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NCERT Solutions for Exercise 5.6 Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Exercise 5.6 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Ramraj Saini | Updated on Dec 03, 2023 05:10 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6

NCERT Solutions for Exercise 5.6 Class 12 Maths Chapter 5 Continuity and Differentiability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In the previous exercises of this NCERT syllabus Class 12 chapter, you have learned about the differentiation of function y = f(x) w.r.t the x, but in exercise 5.6 Class 12 Maths, the function y = f(a) and x = f(a) is given in the parametric form and you need to find the differentiation of function y w.r.t the x. If you have solved the previous exercises by yourself, you won't find any difficulty while solving Class 12th Maths chapter 5 exercise 5.6. You just need to find the differentiation of y and x w.r.t. to its parameter. The differentiation of y w.r.t. x is the same as the ratio of dy/da and dx/da i.e dy/dx = (dy/da)/(dx/da). You can go through the NCERT solutions for Class 12 Maths chapter 5 exercise 5.6 to get detailed solutions to these problems.

12th class Maths exercise 5.6 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6

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Continuity and Differentiability Exercise: 5.6

Question:1 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .

x = 2at^2, y = at^4

Answer:

Given equations are
x = 2at^2, y = at^4
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(2at^2)}{dt}= 4at
Similarly,
\frac{dy}{dt}=\frac{d(at^4)}{dt}= 4at^3
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{4at^3}{4at} = t^2
Therefore, the answer is \frac{dy}{dx}= t^2

Question:2 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx .

x= a \cos \theta , y = b \cos \theta

Answer:

Given equations are
x= a \cos \theta , y = b \cos \theta
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a\cos \theta)}{d\theta}= -a\sin \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(b\cos \theta)}{d\theta}= -b\sin \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-b\sin \theta}{-a\sin \theta} = \frac{b}{a}
Therefore, answer is \frac{dy}{dx}= \frac{b}{a}

Question:3 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx . x = \sin t , y = \cos 2 t

Answer:

Given equations are
x = \sin t , y = \cos 2 t
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(\sin t)}{dt}= \cos t
Similarly,
\frac{dy}{dt}=\frac{d(\cos 2t)}{dt}= -2\sin 2t = -4\sin t \cos t \ \ \ \ \ (\because \sin 2x = \sin x\cos x)
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{-4\sin t \cos t }{\cos t} = -4\sin t
Therefore, the answer is \frac{dy}{dx} = -4\sin t

Question:4 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx

x = 4t , y = 4/t

Answer:

Given equations are
x = 4t , y = 4/t
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(4 t)}{dt}= 4
Similarly,
\frac{dy}{dt}=\frac{d(\frac{4}{t})}{dt}= \frac{-4}{t^2}
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{ \frac{-4}{t^2} }{4} = \frac{-1}{t^2}
Therefore, the answer is \frac{dy}{dx} = \frac{-1}{t^2}

Question:5 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta

Answer:

Given equations are
x = \cos \theta - \cos 2\theta , y = \sin \theta - \sin 2 \theta
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(\cos \theta-\cos 2\theta)}{d\theta}= -\sin \theta -(-2\sin 2\theta) = 2\sin 2\theta - \sin \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(\sin \theta - \sin 2\theta)}{d\theta}= \cos \theta -2\cos2 \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}
Therefore, answer is \frac{dy}{dx}= \frac{\cos \theta -2\cos 2 \theta}{2\sin2\theta-\sin \theta}


Question:6 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )

Answer:

Given equations are
x = a ( \theta - \sin \theta ) , y = a ( 1+ \cos \theta )
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a(\theta- \sin \theta))}{d\theta}= a(1-\cos \theta)
Similarly,
\frac{dy}{d\theta}=\frac{d(a(1+\cos \theta))}{d\theta}=-a\sin \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{-a\sin \theta}{a(1-\cos \theta)} = \frac{-\sin }{1-\cos \theta} =- \cot \frac{\theta}{2} \ \ \ \ \ \ \ (\cot \frac{x}{2}=\frac{\sin x}{1-\cos x})
Therefore, the answer is \frac{dy}{dx}=-\cot \frac{\theta}{2}

Question:7 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}

Answer:

Given equations are
x = \frac{\sin ^3 t }{\sqrt {\cos 2t }} , y = \frac{\cos ^3 t }{\sqrt {\cos 2t }}
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(\frac{\sin ^3 t }{\sqrt {\cos 2t }})}{dt}=\frac{\sqrt{\cos 2t}.\frac{d(\sin^3t)}{dt}-\sin^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2} =\frac{3\sin^2 t\cos t.\sqrt{\cos 2t}-\sin^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{{\cos 2t}}
=\frac{3\sin^2t\cos t . \cos 2t+sin^3t\sin 2t}{\cos2t\sqrt{\cos2t}}
=\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}} \ \ \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)
Similarly,
\frac{dy}{dt}=\frac{d( \frac{\cos ^3 t }{\sqrt {\cos 2t }})}{dt}=\frac{\sqrt{\cos 2t}.\frac{d(\cos^3t)}{dt}-\cos^3t.\frac{d(\sqrt{\cos2t})}{dt}}{(\sqrt{\cos2t})^2} =\frac{3\cos^2 t(-\sin t).\sqrt{\cos 2t}-\cos^3t.\frac{1}{2\sqrt{\cos 2t}}.(-2\sin 2t)}{(\sqrt{\cos 2t})^2}
=\frac{-3\cos^2t\sin t\cos2t+\cos^3t\sin 2t}{\cos2t\sqrt{\cos2t}}
=\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}}
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{\frac{ \sin2t\cos^3t(1-3\tan t \cot 2t)}{\cos2t\sqrt{\cos2t}} }{\frac{\sin^3t\sin2t(3\cot t \cot 2t+1)}{\cos2t\sqrt{\cos2t}}} = \frac{\cot^3t(1-3\tan t \cot 2t)}{(3\cot t \cot 2t+1)}
= \frac{\cos^3t(1-3.\frac{\sin t}{\cos t}.\frac{\cos2t}{\sin 2t})}{\sin^3t(3.\frac{\cos t}{\sin t}.\frac{\cos 2t}{\sin 2t}+1)} = \frac{\cos^2t(\cos t\sin2t -3\sin t \cos 2t)}{\sin^2t(3\cos t \cos2t+\sin t \sin 2t)}
=\frac{\cos^2t(\cos t .2\sin t \cos t - 3\sin t (2\cos^2t-1))}{\sin^2t(3\cos t(1-2\sin^2 2t)+\sin t.2\sin t \cos t)}
(\because \sin 2x = 2\sin x\cos x \ and \ \cos 2x = 2\cos^2x-1 \ and \ \cos 2x = 1-2\sin^2x)
=\frac{\cos^2t(2\sin t\cos^2 t-6\sin t\cos^2t+3\sin t)}{\sin^2t(3\cos t-6\cos t \sin^2t+2\sin^2\cos t)}\\=\frac{sint cost(-4cos^3t+3cost)}{sintcost(3sint-4sin^3t)}

\frac{dy}{dx} = \frac{-4\cos^3t+3\cos t}{3\sin t -4\sin^3 t}= \frac{-\cos 3t}{\sin 3t} = -\cot 3t \left ( \because \sin3t = 3\sin t-4\sin^3t \\ \ and \ \cos3t = 4\cos^3t - 3\cos t \right )

Therefore, the answer is \frac{dy}{dx} = -\cot 3t

Question:8 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a ( \cos t + \log \tan t/2 ),y = a \sin t

Answer:

Given equations are
x = a ( \cos t + \log \tan \frac{t}{2} ),y = a \sin t
Now, differentiate both w.r.t t
We get,
\frac{dx}{dt}=\frac{d(a ( \cos t + \log \tan \frac{t}{2} ))}{dt}= a(-\sin t + \frac{1}{\tan\frac{t}{2}}.\sec^2\frac{t}{2}.\frac{1}{2})
= a(-\sin t+\frac{1}{2}.\frac{\cos \frac{t}{2}}{\sin\frac{t}{2}}.\frac{1}{\cos^2\frac{t}{2}}) = a(-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}})
=a(-\sin t+\frac{1}{\sin 2.\frac{t}{2}} ) = a(\frac{-\sin^2t+1}{\sin t})= a(\frac{\cos^2t}{\sin t})
Similarly,
\frac{dy}{dt}=\frac{d(a\sin t)}{dt}= a\cos t
Now, \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{a \cos t }{ a(\frac{\cos^2t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t
Therefore, the answer is \frac{dy}{dx} = \tan t

Question:9 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a \sec \theta , y = b \ tan \theta

Answer:

Given equations are
x = a \sec \theta , y = b \ tan \theta
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a\sec \theta)}{d\theta}= a\sec \theta \tan \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(b\tan \theta)}{d\theta}= b\sec^2 \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{b\sec^2 \theta}{a\sec\theta\tan \theta} = \frac{b\sec\theta}{a\tan \theta}= \frac{b\frac{1}{\cos\theta}}{a\frac{\sin \theta}{\cos \theta}} = \frac{b }{a\sin \theta} = \frac{b cosec \theta}{a}
Therefore, the answer is \frac{dy}{dx} = \frac{b cosec \theta}{a}

Question:10 If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )

Answer:

Given equations are
x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )
Now, differentiate both w.r.t \theta
We get,
\frac{dx}{d\theta}=\frac{d(a(\cos \theta+ \theta\sin \theta))}{d\theta}= a(-\sin \theta+\sin \theta+ \theta\cos \theta)= a \theta\cos \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(a(\sin \theta- \theta\cos \theta))}{d\theta}= a(\cos \theta-\cos \theta+ \theta\sin \theta) = a \theta\sin \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{a \theta\sin \theta}{a \theta\cos \theta} = \tan \theta
Therefore, the answer is \frac{dy}{dx}= \tan \theta

Question:11 If x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}, show that dy/dx = - y /x

Answer:

Given equations are
x = \sqrt {a ^{\sin ^{-1}t}} , y = \sqrt { a ^{ \cos ^{-1}t}}

xy=\sqrt{a^{sin^{-1}t+cos^{-1}t}}\\since\ sin^{-1}x+cos^{-1}x=\frac{\pi}{2}\\xy=a^{\frac{\pi}{2}}=constant=c

differentiating with respect to x

x\frac{dy}{dx}+y=0\\\frac{dy}{dx}=\frac{-y}{x}

More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6:-

In Class 12th Maths chapter 5 exercise 5.6, there are 11 questions in which you need to find the differentiation of y w.r.t x where x and y are given in the parametric form and you can't use the elimination of parameter method to these differentiation. The three examples given in the NCERT book prior to Class 12 Maths ch 5 ex 5.6 are related to differentiation in parametric form also. You can solve these examples to get more clarity.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6:-

  • NCERT solutions for Class 12 Maths Chapter 5 exercise 5.6 are beneficial not only for this exercise but also to learn differentiation of functions in parametric function.
  • Class 12 Maths chapter 5 exercise 5.6 solutions can be used as the revision notes
  • Class 12 Maths ch 5 ex 5.6 solutions are designed in a descriptive manner so you use these solution to understand the concept easily.
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Key Features Of NCERT Solutions for Exercise 5.6 Class 12 Maths Chapter 5

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 5.6 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 5.6, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 5.6 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 5.6 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 5.6 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 5.6 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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Frequently Asked Questions (FAQs)

1. If x = sin (t) then find differentiation of x w.r.t t ?

Given x = sin (t)

dx/dt = cos(t)

2. If y = sin (t) then find the differentiation of y w.r.t x. ?

y=sin(t)

As y is not dependent on the x.

dy/dx = 0

3. Which book is best book for NCERT Class 12 Maths ?

NCERT book is best for CBSE Class 12 Maths. You don't need other books for the CBSE board exams.

4. Which book should i refer for Maths JEE main ?

Mathematics book by M.L. khana is considered to be good book for the Maths JEE main.

5. Do I need to but CBSE chapter wise solution book Class 12 Maths ?

You don't need to buy any solution book for CBSE Class 12 Maths. Can follow NCERT book, solutions, NCERT exemplar and previous year solutions.

6. Can i get chapter-wise solutions for Class 12 Maths ?
7. Can I get marks distribution for Class 12 Maths ?
8. Can I get NCERT solutions for Class 11 Maths ?

Here you will get NCERT Solutions for Class 11 Maths. Solutions to each chapter are available with all the necessary steps.

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i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

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kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
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Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

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kumardurgesh1802 10 July,2024
I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

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satyamercedes0626 23 July,2024
I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

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Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.


Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.


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Submit list of candidates for CBSE board exams 2025 correctly, on time: Schools reminded
September 12, 2024 - 01:16 PM IST
CBSE sample paper 2024-25, marking scheme out for skill education subjects
September 02, 2024 - 12:13 PM IST
CBSE to overhaul internal assessment system, develop framework with British Council
August 28, 2024 - 04:58 PM IST
65 lakh students didn't clear Class 10, 12 board exams last year: Education ministry
August 21, 2024 - 10:24 PM IST
CBSE Compartment Result 2024: Marks verification, answer book photocopy, revaluation dates out
August 06, 2024 - 12:37 PM IST
CBSE Class 12 compartment results 2024 declared at cbseresults.nic.in; 29.78% pass
August 02, 2024 - 04:46 PM IST
CBSE best, Kerala board worst in curriculum, infrastructure: NCERT PARAKH report
July 30, 2024 - 04:48 PM IST

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