CBSE Class 9

CBSE Class 9

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

Other Exercises

Question 1.
If f(x) = 2x3 – 13x2 + 17x + 12, find
(i) f (2)
(ii) f (-3)
(iii) f(0)
Solution:
f(x) = 2x3 – 13x2 + 17x + 12
(i) f(2) = 2(2)3 – 13(2)2 + 17(2) + 12
= 2 x 8-13 x 4+17 x 2+12
= 16-52 + 34 + 12
= 62 – 52
= 10
(ii) f(-3) = 2(-3)3 – 13(-3)2 + 17 x (-3) + 12
= 2 x (-27) – 13 x 9 + 17 x (-3) + 12
= -54 – 117 -51 + 12
= -222 + 12
= -210
(iii) f(0) = 2 x (0)3 – 13(0)2 + 17 x 0 + 12
= 0-0 + 0+ 12 = 12

Question 2.
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases: [NCERT]
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.1
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.2
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.3
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.4
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.5
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.6

Question 3.
If x = 2 is a root of the polynomial f(x) = 2x2-3x + la, find the value of a.
Solution:
p(x) = 2x2 – 3x + 7a
∵ x = 2 is its zero, then
p(0) = 0
∴ p( 2) = 2(2)2 – 3×2 + la = 0
⇒2 x 4-3 x2 + 7a = 0
⇒ 8 – 6 + 7o = 0
⇒2 + 7a = 0
⇒ 7a = -2 ⇒ a =\(\frac { -2 }{ 7 }\)
∴ Hence a = \(\frac { -2 }{ 7 }\)

Question 4.
If x = –\(\frac { 1 }{ 2 }\) is a zero of the polynomial p(x) = 8x3 – ax2 – x + 2, find the value of a.
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q4.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q4.2

Question 5.
If x = 0 and x = -1 are the roots of the polynomial f(x) = 2x3 – 3x2 + ax + b, find the value of a and b.
Solution:
f(x) = 2x3 – 3x2 + ax + b
∵ x = 0 and x = -1 are its zeros
∴ f(0) = 0 and f(-1) = 0
Now, f(0) = 0
⇒  2(0)3 – 3(0)2 + a x 0 + b = 0
⇒ 0-0 + 0 + b= 0
∴ b = 0
and f(-1) = 0
⇒ 2(-1)3 – 3(-1)2 + a(-1) + b = 0
⇒  2 x (-1) – 3 x 1 + a x (-1) + b = 0
⇒ -2 -3-a + b = 0
⇒ -2-3-a + 0 = 0
⇒ -5- a = 0=>a =-5
Hence a = -5, b = 0

Question 6.
Find the integral roots of the polynomial f(x) = x3 + 6x2 + 11x + 6.
Solution:
f(x) = x3 + 6x2 + 11x + 6
Construct = 6 = ±1, ±2, +3, ±6
If x = 1, then
f(1) = (1)3 + 6(1)2 + 11 x 1 + 6
= 1+ 6+11+ 6 = 24
∵  f(x) ≠ 0, +0
∴ x = 1 is not its zero
Similarly, f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6
= -1 + 6 x 1-11+6
=-1+6-11+6
= 12-12 = 0
∴  x = -1 is its zero
f(-2) = (-2)3 + 6(-2)2 + 11 (-2) + 6
= -8 + 24 – 22 + 6
= -30 + 30 = 0
∴ x = -2 is its zero
f(-3) = (-3)3 + 6(-3)2 + 11 (-3) + 6
= -27 + 54 – 33 + 6 = 60 – 60 = 0
∴  x = -3 is its zero
x = -1, -2, -3 are zeros of f(x)
Hence roots of f(x) are -1, -2, -3

Question 7.
Find the rational roots of the polynomial f(x) = 2x3 + x2 – 7x – 6.
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q7.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q7.2
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q7.3

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A

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RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A.

Question 1.
Solution:
Co-ordinates of
A are ( – 6, 5)
B are (5, 4)
C are ( – 3, 2)
D are (2, – 2)
E are ( – 1, – 4)
Ans.
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q1.1

Question 2.
Solution:
The given points have been plotted as shown in the adjoining graph.
Where X’OX and YOY’ are the axis:
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q2.1

Question 3.
Solution:
(i) (7, 0) lies on x-axis as its ordinate is (0)
(ii) (0, – 5) lies on y-axis as its abscissa is (0)
(iii) (0, 1) lies on y-axis as its abscissa is (0)
(iv) ( – 4, 0) lies on jc-axis as its ordinate is (0)
Ans.

Question 4.
Solution:
(i) ( – 6, 5) lies in second quadrant because A is of the type (-, +)
(ii) ( – 3, – 2) lies in third quadrant because A is of the type (-, -)
(iii) (2, – 9) lies in fourth quadrant because it is of the type (+, -).

Question 5.
Solution:
In the given equation, .
y = x+1
Put x = 0, then y = 0 + 1 = 1
x = 1, then, y = 1 + 1=2
x = 2, then, y = 2 + 1 = 3
Now, plot the points as given in the table given below on the graph, and join them as shown.
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q5.1

Question 6.
Solution:
In the given equation
y = 3x + 2
Put x = 0,
then y = 3 x 0 + 2 = 0 + 2 = 2
x = 1, then
y = 3 x 1 + 2 = 3 + 2 = 5
and x = – 2, then
y = 3 x ( – 2) + 2 = – 6 + 2 = – 4
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q6.1

Question 7.
Solution:
In the given equation,
y = 5x – 3
Put x = 1,y = 5 x 1 – 3 = 5 – 3 = 2
x = 0 then,
y = 5 x 0 – 3 = 0 – 3 = – 3
and x = 2, then
y = 5 x 2 – 3 = 10 – 3 = 7
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q7.1

Question 8.
Solution:
In the given equation,
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q8.1
y = 3x,
Put x = 0,
then y = 3 x 0 = 0
Put x = 1, then
y = 3 x 1 = 3
Put x = – 1, then
y = 3 ( – 1) = – 3
Now, plot the points as given in the table below
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q8.2
and join them as shown.

Question 9.
Solution:
In the given equation, y = – x,
Put x = 1,
then y = – 1
Put x = 2, then
y = – 2
Put x = – 2, then
y = – ( – 2) = 2
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q9.1

Hope given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A are helpful to complete your math homework.

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Value Based Questions in Science for Class 9 Chapter 2 Is Matter Around Us Pure

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Value Based Questions in Science for Class 9 Chapter 2 Is Matter Around Us Pure

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 2 Is Matter Around Us Pure

Question 1.
A student was asked by his teacher to separate an impure sample of sulphur containing sand as the impurity. He tried to purify it with the help of sublimation. But he was not successful. Particles of sulphur could not be separated completely from sand.

  1. Why did not the sublimation process succeed ?
  2. Suggest an alternate method to affect the separation.
  3. What is the value based information associated with this ?

Answer:

  1. Sulphur is not of volatile nature. Upon heating, it melts. Therefore, sublimation process could not succeed.
  2. The student should have dissolved the impure sample in carbon disulphide. It is a liquid in which sulphur completely dissolves while iron does not. From the solution, sulphur can be recovered with the help of crystallization process.
  3. The process of sublimation is useful only if one of the constituents present in the mixture can sublime while the others do not undergo sublimation.

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Question 2:
Amit was asked by his teacher to separate a liquid mixture of acetone and ethyl alcohol. He set up a distillation apparatus and tried to distil the mixture. To his surprise, both the liquids got distilled. Teacher told Amit to repeat the experiment by using a fractionating column in the distillation flask. Amit followed the advice of the teacher and he was able to separate the two liquids.

  1. Why was Amit not successful in separating the liquid mixture earlier ?
  2. Why did teacher ask him to use the fractionating column ?
  3. Which liquid was distilled first ?
  4. As a student of chemistry, what value based information you have gathered ?

Answer:

  1. The difference in boiling point temperatures of acetone (56°C) and ethyl alcohol (78°C) is only 22°C. Therefore, process of simple distillation fails in this case.
  2. Fractionating column is quite effective in this case because it obstructs the distillation of ethyl alcohol which is high boiling and at the same time helps in the distillation of acetone which is low boiling.
  3. Acetone was distilled first since it has comparatively low boiling point.
  4. The process of simple distillation can be used only in case, the liquids present in the mixture differ in their boiling point by 25°C or more.

Question 3.
A housewife got a cut on her finger while working in the Kitchen. She tried to stop the bleeding by applying dettol on it but it was not effective. By chance, her friend was also there. She asked her to rub alum on the cut which she did. The bleeding immediately stopped.

  1. Why was not dettol effective in stopping bleeding ?
  2. Why was alum effective ?
  3. What valuable service was done by the friend to the housewife ?

Answer:

  1. Blood is a colloidal solution and the colloidal particles of RBCs carry charge. Dettol is an organic compound. It could not neutralise the charge on the colloidal particles. Therefore, bleeding did not stop.
  2. Alum is a salt in which charged ions are present. When alum was applied on the cut, the charged ions neutralised the charge on the colloidal particles. In the absence of charge, blood became thick and bleeding stopped. In otherwords, blood got coagulated.
  3. The friend has a proper knowledge of colloidal solutions. She must be a student of chemistry. Timely help by her stopped the bleeding. Otherwise, there would have been a further loss of blood.

Question 4.
Mallika’s mother was suffering from cold and cough. Mallika prepared tea for her mother. She boiled water in a pan, then she added tea leaves, sugar and milk to it. She filtered the tea in a cup and served to her mother.

  1. Explain the values shown by Mallika.
  2. Identify solute, solvent, residue and filtrate in this activity. (CBSE 2013)

Answer:

  1. Mallika used the knowledge of chemistry to provide relief to her mother. Actually she prepared an extract of tea leaves which is helpful in curing cold and cough and gives warmth to the body.
  2. Solute : tea leaves and sugar ; Solvent : water and milk; Filtrate : homogeneous mixture of water, milk, sugar and extract of tea leaves.

Question 5.
Nikhil’s father was suffering from high blood pressure and cardiac problems. Doctor suggested him to take low fat milk. Nikhil churned the milk and separated the butter (fat) from it and then he served that milk to his father. Answer the following questions based on above information :

  1. Name the technique by which Nikhil separated butter from milk.
  2. Write any other application of that technique.
  3. Which values are reflected in Nikhil’s behaviour ?

Answer:

  1. It is called centrifugation carried in a centrifugal machine.
  2. Washing machines that are used for washing dirty clothes are centrifugal machines.
  3. Nikhil had the proper knowledge of chemistry. He used this to separate fat from milk with the help of the centrifugal machine. In this way, he rendered proper service to his father.

Hope given Value Based Questions in Science for Class 9 Chapter 2 Is Matter Around Us Pure are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1

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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1

Other Exercises

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 Q1.1
Solution:
(i) 3x2 – 4x + 15,
(ii) y2 + 2\(\sqrt { 3 } \) are polynomial is one variable. Others are not polynomial or polynomials in one variable.

Question 2.
Write the coefficient of x2 in each of the following:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 Q2.1
Solution:
Coefficient of x2,
in (i) is 7
in (ii) is 0 as there is no term of x2 i.e. 0 x2
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 Q2.2

Question 3.
Write the degrees of each of the following polynomials:
(i) 7x3 + 4x2 – 3x + 12
(ii) 12 – x + 2x3
(iii) 5y – \(\sqrt { 2 } \)
(iv) 7
(v) 0
Solution:
(i) Degree of the polynomial 7x3 + 4x2 – 3x + 12 is 3
(ii) Degree of the polynomial 12 – x + 2x3 is 3
(iii) Degree of the polynomial 5y – \(\sqrt { 2 } \)is 1
(iv) Degree of the polynomial 7 is 0
(v) Degree of the polynomial 0 is 0 undefined.

Question 4.
Classify the following polynomials as linear, quadratic, cubic and biquadratic polynomials:
(i) x + x2 + 4
(ii) 3x – 2
(iii) 2x + x2 [NCERT]
(iv) 3y
(v) t2 + 1
(v) 7t4 + 4t3 + 3t – 2
Solution:
(i)  x + x2 + 4 It is a quadratic polynomial.
(ii) 3x – 2 : It is a linear polynomial.
(iii) 2x + x2: It is a quadratic polynomial.
(iv) 3y It is a linear polynomial.
(v) t2+ 1 It is a quadratic polynomial.
(vi) 7t4 + 4t3 + 3t – 2 It is a biquadratic polynomial.

Question 5.
Classify the following polynomials as polynomials in one-variable, two-variables etc.
(i) x2-xy +7y2
(ii) x2 – 2tx + 7t2 – x + t
(iii) t3 -3t2 + 4t-5
(iv) xy + yz + zx
Solution:
(i) x2 – xy + 7y2: It is a polynomial in two j variables x, y.
(ii) x2 – 2tx + 7t2 – x + t: It is a polynomial in two variables in x, t.
(iii) t3 – 3t2 + 4t – 5 : It is a polynomial in one variable in t.
(iv) xy +yz + zx : It is a polynomial in 3 variables in x, y and

Question 6.
Identify polynomials in the following:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 Q6.1
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 Q6.2

Question 7.
Identify constant, linear, quadratic and cubic polynomials from the following polynomials:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 Q7.1
Solution:
(i) f(x) = 0 : It is a constant polynomial as it has no variable.
(ii) g(x) = 2x3 – 7x + 4 : It is a cubic polynomial.
(iii) h(x) = -3x + \(\frac { 1 }{ 2 }\) : It is a linear polynomial.
(iv) p(x) = 2x2 – x + 4 : It is a quadratic polynomial.
(v) q(x) = 4x + 3 : It is linear polynomial.
(vi) r(x) = 3x3 + 4x2 + 5x – 7 : It is a cubic polynomial.

Question 8.
Give one example each of a binomial of degree 35 and of a monomial of degree 100.   [NCERT]
Solution:
Example of a binomial of degree 35 = 9x35 + 16
Example of a monomial of degree 100 = 2y100

 

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure

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HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure

These Solutions are part of HOTS Questions for Class 9 Science. Here we have given HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure.

Question 1.
A house wife churned full cream with a milk churner

  1. What will she observe after churning the milk ?
  2. What could be the possible reason for the observation ?

Answer:

  1. Churning of milk is a centrifugation process. As a result, lighter particles of cream or butter will move upwards and collect at the top. Heavier residual particles will remain at the bottom. This is a very common process used to separate cream from milk or butter from yogurt.
  2. The separation is based on the principle that lighter particles move upwards while denser particle downwards upon centrifugation.

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Question 2.
Based on separation techniques, complete the following. The first one is done for you.
HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 1
Answer:
2. Homogeneous, Evaporation, Difference in nature.
3. Hetereogeneous, Filtration, Difference in solubility in water
4. Homogeneous, Chromatography, Difference in adsorption of different components.

Question 3.
The table given below shows number of grams of five different solids dissolving in 100 g of the solvents : water, alcohol and chloroform (all at 20°C).
HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 2

  1. Which solid dissolves best in water at 20°C ?
  2. Which solid is maximum soluble in alcohol ?
  3. Which solid is insoluble in all the three solvents ?

Answer:

  1. Sugar is best soluble in water at 20°C
  2. Iodine is maximum soluble in alcohol.
  3. Chalk is insoluble in all the three solvents.

Question 4.
Some solids dissolve easily in liquids while the others donot

  1. What is the name given to the liquids which dissolve solids ?
  2. What is the name given to the clear liquid formed when a solid dissolves in a liquid ?
  3. What is the name given to the liquid which contains in it some suspended particles ?

Answer:

  1. The liquids are known as solvents
  2. The clear liquid is called solution or true solution. .
  3. The liquid is known as dispersion medium or dispersing medium.

Question 5.
Butter is an example of one type of colloidal solution. Name it. Give a reason for your choice.
Answer:
The colloidal solution is an example in which solid acts as the dispersion medium while liquid as the dispersed phase. It is also called gel.
Reason for the choice. On pressing butter, liquid drops come out of it leaving behind a solid. This clearly shows that butter is a gel.

Question 6.

  1. The solubility of sodium chloride in water increases with rise in temperature while that of lithium carbonate decreases. Assign reason.
  2. Water containing 88-8% oxygen and 11-2% hydrogen is often used as a fire extinguisher. Can a mixture containing the two gases in the same ratio by mass be used for extinguishing fire ?
  3. The melting point of a solid when determined experimentally comes out to be 160°C. But its actual melting point as given in standard books is 150°C. Predict the nature of the solid.

Answer:

  1. When sodium chloride is dissolved in water, the process is endothermic in nature. This means that heat energy is absorbed in the process. Therefore, solubility increases with rise in the temperature. In case of lithium carbonate, the process of dissolution is exothermic. This means that heat is evolved in the process. Therefore, its solubility in water decreases with rise in temperature.
  2. No, it cannot be used. Actually, in water the two elements are chemically combined with each other. They therefore, lose their identity. But in the mixture, no chemical combination between hydrogen and oxygen has taken place. Even water cannot be formed on mixing the gases. Therefore, the mixture does not extinguish any fire.
  3. Since the experimentally determined melting point of the solid is more than the standard value of the melting point, this means that the solid is not in pure state. It has some impurities present. Please note that the purity of a solid can be determined by finding its melting point and comparing it with the standard value.

Hope given HOTS Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.