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RD Sharma Solution for Class 12 Science Mathematics Chapter 15 - Mean Value Theorems

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 12 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 15 - Mean Value Theorems.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 12 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 12 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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RD Sharma Solution for Class 12 Science Mathematics Chapter 15 - Mean Value Theorems Page/Excercise 15.1

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 2(v)

Solution 2(vi)

Solution 2(vii)

Solution 2(viii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 3(v)

Solution 3(vi)

Solution 3(vii)

Here,

            f open parentheses x close parentheses equals fraction numerator sin x over denominator e to the power of x end fraction space o n space x space element of open square brackets 0 comma space straight pi close square brackets
W e space k n o w space t h a t comma space e x p o n e n t i a l space a n d space sin e space b o t h space f u n c t i o n s space a r e space c o n t i n u o u s space a n d space d i f f e r e n t i a b l e
e v e r y space w h e r e comma space s o space f open parentheses x close parentheses space i s space c o n t i n u o u s space i s space open square brackets 0 comma space straight pi close square brackets space a n d space d i f f e r e n t i a b l e space i s space open square brackets 0 comma space straight pi close square brackets

Now comma space

space space space space space space space space space space space space space straight f open parentheses 0 close parentheses equals fraction numerator sin space 0 over denominator straight e to the power of 0 end fraction equals 0
space space space space space space space space space space space space space space
space space space space space space space space space space space space space space space straight f open parentheses straight pi close parentheses equals fraction numerator sin space straight pi over denominator straight e to the power of straight pi end fraction equals 0

rightwards double arrow straight f open parentheses 0 close parentheses equals straight f open parentheses straight pi close parentheses
Since space Rolle apostrophe straight s space theorem space applicable comma space therefore space there space must space exist space straight a space point space straight c element of open square brackets 0 comma space straight pi close square brackets
such space that space straight f apostrophe open parentheses straight c close parentheses equals 0

Now comma
space space space space space space space space space space straight f open parentheses straight x close parentheses equals sinx over straight e to the power of straight x

space space space space space space space space space space rightwards double arrow straight f apostrophe open parentheses straight x close parentheses equals fraction numerator straight e to the power of straight x open parentheses cosx close parentheses minus straight e to the power of straight x open parentheses sinx close parentheses over denominator open parentheses straight e to the power of straight x close parentheses squared end fraction
Now comma
space space space space space space space space space space space space space space space space straight f apostrophe open parentheses straight c close parentheses equals 0
space space space space space space space space space space space rightwards double arrow straight e to the power of straight c open parentheses cosc minus sinc close parentheses equals 0
space space space space space space space space space space space rightwards double arrow space straight e to the power of straight c not equal to 0 space and space cosc minus sinc equals 0
space space space space space space space space space space space rightwards double arrow space tanc equals 1
space space space space space space space space space space space space space space space straight c equals straight pi over 4 element of open square brackets 0 comma straight pi close square brackets
Hence comma space Rolle apostrophe straight s space theorem space is space verified.

Solution 3(viii)

Solution 3(ix)

Solution 3(x)

Solution 3(xi)

Solution 3(xii)

Solution 3(xiii)

Solution 3(xiv)

Solution 3(xv)

Solution 3(xvi)

Solution 3(xvii)

Solution 3(xviii)

Solution 7

x = 0 then y = 16

Therefore, the point on the curve is (0, 16) 

Solution 8(i)

x = 0, then y = 0

Therefore, the point is (0, 0)

Solution 8(ii)

Solution 8(iii)

x = 1/2, then y = - 27

Therefore, the point is (1/2, - 27)

Solution 9

Solution 10

Solution 11

RD Sharma Solution for Class 12 Science Mathematics Chapter 15 - Mean Value Theorems Page/Excercise 15.2

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)


Solution 1(vii)

Solution 1(viii)

Solution 1(ix)

Solution 1(x)

Solution 1(xi)

Solution 1(xii)


Solution 1(xiii)

Solution 1(xiv)

Solution 1(xv)

Solution 1(xvi)


Solution 2

Solution 3

Solution 4

Solution 5

Solution 6


Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

RD Sharma Solution for Class 12 Science Mathematics Chapter 15 - Mean Value Theorems Page/Excercise MCQ

Solution 1

Correct option: (c)

 

  

Solution 2

Correct option: (c)

 

  

Solution 3

Correct option: (b)

Solution 4

Correct option: (c)

 

Using statement of Lagrange's mean value theorem function is continuous on [a,b], differentiable on (a,b) then there exists c such that a < x1

Solution 5

Correct option: (b)

 

ϕ(x) is continuous and differentiable function then using statement of Rolle's theorem f(a)=f(b). Hence, here sin 0=0 also sin п=0. The answer is [0,  ].

Solution 6

Correct option: (a)

 

  

Solution 7

Correct option: (a)

  

Solution 8

Correct answer: (c)

 

  

Solution 9

Correct option: (d)

 

  

Solution 10

Correct option: (a)

 

  

Solution 11

Correct option: (d)

 

  

RD Sharma Solution for Class 12 Science Mathematics Chapter 15 - Mean Value Theorems Page/Excercise 15VSAQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 15 - Mean Value Theorems for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 15 - Mean Value Theorems.

Text Book Solutions

CBSE XII Science - Mathematics

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