Yearly Archives: 2020

NCERT Important Questions & Solutions for Class 9 Maths Chapter 15 Probability

You can find Chapter 2 Probability Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important Questions & Solutions for Chapter 15 Probability Ex 15.1

1.In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Here, the total number of trials = 30
∵ Number of times, the ball touched the boundary =6
∴ Number of times, the ball missed the boundary = 30 – 6 = 24
Let the event of not hitting the boundary be represented by E, then

P(E)= 24/30= 4/5

Thus, the required probability =4/5

2.In a particular section of class IX, 40 students were asked about the month of their birth and the following graph was prepared for the data so obtained.

Find the probability that a student of the class was born in August.
Solution:
From the graph, we have
Total number of students born in various months = 40
Number of students born in August = 6
∴ Probability of a student of the Class-IX who was born in August = 6 / 40 = 3/20

3.An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below.

Suppose a family is chosen. Find the probability that the family has chosen is
(i) earning ₹ 10000-13000 per month and owning exactly 2 vehicles.
(ii) earning ₹16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not own any vehicle.
(iv) earning ₹13000-16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Here, the total number of families = 2400
(i) ∵ Number of families earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles = 29
∴ Probability of a family earning Rs. 10000 – Rs. 13000 per month and owning exactly 2 vehicles =29 / 2400

(ii) ∵ Number of families earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579
∴ Probability of a family earning Rs. 16000 or more per month and owning exactly 1 vehicle = 579 /2400

(iii) ∵ Number of families earning less than Rs. 7000 per month and do not own any vehicle = 10
∴ Probability of a family earning less than Rs. 7000 per month and does not own any vehicle = 10 / 2400 =1 /240

(iv) ∵ Number of families earning Rs. 13000 – Rs. 16000 per month and owning more than 2 vehicles = 25
∴ Probability of a family earning Rs. 13000 – Rs. 16000 per month and owning more than 2 vehicles = 25/2400 = 1/96

(v) ∵ Number of families owning not more than 1 vehicle
= [Number of families having no vehicle] + [Number of families having only 1 vehicle]
= [10 + 1 + 2 + 1] + [160 + 305 + 535 + 469 + 579] = 14 + 2048 = 2062
∴ Probability of a family owning not more than 1 vehicle = 2062 /2400 = 1031/ 1200

4.The distance (in km) of 40 engineers from their residence to their place of work were found as follows

What is the empirical probability that an engineer lives
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within 1/2 km from her place of work?
Solution:
Here, total number of engineers = 40

(i) ∵ Number of engineers who are living less than 7 km from their work place = 9
∴ Probability of an engineer who is living less than 7 km from her place of work = 9/40

(ii) ∵ Number of engineers living at a distance more than or equal to 7 km from their work place = 31
∴ Probability of an engineer who is living at distance more than or equal to 7 km from her place of work = 31/40

(iii) ∵ The number of engineers living within 1/2km from their work place = 0
∴ Probability of an engineer who is living within 1/2km from her place of work = 0/40 = 0

5.Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg)
4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags,chosen at random contains more than 5 kg of flour.
Solution:
Here, total number of bags = 11
∵ Number of bags having more than 5 kg of flour = 7
∴ Probability of a bag having more than 5 kg of flour = 7/11

NCERT Quick revision Notes of Chapter-15 Probability

NCERT Solutions of Chapter-15 Probability

NCERT MCQs of Chapter-15 Probability

NCERT Important Questions & Solutions for Class 9 Maths Chapter 14 Statistics

You can find Chapter 2 Statistics Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.2

1.The blood groups of 30 students of class VIII are recorded as follows
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
Represent this data in the form of a frequency distribution table. Which is the most common and which is the rarest blood group among these students?
Solution:
The required frequency distribution table is

From the above table, we have The most common blood group is O. The rarest blood group is AB.

2.The relative humidity (in %) of a certain city for a month of 30 days was as follows

(i) Construct a grouped frequency distribution table with classes 84-86, 86-88 etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
Here, the lowest value of observation = 84.9
The highest value of observation = 99.2
So, class intervals are 84 – 86, 86 – 88, 88 – 90, ……. , 98 – 100

(i) Thus, the required frequency distribution table is

(ii) Since, the relative humidity is high during the rainy season, so, the data appears to be taken in the rainy season.
(iii) Range = (Highest observation) – (Lowest observation) = 99.2 – 84.9 = 14.3

3.A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 – 0.04, 0.04 – 0.08 and so on.
(ii) For how many day’s was the the concentration of sulphur dioxide more than 0.11 parts per million ?
Solution:
(i) Here, the lowest value of the observation = 0.01
The highest value of the observation = 0.22
∴ Class intervals are 0.00 – 0.04, 0.04 – 0.08,……., 0.20 – 0.24
The required frequency distribution table is

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 8 days.

4.The value of π upto 50 decimal places is given below
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Solution:
(i) The required frequency distribution table

(ii) The most frequently occurring digits are 3 and 9 and the least frequently occurring digit is 0.

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.3

1.A survey conducted by an organisation for the cause of illness and death among the women between the ages 15-44 (in years) worldwide, found the following figures (in %)

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
Solution:
(i) The required graphical representation is shown as follows:

(ii) The major cause of women’s ill health and death worldwide is ‘reproductive health conditions’.
(iii) Two factors may be un education and poor background.

2.Given below are the seats won by different political parties in the polling outcome of a state assembly elections

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution:
(i) The required bar graph is shown below:

(ii) The political party A won the maximum number of seats.

3.The length of 40 leaves of a plant measured correct to one millimetre and the obtained data is represented in the following table

(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves 153 mm long and Why?
Solution:
(i) The given frequency distribution table is not continuous. Therefore, first we have to modify it to be continuous distribution.

For modification we have to find the difference between 1st row upper limit and 2nd row lower limit and divide them by 2 ( Ex. 126-127= 1 =1/2 = 0.5

Now subtract 0.5 from lower limit and add 0.5 in upper limit of rows ( Ex. 118-0.5=117.5 & 126+0.5=126.5 so the new modified row is (117.5 – 126.5 ) so continue this in all the row of table.

Thus, the modified frequency distribution table is:

Now, the required histogram of the frequency distribution is shown below :

(ii) Yes, other suitable graphical representation is a ‘frequency polygon’.
(iii) No, it is not a correct statement. The maximum number of leaves lie in the class interval 145 – 153.

4.The runs scored by two teams A and B on the first 60 balls in a cricket match are given below

Represent the data of both the teams on the same graph by frequency polygons.
Solution:
The given class intervals are not continuous. Therefore, we first modify the distribution as continuous.

For modification we have to find the difference between 1st row upper limit and 2nd row lower limit and divide them by 2 ( Ex.6-7= 1 =1/2 = 0.5

Now subtract 0.5 from lower limit and add 0.5 in upper limit of rows ( Ex. 1-0.5=0.5 & 6+0.5=6.5 so the new modified row is (0.5 – 6.5 ) so continue this in all the row of table.

Now also Find Class Mark for the given Number of balls. Class Mark= Upper limit+Lower limit/2 ( Ex. 1+6/2=7/2=3.5 )

Now, the required frequency polygons are as shown below:

5.100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Solution:
(i) Since, class intervals of the given frequency distribution are unequal, and the minimum class size = 6 – 4 = 2.
Therefore, we have the following table for length of rectangles.

The required histogram is shown below:

(ii) The maximum frequency is 44, which is corresponding to the class interval 6 – 8.
∴ Maximum number of surnames lie in the class interval 6 – 8.

NCERT Important Questions & Solutions for Chapter 14 Statistics Ex 14.3

1.The following number of goals were scored by a team in a series of 10 matches
2, 3, 4, 5, 0, 1, 3, 3, 4, 3.
Find the mean, median and mode of these scores.
Solution:
To find the mean :
Here, n = 10

Thus, mean = 2.8

To find median:
Now arranging the given data in ascending order,
we have 0,1, 2, 3, 3, 3, 3, 4, 4, 5
∵ n = 10, an even number

Thus, median = 3

To find mode:
In the given data, the observation 3 occurs 4 times,
i.e., maximum number of times.
Thus, mode = 3

2.The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Here, the given observations are in ascending order.
Since, n = 10 (an even number)

Since, median = 63 [Given]
∵ x + 1 = 63 ⇒ x = 63 – 1 = 62
Thus, the required value of x is 62.

3.Find the mean salary of 60 workers of a factory from the following table

Solution:

Thus, the required mean salary = Rs. 5083.33

NCERT Quick revision Notes For Chapter-4 Statistics

NCERT Solutions For Chapter-4 Statistics

NCERT MCQs For Chapter-4 Statistics

NCERT Important questions & Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

You can find Chapter 2 Linear Equations in Two Variables Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.1

1. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35     (ii) 5 = 2x

Solutions

(i) We have 2x + 3y = 9.35
or (2)x + (3)y + (−9.35 ) = 0
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= –9.35

(ii) We have 5 = 2x ⇒ 5 – 2x = 0
or -2x + 0y + 5 = 0
or (-2)x + (0)y + (5) = 0
Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.2

1.Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9

Solutions:-

(i) 2x + y = 7
When x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)
When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)
When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9
When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
When x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
When x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))

2.Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k.
Solution:
We have 2x + 3y = k
putting x = 2 and y = 1 in 2x+3y = k,we get
2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k
Thus, the required value of k is 7.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.3

1.Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2

Solution:
(i) x + y = 4 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y =4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:

Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.

Thus, the line AB is the required graph of x + y = 4

(ii) x – y = 2 ⇒ y = x – 2
If we have x = 0, then y = 0 – 2 = -2
x = 1, then y = 1 – 2 = -1
x = 2, then y = 2 – 2 = 0
∴ We get the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.

Thus, the ime is the required graph of x – y = 2

2.Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution:
(2, 14) means x = 2 and y = 14
Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0
There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.

3.The taxi fare In a city Is as follows: For the first kilometre, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs.y, write a linear equation for this Information, and draw Its graph.

Solution:
Here, total distance covered = x km and total taxi fare = Rs. y
Fare for 1km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1)km = Rs.5 x(x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to question,
y = 8 + 5(x – 1) = y = 8 + 5x – 5
⇒ y = 5x + 3,
which is the required linear equation representing the given information.
Graph: We have y = 5x + 3
Wben x = 0, then y = 5(0) + 3 ⇒ y = 3
x = -1, then y = 5(-1) + 3 ⇒ y = -2
x = -2, then y = 5(-2) + 3 ⇒ y = -7
∴ We get the following table:

Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation y = 5x + 3.

4.In countries like USA and Canada, temperature is measured In Fahrenheit, whereas in countries like India, it is measured in Celsius. Here Is a linear equation that converts Fahrenheit to Celsius:
F = (9/5 )C + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature Is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what Is the temperature In Fahrenheit and If the temperature is 0°F, what Is the temperature In Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find It.
Solution:
(i) We have
F = (9/5 )C + 32
When C = 0 , F = (9/5 ) x 0 + 32 = 32
When C = 15, F = (9/5 )(-15) + 32= -27 + 32 = 5
When C = -10, F = 9/5 (-10)+32 = -18 + 32 = 14
We have the following table:

Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.

(ii) From the graph, we have 86°F corresponds to 30°C.
(iii) From the graph, we have 95°F corresponds 35°C.
(iv) We have, C = 0
From (1), we get
F = (9/5)0 + 32 = 32
Also, F = 0
From (1), we get
0 = (9/5)C + 32 ⇒ −32×59/9 = C ⇒ C = -17.8
(V) When F = C (numerically)
From (1), we get
F = 9/5F + 32 ⇒ F – 9/5F = 32
⇒ −4/5F = 32 ⇒ F = -40
∴ Temperature is – 40° both in F and C.

NCERT Important questions Chapter 4 Linear Equations in Two Variables Ex 4.4

1.Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Solution:
(i) y = 3
∵ y = 3 is an equation in one variable, i.e., y only.
∴ y = 3 is a unique solution on the number line as shown below:

(ii) y = 3
We can write y = 3 in two variables as 0.x + y = 3
Now, when x = 1, y = 3
x = 2, y = 3
x = -1, y = 3
∴ We get the following table:

Plotting the ordered pairs (1, 3), (2, 3) and (-1, 3) on a graph paper and joining them, we get aline AB as solution of 0. x + y = 3,
i.e. y = 3.

NCERT Quick revision Notes For Chapter-4 Linear Equation in Two Variables

NCERT Solutions For Chapter-4 Linear Equation in Two Variables

NCERT MCQs For Chapter-4 Linear Equation in Two Variables

NCERT Important Questions & Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

You can find Chapter 2 Coordinate Geometry Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

1. Write the answer of each of the following questions:
   (i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
   (ii) What is the name of each part of the plane formed by these two lines?
   (iii) Write the name of the point where these two lines intersect.
   Solution:
   (i) The horizontal line: x – axis and the vertical line: y – axis.
   (ii) Each part is called “Quadrant”.
   (iii) Origin

2.See the given figure and write the following:
(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3,-5).
(iv) The point identified by the coordinates (2,-4).
(v) The abscissa of point D.
(vi) The ordinate of the point H.
(vii) The coordinates of the point L.
(viii) The coordinates of the point M.

NCERT Quick revision Notes Chapter-3 Coordinate Geometry

NCERT Solutions Chapter-3 Coordinate Geometry

NCERT MCQ Chapter-3 Coordinate Geometry

NCERT Important Question & Solutions for Class 9 Maths Chapter 2 Polynomials

You can find Chapter 2 Polynomials Class 9 Maths NCERT Important Questions here that will help  Chapter 2 easily without wasting your precious time. This will help in developing your problem solving skills and be aware of the concepts. By taking help from these NCERT Important Questions, you can build your own answers for homework and get good marks in the examination. These Important question & solutions are updated according to the latest NCERT Maths textbook. These solutions are prerequisites before solving exemplar problems and going for advance Maths Books.

NCERT Important Questions & Solutions For Exercise 2.1 Polynomials

1.Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x² – 3x + 7                                (ii) 3 √t + t√2                                    (iii) y+ 2y

Solution:
(i) We have 4x² – 3x + 7 = 4x² – 3x + 7x⁰
It is a polynomial in one variable i.e., x
because each exponent of x is a whole number.

(ii) We have 3 √t + t√2 = 3 √t^1/2 + √2.t
It is not a polynomial, because one of the exponents of t is 12,
which is not a whole number.

(iii) We have y + y+2y = y + 2.y^-1
It is not a polynomial, because one of the exponents of y is -1,
which is not a whole number.

2.Write the coefficients of x² in each of the following
(i) 2 + x² + x
(ii) 2 – x² + x³

Solution:
(i) The given polynomial is 2 + x² + x.
The coefficient of x² is 1.
(ii) The given polynomial is 2 – x² + x³.
The coefficient of x² is -1.

3.Write the degree of each of the following polynomials.
(i) 5x³+4x² + 7x
(ii) 4 – y²
(iii) 5t – √7
(iv) 3
Solution:
(i) The given polynomial is 5x³ + 4x² + 7x.
The highest power of the variable x is 3.
So, the degree of the polynomial is 3.

(ii) The given polynomial is 4- y². The highest
power of the variable y is 2.
So, the degree of the polynomial is 2.

(iii) The given polynomial is 5t – √7 . The highest power of variable t is 1. So, the degree of the polynomial is 1.
(iv) Since, 3 = 3x° [∵ x°=1]
So, the degree of the polynomial is 0.

4.Classify the following as linear, quadratic, and cubic polynomials.    (i) x²+ x    (ii) r²     (iii) 7x³

Solution:
(i) The degree of x² + x is 2. So, it is a quadratic polynomial.

(ii) The degree of r² is 2. So, it is a quadratic polynomial.

(iii) The degree of 7x³ is 3. So, it is a cubic polynomial.

NCERT Important Questions & Solutions For Exercise 2.2 Polynomials

1.Find the zero of the polynomials in each of the following cases

(i) p(x) = x + 5                  (ii) p(x) = x – 5                        (iii) p(x) = ax

 

2.Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) x – 12
(iii) x
(iv) x + π
(v) 5 + 2x

Solution:
Let p(x) = x³ + 3x² + 3x +1
(i) The zero of x + 1 is -1.
∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1
= -1 + 3- 3 + 1 = 0
Thus, the required remainder = 0

(ii) The zero of x−12 is 12

Thus, the required remainder = 278

(iii) The zero of x is 0.
∴ p(0) = (0)³ + 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1 = 1
Thus, the required remainder = 1.

(iv) The zero of x + π is -π.
p(-π) = (-π)³ + 3(- π)²2 + 3(- π) +1
= -π³ + 3π² + (-3π) + 1
= – π³ + 3π² – 3π +1
Thus, the required remainder is -π³ + 3π² – 3π+1.

(v) The zero of 5 + 2x is −52 .


Thus, the required remainder is −278 .

NCERT Important Questions & Solutions For Exercise 2.3 Polynomials

1. Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Solution:
We have, p(x) = x³ – ax² + 6x – a and zero of x – a is a.
∴ p(a) = (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a = 5a
Thus, the required remainder is 5a.

2. Check whether 7 + 3x is a factor of 3x³+7x.
Solution:
We have, p(x) = 3x³+7x. and zero of 7 + 3x is −73.

Since,( −4909) ≠ 0
i.e. the remainder is not 0.
∴ 3x³ + 7x is not divisib1e by 7 + 3x.
Thus, 7 + 3x is not a factor of 3x³ + 7x.

NCERT Important Questions & Solutions For Exercise 2.4 Polynomials

1. Determine which of the following polynomials has (x +1) a factor. (i) x³+x²+x +1 (ii) x⁴ + x³ + x² + x + 1

Solution:
The zero of x + 1 is -1.
(i) Let p (x) = x³ + x² + x + 1
∴ p (-1) = (-1)³ + (-1)² + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, (x+ 1) is a factor of x³ + x² + x + 1.

(ii) Let p (x) = x⁴ + x³ + x² + x + 1
∴ P(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1)+1
= 1 – 1 + 1 – 1 + 1
⇒ P (-1) ≠ 1
So, (x + 1) is not a factor of x⁴ + x³ + x² + x+ 1.

2.Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases
(i) p (x)= 2x³ + x² – 2x – 1, g (x) = x + 1
(ii) p(x)= x³ + 3x² + 3x + 1, g (x) = x + 2

Solution:
(i) We have, p (x)= 2x³ + x² – 2x – 1 and g (x) = x + 1
∴ p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= 2(-1) + 1 + 2 – 1
= -2 + 1 + 2 -1 = 0
⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) x³ + 3x² + 3x + 1 and g(x) = x + 2
∴ p(-2) = (-2)³ + 3(-2)²+ 3(-2) + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

3. Find the value of k, if x – 1 is a factor of p (x) in each of the following cases
(i) p (x) = x² + x + k
(ii) p (x) = 2x² + kx + √2

Solution:
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x² + x + k
Since, p(1) = (1)² +1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.

(ii) Here, p (x) = 2x² + kx + √2
Since, p(1) = 2(1)² + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)

4.Factorise
(i) 12x² – 7x +1
(ii) 2x² + 7x + 3

Solution:
(i) We have,
12x² – 7x + 1 = 12x² – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x² -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x² + 7x + 3 = 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

5. Factories

(i) x³ + 13x² + 32x + 20
(ii) 2y³ + y² – 2y – 1

(i) We have, x³ + 13x² + 32x + 20
= x³ + x² + 12x² + 12x + 20x + 20
= x²(x + 1) + 12x(x +1) + 20(x + 1)
= (x + 1)(x² + 12x + 20)
= (x + 1)(x² + 2x + 10x + 20)
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)
Thus, x³ + 13x² + 32x + 20
= (x + 1)(x + 2)(x + 10)

(ii) We have, 2y³ + y² – 2y – 1
= 2y³ – 2y² + 3y² – 3y + y – 1
= 2y²(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)(2y² + 3y + 1)
= (y – 1)(2y² + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)
Thus, 2y³ + y² – 2y – 1
= (y – 1)(y + 1)(2y +1)

NCERT Important Questions & Solutions For Exercise 2.4 Polynomials

1. Use suitable identities to find the following products

(i) (y²+ 3/2) (y²– 3/2)

2.Evaluate the following products without multiplying directly
(i) 103 x 107

Solution:
(i)We have, 103 x 107 = (100 + 3) (100 + 7)
= ( 100)² + (3 + 7) (100)+ (3 x 7)
[Using (x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + (10) x 100 + 21
= 10000 + 1000 + 21=11021

3.Expand each of the following, using suitable identity

(i) (- 2x + 5y – 3z)²
(ii) [ 1/4a –1/4b + 1] ²

(i)(- 2x + 5y- 3z)² = (- 2x)² + (5y)² + (- 3z)² + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

4. Factorise
(i) 4 x² + 9y² + 16z² + 12xy – 24yz – 16xz

Solution:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
= (2x)² + (3y)² + (- 4z)² + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x)
= (2x + 3y – 4z)² = (2x + 3y + 4z) (2x + 3y – 4z)

5.

Solution:
We have, (x + y)³ = x³ + y³ + 3xy(x + y) …(1)
and (x – y)³ = x³ – y³ – 3xy(x – y) …(2)

6.Factorise each of the following

7. Verify that
x³ +y³ +z³ – 3xyz = 1/2 (x + y+z)[(x-y)² + (y – z)² +(z – x)²]
Solution:
R.H.S
= 1/2(x + y + z)[(x – y)²+(y – z)²+(z – x)²]
= 1/2 (x + y + 2)[(x² + y² – 2xy) + (y² + z² – 2yz) + (z² + x² – 2zx)]
= 1/2 (x + y + 2)(x² + y² + y² + z² + z² + x² – 2xy – 2yz – 2zx)
= 1/2 (x + y + z)[2(x² + y² + z² – xy – yz – zx)]
= 2 x 1/2 x (x + y + z)(x² + y² + z² – xy – yz – zx)
= (x + y + z)(x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz = L.H.S.
Hence, verified.

8. If x + y + z = 0, show that x³ + y³ + z³ = 3 xyz.
Solution:
Since, x + y + z = 0
⇒ x + y = -z (x + y)³ = (-z)³
⇒ x³ + y³ + 3xy(x + y) = -z³
⇒ x³ + y³ + 3xy(-z) = -z³ [∵ x + y = -z]
⇒ x³ + y³ – 3xyz = -z³
⇒ x³ + y³ + z³ = 3xyz
Hence, if x + y + z = 0, then
x³ + y³ + z³ = 3xyz

9.Without actually calculating the cubes, find the value of each of the following
(i) (- 12)³ + (7)³ + (5)³

Solution:
(i) We have, (-12)³ + (7)³ + (5)³
Let x = -12, y = 7 and z = 5.
Then, x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x³ + y³ + z³ = 3xyz
∴ (-12)³ + (7)³ + (5)³ = 3[(-12)(7)(5)]
= 3[-420] = -1260

NCERT Quick revision Notes Chapter-2 Polynomials

NCERT Solutions Chapter-2 Polynomials

NCERT MCQ Chapter-2 Polynomials

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