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# 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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## 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:

$\boldsymbol{\left ( a \right )\: \sqrt{25.3}}$

$\boldsymbol{\left ( b\right )\: \sqrt{49.5}}$

$\boldsymbol{\left ( c\right )\: \sqrt{0.6}}$

$\boldsymbol{\left ( d \right )\: \left ( 0.009 \right )^{\frac{1}{3}}}$

$\boldsymbol{\left ( e \right )\: \left ( 0.999 \right )^{\frac{1}{10}}}$

$\boldsymbol{\left ( f \right )\: \left ( 15 \right )^{\frac{1}{4}}}$

$\boldsymbol{\left ( g \right )\: \left ( 26 \right )^{\frac{1}{3}}}$

$\boldsymbol{\left ( h \right )\: \left ( 255 \right )^{\frac{1}{4}}}$

$\boldsymbol{\left ( j \right )\: \left ( 82 \right )^{\frac{1}{4}}}$

$\boldsymbol{\left ( l \right )\: \left ( 401\right )^{\frac{1}{4}}}$

$\boldsymbol{\left ( m \right )\: \left ( 0.0037\right )^{\frac{1}{2}}}$

$\boldsymbol{\left ( n \right )\: \left ( 26.5\right )^{\frac{1}{2}}}$

$\boldsymbol{\left ( o\right )\: \left ( 26.57\right )^{\frac{1}{3}}}$

Ans. (a) We have to get the value of $\sqrt{25.3}$

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  $y=\sqrt{x}.....(i)$

Differentiating (i) with respect to x, we get

$\therefore \frac{dy}{dx}=\frac{1}{2}x^{\frac{-1}{ \, \, \, 2}}=\frac{1}{2\sqrt{x}}$

We have differential of y

$dy=\frac{dx}{2\sqrt{x}}.....(ii)$

Subtituting the value of  x = 25 and dx = 0.3

$\boldsymbol{dy=\frac{0.3}{2\sqrt{25}}=\frac{0.3}{2\times 5}=0.03}$……..(ii)

Now, we have

$\boldsymbol{\sqrt{25.3}= y + dy}$

$\boldsymbol{\Rightarrow \sqrt{25.3}= \sqrt{x} + dy}$

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

$\boldsymbol{\therefore \sqrt{25.3}= \sqrt{25} + 0.03=5+0.03 =5.03}$

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

Therefore, approximately value of $\sqrt{25.3}$ is  = 5.03.

(b)

We have to get the value of $\boldsymbol{\sqrt{49.5}}$

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 $y=\sqrt{x}.....(i)$

$\therefore \frac{dy}{dx}=\frac{1}{2}x^{\frac{-1}{ \, \, \, 2}}=\frac{1}{2\sqrt{x}}$

$dy=\frac{dx}{2\sqrt{x}}.....(ii)$

$\boldsymbol{dy=\frac{dx}{2\sqrt{x}}=\frac{0.5}{2\sqrt{49}}=\frac{0.5}{2\times 7}=\frac{0.5}{14}=0.0357}$

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

$\boldsymbol{\sqrt{49.5}=y+dy=\sqrt{x}+dy=\sqrt{49}+0.0357=7+0.0357=7.0357}$

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

Therefore, approximately value of $\sqrt{49.5}$ 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  $y=\sqrt{x}.....(i)$

$\therefore \frac{dy}{dx}=\frac{1}{2}x^{\frac{-1}{ \, \, \, 2}}=\frac{1}{2\sqrt{x}}$

$dy=\frac{dx}{2\sqrt{x}}.....(ii)$

Putting the value dx= -0.04 and x = 0.64

$\boldsymbol{dy=\frac{-0.04}{2\sqrt{0.64}}=\frac{-0.04}{2\times 0.8}=\frac{-0.04}{1.6}=-0.025}$

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

$\boldsymbol{\sqrt{0.6}=y +dy=\sqrt{x}+dy=\sqrt{0.64}-0.025=0.8-0.025=0.775}$

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

Therefore, approximately value √0.6  of  is  = 7.0357.

$\boldsymbol{\left ( d \right )\: \left ( 0.009 \right )^{\frac{1}{3}}}$

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

Let

$\boldsymbol{y=\left ( x \right )^{\frac{1}{3}}}$

$\boldsymbol{\frac{dy}{dx}=\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{\frac{-1}{3}}}$

$\boldsymbol{dy=\frac{1}{3}x^{\frac{-1}{3}}dx}$

Putting the value dx= 0.001 and x = 0.008

$\boldsymbol{dy=\frac{1}{3}\left ( 0.008 \right )^{\frac{-2}{3}}\times 0.001}$

$\boldsymbol{dy=\frac{1}{3}\left ( \frac{8}{1000} \right )^{\frac{-2}{3}}\times \frac{1}{1000}}$

$\boldsymbol{dy=\frac{1}{3}\left ( \frac{2}{10} \right )^{\frac{-2}{3}\times 3}\times \frac{1}{(10)^{3}}}$

$\boldsymbol{=\frac{1}{3}\times \frac{\left ( 10 \right )^{2}}{2^{2}}\times \frac{1}{10^{3}}=\frac{1}{120}=0.0083}$

dy = 0.0083

$\boldsymbol{\left ( 0.009 \right )^{\frac{1}{3}}=y +dy=x^{\frac{1}{3}}+dy=\left ( 0.008 \right )^{\frac{1}{3}}+0.0083=0.2+0.0083=0.208}$

Therefore, approximately the value  of  $\boldsymbol{\left ( 0.009 \right )^{\frac{1}{3}}}$  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

$\boldsymbol{\frac{dy}{dx}=8x +5}$

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

$\boldsymbol{\frac{dy}{dx}=3x^{2}-14x}$

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

$\boldsymbol{dv=\frac{dv}{dx}.\delta x}$…….(i)

V = x³

$\boldsymbol{\frac{dv}{dx}=3x^{2}}$……(ii)

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

$\boldsymbol{dv}\boldsymbol{=3x^{2}\times 0.01x=0.03x^{3}}$

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

$\boldsymbol{\delta s=\frac{ds}{dx}.\delta x.......\left ( i \right )}$

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

S = 6x²

$\boldsymbol{\frac{ds}{dx}=12x.....\left ( ii \right )}$

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

$\boldsymbol{\delta s=12x^{2}\times -0.01=-0.12x^{2}}$

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

$\boldsymbol{\delta v=\frac{dv}{dr}.\delta r}$………(i)

Applying the formula for the volume of sphere

$\boldsymbol{V=\frac{4}{3}\pi r^{2}}$

$\boldsymbol{\frac{dv}{dr}=\frac{4\pi }{3}.\frac{d}{dr}\left ( r^{3} \right )}$

$\boldsymbol{\frac{dv}{dr}=\frac{4\pi }{3}.3r^{2}=4\pi r^{2}}$……..(ii)

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

$\boldsymbol{\delta v=4\pi r^{2}.\delta r}\boldsymbol{=4\times \frac{22}{7}\times 7^{2}\times 0.02}$

$\boldsymbol{\delta v=4\times 22\times 7\times 0.02=12.32}$

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

$\boldsymbol{\delta s=\frac{ds}{dr}.\delta r.....\left ( i \right )}$

The surface area of the sphere is given by

S = 4πr²

$\boldsymbol{\frac{ds}{dr}=4\pi .\frac{d}{dr}\left ( r^{2} \right )=8\pi r}$

$\boldsymbol{\frac{ds}{dr}=8\pi r......\left ( ii \right )}$

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

δs = 8πr × 0.03

$\boldsymbol{\delta s=8\times \frac{22}{7}\times 9\times 0.03=6.79}$

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

$\boldsymbol{\frac{dy}{dx}=6x+15}$

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

$\boldsymbol{\delta v=\frac{dv}{dx}.\delta x.....\left ( i \right )}$

The volume of cube is given as follows

V = x³

$\boldsymbol{\frac{dv}{dx}= 3x^{2}....\left ( ii \right )}$

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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