# 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**** m**^{3}** **

**(B) 0.6 **** x³ m**^{3}** **

**(C) 0.09 x³**** m**^{3}

**(D) 0.9 **** x³ m**^{3}

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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