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Question And Answer:

Q.1A measure of dispersion is a good supplement to the central value in understanding a frequency distribution. Comment.
ANSWER:
Dispersion is the extent to which values in a distribution differ from the avarage of the distribution. Knowledge of only average is insufficient as it does not reflect the quantum of variation in values.

Measures of dispersion enhance the understanding of a distribution considerably by providing information about how much the actual value of items in a series deviate from the central value, e.g., per capita income gives only the average income but a measure of dispersion can tell you about income inequalities, thereby improving the understanding of the relative living standards of different sections of the society. Through value of dispersion one can better understand the distribution.

Thus a measure of dispersion is a good supplement to the central value in understanding a frequency distribution.

Q.2Which measure of dispersion is the best and how?
ANSWER:
Standard Deviation is considered to be the best measure of dispersion and is therefore the most widely used measure of dispersion.

• It is based on all values and thus provides information about the complete series. Because of this reason, a change in even one value affects the value of standard deviation.
• It is independent of origin but not of scale.
• It is us’eful in advanced statistical calculations like comparison of variability in two data sets.
• It can be used in testing of hypothesis.
• It is capable of further algebraic treatment.

Q.3Some measures of dispersion depend upon the spread of values whereas some calculate the variation of values from a central value. Do you agree?
ANSWER:
Yes, it is true that some measures of dispersion depend upon the spread of values, whereas some calculate the variation of values from the central value. Range and Quartile Deviation measure the dispersion by calculating the spread within which the value lie. Mean Deviation and Standard Deviation calculate the extent to which the values differ from the average or the central value.

Question 4.
Q.4In town, 25% of the persons earned more than ₹ 45,000 whereas 75% earned more than 18,000. Calculate the absolute and relative values of dispersion.
ANSWER:
25% of the persons earned more than ₹ 45,000. This implies that upper quartile Q₃ = 45,000 75% earned more than 18,000. This implies that lower quartile Q₁ =18,000
Absolute Measure of Dispersion = Q₃ – Q₁ = 45,000 – 18,000 = 27,000
Relative Measure of Dispersion
Co-efficient of Quartile Deviation

Q.5 The yield of wheat and rice per acre for 10 districts of a state is as under

Calculate for each crop,
(i) Range
(ii) QD
(iii) Mean’Deviation about Mean
(iv) Mean Deviation about Median
(v) Standard Deviation
(vi) Which crop has greater variation?
(vii) Compare the value of different measures for each crop.
ANSWER:
(i) Range
(a) Wheat Highest value of distribution (H) = 25
Lowest value of distribution (L) = 9
Range = H – L = 25 – 9 = 16
(b) Rice Highest value of distribution (H) = 34
Lowest value of distribution (L)=12
Range = H – L = 34 – 12 = 22
(ii) Quartile Deviation
(a) Wheat Arranging the production of wheat in increasing order 9, 10, 10, 12, 15, 16, 18, 19, 21, 25
Q₁ = N+14th item = 10+14th item = 114th item
= 2.75th item
= Size of 2nd item + 0.75 (size of 3rd item – size of 2nd item)
= 10 + 0.75(10 – 10)
= 10 + 0.75 × 0
= 10
Q₃ = 3(N+1)4th item = 3(10+1)4th item
= 334th item = 8.25th
= Size of 8th item + 0.25 (size of 9th item – size of 8th item)
= 19 + 0.25(21 – 19)
= 19 + 0.25 × 2
= 19 + 0.50 = 19.50
Quartile Deviation = Q3−Q12=19.50−102=9.502 = 4.75
(b) Rice Arranging the data of production of rice
12, 12, 12, 15, 18, 18, 22, 23, 29, 34 item
Q₁ = N+14th item = 10+14th item
= 2.75 th item
= Size of 2nd item + 0.75 (size of 3rd item – size of 2nd item)
= 12 + 0.75(12 – 12) = 12 + 0.75 × 0
= 12

= 8.25th item
= Size of 8th item + 0.25 (size of 9th item – size of 8th item)
= 23 + 0.25(29 – 23)
= 23 + 0.25 × 6
= 23 + 1.5
= 24.5
Quartile Deviation = Q3−Q12=24.5−122=12.502 = 6.25

(iii) Mean Deviation about Mean
(a) Wheat

(b) Rice

(iv) Mean Deviation about Median
(a) Wheat

(b) Rice

(v) Standard Deviation
(a) Wheat

(b) Rice

(vi) Coefficient of Variation
(a) Wheat
CV =σX¯¯¯¯¯×100=5.0415.5×100 = 32.51
(b) Rice
CV =σX×100=7.1619.5×100 = 36.71
Rice crop has greater variation as the coefficient of variation is higher for rice as compared to that of wheat.
(vii) Rice crop has higher Range, Quartile Deviation, Mean Deviation about Mean, Mean Deviation about Median, Standard Deviation and Coefficient of Variation.

Q.6In the previous question, calculate the relative measures of variation and indicate the value which , in your opinion, is more reliable.
ANSWER:
(i) Coefficient of Range
(a) Wheat

The coefficient of variation is more reliable than all other measures.

Q.7A batsman is to be selected for a cricket team. The choice is between X and Y on the basis of their scores in five previous scores which are

Which batsman should be selected if we want,
(i) a higher run-getter, or
(ii) a more reliable batsman in the team?
ANSWER:
Batsman X

(i) Average of Batsman X is higher than that of Batsman Y, so he should be selected if we want a high scorer.
(ii) The Batsman Y is more reliable than Batsman X. This is because the coefficient of variation of Batsman X is higher than that of Batsman Y.

Q.8To check the quality of two brands of light bulbs, their life in burning hours was estimated as under for 100 bulbs of each brand.

(i) Which brand gives higher life?
(ii) Which brand is more dependable?
ANSWER:
For Brand A

For Brand B

(i) The average life of bulb of Brand B is comparatively higher than that of Brand A.
(ii) The bulbs of Brand B are more dependable as CV of Brand B is lesser than CV of Brand A.
Q.9 Average daily wage of 50 workers of a factory was ₹ 200 with a Standard Deviation of ₹ 40. Each worker is given a raise of ₹ 20. What is the new average daily wage and standard deviation? Have wages become more or less uniform?
ANSWER:
N = 50
x¯¯¯ = 200
σ = 40
Average wage =  Total wages  Number of workers 
200 =  Total wages 50
So, total wages = 200 × 50 = ₹ 10,000
Now, increase in wage rate = ₹20
Total raise = 50 × 20= ₹ 1,000
Total wage after raise = ₹ 10,000 + 1,000 = ₹ 11,000
New average wage =  New total wages  Number of workers =11,00050 = ₹220
Thus, Mean increases by the amount of increase in wage of each worker as the absolute increase was equal for all.
Standard Deviation will remain the same that is ₹40 as Standard Deviation is independent of origin and hence addition of equal amount in all the values will not cause any change in the Standard Deviation.
Uniformity of wages can be seen by coefficient of variation.
Previously, the coefficient of variation was
CV = σX¯¯¯¯¯×100 = (40/200) × 100 = 20
The new coefficient of variation after wage increase is given by
CV = σx×100 = (40/220) × 100 = 18.18
This shows that wages have become more uniform now as the new CV is lower.

Q.10If in the previous question, each worker is given a hike of 10% in wages, how are the Mean and Standard Deviation values affected?
ANSWER:
Average wage = ₹ 200
Hike in wages = 10%
Since arithmetic mean is not independent of scale, the mean will also increase by 10%.
= 10100×200 = ₹ 20
Hence, the new Mean will be
200 + 20 = ₹ 220
Standard Deviation is also not independent of scale, hence, the Standard Deviation will also increase by 10% Initial Standard Deviation = ₹ 40
So, New Standard Deviation = ₹40 +10% of 40
= ₹ (40 + 4)
= ₹ 44

Q.11Calculate the Mean Deviation using Mean and Standard Deviation for the following distribution.

ANSWER:

Q.12The sum of 10 values is 100 and the sum of their squares is 1090. Find out the coefficient of variation.
ANSWER:

Question And Answer:

Q.1 Which average would be suitable in the following cases?
(i) Average size of readymade garments.
(ii) Average intelligence of students in a class.
(iii) Average production in a factory per shift.
(iv) Average wages in an industrial concern.
(v) When the sum of absolute deviations from average is least.
(vi) When quantities of the variable are in ratios.
(vii) In case of open-ended frequency distribution.
ANSWER:
(i) Mode Average size of any ready made garments should be the size for which demand is the maximum. Hence, the modal value which represents the value with the highest frequency should be taken as the average size to be produced.
(ii) Median It is the value that divides the series into two equal parts. Therefore, Median will be the best measure for calculating the average intelligence of students in a class as it will give the average intelligence such that there are equal number of students above and below this average. It will not be affected by extreme values.
(iii) Arithmetic Mean The average production in a factory per shift is best calculated by Arithmetic Mean as it will capture all types of fluctuations in production during the shifts.
(iv) Arithmetic Mean Arithmetic Mean will be the most suitable measure. It is calculated by dividing the sum of wages of all the workers by the total number of workers in the industrial concern. It gives a fair idea of average wage bill taking into account all the workers.
(v) Arithmetic Mean The algebraic sum of the deviations of values about Arithmetic Mean is zero. Hence, when the sum of absolute deviations from average is the least, then mean could be used to calculate the average.
(vi) Median Median will be the most suitable measure in case the variables are in ratios as it is least affected by the extreme values.
(vii) Median Median is the most suitable measure as it can be easily computed even in case of open ended frequency distribution and will not get affected by extreme values.

Q.2 Indicate the most appropriate alternative from the multiple choices provided against each question.
(i) The most suitable average for qualitative measurement is
(a) Arithmetic mean
(b) Median
(c) Mode
(d) Geometric mean
(e) None of these
ANSWER:
(b) Median is the most suitable average for qualitative measurement because Median divides a series in two equal parts thus representing the average qualitative measure without being affected by extreme values.

(ii) Which average is affected most by the presence of extreme items?
(a) Median
(b) Mode
(c) Arithmetic Mean
(d) Geometric Mean
(e) Harmonic Mean
ANSWER:
(c) It is defined as the sum of the values of all observations divided by the number of observations and therefore it is. affected the most by extreme values

(iii) The algebraic sum of deviation of a set of n values from AM is
(a) n
(b) 0
(c) 1
(d) None of these
ANSWER:
(b) This is one of the mathematical properties of arithmetic mean that the algebraic sum of deviation of a set of n values from AM is zero.

Q.3 Comment whether the following statements are true or false.
(i) The sum of deviation of items from median is zero.
(ii) An average alone is not enough to compare series.
(iii) Arithmetic mean is a positional value.
(iv) Upper quartile is the lowest value of top 25% of items.
(v) Median is unduly affected by extreme observations.
ANSWER:
(i) False
This mathematical property applies to the arithmetic mean and not to median.
(ii) True
Average is not enough to compare the series as it does not explain the extent of deviation of different items from the central tendency and the difference in the frequency of values. These are measured by measures of dispersion and kurtosis.
(iii) False
Median is a positional value.
(iv) True
The upper quartile also called the third quartile, has 75 % of the items below it and 25 % of items above it.
(v) False
Arithmetic mean is unduly affected by extreme observations.

Q.4If the arithmetic mean of the data given below is 28, find (a) the missing frequency and (b) the median of the series

ANSWER:
(a) Let the missing frequency br f₁.
Arithmetic Mean = 28

or 2240 -2100 = 35f₁ = 28f₁
or 140 = 7f₁
f₁ = 20
Hence, the missing frequency is 20.
(b)

So, the Median class = Size of (N2)th item = 50th term.
50th item lies in the 57th cumulative frequency and the corresponding class interval is 20-30.

Q.5The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

ANSWER:

N = 10
X¯¯¯¯=ΣXN=240010=240
Arithmetic Mean = ₹ 240

Q.6 Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

ANSWER:

Q.7 The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

ANSWER:

So, the median class = Size of (N2) th item = 190 item
190th lies in the 129 th cumulative frequency and the corresponding class interval is 200-300.

Median size of land holdings = 241.22 acres

Q.8The following series relates to the daily income of workers employed in a firm. Compute (a) highest income of lowest 50% workers, (b) minimum income earned by the top 25% workers and (c) maximum income earned by lowest 25% workers.

ANSWER:

(a) Highest income of lowest 50% workers will be given by the median. Σf = N = 65
Median class = Size of (N2)th item = Size of (652)th item=325 th item
32.5th item lies in the 50th cumulative frequency and the corresponding class interval is 24.5 – 29.5.

(b) Minimum income earned by top 25% workers will be given by the lower quartile Q₁.
Class interval of Q₁ = (N4)th item
= (654)th item = 1625th item
16.25th item lies in the 30th cumulative frequency and the corresponding class interval is 19.5 – 24.5

(c) Maximum income earned by lowest 25% workers will be given by the upper quartile Q₃.
Class interval of Q₃ = (N4)th item
= 3(654)th item
= 3 × 1625th item
= 48.75th item
48.75th item lines in 50th item and the corresponding class interval is 24.5-29.5

Q.9The following table gives production yield in kg per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode production yield.

ANSWER:
(i) Mean

(ii) Median

(iii) Mode
Grouping Table

Analysis Table

(i) Mean

(ii) Median

(iii) Mode
Grouping Table

Analysis Table

In This Post we are  providing Chapter-8 BINOMINAL THEOREM NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON BINOMINAL THEOREM

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1. Find Hence evaluate 

Ans.

Put 

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2. Show that is divisible by 64, whenever n is positive integer.

Ans. 

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3. Find the general term in the expansion of  

Ans. 

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4. In the expansion of prove that coefficients of and are equal.

Ans. 

Put  and  respectively

Coeff of is 

Coeff of is H.P

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5. Expand 

Ans. 

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6. Find the sixth term of the expansion if the binomial coefficient of the third term from the end is 45.

Ans. The binomial coeff of the third term from end = binomial coeff of the third term from beginning = 

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7. Find a if the 17th and 18th terms of the expansion are equal.

Ans. 

ATQ put  and 17

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8. Find the term independent of in the expansion of 

Ans. 

Put 

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9. If the coeff of and terms in the expansion of are equal find 

Ans.

Coeff are

and 

ATQ 

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10. Show that the coeff of the middle term in the expansion of is equal to the sum of the coeff of two middle terms in the expansion of 

Ans. As is even so the expansion has only one middle term which is  term

Coeff of is 

Similarly being odd the other expansion has two middle term i.e

and term

 i.e and 

The coeff are  and 

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11. Find the value of if the coeff of and terms in the expansion of  are equal.

Ans. 

Put 

And 

ATQ 

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12. Find the 13th term in the expansion of 

Ans. 

Put 

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13.Find , if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of is 

Ans.Fifth term from the beginning in the expansion of is

How fifth term from the end would be equal to in term from the beginning

ATQ 

14. The sum of the coeff. 0f the first three terms in the expansion of being natural no. is 559. Find the term of expansion containing 

Ans.The coeff. Of the first three terms of are and 

Therefore, by the given condition

On solving we get 

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15.Show that the middle term in the expansion of is 

Ans.As is even, the middle term of the expansion term

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In This Post we are  providing Chapter-7 PERMUTATIONS AND COMBINATIONS NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON PERMUTATIONS AND COMBINATIONS

1.In how many ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colours.

Ans.No. of ways of selecting 9 balls

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2.Find if 

Ans.

 Rejected. Because if we put the no. in the factorial is –ve.

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3.If find 

Ans.

4.What is 

Ans.Is multiplication of consecutive natural number

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5.If what is the value of 

Ans. then 

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6.How many words, with or with not meaning each of 2 vowels and 3 consonants can be flamed from the letter of the word DAUGHTER?

Ans.In the word DAUGHTER There are 3 vowels and 5 consonants out of 3 vowels, 2 vowels can be selected in ways and 3 consonants can be selected in ways 5 letters 2 vowel and 3 consonant can be arranged in 5! Ways

Total no. of words 

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7.Convert the following products into factorials 

Ans.

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8.Evaluate 

Ans.

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9.Evaluate 

Ans.

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10.Find if 

Ans.Given 

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11.Evaluate 

Ans.

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12.Convert into factorial 2.4.6.8.10.12

Ans. 

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13.How many words with or without meaning can be formed using all the letters of the word ‘EQUATION’ at a time so that vowels and consonants occur together

Ans. In the word ‘EQUATION’ there are 5 vowels [A.E.I.O.U.] and 3 consonants [Q.T.N]

Total no. of letters = 8

Arrangement of 5 vowels = 

Arrangements of 3 consonants = 

Arrangements of vowels and consonants = 

Total number of words 

14. How many words, with or without meaning can be made from the letters of the word MONDAY. Assuming that no. letter is repeated, it

(i) 4 letters are used at a time

(ii) All letters are used but first letter is a vowel?

Ans. Part-I In the word MONDAY there are 6 letters

4 letters are used at a time

Total number of words 

Part-II  All letters are used at a time but first letter is a vowel then OAMNDY

2 vowels can be arranged in 2! Ways

4 consonants can be arranged in 4! Ways

Total number of words   

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15.Prove that 

Ans.Proof L.H.S.

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In This Post we are  providing Chapter-6 LINEAR INEQUALITIES NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON LINEAR INEQUALITIES

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1. Solve the inequality 

Ans. 

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2. Solve the inequality 

Ans. 

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3. Solution set of the in inequations and is.

Ans. 

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4. Solve. Show the graph of the solution on number line.

Ans.

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5. Solve the inequality. 

Ans. 

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6. Solve when is an integer.

Ans. 

7. Ravi obtained 70 and 75 mark in first unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Ans. Let Ravi secure marks in third test

ATQ 

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8. Find all pairs of consecutive odd natural no. both of which are larger than 10 such that their sum is less than 40.

Ans. Let and be consecutive odd natural no.

ATQ

From (i) and (ii)

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9. A company manufactures cassettes and its cost equation for a week is C=300+1.5and its revenue equation is R=2, where is the no. of cassettes sold in a week. How many cassettes must be sold by the company to get some profit?

Ans. Profit = revenue-cost

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10. The longest side of a is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the is at least 61 cm find the minimum length of the shortest side.

Ans. Let shortest side be cm then the longest side is cm and the third side cm.

ATQ 

Length of shortest side is 9 cm.

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11. In drilling world’s deepest hole it was found that the temperature T in degree Celsius, km below the surface of earth was given by At what depth will the tempt. Be between c and 

Ans. Let km is the depth where the tempt lies between and 

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12. A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second.

Ans. Let the shortest length be cm, then second length is (+3) cm and the third length is 2 cm.

ATQ 

Again ATQ

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13. The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between 7.2 and 7.8 If the first pH reading are 7.48 and7.85, find the range of pH value for the third reading that will result in the acidity level being normal.

Ans. Let third reading be  then

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14. Solve graphically 

Ans.  

Put (1,0) in eq. (i)

false

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15. Solve graphically

Ans. 

	0	2
	3	0

Put in eq. ……….

 which is false

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In This Post we are  providing Chapter-5 COMPLEX NUMBERS AND QUADRATIC EQUATIONS NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON COMPLEX NUMBERS AND QUADRATIC EQUATIONS

1. Evaluate i^-39

Ans. 

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2. Solved the quadratic equation 

Ans. 

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3. If = 1, then find the least positive integral value of m.

Ans. 

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4. Evaluate (1+ i)⁴

Ans. 

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5. Find the modulus of 

Ans. Let z = 

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6. Express in the form of a + ib. (1+3i)^-1

Ans. 

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7. Explain the fallacy in -1 = i. i. = 

Ans.  is okay but

 is wrong.

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8. Find the conjugate of 

Ans. Let z = 

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9. Find the conjugate of – 3i – 5.

Ans. Let z = 3i – 5

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10. Let z₁ = 2 – i, z₂ = -2+i Find Re 

Ans. z₁ z₂ = (2 – i)(-2 + i)

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11. Express in the form of a + ib (3i-7) + (7-4i) – (6+3i) + i²³

Ans. Let

Z = 

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12. Find the conjugate of 

Ans. 

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13. Solve for x and y, 3x + (2x-y) i= 6 – 3i

Ans. 3x = 6

x = 2

2x – y = – 3

2 × 2 – y = – 3

– y = – 3 – 4

y = 7

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14. Find the value of 1+i² + i⁴ + i⁶ + i⁸ + —- + i²⁰

Ans.

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15. Multiply 3-2i by its conjugate.

Ans.Let z = 3 – 2i

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In This Post we are  providing Chapter-4 PRINCIPLE ON MATHEMATICAL INDUCTION NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON PRINCIPLE ON MATHEMATICAL INDUCTION

Q1. Give an example of a statement P(n) which is true for all n≥ 4 but P(l), P(2) and P(3) are not true. Justify your answer

Sol. Consider the statement P(n): 3n < n!

For n = 1, 3 x 1 < 1!, which is not true
For n = 2, 3 x 2 < 2!, which is not true
For n = 3, 3 x 3 < 3!, which is not true
For n = 4, 3 x 4 < 4!, which is true
For n = 5, 3 x 5 < 5!, which is true

Q2. Prove that number of subsets of a set containing n distinct elements is 2″, for all n ∈
Sol: Let P(n): Number of subset of a set containing n distinct elements is 2″, for all ne N.
For n = 1, consider set A = {1}. So, set of subsets is {{1}, ∅}, which contains 2¹ elements.
So, P(1) is true.
Let us assume that P(n) is true, for some natural number n = k.
P(k): Number of subsets of a set containing k distinct elements is 2^k To prove that P(k + 1) is true,
we have to show that P(k + 1): Number of subsets of a set containing (k + 1) distinct elements is 2^k+1
We know that, with the addition of one element in the set, the number of subsets become double.
Number of subsets of a set containing (k+ 1) distinct elements = 2×2^k = 2^k+1
So, P(k + 1) is true. Hence, P(n) is true.

Q3. 4ⁿ – 1 is divisible by 3, for each natural number
Sol: Let P(n): 4ⁿ – 1 is divisible by 3 for each natural number n.
Now, P(l): 4¹ – 1 = 3, which is divisible by 3 Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 4^k – 1 is divisible by 3
or               4^k – 1 = 3m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 4^k+1 – 1
= 4^k-4-l
= 4(3m + 1) – 1  [Using (i)]
= 12 m + 3
= 3(4m + 1), which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q4. 2³ⁿ – 1 is divisible by 7, for all natural numbers
Sol: Let P(n): 2³ⁿ – 1 is divisible by 7
Now, P( 1): 2³ — 1 = 7, which is divisible by 7.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 2^3k – 1 is divisible by 7.
or               2^3k -1 = 7m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 2^3(k+1)– 1
= 2^3k.2³– 1
= 8(7 m + 1) – 1
= 56 m + 7
= 7(8m + 1), which is divisible by 7.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q5. n³ – 7n + 3 is divisible by 3, for all natural numbers
Sol: Let P(n): n³ – 7n + 3 is divisible by 3, for all natural numbers n.
Now P(l): (l)³ – 7(1) + 3 = -3, which is divisible by 3.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) = K³ – 7k + 3 is divisible by 3
or K³ – 7k + 3 = 3m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1 ):(k + l)³ – 7(k + 1) + 3
= k³ + 1 + 3k(k + 1) – 7k— 7 + 3 = k³ -7k + 3 + 3k(k + l)-6
= 3m + 3[k(k+l)-2]  [Using (i)]
= 3[m + (k(k + 1) – 2)], which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n

Q6. 3²ⁿ – 1 is divisible by 8, for all natural numbers
Sol: Let P(n): 3²ⁿ – 1 is divisible by 8, for all natural numbers n.
Now, P(l): 3² – 1 = 8, which is divisible by 8.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): 3^2k – 1 is divisible by 8
or               3^2k -1 = 8m, m ∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 3^2(k+1)– l
= 3^2k • 3² — 1
= 9(8m + 1) – 1     (using (i))
= 72m + 9 – 1
= 72m + 8
= 8(9m +1), which is divisible by 8 Thus P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q7. For any natural number n, 7ⁿ – 2ⁿ is divisible by 5.
Sol: Let P(n): 7ⁿ – 2ⁿ is divisible by 5, for any natural number n.
Now, P(l) = 7¹-2¹ = 5, which is divisible by 5.
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.

.’.  P(k) = 7^k -2^k is divisible by 5
or  7^k – 2^k = 5m, m∈ N                                                                           (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 7^k+1 -2^k+1
= 7^k-7-2^k-2
= (5 + 2)7^k -2^k-2
= 5.7^k + 2.7^k-2-2^k
= 5.7^k + 2(7^k – 2^k)
= 5 • 7^k + 2(5 m)     (using (i))
= 5(7^k + 2m), which divisible by 5.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Q8. For any natural number n, xⁿ -yⁿ is divisible by x -y, where x and y are any integers with x ≠y
Sol: Let P(n) : xⁿ – yⁿ is divisible by x – y, where x and y are any integers with x≠y.
Now, P(l): x¹ -y¹ = x-y, which is divisible by (x-y)
Hence, P(l) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): x^k -y^k is divisible by (x – y)
or   x^k-y^k = m(x-y),m ∈ N …(i)
Now, we have to prove that P(k + 1) is true.
P(k+l):x^k+l-y^k+l
= x^k-x-x^k-y + x^k-y-y^ky
= x^k(x-y) +y(x^k-y^k)
= x^k(x – y) + ym(x – y)  (using (i))
= (x -y) [x^k+ym], which is divisible by (x-y)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q9. n³ -n is divisible by 6, for each natural number n≥
Sol: Let P(n): n³ – n is divisible by 6, for each natural number n> 2.
Now, P(2): (2)³ -2 = 6, which is divisible by 6.
Hence, P(2) is true.
Let us assume that, P(n) is true for some natural number n = k.
P(k): k³ – k is divisible by 6
or    k³ -k= 6m, m∈ N       (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): (k+ l)³-(k+ 1)
= k³+ 1 +3k(k+ l)-(k+ 1)
= k³+ 1 +3k² + 3k-k- 1 = (k³-k) + 3k(k+ 1)
= 6m + 3 k(k +1)  (using (i))
Above is divisible by 6.   (∴ k(k + 1) is even)
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 2.

Q10. n(n² + 5) is divisible by 6, for each natural number
Sol: Let P(n): n(n² + 5) is divisible by 6, for each natural number.
Now P(l): 1 (l² + 5) = 6, which is divisible by 6.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): k( k² + 5) is divisible by 6.
or K (k²+ 5) = 6m, m∈ N         (i)
Now, we have to prove that P(k + 1) is true.
P(K+l):(K+l)[(K+l)² + 5]
= (K + l)[K² + 2K+6]
= K³ + 3 K² + 8K + 6
= (K² + 5K) + 3 K² + 3K + 6 =K(K² + 5) + 3(K² + K + 2)
= (6m) + 3(K² + K + 2)        (using (i))
Now, K² + K + 2 is always even if A is odd or even.
So, 3(K² + K + 2) is divisible by 6 and hence, (6m) + 3(K² + K + 2) is divisible by 6.
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q11. n² < 2ⁿ, for all natural numbers n ≥
Sol: Let P(n): n² < 2ⁿ for all natural numbers n≥ 5.
Now P(5): 5² < 2⁵ or 25 < 32, which is true.
Hence, P(5) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): k² < 2^k  (i)
Now, to prove that P(k + 1) is true, we have to show that P(k+ 1): (k+ l)² <2^k+1
Using (i), we get
(k + l)² = k² + 2k + 1 < 2^k + 2k + 1         (ii)
Now let, 2^k + 2k + 1 < 2^k+1     (iii)
∴ 2^k + 2k + 1 < 2 • 2^k
2k + 1 < 2^k, which is true for all k > 5 Using (ii) and (iii), we get (k + l)² < 2^k+1 Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 5.

Q12. 2n<(n + 2)! for all natural numbers
Sol: Let P(n): 2n < (n + 2)! for all natural numbers n.
P( 1): 2 < (1 + 2)! or 2 < 3! or 2 < 6, which is true.
Hence,P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k) :2k<(k + 2)!  (i)
To prove that P(k + 1) is true, we have to show that
P(k + 1): 2(k+ 1) < (k + 1 + 2)!
or 2(k+ 1) < (k + 3)!
Using (i), we get
2(k + 1) = 2k + 2<(k+2) !  +2  (ii)
Now let, (k + 2)! + 2 < (k + 3)!  (iii)
=>  2 < (k+ 3)! – (k+2) !
=> 2 < (k + 2) ! [k+ 3-1]
=>2<(k+ 2) ! (k + 2), which is true for any natural number.
Using (ii) and (iii), we get 2(k + 1) < (k + 3)!
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q13. 2 + 4 + 6+… + 2n = n² + n, for all natural numbers
Sol: Let P(n) :2 + 4 + 6+ …+2 n = n² + n
P(l): 2 = l² + 1 = 2, which is true
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): 2 + 4 + 6 + .,.+2k = k² + k  (i)
Now, we have to prove that P(k + 1) is true.
P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)
= k² + k + 2(k+ 1)  [Using (i)]
= k² + k + 2k + 2
= k² + 2k+1+k+1
= (k + 1)² + k+ 1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q14. 1 + 2 + 2² + … + 2ⁿ = 2^n +1 – 1 for all natural numbers
Sol: Let P(n): 1 + 2 + 2² + … + 2ⁿ = 2^n +1 – 1, for all natural numbers n
P(1): 1 =2^{0 + 1} — 1 = 2 — 1 = 1, which is true.
Hence, ,P(1) is true.
Let us assume that P(n) is true for some natural number n = k.

P(k): l+2 + 2²+…+2^k = 2^k+1-l              (i)

Now, we have to prove that P(k + 1) is true.

P(k+1): 1+2 + 2²+ …+2^k + 2^k+1
= 2^{k +1} – 1 + 2^k+1  [Using (i)]
= 2.2^k+l– 1 = 1
₌ 2^(k+1)+1-1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

Q15. 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers
Sol: Let P(n): 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), for all natural numbers n.
P(1): 1 = 1(2 x 1 – 1) = 1, which is true.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k):l+5 + 9 +…+(4k-3) = k(2k-1)  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 1 + 5 + 9 + … +  (4k- 3) + [4(k+ 1) – 3]
= 2k² -k+4k+ 4-3
= 2k² + 3k + 1
= (k+ 1)( 2k + 1)

= (k+l)[2(k+l)-l]

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n.

In This Post we are  providing Chapter- 2 TRIGNOMETRIC FUNCTIONS NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON TRIGNOMETRIC FUNCTIONS

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1.Convert into radian measures. -37⁰ 30’

Ans. – 37⁰ 30’ = –

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2.Prove Sin (n+1) x Sin (n+2) x + Cos (n+1) x. Cos (n+2) x = Cos x 

Ans.L.H.S = Cos (n+1) x Cos (n+2) x + Sin (n+1) x Sin (n+2) x

=Cos 

=Cos x

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3.Find the value of Sin 

Ans. Sin = Sin 

= sin 

= Sin 

=

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4.Find the principal solution of the eq. tan x = 

Ans. tan x = 

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5.Convert into radian measures. 

Ans. 5⁰ 37¹ 30¹¹ = 5⁰ + 

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6.Evaluate 2 Sin 

Ans.2 Sin 

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6.Find the solution of Sin x = 

Ans.Sin x = 

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7.Prove that 

Ans.L. H. S = tan 36⁰

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8.Find the value of tan 

Ans.

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9.Prove Cos 4x = 1 – 8 Sin² x. Cos²x

Ans. L. H. S = Cos 4x

10.Find the value of Cos (- 1710⁰).

Ans. Cos (-1710⁰) = Cos (1800-90)[Cos (-θ) = Cos θ

= Cos [5 360 +90]

= Cos  = 0

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11.A wheel makes 360 revolutions in 1 minute. Through how many radians does it turn in 1 second.

Ans.N. of revolutions made in 60 sec. = 360

N. of revolutions made in 1 sec = 

Angle moved in 6 revolutions = 2 π 6 = 12 π

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12.If in two circles, arcs of the same length subtend angles 60⁰ and 75⁰ at the centre find the ratio of their radii.

Ans.

(1)÷ ( 2)

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13.Prove that Cos 6x= 32 Cos⁶x – 48 Cos⁴ x + 18 Cos² x-1

Ans.L.H.S. = Cos 6x

= 

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14.Solve Sin2x-Sin4x+Sin6x=o

Ans.

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15.Prove that (Cos x + Cos y)² + (Sin x – Sin y)² = 4 Cos² 

Ans. L. H. S = (Cos x + Cos y)² + (Sin x – Sin y)²

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In This Post we are  providing Chapter-2 RELATIONS AND FUNCTIONS NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON RELATIONS AND FUNCTIONS

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1. If P = {1,3}, Q = {2,3,5}, find the number of relations from A to B

Ans. = 64

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2. If A = {1,2,3,5} and B = {4,6,9}, R = {(x, y) : |x – y| is odd, x ∈ A, y ∈ B} Write R in roster form

Ans. Function

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3.

Let f and g be two real valued functions, defined by, f(x) = x2, g(x) = 3x +2.

Ans. Not a function

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4. Find the domain of the relation, R = { (x, y) : x, y ϵ Z, xy = 4} Find the range of the following relations : (Question-23, 24)

Ans. {–4, –2, –1,1,2,4}

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5.If and form the sets and are these two Cartesian products equal?

Ans. Given and by definition of cartesion product, we set

and 

By definition of equality of ordered pains the pair is not equal to the pair therefore 

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6.If and are finite sets such that and find the number of relations from to 

Ans. Linen 

the number of subsets of 

then the number of subsets of 

Since every subset of is a relation from A to B therefore the number of relations from A to B = 2^mk

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7.Let be a function from z to z defined by for same integers a and b determine a and b.

Ans. Given 

Since 

Subtracting (i) from(ii) we set a=2

Substituting a=2 is (ii) we get 2+b=1

b = -1

Hence a = 2, b = -1

8.Express as the set of ordered pairs

Ans. Since and 

Put 

For anther values of we do not get 

Hence the required set of ordered peutes is 

9.Function is defined by find 

Ans. 

10.Let and be the relation, is one less than from to then find domain and Range of 

Ans. Given and is the relation ‘is one less than’ from to therefore 

Domain of and range of 

11.Let be a relation from to define by.

Is the following true implies 

Ans. No; let As so but so 

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12.Let be the set of natural numbers and the relation be define in by =what is the domain, co domain and range of? Is this relation a function?

Ans. Given 

Domain of co domain of and Range of is the set of even natural numbers.

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13. Let A = {1,2,3,4}, B = {1,4,9,16,25} and R be a relation defined from A to B as, R = {(x, y) : x ϵ A, y ϵ B and y = x²}

(a) Depict this relation using arrow diagram.

(b) Find domain of R.

(c) Find range of R.

(d) Write co-domain of R.

Ans.

(b) {1,2,3,4}

(c) {1,4,9,16}

(d) {1,4,9,16,25}

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14. Let R = { (x, y) : x, y ϵ N and y = 2x} be a relation on N. Find :

(i) Domain

(ii) Codomain

(iii) Range

Is this relation a function from N to N

Ans. (i) N

(ii) N

(iii) Set of even natural numbers

yes, R is a function from N to N.

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15. Find the domain and range of, f(x) = |2x – 3| – 3

Ans. Domain is R

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In This Post we are  providing Chapter-1 SETS NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON SETS

Q1.If A = { 1,2,3,4,5,6}, B = {2,4,6, 8} then find A – B

Ans. We are given the sets A ={ 1,2,3,4,5,6}, B = {2,4,6, 8}

A – B = { 1,2,3,4,5,6}- {2,4,6, 8} = {1, 3, 5}

Q2. Let A and B be two sets containing 3 and 6 elements respectively. Find the maximum and the minimum number of elements in A ∪ B.

Ans.There may be the case when atleast 3 elements are common between both sets

Let a set A = {a, b, c} and B = {a, b, c, d, e, f}

∴ A ∪ B = {a, b, c, d, e, f} implies that the minimum number of elements in A ∪ B are = 6

There may be the case when there are no any elements are common between both sets

lf A = {a, b, c}, B = { d, e, f, g, h, i}

A ∪ B = {a, b, c, d, e, f,g,h,i} implies that the maximum number of elements in A ∪ B are = 9

Q3.If A = {(x,y) : x² + y²= 25  where x, y ∈ W } write a set of all possible ordered pair .

Ans.  We are given the set A = {(x,y) : x² + y²= 25  where x, y ∈ W }

All possible ordered pair of set A are following

For x = 0,y =5, x=3,y=4,for x =4, y =3,for x=5,y =0

A = {(0,5),(3,4),(4,3),(5,0)}

Q4.If A = {1,2,3}, B = {4, 5, 6} and C ={5} verify that A ∪ ( B ∩ C) = (A ∪ B) ∩ A ∪ C.

We are given the sets A = {1,2,3}, B = {4, 5, 6} ,C = {5}

B ∩ C = {5}

LHS

A ∪ ( B ∩ C) = {1,2,3,5}

(A ∪ B) and A ∪ C

A ∪ B = {1,2,3,4,5,6} and A ∪ C = {1,2,3,5}

RHS

(A ∪ B) ∩ A ∪ C = {1,2,3,5}

Therefore

A ∪ ( B ∩ C) = (A ∪ B) ∩ A ∪ C, Hence proved

Q5.From the adjoining Venn diagram, write the value of the following.

(a) A ‘

(b) B’

(c) (A ∩ B)’

Ans. From the venn diagram ,we have

U = {1,2,3,4………15}

A = {7,9.11}. B ={11,12,13,14}

A’ = U – A = {1,2,3,4………15} – {7,9.11} = {1,2,3,4,5,6,8,10,12,13,14,15}

B’ = U – B = {1,2,3,4………15} – {11,12,13,14}= {1,2,3,4,5,6,7,8,9,10,,15}

We have,(A ∩ B) = 11

(A ∩ B)’ = U – (A ∩ B) = {1,2,3,4………15} – {11} = (1,2,3,4,5,6,7,8,9,10,12,13,14,15}

Q6. If P(A) = P(B) show that A = B.

Ans. P(A} and P(B) implies that both are power sets of A and B respectively

Every set is an element of its power set , so A ∈ P(A)

Since, we are given that

P(A) = P(B)

Therefore, A ∈ P(B)

Indicates that every element of A belongs to the set B

So, A ⊂ B……(i)

Similarly B ∈ P(B)

Since, we are given that

P(A) = P(B)

Indicates that every element of B belongs to the set A

So, B ⊂ A……(ii)

From (i) and (ii), we get

A = B, Hence proved

Q7. Let A  and B be sets ; if A∩X = B∩X = ∅ and A∪X = B∪X for some set X. Show that A=B.

Ans. We are given that A∩X = B∩X = ∅ and A∪X = B∪X

To prove   A=B

Proof.  A∪X = B∪X (given)

Multiplying both sides by A∩

A∩ (A∪X) =A∩ (B∪X)

Using distributive property

(A∩ A) ∪ (A ∩ X) = (A∩ B )∪ (A∩ X)

A∩Φ = (A∩ B) ∪ Φ

A = (A∩ B) …………(i)

A∪X = B∪X

Multiplying both sides by B∩

B∩(A∪X) = B∩(B∪X)

(B∩ A) ∪ (B ∩ X) = (B∩ B )∪ (B∩ X)

(B∩ A) ∪φ = B ∪ φ

B = (B∩ A)

B = (A∩ B) …………(ii)

From (i) and (ii)

A = B, Hence proved

Q8.If A ={1,2,3,4,5},then write the proper subsets of A.

Ans. The number of elements in the given sets A ={1,2,3,4,5} are =5

The number of proper subsets of any set are = 2ⁿ – 1

   ^{Where n = number of elements = 5}

  ^{The number of proper subsets of any set are =}  2⁵ – 1 =32 – 1 = 31

Q9.Write the following sets in the Roster form

(i) A={x : x ∈ R, 2x+11 =15}

(ii)B={x |x² =x, x ∈ R}

(iii)C={x = x is a positive factor of the prime number p}

Ans.(i)We have, A={x : x ∈ R, 2x+11 =15}

2x+1= 15 ⇒x= 2

∴ A = {2}

(ii)We have,B={x |x² =x, x ∈ R}

∴ x² = x ⇒ x²-x= 0 ⇒x(x-1)= 0 ⇒x=0,1

∴ B ={0,1}

(iii)We have, C={x = x is a positive factor of the prime number p}

Sice positive factors of a prime u∪mer are 1 ad the number itself,we have

C={1,p}

Q10.For all sets A,B and C show that (A – B) ∩(A – C) = A – (B ∪ C).

Ans. Considering that

x ∈ (A – B)∩(A – C)

⇒ x ∈ (A – B) and x ∈ (A – C)

⇒ (x ∈ A and x∉ B) and (x ∈ A and x∉ C)

⇒ (x ∈ A ) and (x ∉B and  x∉ C )

⇒(x ∈ A ) x∉ (B ∪C)∈

⇒x ∈ A – (B ∪C)

⇒(A – B) ∩(A – C)⊂A – (B ∪ C)…….(i)

Now,Considering that

y ∈A – (B ∪ C)

⇒ y ∈A and y ∉  (B ∪ C)

⇒y ∈A and (y ∉B and y ∉ C)

⇒(y ∈A and y ∉B) and (y ∈A and y ∉ C)

⇒y ∈ (A – B) and y ∈ (A – C)

⇒y ∈ (A – B) ∩ y ∈ (A – C)

⇒A – (B ∪ C) ⊂(A – B) ∩(A – C)………(ii)

From (i) and (ii)

(A – B) ∩(A – C)= A – (B ∪ C), Hence proved

Q11.Let A,B and C be the sets such that A∪B = A∪C and A ∩B = A ∩C,show that B = C.

Ans. According to question, A ∪ B = A ∪ C and A ∩ B = A ∩ C

To show, B = C

Let us assume, x ∈ B So, x ∈ A ∪ B

x ∈ A ∪ C

Hence, x ∈ A or x ∈ C

when x ∈ A, then x ∈ B

∴ x ∈ A ∩ B

As, A ∩ B = A ∩ C

So, x ∈ A ∩ C

∴ x ∈ A or x ∈ C

x ∈ C

∴ B ⊂ C

similarly, it can be shown that C ⊂ B

Hence, B = C

Q12.Show that for any sets A and B

A = (A∩B)∪(A-B) 

Ans.We have to prove

A = (A∩B)∪(A-B)

Taking RHS and solving it

(A∩B)∪(A-B)

Using the property

A -B = A -(A∩B)

A-B = A∩B’

=  (A∩B) ∪ (A∩B’)

Applying distributive property

A∩(B∪C) = (A∩B) ∪(A∩C)

Replacing C = B’ in LHS

A-B = A∩(B∪B’)

= A∩ (U) [since B∪B’ = U]

= A [since A∩ U = A]

= RHS, Hence proved

Q13. Write the following sets in the roster form:

(i) A = {x : x ∈ R, 2x + 15 = 15}

(ii) B = {x : x² = x, x ∈ R}

Ans(i) It is given to us that set

A = {x : x ∈ R, 2x + 11 = 15}

2x + 11 = 15

2x = 15 – 11

2x = 4

x = 2

Therefore in roster form it is written as  A = {2}

(ii) It is given to us that

B = {x : x² =x, x ∈ R}

x² = x

x² – x = 0

x(x – 1) = 0

x = 0, x = 1

Therefore in roster form it is written as

B = {0, 1}

Q14. If  A and B are subsets of the universal set U, then show that 

(i) A ⊂ A ∪ B

Ans. Let’s prove that

A ⊂ A ∪ B

Let  x ∈ A or x ∈ B

If x ∈ A then x ∈ A ∪ B

Hence A ⊂ A ∪ B

Q15. A,B and C are subsets of universal set if A = {2,4,6,8,12,20}, B ={3,6.9.12.15},C ={5,10,15,20} and U is the set of all whole numbers, draw a venn diagram showing the relation of  U,A,B and C.

Ans.