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NCERT Solutions for Exercise 5.2 Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Exercise 5.2 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Ramraj Saini | Updated on Dec 03, 2023 04:55 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.2

NCERT Solutions for Exercise 5.2 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 NCERT syllabus Class 11 Maths, you have already learned about the derivatives of real value functions. Differentiation is defined as the process of finding the derivative of a function. In this article, you will get NCERT solutions for Class 12 Maths chapter 5 Exercise 5.2. This NCERT book exercise consists of questions related to finding derivatives of different types of functions.

Mainly derivatives of composite functions using chain rule are covered in the Class 12 Maths ch 5 ex 5.2. The chain rule is a very important concept to find the derivative of a composite function of two differentiable functions using derivatives of these functions. Solve problems from exercise 5.2 Class 12 Maths to get an understanding of the chain rule. Also, you can try to prove this theorem to get conceptual clarity. 12th class Maths exercise 5.2 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 5.2

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Continuity and Differentiability Exercise: 5.2

Question:1.Differentiate the functions with respect to x in

\sin (x^2 +5 )

Answer:

Given function is
f(x)=\sin (x^2 +5 )
when we differentiate it w.r.t. x.
Lets take t = x^2+5 . then,
f(t) = \sin t
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} (By chain rule)
\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (x^2+5)
\frac{dt}{dx} = \frac{d(x^2+5 )}{dx} = 2x
Now,
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (x^2+5).2x
Therefore, the answer is 2x \cos (x^2+5)

Question:2. Differentiate the functions with respect to x in

\cos ( \sin x )

Answer:

Given function is
f(x)= \cos ( \sin x )
Lets take t = \sin x then,
f(t) = \cos t
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} ( By chain rule)
\frac{df(t)}{dt} = \frac{d(\cos t)}{dt} = -\sin t = -\sin (\sin x)
\frac{dt}{dx} = \frac{d(\sin x)}{dt} = \cos x
Now,
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = -\sin(\sin x).\cos x
Therefore, the answer is -\sin(\sin x).\cos x

Question:3. Differentiate the functions with respect to x in

\sin (ax +b )

Answer:

Given function is
f(x) = \sin (ax +b )
when we differentiate it w.r.t. x.
Lets take t = ax+b . then,
f(t) = \sin t
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} (By chain rule)
\frac{df(t)}{dt} = \frac{d(\sin t )}{dt} = \cos t = \cos (ax+b)
\frac{dt}{dx} = \frac{d(ax+b )}{dx} = a
Now,
\frac{df(t)}{dx} = \frac{df(t)}{dt}.\frac{dt}{dx} = \cos (ax+b).a
Therefore, the answer is a \cos (ax+b)

Question:4. Differentiate the functions with respect to x in

\sec (\tan (\sqrt x) )

Answer:

Given function is
f(x)=\sec (\tan (\sqrt x) )
when we differentiate it w.r.t. x.
Lets take t = \sqrt x . then,
f(t) = \sec (\tan t)
take \tan t = k. then,
f(k) = \sec k
\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx} (By chain rule)
\frac{df(k)}{dk} = \frac{d(\sec k )}{dk} = \sec k \tan k = \sec(\tan\sqrt x)\tan(\tan\sqrt x)
(\because k = \tan t \ and \ t = \sqrt x)
\frac{df(t)}{dt} = \frac{d(\tan t )}{dt} = \sec^2 t =\sec^2 (\sqrt x) \ \ \ \ \ \ (\because t = \sqrt x)
\frac{dt}{dx} = \frac{d(\sqrt x)}{dx} = \frac{1}{2\sqrt x}
Now,
\frac{df(k)}{dx} = \frac{df(k)}{dk}.\frac{dk}{dt}.\frac{dt}{dx} =\sec(\tan \sqrt x)\tan(\tan \sqrt x).\sec^2 (\sqrt x) . \frac{1}{2\sqrt x}
Therefore, the answer is \frac{\sec(\tan \sqrt x).\tan(\tan \sqrt x).\sec^2 (\sqrt x)}{2\sqrt x}

Question:5. Differentiate the functions with respect to x in

\frac{\sin (ax +b )}{\cos (cx + d)}

Answer:

Given function is
f(x) = \frac{\sin (ax +b )}{\cos (cx + d)} = \frac{g(x)}{h(x)}
We know that,
f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)}
g(x) = \sin(ax+b) and h(x) = \cos(cx+d)
Lets take u = (ax+b) \ and \ v = (cx+d)
Then,
\sin (ax+b) = \sin u \ and \ \cos(cx+d) = \cos c
g^{'}(x)=\frac{d(g(x))}{dx} = \frac{d(g(x))}{du}.\frac{du}{dx} (By chain rule)
\frac{d(g(x))}{du} = \frac{d(\sin u)}{du} = \cos u = \cos(ax+b) \ \ \ \ \ \ \ \ \ (\because u = ax +b)
\frac{du}{dx} = \frac{d(ax+b)}{dx} = a
g^{'}(x)=a\cos (ax+b) -(i)
Similarly,
h^{'}(x)=\frac{d(h(x))}{dx} = \frac{d(h(x))}{dv}.\frac{dv}{dx}
\frac{d(h(x))}{dv}= \frac{d(\cos v)}{dv} = -\sin v = -\sin (cx+d) \ \ \ \ \ \ \ (\because v = (cx+d))
\frac{dv}{dx}= \frac{d(cx+d)}{dv} = c
h^{'}(x)=-c\sin(cx+d) -(ii)
Now, put (i) and (ii) in
f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)} = \frac{a\cos(ax+b).\cos(cx+d)-\sin(ax+b).(-c.\sin(cx+d))}{\cos^2(cx+d)}
= \frac{a\cos(ax+b).\cos(cx+d)}{\cos^2(cx+d)}+\frac{\sin(ax+b).c.\sin(cx+d)}{\cos^2(cx+d)}
= a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)
Therefore, the answer is a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)

Question:6. Differentiate the functions with respect to x in

\cos x^3 . \sin ^ 2 ( x ^5 )

Answer:

Given function is
f(x)=\cos x^3 . \sin ^ 2 ( x ^5 )
Differentitation w.r.t. x is
f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x)
g(x) = \cos x^3 \ and \ h(x) = sin^2(x^5)
Lets take u = x^3 \ and \ v = x^5
Our functions become,
\cos x^3 = \cos u and \sin^2(x^5) = \sin^2v
Now,
g^{'}(x) = \frac{d(g(x))}{dx} =\frac{d(g(u))}{du}.\frac{du}{dx} ( By chain rule)
\frac{d(g(u))}{du} = \frac{d(\cos u)}{du} = -\sin u =- \sin x^3 \ \ \ \ (\because u = x^3)
\frac{du}{dx} = \frac{d(x^3)}{dx} = 3x^2
g^{'}(x) = -\sin x^3.3x^2 -(i)
Similarly,
h^{'}(x) = \frac{d(h(x))}{dx} =\frac{d(h(v))}{dv}.\frac{dv}{dx}
\frac{d(h(v))}{dv}= \frac{d(\sin^2v)}{dv} =2\sin v \cos v =2\sin x^5\cos x^5 \ \ \ (\because v = x^5)

\frac{dv}{dx} = \frac{d(x^5)}{dx} = 5x^4
h^{'}(x) = 2\sin x^5\cos x^5.5x^4 = 10x^4\sin x^5\cos x^5 -(ii)
Put (i) and (ii) in
f^{'}(x) = g^{'}(x).h(x) + g(x).h^{'}(x) = -3x^2\sin x^3.\sin^2 x^5+\cos x^3.10x^4\sin x^5\cos x^5
Therefore, the answer is 10x^4\sin x^5\cos x^5.\cos x^3 -3x^2\sin x^3.\sin^2 x^5

Question:7. Differentiate the functions with respect to x in

2 \sqrt { \cot ( x^2 )}

Answer:

Give function is
f(x)=2 \sqrt { \cot ( x^2 )}
Let's take t = x^2
f(t) = 2\sqrt{\cot t}
Now, take \cot t = k^2
f(k) = 2k
Differentiation w.r.t. x
\frac{d(f(k))}{dx} = \frac{d(f(k))}{dk}.\frac{dk}{dt}.\frac{dt}{dx} -(By chain rule)
\frac{d(f(k))}{dk} = \frac{d(2k)}{dk} = 2
\frac{dk}{dt} = \frac{d(\sqrt{\cot t})}{dt} = \frac{1}{2\sqrt{cot t}}.(-cosec^2 t) = \frac{-cosec^2 x^2}{2\sqrt{cot x^2}} \ \ \ (\because t = x^2)
\frac{dt}{dx} = \frac{d(x^2)}{dx} = 2x
So,
\frac{d(f(k))}{dx} = 2.\frac{-cosec^2 x^2}{2\sqrt{cot x^2}}.2x = \frac{-2\sqrt2x}{\sin^2x^2\sqrt{\frac{2\sin x^2\cos x^2}{\sin^2x^2}} } ( Multiply and divide by \sqrt 2 and multiply and divide \sqrt {\cot x^2} by \sqrt{\sin x^2)
(\because \cot x = \frac{\cos x}{\sin x} \ and \ cosec x = \frac{1}{\sin x } )
=\frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}} \ \ \ \ (\because 2\sin x\cos x=\sin2x)
There, the answer is \frac{-2\sqrt2x}{\sin x^2\sqrt{\sin2x^2}}

Question:8 Differentiate the functions with respect to x in

\cos ( \sqrt x )

Answer:

Let us assume : y\ =\ \cos ( \sqrt x )

Differentiating y with respect to x, we get :

\frac{dy}{dx}\ =\ \frac{d(\cos ( \sqrt x ))}{dx}

or =\ - \sin \sqrt{x}.\frac{d( \sqrt x )}{dx}

or =\ \frac{- \sin \sqrt{x}}{2\sqrt{x}}



Question:9. Prove that the function f given byf (x) = |x-1 | , x \epsilon R is not differentiable at x = 1.

Answer:

Given function is
f (x) = |x-1 | , x \epsilon R
We know that any function is differentiable when both
\lim_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h} and \lim_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h} are finite and equal
Required condition for function to be differential at x = 1 is

\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}
Now, Left-hand limit of a function at x = 1 is
\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^-}\frac{|1+h-1|-|1-1|}{h} = \lim_{h\rightarrow 0^-}\frac{|h|-0}{h}
= \lim_{h\rightarrow 0^-}\frac{-h}{h} = -1 \ \ \ \ (\because h < 0)
Right-hand limit of a function at x = 1 is
\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{|1+h-1|-|1-1|}{h} = \lim_{h\rightarrow 0^+}\frac{|h|-0}{h}
=\lim_{h\rightarrow 0^-}\frac{h}{h} = 1
Now, it is clear that
R.H.L. at x= 1 \neq L.H.L. at x= 1
Therefore, function f (x) = |x-1 | is not differentiable at x = 1

Question:10. Prove that the greatest integer function defined by f (x) = [x] , 0 < x < 3 is not differentiable at

x = 1 and x = 2.

Answer:

Given function is
f (x) = [x] , 0 < x < 3
We know that any function is differentiable when both
\lim_{h\rightarrow 0^-}\frac{f(c+h)-f(c)}{h} and \lim_{h\rightarrow 0^+}\frac{f(c+h)-f(c)}{h} are finite and equal
Required condition for function to be differential at x = 1 is

\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h}
Now, Left-hand limit of the function at x = 1 is
\lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^-}\frac{[1+h]-[1]}{h} = \lim_{h\rightarrow 0^-}\frac{0-1}{h}
=\lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ \ (\because h < 0 \Rightarrow 1+h<1, \therefore [1+h] =0)
Right-hand limit of the function at x = 1 is
\lim_{h\rightarrow 0^+}\frac{f(1+h)-f(1)}{h } = \lim_{h\rightarrow 0^+}\frac{[1+h]-[1]}{h} = \lim_{h\rightarrow 0^+}\frac{1-1}{h}
=\lim_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \Rightarrow 1+h>1, \therefore [1+h] =1)
Now, it is clear that
R.H.L. at x= 1 \neq L.H.L. at x= 1 and L.H.L. is not finite as well
Therefore, function f(x) = [x] is not differentiable at x = 1
Similary, for x = 2
Required condition for function to be differential at x = 2 is

\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h}
Now, Left-hand limit of the function at x = 2 is
\lim_{h\rightarrow 0^-}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^-}\frac{[2+h]-[2]}{h} = \lim_{h\rightarrow 0^-}\frac{1-2}{h}
=\lim_{h\rightarrow 0^-}\frac{-1}{h} = -\infty \ \ \ \ (\because h < 0 \Rightarrow 2+h<2, \therefore [2+h] =1)
Right-hand limit of the function at x = 1 is
\lim_{h\rightarrow 0^+}\frac{f(2+h)-f(2)}{h } = \lim_{h\rightarrow 0^+}\frac{[2+h]-[2]}{h} = \lim_{h\rightarrow 0^+}\frac{2-2}{h}
=\lim_{h\rightarrow 0^-}\frac{0}{h} = 0 \ \ \ \ \ (\because h > 0 \Rightarrow 2+h>2, \therefore [2+h] =2)
Now, it is clear that
R.H.L. at x= 2 \neq L.H.L. at x= 2 and L.H.L. is not finite as well
Therefore, function f(x) = [x] is not differentiable at x = 2

More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2:-

Class 12 Maths chapter 5 exercise 5.2 solutions is all about using the chain rule to find derivatives of composite functions. There are a total of 10 questions out of which 8 questions are from chain rule in Class 12 Maths ch 5 ex 5.2. There are 3 examples given before this exercise in the NCERT textbook, that you can solve to understand this concept fully. This concept is very easy as well useful for composite functions. You will find the use of this concept not only in this exercise but in physics, chemistry, and in other chapters also.

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

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2:-

  • NCERT Solutions for Class 12 Maths Chapter 5 exercise 5.2 is useful in solving this exercise as well as other subjects for finding derivatives of composite functions.
  • There are other methods to find the derivative of a composite function, but the chain rule is more simple to use.
  • The last two questions in exercise 5.2 Class 12 Maths are related to checking the differentiability of given functions
  • You can use Class 12th Maths chapter 5 exercise 5.2 for quick revision of chain rule.
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Key Features Of NCERT Solutions for Exercise 5.2 Class 12 Maths Chapter 5

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 5.2 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 5.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 5.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 5.2 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 5.2 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 5.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. What is the definition of the composite function ?

A composite function is obtained from other functions where the output of one function is the input of another function.

2. Give an example of composite function ?

sin (x/2)  is an example of a composite function.

3. What is derivative of sin(x/2) ?

d(sin(x/2)/dx = cos(x/2)/2 

4. What is derivative of cos(x/2) ?

d(cos(x/2)/dx = -sin(x/2)/2

5. what is a limit of a function ?

Limit of function is defined as the value of functions reaches when the limit reaches.

6. Does calculus is so hard ?

Some people consider it hard as it is a new concept included in class 11 maths and the level of maths till class 10 is too easy.

7. Why calculus is so hard ?

As the foundation of your maths is low, you may find it hard to grasp but with more practice, you will grasp the concept easily.

8. Can I get free NCERT Syllabus for Class 12 Maths ?

Yes, You can check here for the NCERT Syllabus for Class 12 Maths. NCERT book exercise questions, NCERT exemplar problems and solutions are helpful to practice questions for the CBSE board exam.

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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.

Possible steps:

  1. Re-evaluate Your Study Strategies:

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  5. 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
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.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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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