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Class 9 NCERT Solutions of Chapter 2-Polynomial
The NCERT solutions of class 9 maths chapter 2- Polynomial is a basic chapter of algebra. The NCERT solutions of chapter 2-Polynomial will help you in understanding the questions based on the different structures of polynomials. The NCERT solutions of chapter 2-Polynomial provide the use of all algebraic identities which are used in higher classes and also in other branches of mathematics. All the NCERT solutions of chapter 2-Polynomial are solved by an expert teacher of maths as per the CBSE norms.
Polynomial is an algebraic expression in which the degree of the variable is a whole number as an example x, x +1,x²+1, etc but constants like 1,2,3 3/2, etc are also supposed as a polynomial because we can write them x0,2x0,3x0,3x0/2etc.In terms of the number of terms, polynomials are classified as monomial(single), binomial(two terms), trinomial(three terms), etc. In terms of degree, the polynomials are classified as linear polynomials (degree 1), quadratic polynomials (degree 2), cubic polynomials (degree 3), etc.
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Chapter 2 Polynomial of Class 9 contains 5 exercises
Exercise 2.1
Exercise 2.2
Exercise 2.3
Exercise 2.4
Exercise 2.5
NCERT Solutions of all chapters of Maths for Class 10 from Chapters 1 to 15
NCERT Solutions of Class 9 Maths : from chapter 1 to 15
You can also see the video NCERT Solutions for Class 10 Maths Chapter 2 -Polinomial updated for 2023-24 CBSE Board Exams,these NCERT solutions are referred to Chapter- 2 Current maths NCERT text book.
Class 9 maths NCERT Solutions of the chapter 2-Polynomial
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Exercise 2.1
Q1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x² –3x + 7
(ii) y² + √2
(iii) 3√t + t√2
Ans. (i) 4x² –3x + 7
Hint: Polynomial is the algebraic expression in which the power of the variable is a whole number
The given expression is a polynomial in one variable x since each power of x (2,1,0) is a whole number
(ii) y² + √2
The given expression is polynomial in one variable y since each power of y (2,0) is a whole number
(iii) 3√t + t√2
The given expression is not a polynomial since the power of variable t is (1/2, 1) in which 1/2 is not a whole number
The given expression is not a polynomial since the power of variable y is (1, –1) in which –1 is not a whole number.
The given expression is a polynomial in three variables x,y, and t since each exponent of the variable is 10, 3, and 50 in which all are whole numbers.
Q2.Write the coefficient of x² in each of the following.
(i) 2 + x² + x
(ii) 2 – x² + x³
Ans.
(i) 2 + x² + x
In the given expression the term containing x² has the coefficient 1
(ii) 2 – x² + x³
In the given expression the term containing x² has the coefficient –1
In the given expression the term containing x² has the coefficient π/2.
In the given expression there is no any term x², so the coefficient of x² is 0.
Q3. Give one example each of a binomial of degree 35, and a monomial of degree 100.
Ans. Binomial of degree 35 is following
Monomial of degree 100 is following
Q4. Write the degree of the following polynomial
(i) 5x³ + 4x² + 7x
(ii) 4 – y²
(iii)5t –√7
(iv) 3
Ans.
(i) 5x³ + 4x² + 7x
The highest exponent of the variable x is 3, so the degree of the given polynomial is 3
(ii) 4 – y²
The highest exponent of the variable y in the expression is 2, so the degree of the polynomial is 2
(iii)5t –√7
The highest exponent of the variable t in the expression is 1, so the degree of the polynomial is 1
(iv) 3
There is no variable in the given term, so the degree of the polynomial is 0
Q5. Classify the following as linear, quadratic, and cubic polynomial.
(i) x² + x
(ii) x – x³
(iii) y + y² + 4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x³
Ans.
(i) x² + x
The given polynomial is of the degree 2, so it is a quadratic polynomial
(ii) x – x³
The given polynomial is of the degree 3, so it is a cubic polynomial
(iii) y + y² + 4
The given polynomial is of the degree 2, so the given polynomial is a quadratic polynomial
(iv) 1 + x
The given polynomial is of the degree 1, so the given polynomial is a linear polynomial
(v) 3t
The given polynomial is of the degree 1, so the given polynomial is a linear polynomial
(vi) r²
The given polynomial is of the degree 2, so the given polynomial is a quadratic polynomial
(vii) 7x³
The given polynomial is of the degree 3, so the given polynomial is a cubic polynomial
See the video for exercise 2.1 solutions
Class 9 Maths NCERT Solutions of Chapter 2-Polynomial
Exercise 2.2
Q1. Find the value of polynomial 5x –4x² +3 at
(i) x = 0 (ii)x = –1 (iii) x = 2
Ans. Let the given polynomial is expressed as p(x)
(i) x = 0
p(x) = 5x –4x² +3
p(0) = 5× 0 –4 × 0 + 3 = 3
(ii)x = –1
p(x) = 5x –4x² +3
p(–1) = 5 ×–1 – 4 (–1) ² + 3 = –5 –4 + 3 = –9 + 3 = –6
(iii) x = 2
p(x) = 5x –4x² +3
p(2) = 5 × 2 – 4 ×2² + 3 = 10 – 16 + 3 = –3
Q2.Find p(0), p(1) and p(2) for each of the following polynomials
(i) p(y) = y² –y + 1
(ii) p(t) = 2 +t +2t² – t³
(iii) p(x) = x³
(iv) p(x) = (x +1)(x – 1)
Ans.
(i) p(y) = y² –y + 1
p(0) = 0 – 0 +1 = 1
p(1) = 1² –1 + 1 = 1
p(2) = 2² –2 + 1 = 3
(ii) p(t) = 2 +t +2t² – t³
p(0) = 2 +0 +2×0² – 0³= 2
p(1) = 2 +1 +2×1² – 1³ = 3 +2 – 1 = 4
p(2)=2 +2 +2×2² – 2³ = 4 + 8 – 8 = 4
(iii) p(x) = x³
p(0) = 0³ = 0
p(1) = 1³ = 1
p(2) = 2³ = 8
(iv) p(x) = (x +1)(x – 1)
p(0) = (0+1)(0 – 1) = –1
p(1) = (1 +1)(1 – 1) = 0
p(2) = (2 +1)(2 – 1) = 3
Q3. Verify whether the following are zeroes of the polynomial , indicated against them.
Ans.
Since the value of the polynomial at x = – 1/3 is zero, so x = – 1/3 is the zero of the given polynomial
The value of the polynomial is not zero at the value ,x = 4/5, so x = 4/5 is not the zero of the given polynomial
p(1) = 1² – 1 = 0
The value of the polynomial at x = 1, is zero, so x = 1 is the zero of the given polynomial.
The value of the polynomial at x = –1/2, is not zero, so x = –1/2 is not the zero of the given polynomial.
p(0) = 0
The value of the polynomial at x = 0, is zero, so x = 0 is the zero of the given polynomial
The value of the polynomial at x = –m/l is zero, so x = –m/l is the zero of the given polynomial
The value of the polynomial at x = –1/l√3 is zero, so x = –1/√3 is the zero of the given polynomial
The value of the polynomial at x = 2/√3 is not zero, so x = 2/√3 is not the zero of the given polynomial
The value of the polynomial at x = 1/2 is not zero, so x = 1/2 is not the zero of the given polynomial
Q4. Find the zero of the polynomial in each of the given cases.
(i) p(x) = x +5
(ii) p(x) = x – 5
(iii) p(x) = 2 x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0 where c and d are real numbers
Ans.
(i) p(x) = x +5
x +5 = 0
x = – 5
Hence the zero of the given polynomial is x =– 5
(ii) p(x) = x – 5
x – 5 = 0
x = 5
Hence the zero of the given polynomial is x = 5
(iii) p(x) = 2 x + 5
2 x + 5 = 0
Hence the zero of the given polynomial is – 5/2
(iv) p(x) = 3x – 2
3x – 2 = 0
Hence the zero of the given polynomial is 2/3
(v) p(x) = 3x
3x = 0
x = 0
Hence the zero of the given polynomial is 0
(vi) p(x) = ax, a ≠ 0
ax = 0
x = 0
Hence the zero of the given polynomial is 0
vii) p(x) = cx + d
cx + d = 0
Hence zero of the given polynomial is – d/c
Class 9 Maths NCERT Solutions of Chapter 2-Polynomial
EXERCISE-2.3
Q1.Find the remainder when is divided by
(i) x
(iii) x
(iv) x+π
(v) 5+2x
Ans.
(i) We have to divide the polynomial by the polynomial x + 1
Let’s find the zero of the x = 0
x = – 1
Putting the value of x = – 1, in
= – 1 + 3 – 3 + 1 = 0
(ii) Dividing the polynomial by
, we can get the reminder in the following way
Putting the value in
(iii) Dividing the polynomial by x, we can get the reminder by the following way
Putting x = 0 the zero of the polynomial x, in the given polynomial
0 + 0 + 0 +1 = 1
(iv) Dividing the polynomial by x + π, we can get the reminder by the following way
x + π = 0
x = – π , is the zero of the polynomial x + π
Putting the value x = – π, in the given polynomial
The reminder is
(v) Dividing the polynomial by 5 + 2x, we can get the reminder by the following way
5+2x = 0
, is the zero of the polynomial 5+2x
Putting the value of in the polynomial
, we get the reminder
Q2.Find the remainder when is divided by
.
The zero the polynomial x – a, is
x – a = 0
x = a
Putting this value in
a³ – a × a² + 6 × a – a
= a³ – a³ +6a – a = 5a
Therefore the required reminder is 5a
Q3.Check whether is a factor of
.
The zero of the polynomial 7 + 3x is as follows
7 + 3x = 0
Putting this value of x in , we shall get the required reminder
Class 9 Maths NCERT Solutions of Chapter 2-Polynomial
EXERCISE-2.4
Q1.Determine which of the following polynomials has a factor.
(i)
(ii)
(iii)
(iv)
Ans.
(i)
The zero of the polynomial x + 1 is as follows
x + 1 = 0
x = – 1
Putting this value of x = – 1, in the given polynomial
(–1)³ + 1² + 1 + 1
– 1 + 1 + 1 + 1 = 2
Since value of the polynomial at x = – 1, is not zero so (x +1) is not the factor of the given polynomial
(ii)
The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1
Putting x = – 1, in the given polynomial
= 1 – 1 + 1 – 1 + 1 = 1
Since value of the given polynomial at x = 1, is not zero,therefore (x +1) is not the factor of the given polynomial
(iii)
The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1
Putting x = – 1, in the given polynomial
= 1 – 3 + 3 –1 +1 = 1
Since value of the given polynomial at x = 1, is not zero,therefore (x +1) is not the factor of the given polynomial
(iv)
The zero of the polynomial (x +1) is x + 1 = 0 ⇒ x = – 1
Putting x = – 1, in the given polynomial
The value of the polynomial at x = – 1 is not zero, so (x + 1) is not the factor of the given polynomial
Q2. 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
(iii) p (x) = x3 – 4x2 + x + 6, g (x) = x – 3
Ans. (i) p (x)= 2x3 + x2 – 2x – 1, g (x) = x + 1
x + 1 = 0
x = – 1
Putting the value of x = – 1 in the given polynomial
p(x) = 2x3 + x2 – 2x – 1
= 2× – 1 + 1 + 2 – 1
p(–1) = – 2 + 3 – 1 = 0
Since value of p(–1) is zero ,therefore g(x) is the factor of the given polynomial
(ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2
x + 2 = 0
x = – 2
p(x)= x3 + 3x2 + 3x + 1
p(– 2) = (– 2)3 + 3(–2)² + 3 (– 2) +1 = -8 +12 -6 +1 = -1
Since the value of p(– 2) is not zero, therefore the given polynomial is not the factor of (x + 2)
(iii) p (x) = x3 – 4x2 + x + 6, g (x) = x – 3
x – 3 = 0
x = 3
Putting the value of x in the given polynomial
p(3) = 3³ –4× 3² + 3 + 6
= 27– 36 + 9
= 0
Since the value of p(3) is zero, so g(x) is the factor of p(x)
Q3.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
(iii) p (x) = kx2 – √2 x + 1
(iv) p (x) = kx2 – 3x + k
Ans.
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)
(iii) Here, p (x) = kx2 – √2 x + 1
Since, p(1) = k(1)2 – (1) + 1
= k – √2 + 1 = 0
⇒ k = √2 -1
(iv) Here, p(x) = kx2 – 3x + k
p(1) = k(1)2 – 3(1) + k
= k – 3 + k
= 2k – 3 = 0
⇒ k =
Q4. Factorise
(i) 12x2 – 7x +1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4
Ans.
(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)
(iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2)
(iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Thus, 3x2 – x – 4 = (3x – 4)(x + 1)
Q5. Factorise
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20
(iv) 2y3 + y2 – 2y – 1
Ans.
(i) We have, x3 – 2x2 – x + 2
Rearranging the terms, we have x3 – x – 2x2 + 2
= x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2)
= [(x)2 – (1)2](x – 2)
= (x – 1)(x + 1)(x – 2)
[∵ (a2 – b2) = (a + b)(a-b)]
Therefore , x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2)
(ii) We have, x3 – 3x2 – 9x – 5
= x3 + x2 – 4x2 – 4x – 5x – 5 ,
= x2 (x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x2 – 4x – 5)
= (x + 1)(x2 – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
Therefore, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1)
(iii) 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)
Therefore, x3 + 13x2 + 32x + 20
= (x + 1)(x + 2)(x + 10)
(iv) 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)
Therefore, 2y3 + y2 – 2y – 1
= (y – 1)(y + 1)(2y +1)
See the video for Question (4) and Question (5)
Class 9 Maths NCERT Solutions of Chapter 2-Polynomial
Exercise 2.5
Q1.Use suitable identities to find the following products
(i)
(ii)
(iii)
(iv)
Ans.
(i) (x + 4)( x+ 10)
Q2.Evaluate the following products without multiplying directly:
(i)103×107
(ii)95×96
(iii)104× 96
Ans.
(i)103×107
Modifying the given expression in form of (x + a)(x +b) as following
(103)×(107) = (100 +3)(100 +7)
(100 +3)(100 +7) = 100² +(3 +7)100 + 3×7 = 10000 +1000 + 21 = 110021
(ii) 95×96
95×96 = (100 – 5)(100 – 4)
(100 – 5)(100 – 4) = 100² +( – 5 – 4)100 +( – 4)( – 5)
10000 – 900 + 20 = 9120
(iii) 104× 96
(100 +4)(100 – 4)
(100 +4)(100 – 4) = 100² – 4² = 10000 – 16 = 9984
Q3.Factorize the following using appropriate identities
(i)
(ii)
(iii)
Ans.
(i)
Using the following identity
(ii)
(iii)
Q4.Expand each of the following, using a suitable identity
(i)
(ii)
(iii)
(iv)
(v)
(iv)
Ans.
(i)
(ii)
(iii)
(iv)
(v)
(iv)
Q5.Factorise:
(i)
(ii)
Ans.
(i)
Since the terms –24yz and –16xz, both are having negative sign, so the term z which is common between them must be negative
(ii)
Q6.Write the following cubes in expanded form
Ans.
Applying the following identity
=
=
Applying the following identity
Using the following identity
Q7.Evaluate the following using suitable identities
(i)99³ (ii)102³ (iii) 998³
Ans.
(i)99³
Modifying 99 into (100– 1) and using the following identity
(x +y)³ = x³ + 3x²y + 3y²x + y³
= 1000000 – 30000 + 300 – 1
= 970299
(ii)102³
Modifying 102 into ( 100 +2) and using the following identity
(x +y)³ = x³ + 3x²y + 3y²x + y³
(100 +2)³ = 100³ + 3× 100² ×2 + 3×2²× 100 + 2³
= 1000000 + 60000 +1200 +8
=1061208
(iii) 998³
Modifying the number 998 into (1000 – 2) and using the following identity
(x +y)³ = x³ + 3x²y + 3y²x + y³
(1000 –2)³ = 1000³ – 3× 1000² ×2 + 3×2²× 1000 – 2³
1000000000 –6000000 +12000 – 8
=994011992
Q8.Factorise each of the following
(1) 8a³ + b³ +12a²b + 6ab²
(ii) 8a³ – b³ – 12a²b + 6ab²
(iii) 27 – 125a³ – 135a + 225a²
(iv) 64a³ – 27b³ – 144a²b + 108b²a
Ans.
(1) 8a³ + b³ +12a²b + 6ab²
Using the following identity
(x + y)³ = x³ + y³ +3x²y + 3y²x
Modifying the given expression as following
(2a)³ + b³ +3×(2a)²×b + 3×b²×2a
= (2a + b)³
(ii) 8a³ – b³ – 12a²b + 6ab²
Using the following identity
(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x
Modifying the given expression as following
(2a)³ – b³ –3×(2a)²×b + 3×b²×2a
= (2a – b)³
(iii) 27 – 125a³ – 135a + 225a²
Using the following identity
(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x
27 – 125a³ – 135a + 225a²
= 3³ – (5a)³ –3×(3)²×5a + 3×(5a)²×5a
= (3 – 5a)³
(iv) 64a³ – 27b³ – 144a²b + 108b²a
(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x
Modifying the given expression as following
(4a)³ – (3b)³ – 3×(4a)²×3b + 3×(3b)²× 4a
(4a – 3b)³
(x – y)³ = x³ – y³ –3×x²×y + 3×y²×x
Modifying the given expression as following
Q9.Verify
Ans.
Using the following identity
Using the following identity
Hence Proved
Q10.Factorise each of the following
Ans.
Modifying the given expression according to the identity
Modifying the given expression according to the identity
Q11.Factorise
Using the following identity
Modifying the given expression according to the identity
Q12.Verify that
Applying the following identity
Dividing and multiplying the RHS by 2
Q13. If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.
Ans.
Using the following identity
Since, x + y + z = 0
x3 + y3 + z3 – 3 xyz = 0
x3 + y3 + z3 = 3 xyz
Hence if x + y + z = 0, then x3 + y3 + z3 = 3 xyz
See the video for Solutions of Question 1 to Question 13
Q14.Without actually calculating the cubes, find the value of each of the following
(i) (– 12)3 + (7)3 + (5)3
(ii) (28)3 + (– 15)3 + (– 13)3
Ans.
(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
= – 1260
(ii) We have, (28)3 + (–15)3 + (–13)3
Let x = 28, y = –15 and z = – 13
Then, x + y + z = 28 – 15 – 13 = 0
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz
∴ (28)3 + (–15)3 + (–13)3 = 3×28×–15×–13
= 16380
Q15.Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given
(i) Area 25a2 – 35a + 12
(ii) Area 35y2 + 13y – 12
Ans.
Area of a rectangle = Length x Breadth
(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).
(ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).
Q16.What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume 3x2 – 12x
(ii) Volume 12ky2 + 8ky – 20k
Ans.
Volume of a cuboid = Length x Breadth x Height
(i) We have, 3x2 – 12x = 3(x2 – 4x)
= 3 x x x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).
(ii) We have, 12ky2 + 8ky – 20k
= 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)]
= 4 x k x (3y2 + 2y – 5)
= 4k[3y2 – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).
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CBSE Class 10-Question paper of maths 2019 with solutions
NCERT Solutions for Class 10 Science
NCERT Solutions for class 11 maths
| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
| Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
| Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
| Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
| Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
| Chapter 8- Binomial Theorem | Chapter 16- Probability |
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
NCERT solutions for class 12 maths
| Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
| Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
| Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
| Chapter 4-Determinants | Chapter 12-Linear Programming |
| Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
| Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
| Chapter 7- Integrals | |
| Chapter 8-Application of Integrals |
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution
