs like farming, scavenging or leatherwork. Because of that, he had very limited chances of getting jobs with higher pay. Such social inequalities were further aggravated by rules like marrying in the same caste. This prevented people from removing the boundary lines through mixed marriages.

c) The ways of faith or ideologies support Social stratification: 

If Social stratification is to continue through generations it must be proved to be inevitable. For example, the Jati system was justified through the ‘PurityPollution’ (suddha-asuddha) concept. This made Brahmins high caste and Dalits low caste merely by birth and profession. Not everyone approves inequality as a legitimate system. People with special privileges in the society support systems like Jati and Varna. But those who suffer contempt and negligence raise their voice against them.

Correct answer is a) Social inequality

Everywhere in the world, there are common concepts about differently-abled people. 

Here are the main points:

• Handicap is a biological phenomenon. 
• The problems of the differently-abled person come from his/her handicaps. 
• Differently-abled people are considered ‘victims’. 
• A person’s handicap is related to his self-respect. 
• The very word handicap suggests that the person needs help.

Let the required number be x.Then,

= 2x – 7 = 45

Transposing -7 to RHS and it becomes 7.

= 2x = 45 + 7

= 2x = 52

Multiplying both side by (1/2)

= 2x × (1/2) = 52 × (1/2)

= x = 26

∴ The required number is 26.

Let width of the original rectangle = x cm

Length of the original rectangle = (2x – 5)cm

Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm

New width of the rectangle = (x + 2) cm

New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm

Given; new perimeter = 74 cm

6x – 16 = 74

=> 6x = 74 + 16

=> 6x = 90

=>x = 15

Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm

Width of the original rectangle = x = 15 cm

Number of first page of a opened book x 

Number of second page = x + 1 

∴ The sum of the numbers of two pages = 373 

⇒ x + (x + 1) = 373 

⇒ 2x + 1 = 373 

2x = 372 

⇒ x = 186 

∴x + 1 = 186 + 1 = 187 

∴ Numbers of two consecutive pages = 186, 187

Let width of the rectangle = x cm 

Length of the rectangle = 2x cm 

Perimeter of the rectangle = 2 [Length + Width] = 2 [2x + x] = 2 x 3x = 6x cm 

Given perimeter = 54 cm 

6x = 54 

=> x = 9 

Length = 2x = 2 x 9 = 18 cm

Given: 

|5 – 2x| ≤ 3, x ϵ R. 

5 – 2x ≥ - 3 or 5 – 2x ≤ 3 

5 – 2x ≥ -3 

Subtracting 5 from both the sides in the above equation 

5 – 2x – 5 ≥ - 3 – 5 

-2x ≥ - 8 

Now, multiplying by -1 on both the sides in the above equation 

-2x(-1) ≥ -8(-1) 

2x ≤ 8 

Now dividing by 2 on both the sides in the above equation

\(\frac{2{\text{x}}}{2} \le \frac{8}{2}\)

x ≤ 4 

5 – 2x ≤ 3 

Subtracting 5 from both the sides in the above equation 

5 – 2x – 5 ≤ 3 – 5

-2x ≤ -2 

Now, multiplying by -1 on both the sides in the above equation 

-2x(-1) ≤ -2(-1) 

2x ≥ 2 

Now dividing by 2 on both the sides in the above equation

\(\frac{2{\text{x}}}{2} \ge \frac{2}{2}\)

x ≥ 1

Therefore,

x є [1, 4]

Given: 

|3x – 7|> 4, x ϵ R. 

3x – 7 < -4 or 3x – 7 > 4 

(Because |x| > a, a>0 then x < -a and x > a) 

3x – 7 < -4 

Now, adding 7 to both the sides in the above equation 

3x – 7 + 7 < -4 +7 

3x < 3 

Now, dividing by 3 on both the sides of above equation

\(\frac{3{\text{x}}}{3} < \frac{3}{3}\)

x < 1 

Now, 3x – 7 > 4 

Adding 7 on both the sides in above equation 

3x – 7 + 7 > 4 + 7 

3x > 11 

Now, dividing by 3 on both the sides in the above equation

\(\frac{3x}{3} > \frac{11}{3}\)

x > \(\frac{11}{3}\)

Therefore,

x ∈ (-∞ , 1) U ( \(\frac{11}{3}, \infty\))

Let x meter be the breadth of rectangle.

Then, length of rectangle = (x + 6) meter

∴Perimeter = 2 (Length + Breadth)

= 2 {x + (x + 6)}

= 2 (2x + 6) meter

According to question,

2(2x + 6) = 64

⇒ 2x + 6 = 64/2

⇒ 2x + 6 = 32

⇒ 2x = 32 – 6

⇒ 2x = 26

⇒ x = 26/2

⇒ x = 13

Hence, the breadth of rectangle = 13 meter length of rectangle = (13 + 6) meter = 19 meter

Let the breadth of the rectangle be x meter

Length of the rectangle be (x + 9) meter

Area of the rectangle length×breadth = x(x +9) m²

When length and breadth increased by 3cm then,

New length = x + 9 + 3 = x + 12

New breadth = x + 3

So, Area is

(x + 12) (x + 3) = x (x + 9) + 84

x² + 15x + 36 = x² + 9x + 84

15x – 9x = 84 – 36

6x = 48

x = 48/6

= 8

∴ Length of the rectangle (x + 9) = (8 + 9) = 17cm and breadth of the rectangle is 8cm.

Breadth of a volleyball court = x feet say. 

∴ Its length = 2 × x = 2 x feet. 

The perimeter of a court = 177 feet. 

⇒ 2(l + b) = 177 

⇒ 2(2x + x) = 177 

⇒ 2 x 3x = 177 

⇒ 6x = 177 

⇒ x = \(\frac{177}6\)

∴ x = 29.5 

∴ The breadth of a volleyball court = 29.5ft 

The length of a volleyball court = 2x = 2 × 29.5 = 59ft

Let the breadth of rectangle be x metre.

∴ Length of rectangle = (3 + x) metre

∵ Perimeter = 2 (L + B)

∴ Perimeter = 2{(3 + x) + x}

= 2(2x + 3) m

According to question,

2(2x + 3) = 54

Therefore 2x + 3 = 54/2

2x + 3 = 27

2x = 27 – 3 = 24

x = 24/2 = 12

Hence the breadth of rectangle =12 meter and length of rectangle = 12 + 3 = 15 meter

A triangle in which the length of the third side = x m. say.

The length of remaining two equal sides = 2 × x – 5 = (2x – 5) m. 

Perimeter of a triangle = 55 m. 

∴ (2x – 5) + (2x -5) + x = 55 

⇒ 5x – 10 = 55 

⇒ 5x = 65 

⇒ x = \(\frac{65}5\)

∴ x = 13m 

2x – 5 = 2 × 13 – 5 = 26 – 5 = 21m. 

∴ The lengths of three sides of a triangle are 13, 21, 21. (in m.)

We have,

10x – 5 – 7x = 5x + 15 – 8

3x – 5 = 5x + 7

Transposing – 5 to RHS it becomes 5 and 5x to LHS it becomes -5x.

3x – 5x = 7 + 5

– 2x = 12

x = -12/2

x = – 6

We have,

0.4(3x –1) = 0.5x + 1

1.2x – 0.4 = 0.5x + 1

Transposing – 0.4 to RHS it becomes 0.4 and 0.5x to LHS it becomes -0.5x.

1.2x – 0.5x = 1 + 0.4

0.7x = 1.4

x = 1.4/0.7

x = 14/7

x = 2