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RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions.

Board	RBSE
Class	Class 9
Subject	Maths
Chapter	Chapter 3
Chapter Name	Polynomial
Exercise	Additional Questions
Number of Questions Solved	36
Category	RBSE Solutions

RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions

Multiple Choice Questions

Question 1.
A polynomial (RBSESolutions.com) of degree n in x has atmost:
(A) n terms
(B) (n – 1) terms
(C) (n + 1) terms
(D) \(\frac { n }{ 2 }\) terms
Solution
C

Question 2.
The zeroes of the (RBSESolutions.com) polynomial p(x) = x(x² – 1) are:
(A) 0, 1
(B) 0, – 1
(C) 0, – 1, 1
(D) ± 1
Solution
C

Question 3.
If

is equal to

Solution
B

Question 4.
√7 is a polynomial (RBSESolutions.com) of degree:
(A) 0
(B) 7
(C) \(\frac { 1 }{ 2 }\)
(D) 2
Solution
A

Question 5.
The coefficient of x in (x – 3)(x – 4) is:
(A) 7
(B) 1
(C) – 7
(D) 12
Solution
C

Question 6.
If p(x) = x² -3√2x + 1, then p(3√2) is equal to:
(A) 3√2
(B) 3√2 – 1
(C) 6√2 – 1
(D) 1
Solution
D

Question 7.
Which of the following is (RBSESolutions.com) a polynomial in one variable:

Solution
A

Question 8.
The degree of the zero polynomial is:
(A) 0
(B) 1
(C) any real number
(D) not exist
Solution
D

Question 9.
If x⁹¹ + 91 is divided by x + 1, then (RBSESolutions.com) the remainder is:
(A) 0
(B) 90
(C) 92
(D) None of these
Solution
B

Question 10.
Zero of the polynomial p(x) = √3x + 3 is

Solution
A

Very Short Answer Type Questions

Question 1.
Show that \(\frac { 1 }{ 2 }\) is zero (RBSESolutions.com) of the polynomial 2x² + 7x – 4.
Solution.
Let p(x) = 2x² + 7x – 4

Question 2.
Find the (RBSESolutions.com) remainder when
x⁴ + x³ – 2x² + x + 1 is divided by x – 1.
Solution.
Let p(x) = x⁴ + x³ – 2x² + x + 1
zero of x – 1 is 1
So, P(1) = (1)⁴ + (1)³ – 2(1)² + (1) + 1 = 1 + 1 – 2 + 2 = 2
Hence, 2 is the remainder, when x⁴ + x³ – 2x² + x + 1 is divided by x – 1.

Question 3.
If x – 1 is a factor of p(x) = 2x² + kx + √2 then find (RBSESolutions.com) the value of k.
Solution.
x – 1 is a factor of p(x) = 2x² + kx + √2
p(1) = 0 (by factor theorem)
⇒ 2(1)² + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
k = -(2 + √2)

Question 4.
Find the (RBSESolutions.com) value of m, if (x + 3) is a factor of 3x² + mx + 6.
Solution.
Let p(x) = 3x² + mx + 6
If x + 3 is a factor of p(x), then
p(-3) = 0
p(-3) = 3(- 3)² + m(-3) + 6 = 0
⇒ 3 x 9 – 3m + 6 = 0
⇒ 27 – 3m + 6 = 0
⇒ 3m = 33
⇒ m = 11

Question 5.
Verify that 1 is not a zero (RBSESolutions.com) of the polynomial 4x⁴ – 3x³ + 2x² – 5x + 1.
Solution.
Let p(x) = 4x⁴ – 3x³ + 2x² – 5x + 1
Now
P(1) = 4(1)⁴ – 3(1)³ + 2(1)² – 5(1) + 1 = 4 – 3 + 2 – 5 + 1 = -1
p(1) ≠ 0
1 is not a zero of p(x).

Remainder Theorem Calculator. Various types of online remainder theorem calculators are available online.

Question 6.
Using remainder theorem, find the (RBSESolutions.com) value of k so that (4x² + kx – 1) leaves the remainder 2 when divided by (x – 3).
Solution.
Let p(x) = 4x² + kx – 1
Here we are given that p(3) = 2
i.e. 4(3)² + k(3) -1 = 2
⇒ 4 x 9 + 3k – 1 = 2
⇒ 36 + 3k = 3
⇒ 3k = -33
⇒ k = -11

Question 7.
If (x – 2) is a factor of the (RBSESolutions.com) polynomial x⁴ – 2x³ + ax – 1, then find the value of a.
Solution.
(x – 2) is a factor of the polynomial p(x) = x⁴ – 2x³ + ax – 1, then according to factor theorem p(2) = 0
⇒ (2)⁴ – 2(2)³ + a(2) – 1 = 0
⇒ 2a – 1 = 0
⇒ a = \(\frac { 1 }{ 2 }\)

Question 8.
Find the value of
(x – y)³ + (y – z)³ + (z – x)³.
Solution.
Let a = x – y, b = y – z, c = z – x
Here a + b + c = x – y + y – z + z – x = 0
a³ + b³ + c³ = 3abc
⇒ (x – y)³ + (y – z)³ + (z – x)³ = 3(x – y)(y – z)(z – x).

Question 9.
If a, b, c are all non-zero and a + b + c = 0 then find the (RBSESolutions.com) value of \(\frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { { c }^{ 2 } }{ ab }\)
Solution.
We know that a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
a + b + c then a³ + b³ + c³ = 3abc …(1)
Dividing equation (1) by abc, we get

Question 10.

Solution.

Short Answer Type Questions

Question 1.
Find the remainder (RBSESolutions.com) obtained on dividing p(x) = x³ + 1 by x + 1.
Solution.
By long division method

Here we find that remainder is zero.
Also p(-1) = (-1)³ + 1 = -1 + 1 = 0,
which is equal to the (RBSESolutions.com) remainder obtained by actual division.

Question 2.
Find the value of
4a² + b² + 25c² + 4ab – 10bc – 20ca when a = 1, b = 2 and c = 3.
Solution.
We have,
4a² + b² + 25c² + 4ab – 10bc – 20ca = (2a)² + (b)² + (-5c)² – 2(2a)(b) + 2(b)(-5c) + 2(-5c)(2a) = (2a + b – 5c)²
It is given that a = 1, b = 2 and c = 3,
so (2a + b – 5c)² = (2 x 1 + 2 – 5 x 3)² = (2 + 2 – 15)² = (4 – 15)² = (-11)² = 121

Question 3.
If 3a – 2b = 11 and ab = 12, then find (RBSESolutions.com) the value of 27a³ – 8b³.
Solution.
We know that
(a – b)³ = a³ – b³ – 3ab(a – b)
Using this identity,
(3a – 2b)³ = (3a)³ – (2b)³ -3 x 3a x 2b (3a – 2b)
⇒ (3a – 2b)³ = 27a³ – 8b³ – 18ab(3a – 2b)
Now substituting 3a – 2b = 11 and ab = 12,
we get
(11)² = 27a³ – 8b³ – 18 x 12 x 11
⇒ 1331 = 27a³ – 8b³ – 2376
⇒ 27a³ – 8b² = 1331 + 2376
⇒ 27a³ – 8b³ = 3707

Question 4.
Use factor (RBSESolutions.com) theorem, show that (x + √2) is a factor of (2√2x² + 5x + √2).
Solution.
Let p(x) = (2√2x² + 5x + √2)
If x + √2 is a factor of p(x), then according to factor theorem p(√2) = 0.
p(-√2) = 2√2(-√2)² + 5(-√2) + √2 = 2√2 x 2 – 5√2 + √2 = 5√2 – 5√2 = 0
p(-√2) = 0
(x + √2) is a factor p(x) i.e. (2√2 x² + 5x + √2).

Question 5.
If a + b + c = 9 and ab + bc + ca = 23 then find (RBSESolutions.com) the value of a² + b² + c².
Solution.
We have, a + b + c = 9 Squaring both sides, we get
(a + b + c)² = (9)²
⇒ a² + b² + c² + 2ab + 2bc + 2ca = 81
⇒ a² + b² + c² + 2(ab + bc + ca) = 81
⇒ a² + b² + c² + 2 x 23 = 81
⇒ a² + b² + c² = 81 – 46
⇒ a² + b² + c² = 35 Thus, a² + b² + c² = 35.

Question 6.
Simplify by using (RBSESolutions.com) suitable identity

Solution.

Question 7.
Factorise
(i) 216a³ – 125
(ii) a³ – b³ – a + b
Solution.
(i) we have,
216a³ – 125 = (6a)³ – (5)³
[a³ – b³ = (a – b)(a² + ab + b²)]
= (6a – 5) {(6a)² + 6a x 5 + (5)²}
= (6a – 5)(36a² + 30a + 25)
(ii) We have,
a³ – b³ – a + b = a³ – b³ – (a – b)
= (a – b)(a² + ab + b²) – (a – b)
[a³ – b³ = (a – b)(a² + ab + b²)]
= (a – b)(a² + ab + b² – 1)

Question 8.
If p = 4 – q show that p³ + q³ + 12pq = 64.
Solution.
We have,
p = 4 – q ⇒ p + q = 4 …(i)
Cubing both sides, (RBSESolutions.com) we get
(p + q)³ = (4)³
⇒ p³ + 3p2q + 3pq² + q³ = 64
⇒ p³ + q³ + 3pq(p + q) = 64
⇒ p³ + q³ + 3pq x 4 = 64 [using (i)]
⇒ p³ + q³ + 12pq = 64

Question 9.

Solution.

Long Answer Type Questions

Question 1.
If (x – 3) and (x – \(\frac { 1 }{ 2 }\)) are both (RBSESolutions.com) factors of ax² + 5x + b, show that a = b.
Solution.
Let p(x) = ax² + 5x + b
x – 3 is a factor of p(x).
p(3) = 0

Question 2.
For what values of a and b so that (RBSESolutions.com) the polynomial x³ + 10x² + ax – 6 is exactly divisible by (x – 1) and (x + 2).
Solution.
Let p(x) = x³ + 10x² + ax + b
p(x) i.e. x³ + 10x² + ax + 6 is exactly divisible by (x – 1) and (x + 2)
Therefore, p(1) and p(-2) must equal to zero.
p(1) = (1)³ + 10(1)² + a x 1 + b = 0
⇒ 1 + 10 + a + b = 0
⇒ a + b = – 11 …(i)
Also, p(-2) = 0
(-2)³ + 10(-2)² + a x (- 2) + b = 0
-8 + 40 – 2a + b = 0
⇒ – 2a + b = – 32 …(ii)
Solving (i) and (ii), we get
a = 7 and b = -18.

Question 3.
If ax³ + bx² + x – 6 has x + 2 as a factor and leaves (RBSESolutions.com) a remainder 4 when divided by (x – 2), find the values of a and b.
Solution.
Let p(x) = ax³ + bx² + x – 6
(x + 2) is a factor of p(x)
⇒ p(-2) = 0 [∴ x + 2 = 0 ⇒ x = -2]
⇒ a(-2)³ + b(-2)² + (-2) – 6 = 0
⇒ -8a + 4b – 2 – 6 = 0
⇒ -8a + 4b = 8
⇒ -2a + b = 2 …(i)
It is given that p(x) leaves the (RBSESolutions.com) remainder 4 when it divided by (x – 2) i.e. p(2) = 4.
⇒ a(2)³ + b(2)² + (2) – 6 = 4
⇒ 8a + 4b – 4 = 4
⇒ 8a + 46 = 8
⇒ 2a + b = 2 …(ii)
Solving (i) and (ii), we get a = 0 and b = 2.

Question 4.
Evaluate

Solution.
In the given expression, we see that both numerator and denominator (RBSESolutions.com) is in the form a³ + b³ + c³ = 3abc because a + b + c = 0.
From numerator, we see that a² – b² + b² – c² + c² – a² = 0
⇒ (a² – b²)³ + (b² – c²)³ + (c² – a²)³ = 3 (a² – b²)(b² – c²)(c² – a²)
Similarly, from denominator,
a – b + b – c + c – a = 0
(a – b)³ + (b – c)³ + (c – a)³ = 3(a – b)(b – c)(c – a)
Value of the expression

Question 5.
If the polynomials (3x³ + ax² + 3x + 5) and (4x³ + x² – 2x + a) leaves the same (RBSESolutions.com) remainder when divided by (x – 2), find the value of a. Also the find the remainder in each-case.
Solution.
Let the given polynomials are p(x) = 3x³ + ax² + 3x + 5 and f(x) = 4x³ + x² – 2x + a
According to question p(2) = f(2)
3(2)³ + a(2)² + 3(2) + 5 = 4(2)³ + (2)² – 2(2) + a
⇒ 3 x 8 + 4a + 6 + 5 = 32 + 4 – 4 + a
⇒ 24 + 4a + 11 = 32 + a
⇒ 4a – a = 32 – 35
⇒ 3a = -3
⇒ a = -1
As the remainder is same, so p(2) or f(2) would (RBSESolutions.com) be same when a = – 1
p(2) = 3(2)³ + (-1)(2)² + 3 x 2 + 5 [∴ a = -1]
= 24 – 4 + 6 + 5
= 35 – 4 = 31
Thus, a = – 1 and p(2) or f(2) = 31

Question 6.
Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a).
Solution.
We have,
L.H.S.
= (a + b + c)³ – a³ -b³ – c³
= {(a + b + c)³ – (a)³} – (b³ + c³)
= (a + b + c – a){(a + b + c)² + a(a + b + c) + a²} – (b + c){b² – bc + c²)
[∵ x³ – y³ = (x – y)(x² + xy + y²) and x³ + y³ = (x + y)(x² – xy + y²)]
= (b + c){a² + b² + c² + 2ab + 2bc + 2ca + a² + ab + ac + a²} – (b + c)(b² – bc + c²)
= (b + c)[3a² + b² + c² + 3ab + 2bc + 3ca – b² + bc – c²]
= (b + c)[3a² + 3ab + 3bc + 3ca]
= 3(b + c)[a² + ab + bc + ca]
= 3(b + c)[a(a + b) + c(a + b)]
= 3(b + c)(a + b)(a + c)
= 3(a + b)(b + c)(c + a)
= R.H.S.

Question 7.
(i) For what (RBSESolutions.com) value of m is x³ – 2mn² + 16 divisible by (x + 2)?
(ii) Show that (2x – 3) is a factor of x + 2x³ – 9x + 12.
Solution.
(i) Let p(x) = x³ – 2mn² + 16
p(x) will be divisible by (x + 2) if
p(-2)= 0
p(-2) = (-2)³ – 2m(-2)² + 16 = -8 – 8m + 16 = – 8m + 8
p(-2) = 0
⇒ -8m + 8 = 0
⇒ 8m = 8
⇒ m = 1
(ii) We know that (2x – 3) will be a factor of x + 2x³ – 9x + 12 if p(x) on dividing by 2x – 3, leaves (RBSESolutions.com) a remainder zero

So, the remainder obtained (RBSESolutions.com) on dividing p(x) by 2x – 3 is zero.
Hence, (2x – 3) is a factor of x + 2x³ – 9x + 12

We hope the given RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 8 Social Science Chapter 6 Industrial Scenario

RBSE Solutions for Class 8 Social Science Chapter 6 Industrial Scenario are part of RBSE Solutions for Class 8 Social Science. Here we have given RBSE Rajasthan Board Solutions for Class 8 Social Science Chapter 6 Industrial Scenario.

Board	RBSE
Class	Class 8
Subject	Social Science
Chapter	Chapter 6
Chapter Name	Industrial Scenario
Number of Questions Solved	13
Category	RBSE Solutions

Rajasthan Board Class 10 Science Textbook Questions Solved

Question 1.
Choose the best option.

1.  Which of the following is minerals based industry?
   (a) Cotton textile
   (b) Cement
   (c) Leather
   (d) Sugar
2. Which industry is known as the seasonal Industry?
   (a) Ornament Industry
   (b) Granite Industry
   (c) Sugar Industry
   (d) Dairy Industry

Answers:

1. (b)
2. (c)

Question 2.
Match the following:

1.  Sugarcane                   (a) Cotton Textile Industry
2.  Limestone                   (b) Hindustan Copper Limited
3.  Copper                         (c) Sugar Industry
4.  Mustard                       (d) Cement Industry
5.  Cotton                          (e) Raw Oil Mill

Answers:

1. (c)
2. (d)
3. (b)
4. (e)
5. (a)

Question 3.
Fill in the blanks: –

1. There is……… corporation for the development of cottage and small scale industries.
2. The first cement factory was set up in………of…………district.
3. Jodhpur is known for its…….. Khadi.
4.  ………………in Jaipur is the world famous centre for the gold and Silver ornaments and
   the jewels.

Answers:

1. Rajasthan Small Scale Industries corporation.
2. (a) Lakheri
   (b) Bundi
3. Marino
4. Johari Bazar

Question 4.
Classify Industries on the basis of size.
Answer :
Classification of Industries based on their size:

• Cottage Industries include baskets making, knitting sweaters, making wooden goods, earthen pots. They need less capital investment, labour and electricity. These industries are mainly run by the family members in their own house. Most of the goods produced here are handmade.
  They are run in small factory areas and employ 10 to 100 labourers. Examples: match-sticks making, dyeing and printing, bricks making etc.
• Large Scale Industries engage a large number of workers, need more capital investment and high level of technology. Some of them are industry cotton-textile, cement factory, Iron and Steel, automobile industry etc.

Question 5.
Which city of Rajasthan is known as ‘City of Garment’ and why?
Answer :
Bhilwara is known as the ‘City of Garment’ or Manchester of Rajasthan’ because Bhilwara district is the biggest centre of manufacturing cloth.

Question 6.
Why is the ‘Make in India’ programme launched?
Answer :
‘Make in India’ means manufacturing or producing all kinds for goods of our use in India and labelling them as ‘Made in India’ because the goods produced within the country will be cheaper and they will add to the national income by their export.

Question 7.
In how many types industries can be classified? Explain their examples.
Answer:
Industries can be classified on the basis of their size, ownership and raw material, such as follows-

1. On the Basis of Size:
   (a) Cottage Industries – basket weaving, knitting sweaters, making wooden goods, earthern pots. etc.
   (b) Small Scale Industries – Match-Stick, bricks, tie and dye-print industries.
   (c) Large Scale Industries – cotton textile, cement, Iron and Steel, and automobile industries.
2.  On the Basis of Ownership:
   (a) Private Sector Industries as – All those which are run by an individual or a group of individuals. Example: Reliance.
   (b) Public Sector Industries as – Hindustan Aeronautic Limited and Hindustan Copper Limited.
   (c) Joint Sector Industries as – Maruti Udyog Limited.
   (d) Cooperative sector: Anand Milk Union Limited and Saras Dairy Cooperative Unit.
3. On the Basis of Raw Material:
   (a) Agricultural as cotton textile and sugar industries.
   (b) Sea based as fishing oil, sea food processing.
   (c) Mineral based as Iron ore and ornaments.
   (d) Forest based as medicines/drugs, paper, furniture making industries.

Question 8.
What are cottage industries? Give few examples of them.
Answer:
Cottage industries are small scale industries run by the family members from their own homes. The architects prepare the goods with their hands rather than the machines. Examples: basket weaving, poultry farming, sweater knitting, wooden tools, earthen pots, rope making etc.

Question 9.
Describe the Large Scale Industries of Rajasthan.
Answer:
Large Scale Industries of Rajasthan:

1.  Cotton – Textile Industry: It is the oldest industry of Rajasthan. The first Cotton Textile mill was established in 1889 at Beawar in the Ajmer district, but Bhilwara district is the main centre of producing readymade clothes in Rajasthan. Hence it is known as the ‘city of Garment’ or Manchester of Rajasthan. Jaipur, Jodhpur, Churu, Bikaner and Nagaur are the main centre of dyeing and printing or Bandhej. Jaipur and Jodhpur are famous for the stitched clothes.
2. Cement Industry: First cement Industry was set up in 1915 at Lakheri in Bundi district. Raw materials such as limestone, gypsum for producing cement are sufficiently available in Rajasthan but Coal as a raw material for cement is brought from Madhya Pradesh and Chattisgarh. Chittorgarh district is considered ideal for the production of cement because of the easy availability of raw materials for making cement. Beside there are cement manufacturing units in Jaisalmer, Kota, Nagaur, Pali, Sawaimadhopur, Ajmer Udaipur and Banswara districts.
3. Sugar Industry: It is agriculture based seasonal industry and is established mainly in the sugarcane producing region because sugarcane is its raw material. The first sugar mill was established in Rajasthan in 1932 in the private sector at Bhupal sagar in Chittorgarh district, presently not in operation. Simitarly another sugar industries not in operation are at Keshoraipatan in the Bundi district and Sri Sriganganagar.
4.  Other Industries are as:
   
   • Engineering and Instruments Industry in Kota, Bharatpur, Jaipur, Ajmer and Alwar.
   • Chemical Industry in Deedwana, Kota, Alwar, Udaipur and Chittaurgarh.
   • Khetri Copper Complex in Jhunjhunu.
   • Hindustan Zinc Limited in Udaipur.
   • Electronics goods Industry in Bhiwari, Pilani and Jodhpur.

RBSE Solutions for Class 10 Additional Questions Solved

Multiple Choice Questions
Question 1.
In a…………..industry the crafstmen manufacture goods by using local raw materials.
(a) cottage
(b) large scale
(c) agro based
(d) small scale

Question 2.
……………is the leading producer ofcotton textile industry of India.

(a) Jaipur
(b) Bhilwara
(c) Kota
(d) Ajmer

Answers:
1. (b)
2. (a)

RBSE Books Solutions Short Answer Type Questions

Question 1.
1.What are Special Economic Zones?
Answer.
Special Economic Zones are established with the objective of speedy industrial development and increase in means  of employment.

(i) Discuss the role of industries in your day to day life.
Answer:
We use many things in our daily life from waking up in the morning to till going to bed at night like – toothbrush and toothpaste soap, clothes, pen, paper, means of transport, cosmetics, computer, biscuit, utensils, watch etc. All these things are manufacturerd in one or another unit (Industry). In this way there is no field of life where Industries do not play.

RBSE Solutions Pdf Download Long Answer Type Questions

Question 1.
What do you know about the Skill Development Scheme? Write a short note.
Answer:
Skill Development Scheme means introducing professional education courses from the school level so as to prepare the students to get employment opportunities at the local level soon after completing education. In such a scheme beside industrial skill more focus is given on special skills related to electronics, electrical, food processing etc. which enable trainers either to start their own industry or to get employment. For this Enterprising Development Programme (EDP), Enterprising Skill Development Programme (ESDP), Business Skill Development Programme (BSDP), Industrial Motivating Compaign (IMC). Business and Educational Training etc. programes being run or introduced.

We hope the given RBSE Solutions for Class 8 Social Science Chapter 6 Industrial Scenario will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 8 Social Science Chapter 6 Industrial Scenario, drop a comment below and we will get back to you at the earliest.