NCERT Solutions for Exercise 5.6 Class 12 Maths Chapter 5

Continuity is like a peaceful river flowing without any gaps, and Differentiability is when each ripple of the river follows a certain pattern smoothly and precisely. Oftentimes, we come across some functions which can not simply be represented as $y = f (x)$ , as both the variables are expressed in the form of a third variable. These types of form of functions are generally known as functions in parametric form. In exercise 5.6 of the chapter Continuity and Differentiability, we will learn about the derivatives of functions in parametric form. These concepts will help the students differentiate expressions involving variables $x$ and $y$ given in the form of a third variable or parameter. This article on the NCERT Solutions for Exercise 5.5 Class 12 Maths Chapter 5 - Continuity and Differentiability offers clear and step-by-step solutions for the problems given in the exercise, so that the students can develop their problem-solving ability and understand the logic behind these solutions. For syllabus, notes, and PDF, refer to this link: NCERT.

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Class 12 Maths Chapter 5 Exercise 5.6 Solutions: Download PDF

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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 .

$x = 2 a t^{2}, y = a t^{4}$

Answer:

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

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 .

$x = a \cos θ, y = b \cos θ$

Answer:

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

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,
$\frac{d x}{d t} = \frac{d (\sin t)}{d t} = \cos t$
Similarly,
$\frac{d y}{d t} = \frac{d (\cos 2 t)}{d t} = - 2 \sin 2 t = - 4 \sin t \cos t (∵ \sin 2 x = \sin x \cos x)$
Now, $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{- 4 \sin t \cos t}{\cos t} = - 4 \sin t$
Therefore, the answer is $\frac{d y}{d x} = - 4 \sin t$

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

$x = 4 t, y = 4 / t$

Answer:

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

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 $x = \cos θ - \cos 2 θ, y = \sin θ - \sin 2 θ$

Answer:

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

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 $x = a (θ - \sin θ), y = a (1 + \cos θ)$

Answer:

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

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 $x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}}, y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}}$

Answer:

Given equations are

$x = \frac{\sin^{3} t}{\sqrt{\cos 2 t}}, y = \frac{\cos^{3} t}{\sqrt{\cos 2 t}}$

Now, differentiate both w.r.t $t$ :

$\frac{d x}{d t} = \frac{d (\frac{\sin^{3} t}{\sqrt{\cos 2 t}})}{d t} = \frac{\sqrt{\cos 2 t} \cdot \frac{d (\sin^{3} t)}{d t} - \sin^{3} t \cdot \frac{d (\sqrt{\cos 2 t})}{d t}}{(\sqrt{\cos 2 t})^{2}}$

$= \frac{3 \sin^{2} t \cos t \cdot \sqrt{\cos 2 t} - \sin^{3} t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot (- 2 \sin 2 t)}{\cos 2 t}$

$= \frac{3 \sin^{2} t \cos t \cdot \cos 2 t + \sin^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}$

$= \frac{\sin^{3} t \sin 2 t (3 \cot t \cot 2 t + 1)}{\cos 2 t \sqrt{\cos 2 t}} (∵ \frac{\cos x}{\sin x} = \cot x)$

Similarly,

$\frac{d y}{d t} = \frac{d (\frac{\cos^{3} t}{\sqrt{\cos 2 t}})}{d t} = \frac{\sqrt{\cos 2 t} \cdot \frac{d (\cos^{3} t)}{d t} - \cos^{3} t \cdot \frac{d (\sqrt{\cos 2 t})}{d t}}{(\sqrt{\cos 2 t})^{2}}$

$= \frac{3 \cos^{2} t (- \sin t) \cdot \sqrt{\cos 2 t} - \cos^{3} t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot (- 2 \sin 2 t)}{(\sqrt{\cos 2 t})^{2}}$

$= \frac{- 3 \cos^{2} t \sin t \cos 2 t + \cos^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}$

$= \frac{\sin 2 t \cos^{3} t (1 - 3 \tan t \cot 2 t)}{\cos 2 t \sqrt{\cos 2 t}}$

Now, $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{\sin 2 t \cos^{3} t (1 - 3 \tan t \cot 2 t)}{\cos 2 t \sqrt{\cos 2 t}}}{\frac{\sin^{3} t \sin 2 t (3 \cot t \cot 2 t + 1)}{\cos 2 t \sqrt{\cos 2 t}}}$

$= \frac{\cot^{3} t (1 - 3 \tan t \cot 2 t)}{3 \cot t \cot 2 t + 1}$

$= \frac{\cos^{3} t (1 - 3 \frac{\sin t}{\cos t} \frac{\cos 2 t}{\sin 2 t})}{\sin^{3} t (3 \frac{\cos t}{\sin t} \frac{\cos 2 t}{\sin 2 t} + 1)}$

$= \frac{\cos^{2} t (\cos t \sin 2 t - 3 \sin t \cos 2 t)}{\sin^{2} t (3 \cos t \cos 2 t + \sin t \sin 2 t)}$

$= \frac{\cos^{2} t (2 \sin t \cos^{2} t - 3 \sin t \cos^{2} t + 3 \sin t)}{\sin^{2} t (3 \cos t - 6 \cos t \sin^{2} t + 2 \sin^{2} \cos t)}$

$= \frac{\sin t \cos t (- 4 \cos^{3} t + 3 \cos t)}{\sin t \cos t (3 \sin t - 4 \sin^{3} t)}$

$\frac{d y}{d x} = \frac{- 4 \cos^{3} t + 3 \cos t}{3 \sin t - 4 \sin^{3} t} = \frac{- \cos 3 t}{\sin 3 t} = - \cot 3 t (∵ \sin 3 t = 3 \sin t - 4 \sin^{3} t and \cos 3 t = 4 \cos^{3} t - 3 \cos t)$
Therefore, the answer is $\frac{d y}{d x} = - \cot 3 t$

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,
$\frac{d x}{d t} = \frac{d (a (\cos t + \log \tan \frac{t}{2}))}{d t} = a (- \sin t + \frac{1}{\tan \frac{t}{2}} . \sec^{2} \frac{t}{2} . \frac{1}{2})$
$= a (- \sin t + \frac{1}{2} . \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} . \frac{1}{\cos^{2} \frac{t}{2}}) = a (- \sin t + \frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}})$
$= a (- \sin t + \frac{1}{\sin 2. \frac{t}{2}}) = a (\frac{- \sin^{2} t + 1}{\sin t}) = a (\frac{\cos^{2} t}{\sin t})$
Similarly,
$\frac{d y}{d t} = \frac{d (a \sin t)}{d t} = a \cos t$
Now, $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a \cos t}{a (\frac{\cos^{2} t}{\sin t})} = \frac{\sin t}{\cos t} = \tan t$
Therefore, the answer is $\frac{d y}{d x} = \tan t$

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 $x = a \sec θ, y = b t a n θ$

Answer:

Given equations are
$x = a \sec θ, y = b t a n θ$
Now, differentiate both w.r.t $θ$
We get,
$\frac{d x}{d θ} = \frac{d (a \sec θ)}{d θ} = a \sec θ \tan θ$
Similarly,
$\frac{d y}{d θ} = \frac{d (b \tan θ)}{d θ} = b \sec^{2} θ$
Now, $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{b \sec^{2} θ}{a \sec θ \tan θ} = \frac{b \sec θ}{a \tan θ} = \frac{b \frac{1}{\cos θ}}{a \frac{\sin θ}{\cos θ}} = \frac{b}{a \sin θ} = \frac{b c o s e c θ}{a}$
Therefore, the answer is $\frac{d y}{d x} = \frac{b c o s e c θ}{a}$

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 $x = a (\cos θ + θ \sin θ), y = a (\sin θ - θ \cos θ)$

Answer:

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

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

Answer:

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

$x y = \sqrt{a^{s i n^{- 1} t + c o s^{- 1} t}} s i n c e s i n^{- 1} x + c o s^{- 1} x = \frac{π}{2} x y = a^{\frac{π}{2}} = c o n s t a n t = c$

differentiating with respect to x

$x \frac{d y}{d x} + y = 0 \frac{d y}{d x} = \frac{- y}{x}$

Topics covered in Chapter 5, Continuity and Differentiability: Exercise 5.6

The main topics covered in Chapter 5 of continuity and differentiability, exercises 5.6 are:

Derivatives of functions in parametric form: Understanding how to differentiate expressions when two variables are expressed as a third variable known as a parameter.
Application of the chain rule: Functions in parametric form can easily be differentiated with the help of the chain rule. For example, when $x$ and $y$ are expressed in terms of $t$ , we can find the derivative as $\frac{d y}{d x} = \frac{d y}{d t} \div \frac{d x}{d t}$ .

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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 ?

Yes, here you will get chapter-wise Solutions for Class 12 Maths.

7. Can I get marks distribution for Class 12 Maths ?

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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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Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

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Seek Support:
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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

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Hope you find this useful!

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Yuvan 30 August,2024

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

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Central Board of Secondary Education Class 12th Examination

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

Class 12 Maths Chapter 5 Exercise 5.6 Solutions: Download PDF

Continuity and Differentiability Exercise: 5.6

Topics covered in Chapter 5, Continuity and Differentiability: Exercise 5.6

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