, cbse class 9 maths chapter wise important questions - free pdf download.
CBSE Important Questions for Class 9 Maths are available in Printable format for Free Download.Here you may find NCERT Important Questions and Extra Questions for Class 9 Mathematics chapter wise with answers also. These questions will act as chapter wise test papers for Class 9 Mathematics. These Important Questions for Class 9 Mathematics are as per latest NCERT and CBSE Pattern syllabus and assure great success in achieving high score in Board Examinations
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10 questions mcq test - test: polynomials- case based type questions, beti bachao, beti padhao (bbbp) is a personal campaign of the government of india that aims to generate awareness and improve the efficiency of welfare services intended for girls. in a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on beti bachao, beti padhao. q. if in the group, there are 10 teachers and 58 girls, then what is the number of boys.
No. of teachers = x + y = 10
⇒ (x + y) 2 = (10) 2
⇒ x 2 + y 2 + 2xy = 100
[Since (a + b) 2 = a 2 + b 2 + 2ab]
No. of students = (x 2 + y 2 ) = 58
⇒ 58 + 2xy = 100
⇒ 2xy = 100 – 58
⇒ xy = 42/2
Now, since (x + y) 3 = [x 3 + y 3 + 3xy(x + y)]
⇒ (10) 3 = [x 3 + y 3 + 3 × 21(10)]
⇒ 1000 = (x 3 + y 3 + 630)
⇒ 1000 – 630 = (x 3 + y 3 )
⇒ (x 3 + y 3 ) = 370
In mathematics, a polynomial is an expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.
Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls.
In a school, a group of (x + y) teachers, (x 2 + y 2 ) girls and (x 3 + y 3 ) boys organised a campaign on Beti Bachao, Beti Padhao.
Q. (x – y) 3 =
(x 2 – y 2 – 3xy(x – y))
(x 3 – y 3 – 2xy(x – y)
(x 3 – y 3 – 3xy(x – y)
(x 3 – y 3 – 3xyx – y)
We know that (x - y) 3 can be written as
(x - y)(x - y)(x - y)
We know that (x - y)(x - y) can be multiplied and written as
= x 2 - xy - yx + y 2 (x - y)
= x 2 - 2xy + y 2 (x - y)
= x 3 - 2x 2 y + xy 2 - yx 2 + 2xy 2 - y 3
= x 3 – y 3 – 2xy(x – y)
Q. Using part (D), find (x 2 – y 2 ) if x – y = 23.
Also, x + y = 10
x 2 – y 2 = (x + y)(x – y)
National Association For The Blind (NAB) aimed to empower and well-inform visually challenged population of our country, thus enabling them to lead a life of dignity and productivity.
Q. Find the amount donated by Ravi.
Q. (x – y) 3 =
x 2 - xy - yx + y 2 (x - y)
= x 3 – y 3 – 3xy(x – y)
Q. (x + a)(x + b) = x 2 + ................ x + ab
Q. Which mathematical concept is involved in the above situation?
Polynomials- case based type questions mcqs with answers, online tests for polynomials- case based type questions, welcome back, create you account for free.
CBSE Expert
Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 2023-24. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams.
Case study questions play a pivotal role in enhancing students’ problem-solving skills. By presenting real-life scenarios, these questions encourage students to think beyond textbook formulas and apply mathematical concepts to practical situations. This approach not only strengthens their understanding of mathematical concepts but also develops their analytical thinking abilities.
Table of Contents
Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked. For Class 9 Maths Case Study Questions, there would be 5 case-based sub-part questions, wherein a student has to attempt 4 sub-part questions.
Checkout: Class 9 Science Case Study Questions
And for mathematical calculations, tap Math Calculators which are freely proposed to make use of by calculator-online.net
The above Class 9 Maths Case Study Question s will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Study Questions have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.
UNIT I: NUMBER SYSTEMS
1. REAL NUMBERS (18 Periods)
1. Review of representation of natural numbers, integers, and rational numbers on the number line. Rational numbers as recurring/ terminating decimals. Operations on real numbers.
2. Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as √2, √3 and their representation on the number line. Explaining that every real number is represented by a unique point on the number line and conversely, viz. every point on the number line represents a unique real number.
3. Definition of nth root of a real number.
4. Rationalization (with precise meaning) of real numbers of the type
(and their combinations) where x and y are natural number and a and b are integers.
5. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)
UNIT II: ALGEBRA
1. POLYNOMIALS (26 Periods)
Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomial, terms of a polynomial and zero polynomial. Degree of a polynomial. Constant, linear, quadratic and cubic polynomials. Monomials, binomials, trinomials. Factors and multiples. Zeros of a polynomial. Motivate and State the Remainder Theorem with examples. Statement and proof of the Factor Theorem. Factorization of ax2 + bx + c, a ≠ 0 where a, b and c are real numbers, and of cubic polynomials using the Factor Theorem. Recall of algebraic expressions and identities. Verification of identities:
and their use in factorization of polynomials.
2. LINEAR EQUATIONS IN TWO VARIABLES (16 Periods)
Recall of linear equations in one variable. Introduction to the equation in two variables. Focus on linear equations of the type ax + by + c=0.Explain that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers, plotting them and showing that they lie on a line.
UNIT III: COORDINATE GEOMETRY COORDINATE GEOMETRY (7 Periods)
The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations.
UNIT IV: GEOMETRY
1. INTRODUCTION TO EUCLID’S GEOMETRY (7 Periods)
History – Geometry in India and Euclid’s geometry. Euclid’s method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/postulates and theorems. The five postulates of Euclid. Showing the relationship between axiom and theorem, for example: (Axiom)
1. Given two distinct points, there exists one and only one line through them. (Theorem)
2. (Prove) Two distinct lines cannot have more than one point in common.
2. LINES AND ANGLES (15 Periods)
1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180O and the converse.
2. (Prove) If two lines intersect, vertically opposite angles are equal.
3. (Motivate) Lines which are parallel to a given line are parallel.
3. TRIANGLES (22 Periods)
1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).
2. (Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence).
3. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).
4. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. (RHS Congruence)
5. (Prove) The angles opposite to equal sides of a triangle are equal.
6. (Motivate) The sides opposite to equal angles of a triangle are equal.
4. QUADRILATERALS (13 Periods)
1. (Prove) The diagonal divides a parallelogram into two congruent triangles.
2. (Motivate) In a parallelogram opposite sides are equal, and conversely.
3. (Motivate) In a parallelogram opposite angles are equal, and conversely.
4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.
5. (Motivate) In a parallelogram, the diagonals bisect each other and conversely.
6. (Motivate) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it and (motivate) its converse.
5. CIRCLES (17 Periods)
1. (Prove) Equal chords of a circle subtend equal angles at the center and (motivate) its converse.
2. (Motivate) The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.
3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.
4. (Prove) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
5. (Motivate) Angles in the same segment of a circle are equal.
6. (Motivate) If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.
7. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.
UNIT V: MENSURATION 1.
1. AREAS (5 Periods)
Area of a triangle using Heron’s formula (without proof)
2. SURFACE AREAS AND VOLUMES (17 Periods)
Surface areas and volumes of spheres (including hemispheres) and right circular cones.
UNIT VI: STATISTICS & PROBABILITY
STATISTICS (15 Periods)
Bar graphs, histograms (with varying base lengths), and frequency polygons.
To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.
Regular practice of CBSE Class 9 Maths case study questions offers several benefits to students. Some of the key advantages include:
Solving case study questions can be challenging, but with the right approach, you can excel. Here are some tips to enhance your problem-solving skills:
Remember, solving case study questions is not just about finding the correct answer but also about demonstrating a logical and systematic approach. Now, let’s explore some resources that can aid your preparation for CBSE Class 9 Maths case study questions.
Yes, case study questions are an integral part of the CBSE Class 9 Maths syllabus. They are designed to enhance problem-solving skills and encourage the application of mathematical concepts to real-life scenarios.
Solving case study questions enhances students’ problem-solving skills, analytical thinking, and decision-making abilities. It also bridges the gap between theoretical knowledge and practical application, making mathematics more relevant and engaging.
Case study questions help in exam preparation by familiarizing students with the question format, improving analytical thinking skills, and developing a systematic approach to problem-solving. Regular practice of case study questions enhances exam readiness and boosts confidence in solving such questions.
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Case study questions for class 9 maths chapter 9 areas of parallelograms and triangles, case study questions for class 9 maths chapter 6 lines and angles, case study questions for class 9 maths chapter 7 triangles, case study questions for class 9 maths chapter 5 introduction to euclid’s geometry, case study and passage based questions for class 9 maths chapter 14 statistics, case study questions for class 9 maths chapter 1 real numbers, case study questions for class 9 maths chapter 4 linear equations in two variables, case study questions for class 9 maths chapter 3 coordinate geometry, case study questions for class 9 maths chapter 15 probability, case study questions for class 9 maths chapter 13 surface area and volume, case study questions for class 9 maths chapter 10 circles, case study questions for class 9 maths chapter 9 quadrilaterals, case study questions for class 9 maths chapter 2 polynomials.
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Home > CBSE > Important Questions Class 9 Maths Chapter 2
Mathematics Chapter 2 of Class 9 is about Polynomials. Polynomial consists of two terms, namely Poly (meaning “many”) and Nominal (meaning “terms.”). A polynomial is explained as an expression which is composed of variables, constants and exponents that are combined using mathematical operations like addition, subtraction, multiplication and division (No division operation by a variable). Based on the number of terms present in the expression, it is classified as monomial, binomial, and trinomial.
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Our Mathematics experts believe that students must practice questions regularly to perform better in exams. For this purpose, they have prepared the Important Questions Class 9 Mathematics Chapter 2 to help students get access to questions from all the topics of the Polynomials. The questions are followed by their step-by-step answers, which will further help students to revise the chapter. The questions are curated from various sources such as the NCERT textbook and exemplar book, CBSE past years’ question papers, and other reference materials.
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Our in-house Mathematics faculty experts have collated a complete list of Important Questions Class 9 Mathematics Chapter 2 by referring to various sources. The subject experts have meticulously prepared illustration for individual questions that will enable students to comprehend the notions used in each question. Furthermore, the questions are selected in a way that would cover all the topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak topics. And enhance their preparation by further concentrating on weaker sections of the chapter.
Given below are a few of the questions and answers from our question bank of Important Questions Class 9 Mathematics Chapter 2:
Question 1: Calculate the value of 9x² + 4y² if xy = 6 and 3x + 2y = 12.
Answer 1: Consider the equation 3x + 2y = 12
Now, square both sides:
(3x + 2y)² = 12²
=> 9x² + 12xy + 4y² = 144
=>9x² + 4y² = 144 – 12xy
From the questions, xy = 6
9x² + 4y² = 144 – 72
Thus, the value of 9x² + 4y² = 72
Question 2:Evaluate the following using suitable identity
Answer 2: We can write 102 as 100+2
Using identity,(x+y) ³ = x ³ +y ³ +3xy(x+y)
(100+2) ³ =(100) ³ +2 ³ +(3×100×2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
Question 3:Without any actual division, prove that the following 2x⁴
– 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.
[Hint: Factorise x² – 3x + 2]
Answer 3: x²-3x+2
x(x-2)-1(x-2)
Therefore,(x-2)(x-1)are the factors.
Considering (x-2),
Then, p(x) becomes,
p(x)=2x⁴-5x³+2x²-x+2
p(2)=2(2)⁴-5(2)³+2(2)²-2+2
Therefore, (x-2) is a factor.
Considering (x-1),
p(1)=2(1)⁴-5(1)³+2(1)²-1+2
Therefore, (x-1) is a factor.
Question 4: Using the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case
(i) p(x) = 2x³+x²–2x–1, g(x) = x+1
Answer 4 :p(x) = 2x³+x²–2x–1, g(x) = x+1
∴Zero of g(x) is -1.
p(−1) = 2(−1)³+(−1)²–2(−1)–1
∴By the given factor theorem, g(x) is a factor of p(x).
Question 5: Obtain an example of a monomial and a binomial having degrees of 82 and 99, respectively.
Answer 5: An example of a monomial having a degree of 82 = x⁸²
An example of a binomial having a required degree of 99 = x⁹⁹ + 7
Question 6: If the two x – 2 and x – ½ are the given factors of px ²
+ 5 x + r , show that p = r .
Answer 6: Given, f(x) = px²+5x+r and factors are x-2, x – ½
Substituting x = 2 in place of the equation, we get
f(x) = px²+5x+r
f(2) = p(2)²+5(2)+r=0
= 4p + 10 + r = 0 … eq.(i)
Substituting x = ½ in place of the equation, we get,
f( ½ ) = p( ½ )² + 5( ½ ) + r =0
= p/4 + 5/2 + r = 0
= p + 10 + 4r = 0 … eq(ii)
On solving eq(i) and eq(ii),
4p + r = – 10 and p + 4r = – 10
the RHS of both equations are the same,
4p + r = p + 4r
Hence Proved.
Question 7: Identify constant, linear, quadratic, cubic and quartic polynomials from the following.
(i) – 7 + x
(iii) – ? ³
(iv) 1 – y – ? ³
(v) x – ? ³ + ?⁴
(vi) 1 + x + ?²
Answer 7: (i) – 7 + x
The degree of – 7 + x is 1.
Hence, it is a linear polynomial.
The degree of 6y is 1.
Therefore, it is a linear polynomial.
We know that the degree of – ? ³ is 3.
Therefore, it is a cubic polynomial.
We know that the degree of 1 – y – ? ³ is 3.
We know that the degree of x – ? ³ + ?⁴ is 4.
Therefore, it is a quartic polynomial.
We know that the degree of 1 + x + ?² is 2.
Therefore, it is a quadratic polynomial.
We know that the degree of -6?² is 2.
We know that -13 is a constant.
Therefore, it is a constant polynomial.
We know that the degree of –p is 1.
Question 8: Observe the value of the polynomial 5x – 4x² + 3 at x = 2 and x = –1.
Answer 8 : Let the polynomial be f(x) = 5x – 4x² + 3
Now, for x = 2,
f(2) = 5(2) – 4(2)² + 3
=> f(2) = 10 – 16 + 3 = –3
Or, the value of the polynomial 5x – 4x² + 3 at x = 2 is -3.
Similarly, for x = –1,
f(–1) = 5(–1) – 4(–1)² + 3
=> f(–1) = –5 –4 + 3 = -6
The value of the polynomial 5x – 4x² + 3 at x = -1 is -6.
Question 9:Expanding each of the following, using all the suitable identities:
(i) (x+2y+4z)²
(ii) (2x−y+z)²
(iii) (−2x+3y+2z)²
(iv) (3a –7b–c)²
(v) (–2x+5y–3z)²
Answer 9: (i) (x+2y+4z)²
Using identity, (x+y+z)² = x²+²+z²+2xy+2yz+2zx
Here, x = x
(x+2y+4z)² = x²+(2y)²+(4z)²+(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x²+4y²+16z²+4xy+16yz+8xz
Using identity, (x+y+z)² = x²+y²+z²+2xy+2yz+2zx
Here, x = 2x
(2x−y+z)² = (2x)²+(−y)²+z²+(2×2x×−y)+(2×−y×z)+(2×z×2x)
= 4x²+y²+z²–4xy–2yz+4xz
Here, x = −2x
(−2x+3y+2z)² = (−2x)²+(3y)²+(2z)²+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)
= 4x²+9y²+4z²–12xy+12yz–8xz
Using identity (x+y+z)²= x²+y²+z²+2xy+2yz+2zx
Here, x = 3a
(3a –7b– c)² = (3a)²+(– 7b)²+(– c)²+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)
= 9a² + 49b² + c²– 42ab+14bc–6ca
Here, x = –2x
(–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
Question 10: If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + leave the same remainder when divided by z – 3, find the value of a.
Answer 10: Zero of the polynomial,
Hence, zero of g(z) = – 2a
Let p(z) = az³+4z²+3z-4
Now, substituting the given value of z = 3 in p(z), we get,
p(3) = a (3)³ + 4 (3)² + 3 (3) – 4
⇒p(3) = 27a+36+9-4
⇒p(3) = 27a+41
Let h(z) = z³-4z+a
Now, by substituting the value of z = 3 in h(z), we get,
h(3) = (3)³-4(3)+a
⇒h(3) = 27-12+a
⇒h(3) = 15+a
As per the question,
The two polynomials, p(z) and h(z), leave the same remainder when divided by z-3
So, h(3)=p(3)
⇒15+a = 27a+41
⇒15-41 = 27a – a
Question 11: Compute the perimeter of a rectangle whose area is 25x² – 35x + 12.
Answer 11: A rea of rectangle = 25x² – 35x + 12
We know the area of a rectangle = length × breadth
So, by factoring 25x² – 35x + 12, the length and breadth can be obtained.
25x² – 35x + 12 = 25x² – 15x – 20x + 12
=> 25x² – 35x + 12 = 5x(5x – 3) – 4(5x – 3)
=> 25x² – 35x + 12 = (5x – 3)(5x – 4)
Thus, the length and breadth of a rectangle are (5x – 3)(5x – 4).
So, the perimeter = 2(length + breadth)
Therefore, the perimeter of the given rectangle = 2[(5x – 3)+(5x – 4)]
= 2(5x – 3 + 5x – 4)
= 2(10x – 7)
= 20x – 14
Hence, the perimeter of the rectangle = 20x – 14
Question 12: 2x²+y²+²–2√2xy+4√2yz–8xz
Answer 12: Using identity, (x +y+z)² = x²+y²+z²+2xy+2yz+2zx
We can say that, x²+²+²+2xy+2yz+2zx = (x+y+z)²
2x²+y²+8z²–2√2xy+4√2yz–8xz
= (-√2x)²+(y)²+(2√2z)²+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
= (−√2x+y+2√2z)²
= (−√2x+y+2√2z)(−√2x+y+2√2z)
Question 13: If ? + 2? is a factor of ? ⁵ – 4?²?³ + 2? + 2? + 3, find a.
Answer 13: According to the question,
Let p(x) = x ⁵ – 4a²x³ + 2x + 2a + 3 and g(x) = x + 2a
⟹ x + 2a = 0
Hence, zero of g(x) = – 2a
As per the factor theorem,
If g(x) is a factor of p(x), then p( – 2a) = 0
So, substituting the value of x in p(x), we get,
p ( – 2a) = ( – 2a) ⁵ – 4a²( – 2a)³ + 2( – 2a) + 2a + 3 = 0
⟹ – 32a ⁵ + 32a ⁵ – 2a + 3 = 0
⟹ – 2a = – 3
Question 14: Find the value of x³+ y ³ + z ³ – 3xyz if x² + y² + z² = 83 and x + y + z = 1
Answer 14: Consider the equation x + y + z = 15
From algebraic identities, we know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)
From the question, x² + y² + z²= 83 and x + y + z = 15
152 = 83 + 2(xy + yz + xz)
=> 225 – 83 = 2(xy + yz + xz)
Or, xy + yz + xz = 142/2 = 71
Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),
x ³ + y ³ + z ³ – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))
x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71
So, x ³ + y ³ + z ³ – 3xyz = 15(83 – 71)
=> x ³ + y ³ + z ³ – 3xyz = 15 × 12
Or, x ³ + y ³ + z ³ – 3xyz = 180
Question 15:Verify that:
(i) x³+y³ = (x+y)(x²–xy+y²)
(ii) x³–y³ = (x–y)(x²+xy+y²)
Answer 15: (i) x³+y³ = (x+y)(x²–xy+y²)
We know that (x+y)³= x³+y³+3xy(x+y)
⇒ x³+y³ = (x+y)³–3xy(x+y)
⇒ x³+y³ = (x+y)[(x+y)²–3xy]
Taking (x+y) common ⇒ x³+y³ = (x+y)[(x²+y²+2xy)–3xy]
⇒ x³+y³ = (x+y)(x²+y²–xy)
(ii) x³–y³ = (x–y)(x²+xy+y²)
We know that (x–y)³ = x³–y³–3xy(x–y)
⇒ x³−y³ = (x–y)³+3xy(x–y)
⇒ x³−y³ = (x–y)[(x–y)²+3xy]
Taking (x+y) common ⇒ x³−y³ = (x–y)[(x²+y²–2xy)+3xy]
⇒ x³+y³ = (x–y)(x²+y²+xy)
Question 16: For what value of m is ?³ – 2??² + 16 divisible by x + 2?
Answer 16: According to the question,
Let p(x) = x³ – 2mx² + 16, and g(x) = x + 2
⟹ x + 2 = 0
Hence, zero of g(x) = – 2
if p(x) is divisible by g(x), then the remainder of p(−2) should be zero.
Thus, substituting the value of x in p(x), we obtain,
p( – 2) = 0
⟹ ( – 2)³ – 2m( – 2)² + 16 = 0
⟹ 0 – 8 – 8m + 16 = 0
Question 17:If a + b + c = 15 and a² + b² + c² = 83, find the value of a³ + b³ + c³ – 3abc.
Answer 17: We know that,
a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca) ….(i)
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca ….(ii)
Given, a + b + c = 15 and a² + b² + c² = 83
From (ii), we have
152 = 83 + 2(ab + bc + ca)
⇒ 225 – 83 = 2(ab + bc + ca)
⇒ 142/2 = ab + bc + ca
⇒ ab + bc + ca = 71
Now, (i) can be written as
a³ + b³ + c³ – 3abc = (a + b + c)[(a² + b² + c² ) – (ab + bc + ca)]
a³ + b³+ c³ – 3abc = 15 × [83 – 71] = 15 × 12 = 180.
Question 18: Factorise: 27x³+y³+z³–9xyz
Answer 18: The expression27x³+y³+z³–9xyz can be written as (3x)³+y³+z³–3(3x)(y)(z)
27x³+y³+z³–9xyz = (3x)³+y³+z³–3(3x)(y)(z)
We know that x³+y³+³–3xyz = (x+y+z)(x²+y²+z²–xy –yz–zx)
= (3x+y+z)[(3x)²+y²+z²–3xy–yz–3xz]
= (3x+y+z)(9x²+y²+²–3xy–yz–3xz)
Question 19: If (x – 1/x) = 4, then evaluate (x² + 1/x²) and (x⁴ + 1/x⁴).
Answer 19: Given, (x – 1/x) = 4
Squaring both sides, we get,
(x – 1/x)² = 16
⇒ x² – 2.x.1/x + 1/x² = 16
⇒ x² – 2 + 1/x² = 16
⇒ x² + 1/x² = 16 + 2 = 18
∴ (x² + 1/x²) = 18 ….(i)
Again, squaring both sides of (i), we get
(x² + 1/x²)² = 324
⇒ x⁴ + 2.x².1/x² + 1/x⁴= 324
⇒ x⁴ + 2 + 1/x⁴ = 324
⇒ x⁴ + 1/x⁴ = 324 – 2 = 322
∴ (x⁴ + 1/x⁴) = 322.
Question 20: Factorise
Answer 20: The expression 64m³–343n³ can be written as (4m)³–(7n)³
64m³–343n³ =(4m)³–(7n)³
We know that x³–y³ = (x–y)(x²+xy+y²)
64m³–343n³ = (4m)³–(7n)³
= (4m-7n)[(4m)²+(4m)(7n)+(7n)²]
= (4m-7n)(16m²+28mn+49n² )
Question 21: Find out the values of a and b so that (2x³ + ax² + x + b) has (x + 2) and (2x – 1) as factors.
Answer 21: Let p(x) = 2x³ + ax² + x + b. Then, p( –2) = and p(½) = 0.
p(2) = 2(2)³ + a(2)² + 2 + b = 0
⇒ –16 + 4a – 2 + b = 0 ⇒ 4a + b = 18 ….(i)
p(½) = 2(½)³ + a(½)² + (½) + b = 0
⇒ a + 4b = –3 ….(ii)
On solving (i) and (ii), we get a = 5 and b = –2.
Hence, a = 5 and b = –2.
Question 22: Explain that p – 1 is a factor of p¹⁰ – 1 and p¹¹ – 1.
Answer 22: According to the question,
Let h(p) = ?¹⁰ − 1,and g(p) = ? – 1
zero of g(p) ⇒ g(p) = 0
Therefore, zero of g(x) = 1
We know that,
According to the factor theorem, if g(p) is a factor of h(p), then h(1) should be zero
h(1) = (1) ¹⁰ − 1 = 1 − 1 = 0
⟹ g (p) is a factor of h(p).
Here, we have h(p) = ?¹¹ − 1, g (p) = ? – 1
Putting g (p) = 0 ⟹ ? − 1 = 0 ⟹ ? = 1
As per the factor theorem, if g (p) is a factor of h(p),
Then h(1) = 0
⟹ (1) ¹¹ – 1 = 0
Hence, g(p) = ? – 1 is the factor of h(p) = ? ¹⁰ – 1
Question 23: Examine whether (7 + 3x) is a factor of (3×3 + 7x).
Answer 23: Let p(x) = 3×3 + 7x and g(x) = 7 + 3x. Now g(x) = 0 ⇒ x = –7/3.
By the remainder theorem, p(x) is divided by g(x), and then the remainder is p(–7/3).
Now, p(–7/3) = 3(–7/3)3 + 7(–7/3) = –490/9 ≠ 0.
∴ g(x) is not a factor of p(x).
Question 24:Prove that:
x³+y³+z³–3xyz = (1/2) (x+y+z)[(x–y)²+(y–z)²+(z–x)²]
Answer 24: We know that,
x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²–xy–yz–xz)
⇒ x³+y³+z³–3xyz = (1/2)(x+y+z)[2(x²+y²+z²–xy–yz–xz)]
= (1/2)(x+y+z)(2×2+2y²+²–2xy–2yz–2xz)
= (1/2)(x+y+z)[(x²+y²−2xy)+(y²+z²–2yz)+(x²+z²–2xz)]
= (1/2)(x+y+z)[(x–y)²+(y–z)²+(z–x)²]
Question 25: Find out which of the following polynomials has x – 2 a factor:
(i) 3?² + 6?−24.
(ii) 4?² + ?−2.
Answer 25: (i) According to the question,
Let p(x) =3?² + 6?−24 and g(x) = x – 2
g(x) = x – 2
zero of g(x) ⇒ g(x) = 0
Hence, zero of g(x) = 2
Thus, substituting the value of x in p(x), we get,
p(2) = 3(2)² + 6 (2) – 24
= 12 + 12 – 24
the remainder = zero,
We can derive that,
g(x) = x – 2 is factor of p(x) = 3?² + 6?−24
(ii) According to the question,
Let p(x) = 4?² + ?−2 and g(x) = x – 2
p(2) = 4(2)² + 2−2
Since the remainder ≠ zero,
We can say that,
g(x) = x – 2 is not a factor of p(x) = 4?² + ?−2
Question 26: Factorise x² + 1/x² + 2 – 2x – 2/x.
Answer 26 : x² + 1/x² + 2 – 2x – 2/x = (x² + 1/x² + 2) – 2(x + 1/x)
= (x + 1/x)² – 2(x + 1/x)
= (x + 1/x)(x + 1/x – 2).
Question 27: Factorise
8a³+b³+12a²b+6ab²
Answer 27: The expression, 8a³+b³+12a²b+6ab² can be written as (2a)³+b³+3(2a)²b+3(2a)(b)²
8a³+b³+12a²b+6ab² = (2a)³+b³+3(2a)²b+3(2a)(b)²
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)³ = x³+y³+3xy(x+y) is used.
Question 28: By Remainder Theorem, find out the remainder when p(x) is divided by g(x), where
(i) p(?) = ?³ – 2?² – 4? – 1, g(?) = ? + 1
(ii) p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3
(iii) p(?) = 4?³ – 12?² + 14? – 3, g(?) = 2? – 1
(iv) p(?) = ?³ – 6?² + 2? – 4, g(?) = 1 – 3/2 ?
Answer 28: (i) Given p(x) = ?³ – 2?² – 4? – 1 and g(x) = x + 1
Here zero of g(x) = – 1
By applying the remainder theorem
P(x) divided by g(x) = p( – 1)
P ( – 1) = ( – 1)³ – 2 ( – 1)² – 4 ( – 1) – 1 = 0
Therefore, the remainder = 0
(ii) given p(?) = ?³ – 3?² + 4? + 50, g(?) = ? – 3
Here zero of g(x) = 3
By applying the remainder theorem p(x) divided by g(x) = p(3)
p(3) = 3³ – 3 × (3)² + 4 × 3 + 50 = 62
Therefore, the remainder = 62
(iii) p(x) = 4x³ – 12x² + 14x – 3, g(x) = 2x – 1
Here zero of g(x) = ½
By applying the remainder theorem p(x) divided by g(x) = p (½)
P( ½ ) = 4( ½ )³ – 12( ½ )² + 14 ( ½ ) – 3
= 4/8 – 12/4 + 14/2 – 3
= ½ + 1
= 3/2
Hence, the remainder = 3/2
so, zero of g(x) = 2/3
By applying the remainder theorem p(x) divided by g(x) = p(2/3)
p(2/3) = (2/3)³ – 6(2/3)² + 2(2/3) – 4
Therefore, the remainder = – 136/27
Question 29:Factorise x² – 1 – 2a – a².
Answer 29: x² – 1 – 2a – a² = x² – (1 + 2a + a²)
= x² – (1 + a)²
= [x – (1 – a)][x + 1 + a]
= (x – 1 – a)(x + 1 + a)
∴ x² – 1 – 2a – a² = (x – 1 – a)(x + 1 + a).
Question 30:Evaluate the following using suitable identity
Answer 30: We can write 99 as 1000–2
Using identity,(x–y)³ = x³ –y³ –3xy(x–y)
(998)³ =(1000–2)³
=(1000)³ –2³ –(3×1000×2)(1000–2)
= 1000000000–8–6000(1000– 2)
= 1000000000–8- 6000000+12000
= 994011992
Question 31: Find the zeroes of the polynomial:
p(?)= (? –2)² −(? + 2)²
Answer 31: p(x) = (? –2)² −(? + 2)²
Zero of the polynomial p(x) = 0
Hence, we get,
⇒ (x–2)² −(x + 2)² = 0
Expanding using the identity, a² – b² = (a – b) (a + b)
⇒ (x – 2 + x + 2) (x – 2 –x – 2) = 0
⇒ 2x ( – 4) = 0
Therefore, the zero of the polynomial = 0
Consistently solving questions is a vital element of mastering Mathematics. By solving Mathematics Class 9 Chapter 2 important questions, students can get a further understanding of the polynomials chapter.
A few other advantages of solving Important Questions Class 9 Mathematics Chapter 2 are:
Extramarks leaves no stone unturned to give the best learning material to students while combining fun and learning activities through its own study materials to enhance their learning experience. It provides comprehensive learning solutions for students from Class 1 to Class 12. Our website has abundant resources, along with important questions and solutions. Students can click on the links given below to access some of these resources:
Q.1 By actual division, find the quotient and the remainder when x 5 + 1 is divided by x 1
Marks: 3 Ans
x 4 + x 3 + x 2 + x + 1 x 1 x 5 + 1 x 5 x 4 + x 4 + 1 x 4 x 3 + x 3 + 1 x 3 x 2 + ¯ x 2 + 1 x 2 x + x + 1 x 1 + 2 ¯ ¯ ¯ ¯ Quotient : x 4 + x 3 + x 2 + x + 1 Remainder : 2
Q.2 Find the value of k if x 5 is a factor of kx 2 + 3x + 7.
Marks: 2 Ans
Zero of x 5 is 5 as x 5 = 0 gives x = 5 . p(x) = kx 2 + 3 x + 7 p ( 5 ) = 0 25 k + 15 + 7 = 0 25 k + 22 = 0 k = 22 25
Q.3 If x + y + z = 6 and xy + yz + zx = 11, then find the value of x 2 + y 2 + z 2 .
Given : x + y + z = 6 and xy + yz + zx = 11 Squaring both sides , we get x + y + z 2 = 6 2 x 2 + y 2 + z 2 + 2 xy + 2 yz + 2 zx = 36 x 2 + y 2 + z 2 + 2 xy + yz + zx = 36 x 2 + y 2 + z 2 + 2 11 = 36 Since xy + yz + zx = 11 x 2 + y 2 + z 2 + 22 = 36 x 2 + y 2 + z 2 = 36 22 = 14
Faqs (frequently asked questions), 1. what are the four types of polynomials.
The 4 types of polynomials are zero polynomial, linear polynomial, quadratic polynomial, and cubic polynomial.
On the Extramarks website, you can find all of the important questions for Class 9 Mathematics Chapter 2, along with their answers. On the website, you can also find important questions and NCERT solutions for all classes from 1 to 12.
The NCERT Mathematics book has 15 chapters. Each chapter is equally important when it comes to learning the fundamentals and taking the test. Additionally, because CBSE does not specify the distribution of marks for each chapter, students are advised to fully study all chapters. Each and every chapter must be completely understood to acquire a good grade in exams.
All the fifteen chapters of CBSE Class 9 Mathematics syllabus are given below:
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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 2 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.
Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.
Below we have listed the topics that have been discussed in this chapter. As this is one of the important topics in maths, It comes under the unit – Algebra which has a weightage of 20 marks in class 9 maths board exams.
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Show Answer Total amount invested by both, if x = 1000 is (a) 301506 (b)370561 (c) 4012005 (d)490621 Show Answer The shares of Ankur and Ranjan invested individually are (a) (2x + 1), (2x + 5) (b) (2x + 3), (x + 1) (c) (x + 1), (x + 3) (d) None of these Show Answer Name the polynomial of amounts invested by each partner. (a) Cubic (b) Quadratic
Some of them are given below. Answer them. (i) Which one of the following is not a polynomial? (a) 4x 2 + 2x - 1 (b) y+ (3/y) (c) x 3 - 1 (d) y 2 + 5y + 1 (ii) The polynomial of the type ax2 + bx + c, a = 0 is called (a) Linear polynomial (b) Quadratic polynomial (c) Cubic polynomial (d) Biquadratic polynomial
CBSE Class 9 Maths exam 2022-23 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions. Each question has five sub-questions, each followed by four options and one correct answer.
myCBSEguide App Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes. Install Now If you're looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter.
Well, you're in luck! In this article, we will provide you with a collection of Case Study Questions for Class 9 Maths specifically designed to help you excel in your exams. These questions are carefully curated to cover various mathematical concepts and problem-solving techniques.
TopperLearning provides a complete collection of case studies for CBSE Class 9 Maths Polynomials chapter. Improve your understanding of biological concepts and develop problem-solving skills with expert advice.
Algebraic Identities Students can refer to the NCERT Solutions for Class 9 while solving exercise problems and preparing for their Class 9 Maths exams. NCERT Class 9 Maths Chapter 2 - Polynomials Summary NCERT Solutions for Class 9 Maths Chapter 2 Polynomials is the second chapter of Class 9 Maths.
Updated for new NCERT - 2023-24 Edition. Solutions to all NCERT Exercise Questions and Examples of Chapter 2 Class 9 Polynomials are provided free at Teachoo. Answers to each and every question is explained in an easy to understand way, with videos of all the questions. In this chapter, we will learn. What is a Polynomial.
Subjective Case Based Questions -🔥Polynomials Class 9 Maths Chapter 2 | CBSE Board Exam 2022 Preparation | NCERT Solutions | #VedantuClass9and10 Are all wor... CBSE Exam, class 10
The important questions of ch 2 Maths class 9 are carefully designed under the CBSE board's rules' strict guidance. To perform well in mathematics, academic success is practice; the students must efficiently practice the polynomials class 9 important questions. To prevent any issues or mistakes in the important questions for class 9 maths ...
Some important questions from polynomials are given below with solutions. These questions will help the 9th class students to get acquainted with a wide variety of questions and develop the confidence to solve polynomial questions more efficiently. 1. Give an example of a monomial and a binomial having degrees of 82 and 99, respectively. Solution:
Case Study 1. Ankur and Ranjan start a new business together. The amount invested by both partners together is given by the polynomial p (x) = 4x 2 + 12x + 5, which is the product of their individual shares. Coefficient of x2 in the given polynomial is (a) 2 (b) 3 (c) 4 (d) 12 Show Answer Total amount invested by both, if x = 1000 is
Welcome to our Class 9 Maths exam preparation series for the academic year 2023-24! In this video, we dive deep into Case Study Questions from Chapter 2: Pol...
x 2 + x + 1 To know more about polynomials, click here. Term In the polynomial, each expression is called a term . Suppose x 2 + 5x + 2 is polynomial, then the expressions x 2, 5x, and 2 are the terms of the polynomial. Coefficient Each term of the polynomial has a coefficient.
Free PDF download of NCERT Exemplar for Class 9 Maths Chapter 2 - Polynomials solved by expert Maths teachers on Vedantu as per NCERT (CBSE) Book guidelines. All Chapter 2 - Polynomials exercise questions with solutions to help you to revise the complete syllabus and score more marks in your examinations. Download free Class 9 Maths to amp up ...
Ex 2.1 Class 9 Maths Question 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer. (i) 4x 2 - 3x + 7 (ii) y 2 + √2 (iii) 3 √t + t√2 (iv) y+ 2 y (v) x 10 + y 3 +t 50 Solution: (i) We have 4x 2 - 3x + 7 = 4x 2 - 3x + 7x 0 It is a polynomial in one variable i.e., x
7 Views 6 Downloads Class 9 Polynomials Case Study Questions - Podar International School Class 9 Mathematics Questions with answers in PDF format for free Class 9 Maths Question Papers with answers Class 9 Podar International School Mathematics CBSE 2024 Download Paper Related Papers
Detailed Solution for Test: Polynomials- Case Based Type Questions - Question 1 No. of teachers = x + y = 10 ⇒ (x + y) 2 = (10) 2 ⇒ x 2 + y 2 + 2xy = 100 [Since (a + b) 2 = a 2 + b 2 + 2ab] No. of students = (x 2 + y 2) = 58 ⇒ 58 + 2xy = 100 ⇒ 2xy = 100 - 58 ⇒ 2xy = 42 ⇒ xy = 42/2 ⇒ xy = 21 Now, since (x + y) 3 = [x 3 + y 3 + 3xy (x + y)]
by experts Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 2023-24. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams.
Case Study Questions for Class 9 Maths Archives - Gurukul of Excellence. Q-104, Sector-38, Gurgaon, Haryana 07065827902 [email protected] M-F: 8 AM- 8 PM; Weekend: 10 AM-2 PM.
Download Important Questions Class 9 Maths Chapter 2 - Polynomials free PDF prepared by subject experts as per the latest NCERT books to score marks in exam ... Important Questions Class 9 Mathematics Chapter 2- Polynomials. Mathematics Chapter 2 of Class 9 is about Polynomials. Polynomial consists of two terms, namely Poly (meaning "many ...
Here you will get complete NCERT Solutions for Class 9 Maths Chapter 2 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.
Here we are providing Polynomials Class 9 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. Extra Questions for Class 9 Maths Polynomials with Answers Solutions. ... CA Foundation Business Economics Study Material - Elasticity of Demand ...