##### VMC VIQ Scholarship Test

ApplyRegister for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 03:32 PM IST

RD Sharma books are considered the best books for CBSE students. They contain detailed information about topics and a variety of questions on each concept. This is why they are preferred by teachers and students all over the country.

RD Sharma Class 12th Exercise 14.2 contains the chapter Mean Value Theorem. It is precisely designed to help students study and cover their syllabus efficiently. RD Sharma Solutions materials are preferred by most schools all over the country as they are very informative and exam-oriented. Moreover, a lot of students consider this book as a Math bible because of its specific content.

**JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days**

**JEE Main 2025: Maths Formulas | Study Materials**

**JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics **

Mean value theorem exercise 14.1 question 2(i)

, Hence, Rolle’s Theorem is verified.

f(x) is continuous for all x and hence continuous in [2,6]

on [2,6]

We have

- Being polynomial f(x) is continuous for all x and hence continuous in [2,6]
- , which exists in [2,6]

3.

Thus all the conditions of Rolle’s Theorem are satisfied.

There exists at least onesuch that

Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.2 question 2 (ii)

, Hence, Rolle’s Theorem is verified.

is continuous for all x and hence continuous in [1,3]

on [1,3]

We have

- Being polynomial f(x) is continuous for all x and hence continuous in [1,3]
- , which exists in (1,3)

3.

Thus all the conditions of Rolle’s Theorem are satisfied.

There exists at least onesuch that

Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.1 question 2(iii)

Using discrimination method,

on [1,2]

We have,

- Being polynomial f(x) is continuous for all x and hence continuous in [1,2]

Which exists in (1.2)

is derivable in (1,2)

3.

Thus all the conditions of Rolle’s Theorem are satisfied.

There exists at least one such that

Now,

Using discrimination method,

[Neglecting the value 2 ]

Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.1 question 2(iv)

, hence Rolle’s Theorem is verified.

Using discrimination method,

on [0,1]

We have,

- Being polynomial f(x) is continuous for all x and hence continuous in [0,1]

3.

Thus all the conditions of Rolle’s Theorem are satisfied.

There exists at least one such that

Now,

Using discrimination method,

Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.1 question 2(v)

, Hence Rolle’s Theorem is verified.

Using discrimination method,

on [1,-2]

We have,

- Being polynomial f(x) is continuous for all x and hence continuous in [-1,2]

3.

Thus all the conditions of Rolle’s Theorem are satisfied.

There exists at least onesuch that

Now,

Using discrimination method,

Hence, Rolle’s Theorem is verified.

Mean value theorem exercise 14.1 question 2(vii)

, hence Rolle’s Theorem is verified.

f(x) is continuous for all x and hence continuous in [-3,-2]

on [-3,-2]

We have,

- Being polynomial f(x) is continuous for all x and hence continuous in [-3,-2]
- , which exists in [-3,-2]

3.

Thus all the conditions of Rolle’s Theorem are satisfied.

There exists at least onesuch thatNow,

Hence, Rolle’s Theorem is verified.

Mean Value Theoram exercise 14.2 question 1(i)

is a polynomial function.

It is continuous in [2, 3]

(Which is defined in [2, 3])

is differentiable in [2, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(ii)

is a polynomial function.

It is continuous in [0, 1]

(Which is defined in [0, 1])

is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Therefore,

Mean Value Theoram exercise 14.2 question 1(iii)

is a polynomial function.

It is continuous in [1, 2]

(Which is defined in [1, 2])

is differentiable in [1, 2]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(iv) maths

is a polynomial function.

It is continuous in [-1, 2]

(Which is defined in [-1, 2])

is differentiable in [-1, 2]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(v)

is a polynomial function.

It is continuous in [1, 3]

(Which is defined in [1, 3])

is differentiable in [1, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(vi)

is a polynomial function.

It is continuous in [1, 5]

(Which is defined in [1, 5])

is differentiable in [1, 5]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(vii)

is a polynomial function.

It is continuous in [0, 1]

(Which is defined in [0, 1])

is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(viii) maths

is a polynomial function.

It is continuous in [0, 4]

(Which is defined in [0, 4])

is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Quadratic Equation,

Both values lie between [0, 4]

Mean Value Theoram exercise 14.2 question 1(ix)

is a polynomial function.

It is continuous in [-3, 4]

(Which is defined in [-3, 4])

is differentiable in [-3, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Squaring on both sides,

Mean Value Theoram exercise 14.2 question 1(x)

is a polynomial function.

It is continuous in [0, 1]

(Which is defined in [0, 1])

is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(xi)

is a polynomial function.

It is continuous in [1, 3]

(Which is defined in [1, 3])

is differentiable in [1, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(xii)

is a polynomial function.

It is continuous in [0, 4]

(Which is defined in [0, 4])

is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

But in [0, 4], only exists.

Mean Value Theoram exercise 14.2 question 1(xiii)

is a polynomial function.

It is continuous in [2, 4]

(Which is defined in [2, 4])

is differentiable in [2, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Squaring on both sides,

Mean Value Theoram exercise 14.2 question 1(xiv) maths

is a polynomial function.

It is continuous in [0, 4]

(Which is defined in [0, 4])

is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(xv)

is a polynomial function.

It is continuous in

(Which is defined in )

is differentiable in . Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Mean Value Theoram exercise 14.2 question 1(xvi)

is a polynomial function.

It is continuous in [1, 3]

(Which is defined in [1, 3])

is differentiable in [1, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

Value exists.

Mean Value Theoram exercise 14.2 question 2

Lagrange’s mean value theorem states if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one point x = c on this interval, such that

Differentiability at x = 0:

LHD =

RHD =

is** **not differentiable at x = 0

Lagrange’s mean value theorem is not applicable for function f (x) = |x| on [-1, 1]

Mean Value Theoram exercise 14.2 question 3

It is clear

f (x) exists for all the values of x except 0

f (x) is the discontinuous at x = 0

So, f (x) is not continuous in [-1, 1]

Thus, the Lagrange’s mean value theorem is not applicable for the function:

Mean Value Theoram exercise 14.2 question 4

Here,

f^{'}(x) has the unique values for all x except

So, f (x) is continuous in [1, 4]

Differentiate:

f (x) is differentiable in (1, 4)

So, there exists a point c (1, 4)

Thus, Lagrange’s Theorem is verified.

Mean Value Theoram exercise 14.2 question 5 maths

Since tangent is parallel to chord joining (3, 0) and (4, 1), so, we get:

Hence, the point is

Mean Value Theoram exercise 14.2 question 6

Slope of tangent

Slope of line joining [0, 0] and [1, 2] = = 2

The tangent is parallel to this line:

Slope is equal

Hence, the points are

Mean Value Theoram exercise 14.2 question 7

Since tangent is parallel to chord joining (3, 0) and (4, 1), so, we get:

Hence, the point is

Mean Value Theoram exercise 14.2 question 8

Slope of tangent

Slope of line joining (1, -2) and (2, 2)

The tangent is parallel to the line:

Slope is equal to

Points are

Mean Value Theoram exercise 14.2 question 9 maths

Slope of tangent = 3x2

Slope of line joining (1, 2) and (3, 28)

The tangent is parallel to the line:

Slope is equal to

Points are

Mean Value Theoram exercise 14.2 question 10

The tangent to C is parallel to the chord joining the points (a, 0) and (0, a).

Now,

Now, slope of chord

Slope of chord

Slope of chord

Now,

Now,

Hence Point P is

Mean Value Theoram exercise 14.2 question 11

f (x) is continuous in [a, b] and differentiable in (a, b)

Now,

By Mean Value Theorem,

Hence, proved.

RD Sharma Class 12th Exercise 14.2 has 26 questions, 24 of which are Level 1 and two are Level 2. The Level 1 sums are simple and can be completed in a short time. This exercise is based on Lagrange's Mean Value theorem, calculating the tangents to curve, parabola, and other essential functional problems.

RD Sharma Class 12th Exercise 14.2 material is created by a group of subject experts who have years of experience on question paper patterns. Moreover, it is also updated to the latest version, which means students don't have to worry about missing or additional questions. As this chapter contains many questions, referring to the material will ease your work and reduce preparation time. As there are many questions on a particular concept, students can practice some questions and refer to the solutions to save time and prepare efficiently.

RD Sharma Class 12 Chapter 14 Exercise 14.2 material can help students understand their class lectures better by practicing beforehand. The solved questions make it much easier to verify and pick out mistakes from the answers. This, in turn, reduces the number of doubts and provides a clear understanding of the chapter. Students can stay ahead of their class by practicing these solutions beforehand and completing the portion ahead of time.

RD Sharma Class 12 solutions Chapter 12 Ex 14.2 contain step-by-step solutions, it is straightforward for students to understand. Even as students do not know the chapter, they can easily pick up from these solutions. Moreover, the answers are available for free on Career360's website, which makes them easily accessible.

Thousands of students have already started using RD Sharma Class 12th Exercise 14.2 material as it is convenient and helps score better marks. However, the ones who didn't know about this should give it a try.

- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download E-book1. Can I study only through this material?

Class 12 RD Sharma Chapter 14 Exercise 14.2 Solutions material covers all topics and contains clear concepts. It is the best choice for students for exam preparation.

2. What are the benefits of this material from an exam perspective?

RD Sharma Class 12th Exercise 14.2 material contains solved questions. It will help students save time and be convenient for revision.

3. What is the mean value theorem?

If an arc is drawn using two endpoints, at least one point where its tangent is parallel to the secant of the points.

4. What is Rolle's theorem?

Any differentiable function with equal values at two points will have at least one value where the first derivative is zero. To get more information about this, topic check RD Sharma Class 12 Solutions Chapter 14 Ex 14.2.

5. What is Lagrange's theorem?

If a subgroup is taken from a group of any order, it will be a divisor of that group. This is Lagrange's theorem. For more info, check RD Sharma Class 12 Solutions Mean Value Theorem Ex 14.2.

Application Date:05 September,2024 - 20 September,2024

Application Date:05 September,2024 - 23 September,2024

Admit Card Date:13 September,2024 - 07 October,2024

Admit Card Date:13 September,2024 - 07 October,2024

Get answers from students and experts

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Register for Tallentex '25 - One of The Biggest Talent Encouragement Exam

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

Accepted by more than 11,000 universities in over 150 countries worldwide

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer! Trusted by 3,500+ universities globally

News and Notifications

Back to top