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.

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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

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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

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