Class 12 maths NCERT solutions of exercise 6.4- Application of Derivatives - Future Study Point

# Class 12 maths NCERT solutions of exercise 6.4- Application of Derivatives

Class 12 maths  NCERT solutions of exercise 6.4-Application of Derivatives are created by a maths expert teacher of future study point. These NCERT solutions of exercise 6.4 help the students in understanding the concept of chapter 6 completely that is required to solve the questions during the examination. Solutions of exercise 6.4 -Application of derivatives are formulated here by a step by step method so it is very easy for the students to understand.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Chapter 6 of class 12 maths NCERT is also useful for the candidates who are pursuing the competitive entrance exams like NDA, CDS, and engineering entrance exams. So, these NCERT solutions are not only very important for your class 12 board exam of 2020-21 but also helpful in qualifying professional and government entrance exams.

## Class 12 maths NCERT solutions of – Application of Derivatives

Exercise 6.1 – Application of Derivatives

Exercise 6.2- Application of Derivatives

Exercise 6.3 – Application of Derivatives

Exercise 6.4-Application of Derivatives

Exercise 6.5 – Application of Derivatives

### You can also Study other NCERT Solutions

NCERT Solutions Class 10 Science from chapter 1 to 16

Class 11 Maths solutions of important questions of chapter 1-Sets

NCERT Solutions Of Chapter 6 Exercise 6.3 Class 12-Application of Derivatives

12 Class Maths NCERT Solutions of exercise 5.2 Chapter- Continuity and Differentiability

Class 11 Physics NCERT solutions of Chapter 1- Physical World

Class 12 Maths NCERT Solutions of the Chapter 1- Relations and functions-Download free pdf

Class 11 Maths NCERT Solutions of the Chapter 1-Sets- PDF

Science and Maths NCERT solutions for Class 9 ,10 and 11 classes

## Class 12 maths NCERT solutions of exercise 6.4- Application of Derivatives

### You can download the pdf of solutions of class 12 maths ex.6.4-  Application of Derivatives

PDF-Class 12 maths NCERT solutions of ex 6.4-Application of Derivatives

Q1.Using differentials, find the approximate value of each of the following up to 3 places of decimal:

Ans. (a) We have to get the value of

Since, 25 is the nearest value of 25.3 whose square root can be achieved.

So, x = 25, δx = dx = 0.3, δy= dy = ?

Let

Differentiating (i) with respect to x, we get

We have differential of y

Subtituting the value of  x = 25 and dx = 0.3

……..(ii)

Now, we have

Here, x= 25 and dx = 0.3 then dy= 0.03

Since, δx and δy is approximately equal to dx and dy respectively.

Therefore, approximately value of is  = 5.03.

(b)

We have to get the value of

Since, 49  is the nearest value of 49.5 whose square root can be achieved.

So, x = 49,and  δx= dx = 0.5 then let’s find what is the value of dy

Let

Here, x= 49 and δx=0.5, then dy = 0.0357

Since, δx and δy is approximately equal to dx and dy respectively.

Therefore, approximately value of is  = 7.0357.

(c)  We have to get the value of √0.6

The near about complete square of 0.6,we can get after squaring near about number 0.5²=.25,0.7²=0.49,0.8² =0.64, since the difference (o.64-0.6)=0.04 is least,so near about complete square of 0.6 is o.64, therefore considering x = 0.64 and dx = -0.04
Let

Putting the value dx= -0.04 and x = 0.64

Here x = 0.64, dx = 0.04 then dy = 0.025

Since, δx and δy is approximately equal to dx and dy respectively.

Therefore, approximately value √0.6  of  is  = 7.0357.

Splitting 0.009 =0.008 + 0.001 =x + δx =x +dx

Let

Putting the value dx= 0.001 and x = 0.008

dy = 0.0083

Therefore, approximately the value  of    is  = 0.208

You can do rest of the parts of question number 1 by the same method.

Q2.Find the approximate value of  f(2.01) where f(x) = 4x² + 5x +2.

Ans. Let f(x) = y, x = 2, δx = dx = 0.01,

y = 4x² + 5x +2

dy = (8x + 5)dx

Putting dx = 0.01

dy = (8x + 5)0.01

f(x +dx) = y + dy = 4x² + 5x +2 + (8x + 5)0.01

f(2 .01)=f(2 +0.01) = 4 × 2² +5×2 +2+ (8×2 +5)0.01

f(2.01) =16 +10 +2+0.21 =28.21

Since δx ≈dx, therefore approximate value of  f(2.01) is 28.21.

Q3.Find the approximate value of where f(5.001) where f(x) =x³ – 7x²+ 15.

Ans. Let f(x) = y, x = 5, δx =dx = 0.001

y=x³ – 7x²+ 15

dy = (3x² – 14x)dx

f(x + dx) =y +dy

f(x + dx) = x³ – 7x²+ 15 + (3x² – 14x)dx

f(5.001) =f(5 + 0.001) =(5³- 7× 5² + 15) + (3×5²-14×5)×0.001

= 125 -175 +15 +(75 -70)0.001

=125 -175 +15 +0.005=-34.995

Since δx ≈dx, therefore approximate value of  f(5.001) is-34.995 .

Q4. Find the approximate change in the volume of a cube of side meters caused by increasing the side by 1%.

Ans. Let the side of cube is = x

The volume of the cube = V

Increase in size = δx = 1% of x = 0.01x

We need to find the approximate change in volume i.e δV

dx≈ δx and δV = dV

…….(i)

V = x³

……(ii)

Substituting (ii) and δx  = 0.01x in (i)

δv =0.03V

Therefore volume of cube increses approximately 0.03 times.

Q5. Find the approximate change in the surface area of a cube of side x meters caused by decreasing the side by 1%.

Ans.We are given the side of cube =x meters and a decrease in side,δx =-1 % = -0.01

As we know δr ≈ dr

The surface area(S ) of the cube is given as follows

S = 6x²

Putting the value of dv/dx from eq.(ii) and δx = -0.01 in eq.(i)

Hence approximate error in the surface area of the cube is = – 0.12 x²(The  surface area of the cube will be decreased by -0.12x².)

Q6. If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

Ans. Let the radius of the sphere = r

and error in measuring the radius δr = 0.02 m and the error in volume =δv

The volume of the sphere = V

As we know dv ≈ δv

………(i)

Applying the formula for the volume of sphere

……..(ii)

Putting the value of dv/dr from eq.(ii) and δr =0.02, r=7  in eq.(i)

Hence the approximate error in the volume of the sphere is the 12.32-meter cube.

Q7.If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.

Ans. We are given the radius of the sphere ,r= 9m with an error,δr = 0.03 m

Let’s find the error in the surface area of the sphere , δs

As we know, δr ≈ dr, ds ≈δs

The surface area of the sphere is given by

S = 4πr²

Putting the value of ds/dr from eq.(ii) and δr =0.03, r=9  in eq.(i)

δs = 8πr × 0.03

Hence the error in the surface area of the sphere is = 6.79 sq.m

Q8.If  f(x) = 3x² +15x +5 then the approximate value of f(3.02)  is:

(A) 47.66

(B) 57.66

(C) 67.66

(D) 77.66

Ans. Let y = f(x), x = 3 and δx = 0.02

y = 3x² +15x +5

dy = (6x + 15) dx

As we know, dx ≈δx = 0.02, therefore sustituting the value dx = 0.02

dy = (6x +15)0.02

f(x + dx) = y + dy=3x² +15x +5 + (6x + 15) dx

f(3.02) = f( 3 + 0.02) = 3×3² +15×3 +5 + (6×3 + 15) 0.02

f(3.02) = 27 +45 + 5 + 33×0.02 = 77.66

Therefore the approximate value of f(3.02) is 77.66, hece the correct aswer is (D)

Q9.The approximate change in the volume of a cube of side x meters caused by increasing the side by 3% is:

(A) 0.06 x³ m3

(B) 0.6  x³ m3

(C) 0.09 x³ m3

(D) 0.9  x³ m3

Ans. We are given the side of cube x meter are an increase in side, δx = 3% of x = 0.03x

As we know δx ≈ dx, δv ≈dv

The volume of cube is given as follows

V = x³

Putting the value of dv/dx from eq.(ii)  and δx = 0.03x in eq.(i)

δv = 3x² × 0.03x = 0.09 x³

Therefore the volume is increased by 0.09 x³

Government jobs after 10 or 12 th pass : Qualify Entrance exams of SSC MTS and SSC CHSL

Online shopping of Top class branded : T-Shirts, Shirts,Jeans, Suits ,Shoes, Watches, Smalt T.V,Laptops,Mobiles , Computers and E-books of science and maths.

The best books for cracking the competitive entrance exams of CPO,NDA, IAS, CDS and Bank PO

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

### NCERT Solutions for class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

### NCERT Solutions for class 9 science

 Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion Chapter 2-Is matter around us pure? Chapter 10- Gravitation Chapter3- Atoms and Molecules Chapter 11- Work and Energy Chapter 4-Structure of the Atom Chapter 12- Sound Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ? Chapter 6- Tissues Chapter 14- Natural Resources Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources Chapter 8- Motion Last years question papers & sample papers

### NCERT Solutions for class 10 maths

 Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT Solutions for Class 10 Science

 Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world Chapter 4- Carbon and its Compounds Chapter 12- Electricity Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current Chapter 6- Life Process Chapter 14-Sources of Energy Chapter 7-Control and Coordination Chapter 15-Environment Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

### NCERT Solutions for class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT Solutions for Class 11 Physics

Chapter 1- Physical World

chapter 3-Motion in a Straight Line

### NCERT Solutions for Class 11 Chemistry

Chapter 1-Some basic concepts of chemistry

Chapter 2- Structure of Atom

### NCERT Solutions for Class 11 Biology

Chapter 1 -Living World

### NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution

Scroll to Top