NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

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NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

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NCERT solutions of maths 11 th class Chapter 5 Complex number and Quadratic Equation.

Exercises 5.1 & 5.2

Exercise 5.3

NCERT solutions of class 11 maths

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

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Exercise 5.1

Q1. Express the given complex number in the form of a + ib.

$(5i) \left (- \frac{3}{5} i\right )$

Solution:

$(5i) \left (- \frac{3}{5} i\right )$

$=5\times -\frac{3}{5}i\times i$

= –3i²

= –3 × –1

= 3 = 3 + 0.i ( a =3, b= 0).

Q2.Express the given complex number in the form a + ib, : I⁹ + i¹⁹.

Solution: I⁹ + i¹⁹

= (I²)⁴i + (i²)⁹i

$=(-1)^{4}i+(-1)^{9}i$

= 1 × i +(-1)i

= i – i =0 = 0 +i0 (a=0, b= 0)

Q3.Express the given complex number in the form of a + ib:i^-39

Solution: $i^{-39}$

$=\frac{(i^{2})^{-19}}{i}$

$=\frac{(-1^{})^{-19}}{i}$

$=\frac{(-1)^{-19}\times i}{i^{2}}$

$=\frac{(-1)^{-19}\times i}{{-1}}$

= i = 0 + i

Q4. Express the given complex number in the form a+ib:3(7 + i7) + i(7 +i7).

Solution:

3(7 + i7) + i(7 +i7)

=21 + 21i + 7i + 7i²

=21 + 28i –7

=14 + 28i

=14 + i28

Q5.Express the given complex number in the form a + ib:(1 – i) – (–1 +i6).

Solution:

(1 – i) – (–1 +i6)

=1 – i + 1 – 6i

=2 – 7i

=2 –i7

Q6. Express the given complex number in the form a + ib:

$\mathbf{\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right )}$

Solution:

$\left ( \frac{1}{5}+i\frac{2}{5} \right )-\left ( 4+i\frac{5}{2} \right )$

$=\frac{1}{5}+i\frac{2}{5}-4-i\frac{5}{2}$

$=\left ( \frac{1}{5}-4 \right )+i\left ( \frac{2}{5}-\frac{5}{2} \right )$

$= -\frac{19}{5}+i\left ( \frac{-21}{10}\right )$

$= -\frac{19}{5}-i\frac{21}{10}$

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Q7.Express the given complex number in the form a + ib:

$\mathbf{\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+1 \right )}$

Solution:

${\left [ \left ( \frac{1}{3}+i\frac{7}{3} \right )+\left ( 4+i\frac{1}{3} \right ) \right ]-\left ( -\frac{4}{3}+1 \right )}$

$=\frac{1}{3}+i\frac{7}{3}+4+i\frac{1}{3}+\frac{4}{3}-i$

$=\left (\frac{1}{3} +4+\frac{4}{3} \right )+i\left ( \frac{7}{3}+\frac{1}{3}-1\right )$

$=\frac{17}{3}+i\frac{5}{3}$

Q8.Express the given complex number in the form a + ib:

$\mathbf{(1-i)^{4}}$

Solution:

$(1-i)^{4}$

=[(1 – i)²]²

=[1 + i² – 2i]²

=(1 – 1 – 2i)²

=(–2i)² =4i²= 4 × –1 = –4[ i² = –1]

= –4 + 0i

Q9.Express the given complex number in the form a + ib:

$\mathbf{\left ( \frac{1}{3}+3i \right )^{3}}$

Solution:

${\left ( \frac{1}{3}+3i \right )^{3}}$

$\therefore (x+y)^{3}=x^{3}+y^{3}+3x^{2}y+3y^{2}x$

${\color{DarkBlue} \mathbf{ = \left ( \frac{1}{3} \right )^{3}+\left ( 3i \right )^{3}+3\left ( \frac{1}{3} \right )^{2}.3i +3\left ( 3i \right )^{2}\frac{1}{3}}}$

${\color{DarkBlue} \mathbf{=\frac{1}{27}+27i^{2}.i+i +9i^{2}}}$

${\color{DarkBlue} \mathbf{=\frac{1}{27}-27.i+i -9}}$

${\color{DarkBlue} \mathbf{=\frac{1}{27} -9}}{\color{DarkBlue} \mathbf{-26i}}$

${\color{DarkBlue} \mathbf{=-\frac{-242}{27}-26i}}$

Q10. Express the given complex number in the form a + ib:

${\color{DarkGreen} \mathbf{\left ( -2-\frac{1}{3}i \right )^{3}.}}$

Answer.

${{\color{DarkBlue} \mathbf{\left ( -2-\frac{1}{3}i \right )^{3}.}}}$

${{\color{DarkBlue} \because \mathbf{\left ( a-b \right )^{3}=a^{3 }-b^{3}-3a^{2}b+3b^{2}a}}}$

${\color{DarkBlue} =\mathbf{\left ( -2 \right )^{3}-\left ( \frac{1}{3}i \right )^{3}-3\left ( -2 \right )^{2}\frac{1}{3}i+3\left ( \frac{1}{3}i \right )^{2}\left ( -2 \right )}}$

${\color{DarkBlue} \mathbf{=-8-\frac{1}{27}i^{2}.i-4i-\frac{2}{3}i^{2}}}$

${\color{DarkBlue} \mathbf{=-8+\frac{1}{27}i-4i+\frac{2}{3}}}$ (i² = –1)

${\color{DarkBlue} \mathbf{=\left ( -8+\frac{2}{3} \right )+i\left ( \frac{1}{27}-4 \right )}}$

${\color{DarkBlue} \mathbf{=\frac{-22}{3}-i\frac{107}{27}}}$

Q11.Find the multiplicative inverse of the complex number 4 – 3i.

Answer.

The multiplicative inverse of 4 – 3i, will be

${\color{DarkBlue} \mathbf{=\frac{1}{4-3i}}}$

Multiplying its denominator and numerator by the conjugate of the denominator: 4 + 3i.

${\color{DarkBlue} \mathbf{{ \mathbf{=\frac{1}{4-3i}\times\frac{4+3i}{4+3i} }}}}$

${\color{DarkBlue} \mathbf{=\frac{4+3i}{4^{2}-\left ( 3i \right )^{2}}}}$

(i² = –1)

${\color{DarkBlue} \mathbf{=\frac{4+3i}{4^{2}-\left ( 3i \right )^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{4+3i}{16+9}}}$ ${\color{DarkBlue} \mathbf{=\frac{4+3i}{25}}}$

${\color{DarkBlue} \mathbf{=\frac{4}{25}+i\frac{3}{25}}}$ (i² = –1)

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Q12. Find the multiplicative inverse of the complex number √5 + 3i.

Answer. The multiplicative inverse of √5 + 3i will be

${\color{DarkBlue} \mathbf{\frac{1}{\sqrt{5}+3i}}}$

Multiplying its denominator and numerator by the conjugate of the denominator: √5 – 3i.

${\color{DarkBlue} {\mathbf{\frac{1}{\sqrt{5}+3i}\times \frac{\sqrt{5}-3i}{\sqrt{5}-3i}}}}$

${\color{DarkBlue} \mathbf{=\frac{\sqrt{5}-3i}{\left ( \sqrt{5} \right )^{2}-\left ( 3i \right )^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{\sqrt{5}-3i}{5-9i^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{\sqrt{5}-3i}{5+9}}}$

${\color{DarkBlue} \mathbf{=\frac{\sqrt{5}}{14}-i\frac{3}{14}}}$

Q13. Find the multiplicative inverse of the complex number: -i.

Answer. Let z = –i

then mulplicative inverse of z will be

${\color{DarkBlue}= \mathbf{\frac{1}{z}}}$

${\color{DarkBlue} \mathbf{=\frac{1}{-i}}}$

${\color{DarkBlue} \mathbf{=\frac{1}{-i}\times \frac{-i}{-i}}}$

${\color{DarkBlue} \mathbf{=\frac{-i}{\left ( - i\right )^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{-i}{i^{2}}}}$ ${\color{DarkBlue} \mathbf{=\frac{-i}{-1}}}$

= i

Q14. Express the following expression in the form of a + ib:

${\color{DarkGreen} \mathbf{\frac{\left ( 3+i\sqrt{5} \right )\left ( 3-i\sqrt{5} \right )}{\left ( \sqrt{3}+\sqrt{2}i \right )-\left ( \sqrt{3} -i\sqrt{2}\right )}.}}$

Answer.

${\color{DarkBlue} { \mathbf{\frac{\left ( 3+i\sqrt{5} \right )\left ( 3-i\sqrt{5} \right )}{\left ( \sqrt{3}+\sqrt{2}i \right )-\left ( \sqrt{3} -i\sqrt{2}\right )}}}}$

${\color{DarkBlue} \mathbf{=\frac{3^{2}-\left ( i\sqrt{5} \right )^{2}}{\sqrt{3}+\sqrt{2}i-\sqrt{3}+i\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{9-5i^{2}}{i2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{9+5}{i2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{14}{i2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{14}{i2\sqrt{2}}\times \frac{i}{i}}}$

${\color{DarkBlue} \mathbf{=\frac{7i}{\sqrt{2}i^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{7i}{\sqrt{2}\times -1}\times \frac{\sqrt{2}}{\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{-7\sqrt{2}i}{2}}}$

${\color{DarkBlue} \mathbf{=0-i\frac{7\sqrt{2}}{2}}}$

Study class 11 th CBSE maths NCERT solutions Trigonometric functions

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Exercise 5.2

Q1.Find the modulus and argument of the complex number z = –1 – i√3.

Answer.

Modulus of the complex number :z=a + ib is given by

${\color{DarkBlue} \mathbf{\left | z \right |=\sqrt{a^{2}+b^{2}}}}$

Comparing the given complex number (–1 – i√3 ) with the standard form (a +ib) of complex number

a = –1, b = –√3

$\boldsymbol{\left | Z \right |=\sqrt{\left ( -1 \right )^{2}+\left ( -\sqrt{3} \right )^{2}}=\sqrt{1+3}=\sqrt{4}=2}$

The argument of the given complex number is given by θ as follows.

${\color{DarkBlue} \mathbf{\Theta = tan^{-}\frac{b}{a}}}$

b = –√3 , a = –1, the complex number will lie in the third quadrant in the argand plane.

${\color{DarkBlue} \mathbf{\Theta = tan^{-}\left ( \frac{-\sqrt{3}}{-1} \right )}}$

${\color{DarkBlue} \mathbf{\Theta = tan^{-}\sqrt{3}}}$

${\color{DarkBlue} \mathbf{\Theta =\frac{\Pi }{3}}}$

As θ lies in lll quadrant so the value of argument will be –(π – θ).

Therefore the argument of the given complex number will be as follows.

${\color{DarkBlue} \mathbf{=-\left ( \Pi -\frac{\Pi }{3} \right )=-\frac{2\Pi }{3}}}$

And modulus = 2

Q2.Find the modulus and the argument of the complex number z= –√3+i.

Answer.

We can calculate the argument and modulus by converting the complex number into the polar form.

z = –√3+i = rcosθ + irsinθ

rcosθ = –√3….(i) rsinθ = 1….(ii)

Squaring and adding both equation

r²(cos²θ +sin²θ) = (–√3)² +1²

Since cos²θ +sin²θ =1,so

r² = 4

r =±2

r = 2[ r is the modulus so avoiding the value, –2

Placing this value of r in eq.(i) and (ii),we get

2cosθ = –√3, 2sinθ = 1

As the value of sinθ is (+) and of cosθ (–), so the argument of the complex number will lie in the second quadrant.

Therefore taking 2cosθ = –√3 [ value of sinθ is + in l and ll quadrant]

${\color{DarkBlue} \mathbf{\Rightarrow cos\Theta =-\frac{\sqrt{3}}{2}}}$

${\color{DarkBlue} \mathbf{\Rightarrow cos\Theta =-cos\frac{\Pi }{6}}}$

${\color{DarkBlue} \mathbf{\Rightarrow cos\Theta =cos\left ( \Pi -\frac{\Pi }{6} \right )}}$

${\color{DarkBlue} \mathbf{cos\Theta =cos\frac{5\Pi }{6}}}$

${\color{DarkBlue} \mathbf{\Theta =\frac{5\Pi }{6}}}$

Therefore the modulus and the argument of the complex number will be 2 and 5π/6.

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Q3.Convert the given complex number into the polar form:1 –i.

Answer.In the given complex number 1 – i

a = 1, b = –1, in the argand plane the position of the complex number is (a, b) ,b(imaginary part) is shown along y-axis and a(real part) is shown along x-axis. The distance from origin(0,0) of the complex number r is called modulus and the angle made by r with the x-axis θ is known as argument.

Therefore

${\color{DarkBlue} \mathbf{cos\Theta =\frac{a}{r}, sin\Theta =\frac{b}{r}}}$

a = rcosθ, b = rsinθ

If z = a + ib

Substituting value of a and b by rcosθ and rsinθ

z = rcosθ + irsinθ

r = √(a² + b²) = √[1² +(–1)²] =√2

${\color{DarkBlue} \mathbf{cos\Theta =\frac{1}{\sqrt{2}},sin\Theta =\frac{-1}{\sqrt{2}}}}$

Value of cosθ is + and of sinθ is – in iv quadrants, therefore

Taking sinθ = –1/√2, as the value of cosθ is + in l and lV quadrant

So,

${\color{DarkBlue} \mathbf{sin\Theta =-sin\frac{\Pi }{4}=sin\left ( -\frac{\Pi }{4} \right )}}$

${\color{DarkBlue} \mathbf{\because sin\Theta =sin\left ( -\Theta \right )}}$

It is clear that that argument will be

${\color{DarkBlue} \mathbf{\Theta = -\frac{\Pi }{4}}}$

The polar form of the given complex number will be

${\color{DarkBlue} \mathbf{z = \sqrt{2}cos\left ( -\frac{\Pi }{4} \right )+\sqrt{2}sin\left ( -\frac{\Pi }{4} \right )}}$

Q4.Convert the given complex number into the polar form:–1+i.

Answer.

Polar form of the complex number is = rcosθ +irsinθ

rcosθ = –1….(i)

rsinθ = 1……(ii)

Squaring and adding both equation (i) and (ii)

r²(cos²θ +sin²θ) = 2

r² = 2 (cos²θ +sin²θ =1)

r = ±√2

r is the modulus of the complex number ,so

r = √2

rcosθ = –1, rsinθ = 1

${\color{DarkBlue} \mathbf{cos\Theta =-\frac{1}{\sqrt{2}},sin\Theta =\frac{1}{\sqrt{2}}}}$

Value of cosθ is – and of sinθ is + in l and ll quadrant,so argument will lie in ll.

Since the value of sinθ is + in l and ll quadrant, therefore taking value of cosθ.

${\color{DarkBlue} \mathbf{cos\Theta =\frac{-1 }{\sqrt{2}}=-cos\frac{\Pi }{4}}}$

cos(π –θ)= –cosθ

${\color{DarkBlue} \mathbf{-cos\frac{\Pi }{4}=cos\left ( \Pi -\frac{\Pi }{4} \right )=cos\frac{3\Pi }{4}}}$

${\color{DarkBlue} \mathbf{cos\Theta =cos\frac{3\Pi }{4}}}$

${\color{DarkBlue} \mathbf{\Theta =\frac{3\Pi }{4}}}$

Placing the value of θ and r in z = rcosθ + irsinθ.

The polar form of the complex number z will be as follows

z = rcosθ + irsinθ

${\color{DarkBlue} \mathbf{z=\sqrt{2}cos\frac{3\Pi }{4}+i\sqrt{2}sin\frac{3\Pi }{4}}}$

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Q5.Convert the given complex number into the polar form:–1–i.

Answer. z = –1–i

z = rcosθ + irsinθ

rcosθ = –1….(i)

rsinθ = –1…..(ii)

Squaring and adding both eq.(i) and (ii)

r²(cos²θ + sinθ) = 2

r = √2 [cos²θ + sin²θ]

√2cosθ = –1, √2sinθ = –1

${\color{DarkBlue} \mathbf{cos\Theta =\frac{-1}{\sqrt{2}},sin\Theta =-\frac{1}{\sqrt{2}}}}$

The value of cosθ and sinθ is – in lll quadrant, taking the value of sinθ or cosθ

${\color{DarkBlue} \mathbf{cos\Theta =-cos\frac{\Pi }{4}=cos\left ( \Pi -\frac{\Pi }{4} \right )}}$

${\color{DarkBlue} \mathbf{cos\Theta =cos\frac{3\Pi }{4}}}$

${\color{DarkBlue} \mathbf{\Theta =\frac{3\Pi }{4}}}$

${\color{DarkBlue} \mathbf{z = \sqrt{2}cos\frac{3\Pi }{4}+i\sqrt{2}sin\frac{3\Pi }{4}}}$

Q6.Convert the given complex number into the polar form:-3.

Answer. The given complex number is =–3

z= –3 + 0.i

The polar form of the complex number is written as follows

z = rcosθ +irsinθ

rcosθ = –3….(i) rsinθ =0…..(ii)

Squaring and adding both equation

r²(cos²θ + sin²θ) =9

r =±3 [cos²θ + sin²θ]

r is the modulus of the complex number, so avoiding – sign.

r = 3.

3cosθ = –3, 3sinθ = 0

cosθ = –1, sinθ = 0

${\color{DarkBlue} \mathbf{cos\Theta =-cos0, sin\Theta =sin0}}$

Value of cosθ is – and of sinθ is+ in ll quadrant

Value of sinθ is + in l and ll quadrant ,so taking the value of cosθ

cosθ = –cos0 =cos(π – 0)

cosθ = cosπ

θ = π

Therefore the polar form of the given complex number will be

z = 3cosπ + i3sinπ

Q7.Convert the given complex number into the polar form:√3 +i.

Answer. The given complex number is z = √3 +i

The polar form of the complex number is written as follows

z = rcosθ + irsinθ

rcosθ = √3….(i) rsinθ = 1…..(ii)

Squaring and adding both equations

r²(cos²θ + sin²θ) = 4 [cos²θ + sin²θ]

r = ±2

r is the modulus of the complex number, so avoiding – sign.

r = 2

2cosθ = √3, 2sinθ = 1

${\color{DarkBlue} \mathbf{cos\Theta =\frac{\sqrt{3}}{2}, sin\Theta =\frac{1}{2}}}$

The value of sinθ and cosθ is + in l quadrant,so taking either of the equation

${\color{DarkBlue} \mathbf{cos\Theta =cos\frac{\Pi }{6}}}$

${\color{DarkBlue} \mathbf{\Theta =\frac{\Pi }{6}}}$

The polar form of the complex number will be as follows

${\color{DarkBlue} \mathbf{z =2cos\frac{\Pi }{6}+i2sin\frac{\Pi }{6}}}$

Q8.Convert the given complex number into the polar form:i

Answer.

The given complex number is =i

z = 0 + i

The polar form of the complex number is written as follows

z = rcosθ + irsinθ

rcosθ = 0….(i) rsinθ = 1…..(ii)

Squaring and adding both equations

r²(cos²θ + sin²θ) = [cos²θ + sin²θ]

r ±1

r is the modulus of the complex number, so avoiding – sign.

r = 1

cosθ = 0, sinθ =1

cosθ and sinθ both are + in l quadrant , so taking either of the equations

cosθ =0

${\color{DarkBlue} \mathbf{cos\Theta = cos\frac{\Pi }{2}}}$

${\color{DarkBlue} \mathbf{\Theta = \frac{\Pi }{2}}}$

The polar form of the complex number will be as follows

${\color{DarkBlue} \mathbf{z = cos\frac{\Pi }{2}+isin\frac{\Pi }{2}}}$

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

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CBSE Class 10 -Question paper of maths 2020 with solutions

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NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

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

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

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

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

NCERT solutions of maths 11 th class Chapter 5 Complex number and Quadratic Equation.

NCERT solutions of class 11 maths

CBSE Class 11-Question paper of maths 2015

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Exercise 5.1

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

NCERT Solutions for class 9 science

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NCERT Solutions for Class 10 Science

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NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

NCERT solutions of maths 11 th class Chapter 5 Complex number and Quadratic Equation.

NCERT solutions of class 11 maths

CBSE Class 11-Question paper of maths 2015

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

Exercise 5.1

NCERT Solutions for Class 11 Maths Chapter 5 Complex numbers and Quadratic Equations

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

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