NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise

The NCERT maths solutions of chapter 3 miscellaneous exercise containing 10 questions is the extract of the lesson ‘Trigonometry’ . The questions of this exercise are very important for the purpose of your academic exams and entrance exams of engineering.The solutions are explained discretely in order to clear the doubts generally faced by 11 th class students .All questions are explained readily by the teacher expertise in maths. We hope that these solutions by Future Study Point will help you in your preparations of the exam. After you go through each solution write your comment in the form of suggestions, doubts and your requirement.

NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise

Exercise 3.1 and Exercise 3.2

Exercise 3.3

Exercise 3.4

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

Class 11 th CBSE NCERT Maths Solution of Chapter 3 Miscellaneous Exercise

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Q1. Prove that.

${\color{DarkBlue} \mathbf{2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+cos\frac{3\Pi }{13}+cos\frac{5\Pi }{13}=0}}$

Answer.

L.H.S

${\color{DarkBlue} \mathbf{=2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+cos\frac{3\Pi }{13}+cos\frac{5\Pi }{13}}}$

cosA + cosB =2cos(A+B)/2.cos(A–B)/2

${\color{DarkBlue} \mathbf{{ \mathbf{=2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+2cos\frac{\frac{3\Pi }{13}+\frac{5\Pi }{13}}{2}cos\frac{\frac{3\Pi }{13}-\frac{5\Pi }{13}}{2}}}}}$

${\color{DarkBlue} { =\mathbf{2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+2cos\frac{4\Pi }{13}.cos\left ( -\frac{\Pi }{13} \right )}}}$

As we know cos(–x) = cosx

${\color{DarkBlue} { =\mathbf{2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+2cos\frac{4\Pi }{13}.cos\left ( \frac{\Pi }{13} \right )}}}$

${\color{DarkBlue} \mathbf{=2cos\frac{\Pi }{13}(cos\frac{9\Pi }{13}+cos\frac{4\Pi }{13})}}$

${\color{DarkBlue} =\mathbf{2cos\frac{\Pi }{13}\left ( 2cos\frac{\frac{9\Pi }{13}+\frac{4\Pi }{13}}{2}.cos\frac{\frac{9\Pi }{13}-\frac{4\Pi }{13}}{2} \right )}}$

${\color{DarkBlue} =\mathbf{2cos\frac{\Pi }{13}\left ( 2cos\frac{\Pi }{2}.cos\frac{ 5\Pi }{26} \right )}}$

${\color{DarkBlue} \mathbf{\because cos\frac{\Pi }{2}=0}}$

${\color{DarkBlue} \mathbf{=2cos\frac{\Pi }{13}\times 0\times cos\frac{5\Pi }{26}}}$

= 0 = R.H.S

Q2. Prove that

(sin3x + sinx)sinx + (cos3x – cosx)cosx = 0

∵sinA + sinB = 2sin(A+B)/2.cos(A–B)/2

∵cosA –cosB = –2sin(A+B)/2.sin(A–B)/2

${\color{DarkBlue} \mathbf{=2sin\frac{3x+x}{2}.cos\frac{3x-x}{2}.sinx-2sin\frac{3x+x}{2}sin\frac{3x-x}{2}.cosx}}$

=2sin2x.cosx.sinx – 2sin2x.sinx.cosx = 0= R.H.S

Q3.Prove that

${\color{DarkBlue} \mathbf{\left ( cosx+cosy \right )^{2}+\left ( sinx-siny \right )^{2}=4cos^{2}\frac{x+y}{2}}}$

Answer.

L.H.S

${\color{DarkBlue} =\mathbf{\left ( cosx+cosy \right )^{2}+\left ( sinx-siny \right )^{2}}}$

=cos²x + cos²y + 2cosx.cosy + sin²x + sin²y –2sinx.siny

=(cos²x + sin²x )+ (cos²y + sin²y) + 2(cosx.cosy – sinx.siny)

=1 + 1 + 2(cosx.cosy – sinx.siny)

=2 + 2(cosx.cosy – sinx.siny)

∵cos(A +B) = cosA.cosB – sinA.sinB

=2 + 2cos(x + y)

=2[1 + cos(x + y)]

∵ cos2A = 2cos²A – 1

⇒1+cos2A = 2cos²A

${\color{DarkBlue} \mathbf{\therefore 1+cos(x+y)=2cos^{2}\left ( \frac{x+y}{2} \right )}}$

${\color{DarkBlue} \mathbf{=4cos^{2}\left ( \frac{x-y}{2} \right )}\mathbf{=R.H.S}}$

Q4.Prove that

${\color{DarkGreen} { \mathbf{\left ( cosx -cosy \right )^{2}+\left ( sinx-siny \right )^{2}=4sin^{2}\left ( \frac{x-y}{2} \right )}}}$

Answer.

L.H.S (cosx – cosy)² + (sinx – siny)²

=cos²x + cos²y – 2cosx.cosy + sin²x + sin²y –2sinx.siny

=(cos²x + sin²x) + (cos²y + sin²y) –2(cosx.cosy + sinx.siny)

= 1 + 1 – 2cos(x – y) [∵ cos(A – B) = cosA.cosB + sinA.sinB]

= 2 – 2cos(x– y)

=2[1 – cos(x –y)]

∵ cos2A = 1 – 2sin²A⇒ 1 – cos2A = 2sin²A

${\color{DarkBlue} \mathbf{\therefore 1- cos\left ( x-y \right )= 2sin^{2}\left ( \frac{x-y}{2} \right )}}$

${\color{DarkBlue} \mathbf{=2\left [ 2sin^{2}\frac{x-y}{2} \right ]}}$

${\color{DarkBlue} \mathbf{=4sin^{2}\left ( \frac{x-y}{2} \right )}}$

= R.H.S

Q5. Prove that sinx + sin3x + sin5x + sin7x = 4cosx.cos2x.sin4x

Answer.

L.H.S

=sinx + sin3x + sin5x + sin7x

= (sinx + sin3x )+ (sin5x + sin7x)

∵ sinA + sinB = 2sin(A + B)/2 .cos(A –B)/2

${\color{DarkBlue} \mathbf{=2sin\frac{x+3x}{2}.cos\frac{x-3x}{2}+2sin\frac{5x+7x}{2}.cos\frac{5x-7x}{2}}}$

=2sin2x.cos(–x) + 2sin6x.cos(–x)

∵ cos(–x) = cosx

=2sin2x.cosx + 2sin6x.cosx

=2cosx(sin2x + sin6x)

∵ sinA + sinB = 2sin(A + B)/2 .cos(A –B)/2

${\color{DarkBlue} \mathbf{=2cosx\left [ 2sin\frac{2x+6x}{2}.cos\frac{2x-6x}{2} \right ]}}$

= 4cosx.sin4x.cos(–2x)

∵ cos(–θ) = cosθ

= 4cosx.cos2x.sin4x = R.H.S

Q6. Prove that

${\color{DarkGreen} \mathbf{\frac{\left ( sin7x+sin5x \right )+\left ( sin9x +sin3x \right )}{\left ( cos7x+cos5x \right )+\left ( cos9x+cos3x \right )}=tan6x}}$

Answer.

L.H.S

${\color{DarkBlue} ={ \mathbf{\frac{\left ( sin7x+sin5x \right )+\left ( sin9x +sin3x \right )}{\left ( cos7x+cos5x \right )+\left ( cos9x+cos3x \right )}}}}$

${\color{DarkBlue} \mathbf{\because sinA +sinB = 2sin\frac{A+B}{2}.cos\frac{A-B}{2}}}$

${\color{DarkBlue} \mathbf{\because cosA +cosB = 2cos\frac{A+B}{2}.cos\frac{A-B}{2}}}$

${\color{DarkBlue} \mathbf{=\frac{\left [ 2sin\left ( \frac{7x+5x}{2}).cos(\frac{7x-5x}{2} \right ) \right ]+\left [ 2sin(\frac{9x+3x}{2}).cos(\frac{9x-3x}{2}) \right ]}{\left [ 2cos\left ( \frac{7x+5x}{2}).cos(\frac{7x-5x}{2} \right ) \right ]+\left [ 2cos\left ( \frac{9x+3x}{2}.cos(\frac{9x-3x}{2} \right ) \right ] }}}$

${\color{DarkBlue} \mathbf{=\frac{2sin6x.cosx+2sin6x.cos3x}{2cos6x.cosx+2cos6x.cos3x}}}$

${\color{DarkBlue} \mathbf{=\frac{2sin6x(cosx +cos3x)}{2cos6x(cosx +cos3x)}}}$

${\color{DarkBlue} \mathbf{=\frac{2sin6x}{2cos6x}}}$

= tan6x = R.H.S

Q7. Prove that

${\color{DarkBlue} \mathbf{sin3x + sin2x-sinx = 4sinx.cos\frac{x}{2}.cos\frac{3x}{2}}}$

Answer. L.H.S = sin3x + (sin2x – sinx)

${\color{DarkBlue} \mathbf{\because sinA-sinB = 2cos\frac{A+B}{2}.sin\frac{A-B}{2}}}$

${\color{DarkBlue}= \mathbf{sin3x+ 2cos\frac{2x+x}{2}.sin\frac{2x-x}{2}}}$

${\color{DarkBlue}= \mathbf{sin3x+ 2cos\frac{3x}{2}.sin\frac{x}{2}}}$

∵sin2A = 2sinA.cosA

${\color{DarkBlue} {\therefore \mathbf{sin3x=}\mathbf{2sin\frac{3x}{2}.cos\frac{3x}{2}}}}$

${\color{DarkBlue}= \mathbf{2sin\frac{3x}{2}.cos\frac{3x}{2}+ 2cos\frac{3x}{2}.sin\frac{x}{2}}}$

${\color{DarkBlue} \mathbf{=2cos\frac{3x}{2}\left ( sin\frac{3x}{2}+sin\frac{x}{2} \right )}}$

${\color{DarkBlue} \mathbf{\because sinA +sinB = 2sin\frac{A+B}{2}.cos\frac{A-B}{2}}}$

${\color{DarkBlue} \mathbf{ = 2cos\frac{3x}{2}.2sin\frac{(\frac{3x}{2}+\frac{x}{2})}{2}.cos\frac{(\frac{3x}{2}-\frac{x}{2})}{2}}}$

${\color{DarkBlue} \mathbf{=4sinx.cos\frac{x}{2}.cos\frac{3x}{2}= R.H.S}}$

${\color{DarkGreen} {\mathbf{Q8.} \mathbf{Find \; sin\frac{x}{2},cos\frac{x}{2}\; and\; \; tan\frac{x}{2},if\; tan\frac{-4}{3},x\; in \; quadrant\; ll.}}}$

Answer.

Since x is in 2nd quadrant, so the value of x would be estimated as follows

${\color{DarkBlue} \mathbf{\frac{\Pi }{2}< x< \Pi }}$

Dividing all the terms of this equation of inequality

${\color{DarkBlue} \mathbf{\frac{\Pi }{4}< \frac{x}{2}< \frac{\Pi }{2}}}$

The value of ${\color{DarkBlue} \mathbf{sin\frac{x}{2},cos\frac{x}{2}\; and \; tan\frac{x}{2}}}$ will lye in l quadrant.

We are given that

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

${\color{DarkBlue} \because \mathbf{tan2x=\frac{2tanx}{1-tan^{2}x}}}$

${\color{DarkBlue} \mathbf{\therefore tanx = \frac{2tan\frac{x}{2}}{1-tan^{2}\frac{x}{2}}}}$

Placing the value of tanx =(–4/5)

${\color{DarkBlue} \mathbf{\Rightarrow \frac{-4}{3}=\frac{2tan\frac{x}{2}}{1-tan^{2}\frac{x}{2}}}}$

${\color{DarkBlue} \mathbf{\Rightarrow 4tan^{2}\frac{x}{2 }-4=6tan\frac{x}{2}}}$

${\color{DarkBlue} \mathbf{\Rightarrow 4tan^{2}\frac{x}{2}-6tan\frac{x}{2}-4 =0}}$

${\color{DarkBlue} \mathbf{\Rightarrow 4tan^{2}\frac{x}{2}-8tan\frac{x}{2}+tan\frac{x}{2}-4=0}}$

${\color{DarkBlue} \mathbf{\Rightarrow 4tan\frac{x}{2}\left ( tan\frac{x}{2}-2 \right )+2\left ( tan\frac{x}{2}-2 \right )=0}}$

${\color{DarkBlue} \mathbf{\Rightarrow \left ( tan\frac{x}{2}-2 \right )\left ( 4tan\frac{x}{2}+2 \right )=0}}$

${\color{DarkBlue} \mathbf{\Rightarrow tan\frac{x}{2 }=2,tan\frac{x}{2}=-\frac{1}{2}}}$

As we have shown above that x/2 lyes in 1st quadrant,tan(x/2) will be positive

${\color{DarkBlue} \mathbf{\therefore tan\frac{x}{2}=2}}$

${\color{DarkBlue} \mathbf{sec\frac{x}{2}=\pm \sqrt{1+tan^{2}\frac{x}{2}}=\pm \sqrt{1+2^{2}}=\pm \sqrt{5}}}$

x/2 lies in 1 st quadrant so sec(x/2) =√5

${\color{DarkBlue} \mathbf{\therefore cos\frac{x}{2}=\frac{1}{\sqrt{5}}}}$

sinx/2 = ±√(1–cos²x/2)

x/2 lies in 1st quadrant so sin(x/2) will be positive

${\color{DarkBlue} \mathbf{sin\frac{x}{2}=\sqrt{1-cos^{2}\frac{x}{2}}=\sqrt{1-\left ( \frac{1}{\sqrt{5}} \right )^{2}}}}$

${\color{DarkBlue} \mathbf{sin\frac{x}{2}= \frac{2}{\sqrt{5}}}}$

${\color{DarkBlue} \mathbf{Therefore \; \; value\; of\; sin\frac{x}{2},cos\frac{x}{2}\; and\; \; tan\frac{x}{2}\; will\; be\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\; and\; 2.}}$

${\color{DarkGreen} {\mathbf{Q9.} \mathbf{Find \; sin\frac{x}{2} , cos\frac{x}{2}\; and\; tan\frac{x}{2}\; for \; cosx=-\frac{1}{3},x\; in\; quadrant\; lll.}}}$

Answer. x is in lll quadrant, so its value would be estimated as follows

π < x < 3π/2, dividing this eq. by 2,we shall have

π/2 < x/2<3π/4, which reveals that x/2 lyes in ll quadrant .

∵cos2x = 2cos²x –1

${\color{DarkBlue} \mathbf{\therefore cosx =2cos^{2}\frac{x}{2}-1}}$

${\color{DarkBlue} \mathbf{\Rightarrow cos\frac{x}{2}=\pm \sqrt{\frac{1-\frac{1}{3}}{2}}}}$

${\color{DarkBlue} \Rightarrow \mathbf{cos\frac{x}{2}=\pm \frac{1}{\sqrt{3}}}}$

As we have shown above that x/2 lies in ll quadrant and it is known to us ,the value of cos is negative in 2nd quadrant.

${\color{DarkBlue} \therefore \mathbf{cos\frac{x}{2}=-\frac{1}{\sqrt{3}}}}$

${\color{DarkBlue} \mathbf{sin\frac{x}{2}=\pm \sqrt{1-cos^{2}\frac{x}{2}}=\pm \sqrt{1-\left ( -\frac{1}{\sqrt{3}} \right )^{2}}=\pm \sqrt{\frac{2}{3}}}}$

As we have shown above that x/2 lies in ll quadrant and it is known to us ,the value of sin x/2 is positive in 2nd quadrant.

${\color{DarkBlue} \mathbf{\therefore sin\frac{x}{2}=\sqrt{\frac{2}{3}}}}$

${\color{DarkBlue} \mathbf{\because tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}}}$

${\color{DarkBlue} \mathbf{\Rightarrow tan\frac{x}{2}= \frac{\sqrt{\frac{2}{3}}}{-\frac{1}{\sqrt{3}}}}}$

${\color{DarkBlue} \Rightarrow \mathbf{tan\frac{x}{2}=-\sqrt{2}}}$

${\color{DarkBlue} Therefore\; \mathbf{The \; value \; of\; sin\frac{x}{2},\; cos\frac{x}{2}\; \; and\; \; tan\frac{x}{2} are\; {-\sqrt{\frac{2}{3}}},-\frac{1}{\sqrt{3}}\; and -\sqrt{2}.}}$

${\color{DarkBlue} \mathbf{Q10.}\mathbf{Find \; sin\frac{x}{2},cos\frac{x}{2}\; and\; \; tan\frac{x}{2}\; for\; sinx=\frac{1}{4},x\; in \; quadrant\; ll.}}$

Answer.

Since ,x is in ll quadrant ,its value would be estimated as follows

π/2 < x < π

⇒π/4 < x/2 < π/2, implies that x/2 lies in l st quadrant, which shows that all sinx/2, cos x/2 and tan x/2 will lye in 1st quadrant.

${\color{DarkBlue} \mathbf{cosx=\pm \sqrt{1-sin^{2}x}=\pm \sqrt{1-\left ( \frac{1}{4} \right )^{2}}=\pm \frac{\sqrt{15}}{4}}}$

x is in ll quadrant ,so cosx will be negative

${\color{DarkBlue} \therefore \mathbf{cosx =-\frac{\sqrt{15}}{4}}}$

cos2x = 2cos²x – 1

${\color{DarkBlue} \mathbf{\therefore cosx = \pm \sqrt{}\frac{1+cos2x}{2}}}$

${\color{DarkBlue} \mathbf{\Rightarrow cos\frac{x}{2}=\pm \sqrt{\frac{1+cosx}{2}}}}$

${\color{DarkBlue} \mathbf{\Rightarrow cos\frac{x}{2}=\pm \sqrt{\frac{1+\frac{\sqrt{15}}{4}}{2}}}}$

We have shown that x/2 lyes in l quadrant and value of cos is positive in l quadrant.

${\color{DarkBlue} \mathbf{\therefore cos\frac{x}{2}=\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{sin\frac{x}{2} =\pm \sqrt{1-cos^{2}\frac{x}{2}}}}$

As it is already known to us that x/2 is in l st quadrant and value of sin x/2 is postitive in lst quadrant.

${\color{DarkBlue} \mathbf{\Rightarrow sin\frac{x}{2}=\sqrt{1-\left ( \frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}} \right )^{2}}}}$

${\color{DarkBlue} \mathbf{\Rightarrow sin\frac{x}{2}= \frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{tan\frac{x}{2} =\frac{sin\frac{x}{2}}{cos\frac{x}{2}}}}$

${\color{DarkBlue} \mathbf{tan\frac{x}{2}= \frac{\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}}{\frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}}}}}$

${\color{DarkBlue} \mathbf{tan\frac{x}{2}=\frac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}}}}$

Multiplying the denominator and numerator by the conjugate of denominator ${\color{DarkBlue} \mathbf{\sqrt{4+\sqrt{15}}}}$ .

${\color{DarkBlue} \mathbf{\Rightarrow \frac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}}\times \frac{\sqrt{4+\sqrt{15}}}{\sqrt{4+\sqrt{15}}}}}$

${\color{DarkBlue} \mathbf{=4 +\sqrt{15}}}$

${\color{DarkBlue} \mathbf{The \; value \; of\; sin\frac{x}{2},cos\frac{x}{2}\; and \; \; tan\frac{x}{2}\; are\; \frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}},\frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}}\; \; and\; 4+\sqrt{15}.}}$

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

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

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