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NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Ramraj Saini | Updated on Dec 03, 2023 05:13 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.7

NCERT Solutions for Exercise 5.7 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, you have learned about finding the first-order differentiation of different types of functions. In NCERT solutions for Class 12 Maths chapter 5 exercise 5.7, you will learn about finding the derivatives of second order. If a function y=f(x) is given, you can find its first-order derivative by differentiation of y y w.r.t x i.e. f'(x) = dy/dx and if you again differentiate f'(x) w.r.t the x, you will get the second-order f''(x).

If you have command on the first-order differentiation then exercise 5.7 Class 12 Maths will be very easy for you as you just need to differentiate the given function two times. The first and second-order derivatives are useful in finding minimum and maximum values of the functions, finding the domain and range of the function, other subjects also. 12th class Maths exercise 5.7 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 5.7

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Continuity and Differentiability Exercise: 5.7

Question:1 Find the second order derivatives of the functions given in Exercises 1 to 10.

x^2 + 3x+ 2

Answer:

Given function is
y=x^2 + 3x+ 2
Now, differentiation w.r.t. x
\frac{dy}{dx}= 2x+3
Now, second order derivative
\frac{d^2y}{dx^2}= 2
Therefore, the second order derivative is \frac{d^2y}{dx^2}= 2

Question:2 Find the second order derivatives of the functions given in Exercises 1 to 10.

x ^{20}

Answer:

Given function is
y=x ^{20}
Now, differentiation w.r.t. x
\frac{dy}{dx}= 20x^{19}
Now, the second-order derivative is
\frac{d^2y}{dx^2}= 20.19x^{18}= 380x^{18}
Therefore, second-order derivative is \frac{d^2y}{dx^2}= 380x^{18}

Question:3 Find the second order derivatives of the functions given in Exercises 1 to 10.

x \cos x

Answer:

Given function is
y = x \cos x
Now, differentiation w.r.t. x
\frac{dy}{dx}= \cos x + x(-\sin x ) = \cos x-x\sin x
Now, the second-order derivative is
\frac{d^2y}{dx^2}= -\sin x-(\sin x+x\cos x) = -2\sin x - x\sin x
Therefore, the second-order derivative is \frac{d^2y}{dx^2}= -2\sin x - x\sin x

Question:4 Find the second order derivatives of the functions given in Exercises 1 to 10.

\log x

Answer:

Given function is
y=\log x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{1}{x}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{x^2}
Therefore, second order derivative is \frac{d^2y}{dx^2}= \frac{-1}{x^2}

Question:5 Find the second order derivatives of the functions given in Exercises 1 to 10.

x ^3 \log x

Answer:

Given function is
y=x^3\log x
Now, differentiation w.r.t. x
\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}= 3x^2.\log x+ x^2
Now, the second-order derivative is
\frac{d^2y}{dx^2}= 6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x = x(6.\log x+5)
Therefore, the second-order derivative is \frac{d^2y}{dx^2} = x(6.\log x+5)

Question:6 Find the second order derivatives of the functions given in Exercises 1 to 10.

e ^x \sin5 x

Answer:

Given function is
y= e^x\sin 5x
Now, differentiation w.r.t. x
\frac{dy}{dx}=e^x.\sin 5x +e^x.5\cos 5x = e^x(\sin5x+5\cos5x)
Now, second order derivative is
\frac{d^2y}{dx^2}= e^x(\sin5x+5\cos5x)+e^x(5\cos5x+5.(-5\sin5x))
= e^x(\sin5x+5\cos5x)+e^x(5\cos5x-25\sin5x)=e^x(10\cos5x-24\sin5x)
=2e^x(5\cos5x-12\sin5x)
Therefore, second order derivative is \frac{dy}{dx}=2e^x(5\cos5x-12\sin5x)

Question:7 Find the second order derivatives of the functions given in Exercises 1 to 10.

e ^{6x}\cos 3x

Answer:

Given function is
y= e^{6x}\cos 3x
Now, differentiation w.r.t. x
\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)
Now, second order derivative is
\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)
= 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)
e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)
Therefore, second order derivative is \frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)

Question:8 Find the second order derivatives of the functions given in Exercises 1 to 10.

\tan ^{-1} x

Answer:

Given function is
y = \tan^{-1}x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{(1+x^2)^2}.2x = \frac{-2x}{(1+x^2)^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}

Question:9 Find the second order derivatives of the functions given in Exercises 1 to 10.

\log (\log x )

Answer:

Given function is
y = \log(\log x)
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\log(\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}= \frac{1}{x\log x}
Now, second order derivative is
\frac{d^2y}{dx^2}= \frac{-1}{(x\log x)^2}.(1.\log x+x.\frac{1}{x}) = \frac{-(\log x+1)}{(x\log x)^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-(\log x+1)}{(x\log x)^2}

Question:10 Find the second order derivatives of the functions given in Exercises 1 to 10.

\sin (\log x )

Answer:

Given function is
y = \sin(\log x)
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(\sin(\log x))}{dx}=\cos (\log x).\frac{1}{x}= \frac{\cos (\log x)}{x}
Now, second order derivative is
Using Quotient rule
\frac{d^2y}{dx^2}=\frac{-\sin(\log x)\frac{1}{x}.x-\cos(\log x).1}{x^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}
Therefore, second order derivative is \frac{d^2y}{dx^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}

Question:11 If y = 5 \cos x - 3 \sin x prove that \frac{d^2y}{dx^2}+y = 0

Answer:

Given function is
y = 5 \cos x - 3 \sin x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}=-5\sin x-3\cos x
Now, the second-order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}=-5\cos x+3\sin x
Now,
\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0
Hence proved

Question:12 If y = \cos ^{-1} x Find \frac{d ^2 y }{dx^2 } in terms of y alone.

Answer:

Given function is
y = \cos ^{-1} x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2} -(i)
Now, we want \frac{d^2y}{dx^2} in terms of y
y = \cos ^{-1} x
x = \cos y
Now, put the value of x in (i)
\frac{d^2y}{dx^2} = \frac{-2\cos y }{1-\cos^2 y } = \frac{-2\cos y}{\sin ^2 y}= -2\cot y cosec y
(\because 1-\cos^2x =\sin^2 x\ and \ \frac{\cos x}{\sin x} = \cot x \ and \ \frac{1}{\sin x}= cosec x)
Therefore, answer is \frac{d^2y}{dx^2} = -2\cot y cosec y

Question:13 If y = 3 \cos (\log x) + 4 \sin (\log x), show that x^2 y_2 + xy_1 + y = 0

Answer:

Given function is
y = 3 \cos (\log x) + 4 \sin (\log x)
Now, differentiation w.r.t. x
y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}
=\frac{4\cos (\log x)-3\sin(\log x)}{x} -(i)
Now, second order derivative is
By using the Quotient rule
y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{4\cos (\log x)-3\sin(\log x)}{x})}{dx^2}= \frac{(-4\sin(\log x).\frac{1}{x}-3\cos(\log x).\frac{1}{x}).x-1.(4\cos (\log x)-3\sin(\log x))}{x^2}
=\frac{-\sin(\log x)+7\cos (\log x)}{x^2} -(ii)
Now, from equation (i) and (ii) we will get y_1 \ and \ y_2
Now, we need to show
x^2 y_2 + xy_1 + y = 0
Put the value of y_1 \ and \ y_2 from equation (i) and (ii)
x^2\left ( \frac{-\sin(\log x)+7\cos (\log x)}{x^2} \right )+x\left ( \frac{4\cos (\log x)-3\sin(\log x)}{x} \right )+ 3 \cos (\log x) +4\sin(\log x)
-\sin(\log x)-7\cos(\log x)+4\cos(\log x)-3\sin(\log x)+3\ cos (\log x) +4\sin(\log x)
=0
Hence proved

Question:14 If y = A e ^{mx} + Be ^{nx} , show that \frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0

Answer:

Given function is
y = A e ^{mx} + Be ^{nx}
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx} -(ii)
Now, we need to show
\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0
Put the value of \frac{d^2y}{dx^2} \ and \ \frac{dy}{dx} from equation (i) and (ii)
m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})
m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx}+mnBe^{nx}
=0
Hence proved

Question:15 If y = 500 e ^{7x} + 600 e ^{- 7x } , show that \frac{d^2 y}{dx ^2} = 49 y
Answer:

Given function is
y = 500 e ^{7x} + 600 e ^{- 7x }
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}
= 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x} -(ii)
Now, we need to show
\frac{d^2 y}{dx ^2} = 49 y
Put the value of \frac{d^2y}{dx^2} from equation (ii)
24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})
= 24500e^{7x}+29400e^{-7x}
Hence, L.H.S. = R.H.S.
Hence proved

Question:16 If e ^y (x+1) = 1 show that \frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2

Answer:

Given function is
e ^y (x+1) = 1
We can rewrite it as
e^y = \frac{1}{x+1}
Now, differentiation w.r.t. x
\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1} -(i)
Now, second order derivative is
\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2} -(ii)
Now, we need to show
\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2
Put value of \frac{d^2y}{dx^2} \ and \ \frac{dy}{dx} from equation (i) and (ii)
\frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2
=\frac{1}{(x+1)^2}
Hence, L.H.S. = R.H.S.
Hence proved

Question:17 If y = (\tan^{-1} x)^2 show that (x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2

Answer:

Given function is
y = (\tan^{-1} x)^2
Now, differentiation w.r.t. x
y_1=\frac{dy}{dx}=\frac{d((\tan^{-1}x)^2)}{dx}= 2.\tan^{-1}x.\frac{1}{1+x^2}= \frac{2\tan^{-1}x}{1+x^2} -(i)
Now, the second-order derivative is
By using the quotient rule
y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{2\tan^{-1}x}{1+x^2})}{dx^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2\tan^{-1}x(2x)}{(1+x^2)^2}=\frac{2-4x\tan^{-1}x}{(1+x^2)^2} -(ii)
Now, we need to show
(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2
Put the value from equation (i) and (ii)
(x^2+1)^2.\frac{2-4x\tan^{-1}x}{(1+x^2)^2}+2x(x^2+1).\frac{2\tan^{-1}x}{x^2+1}\\ \Rightarrow 2-4x\tan^{-1}x+4x\tan^{-1}x = 2
Hence, L.H.S. = R.H.S.
Hence proved

More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7

In Class 12th Maths chapter 5 exercise 5.7, you will get 17 questions related to finding the second-order derivatives. There are four examples given in the NCERT book prior to the ex 5.7 which you can solve to get more familiar with second derivatives before solving the exercise question. After solving examples, you can try to solve Class 12th Maths chapter 5 exercise 5.7 questions. You may not be to solve these exercise 5.7 Class 12 Maths problems by yourself at first. You can go through Class 12 Maths chapter 5 exercise 5.7 solutions to get conceptual clarity.

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

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7

  • Class 12 Maths chapter 5 exercise 5.7 solutions are designed in a descriptive manner that you can understand easily.
  • Don't go through NCERT solutions for Class 12 Maths chapter 5 exercise 5.7 without trying to solve NCERT problems by yourself.
  • You can use exercise 5.7 Class 12 Maths solutions for reference.
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Key Features Of NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 5.7 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 5.7, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 5.7 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 5.7 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 5.7 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 5.7 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

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Happy learning!!!

Frequently Asked Questions (FAQs)

1. If y = c is a function where c is a constant then find dy/dx ?

y = c

dy/dx = 0

2. If y = c is a function where c is a constant then find the second order derivative of y ?

y = c

dy/dx = 0

d(dy/dx)/dx = 0

3. Find the first derivative of y = x ?

Given y = x

dy/dx = 1

4. Find the second order derivative of y = x ?

Given y = x

dy/dx = 1

d(dy/dx)/dx = 0

5. What is the second order derivative of y = e^x ?

y = e^x

dy/dx = e^x

d(dy/dx)/dx = e^x

d^(2)y/dx^2 = e^x

6. Can i get CBSE Class 10 exam pattern ?

Click on the link to get CBSE Class 10 Exam Pattern.

7. Can I get detailed syllabus for CBSE Class 10 ?

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8. Can I get detailed syllabus for CBSE Class 10 Maths ?

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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.

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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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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.

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

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kumardurgesh1802 10 July,2024
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200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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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
Read More

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