GFG App
Open App
Browser
Continue

# Class 12 RD Sharma Solutions – Chapter 14 Differentials, Errors and Approximations – Exercise 14.1 | Set 2

### (xiv)

Solution:

Considering the function as

y = f(x) = cos x

Taking x = Ď€/3, and

x+â–łx = 11Ď€/36

â–łx = 11Ď€/36-Ď€/3 = -Ď€/36

y = cos x

= cos (Ď€/3) = 0.5

= – sin x

= – sin (Ď€/3) = -0.86603

â–ły = dy =  dx

â–ły = (-0.86603) (-Ď€/36)

â–ły = 0.0756

Hence, = 0.5+0.0756 = 0.5755

### (xv)

Solution:

Considering the function as

y = f(x) =

Taking x = 81, and

x+â–łx = 80

â–łx = 80-81 = -1

â–ły = dy =  dx

â–ły =  â–łx

â–ły = (-1) =  = -0.009259

Hence,

= y+â–ły = 3 + (-0.009259) = 2.99074

### (xvi)

Solution:

Considering the function as

y = f(x) =

Taking x = 27, and

x+â–łx = 29

â–łx = 29-27 = 2

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  (2) = 0.074

Hence,

= y+â–ły = 3+0.074 = 3.074

### (xvii)

Solution:

Considering the function as

y = f(x) =

Taking x = 64, and

x+â–łx = 66

â–łx = 66-64 = 2

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  (2) = 0.042

Hence,

= y+â–ły = 4+0.042 = 4.042

### (xviii)

Solution:

Considering the function as

y = f(x) =

Taking x = 25, and

x+â–łx = 26

â–łx = 26-25 = 1

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  (1) = 0.1

Hence,

= y+â–ły = 5 + 0.1 = 5.1

### (xix)

Solution:

Considering the function as

y = f(x) =

Taking x = 36, and

x+â–łx = 37

â–łx = 37-36 = 1

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  (1) = 0.0833

Hence,

= y+â–ły = 6 + 0.0833 = 6.0833

### (xx)

Solution:

Considering the function as

y = f(x) =

Taking x = 0.49, and

x+â–łx = 0.48

â–łx = 0.48-0.49 = -0.01

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  (-0.01) = -0.007143

Hence,

= y+â–ły = 0.7 + (-0.007143) = 0.693

### (xxi)

Solution:

Considering the function as

y = f(x) =

Taking x = 81, and

x+â–łx = 82

â–łx = 82-81 = 1

â–ły = dy =  dx

â–ły = â–łx

â–ły =  = 0.009259

Hence,

= y+â–ły = 3 + 0.009259 = 3.009259

### (xxii)

Solution:

Considering the function as

y = f(x) =

Taking x = 16/81, and

x+â–łx = 17/81

â–łx = 17/81-16/81 = 1/81

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  = 0.01042

Hence,

= y+â–ły = 2/3 + 0.01042 = 0.6771

### (xxiii)

Solution:

Considering the function as

y = f(x) =

Taking x = 32, and

x+â–łx = 33

â–łx = 33-32 = 1

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  = 0.0125

Hence,

= y+â–ły = 2 + 0.0125 = 2.0125

### (xxiv)

Solution:

Considering the function as

y = f(x) =

Taking x = 36, and

x+â–łx = 36.6

â–łx = 36.6-36 = 0.6

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  (0.6) = 0.05

Hence,

= y+â–ły = 6 + 0.05 = 6.05

### (xxv)

Solution:

Considering the function as

y = f(x) =

Taking x = 27, and

x+â–łx = 25

â–łx = 25-27 = -2

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  (-2) = -0.07407

Hence,

= y+â–ły = 3+(-0.07407) = 2.9259

### (xxvi)

Solution:

Considering the function as

y = f(x) =

Taking x = 49, and

x+â–łx = 49.5

â–łx = 49.5-49 = 0.5

â–ły = dy =  dx

â–ły =  â–łx

â–ły =  (0.5) = 0.0357

Hence,

= y+â–ły = 7 + 0.0357 = 7.0357

### Question 10: Find the appropriate value of f(2.01), where f(x) = 4x2+5x+2

Solution:

Considering the function as

y = f(x) = 4x2+5x+2

Taking x = 2, and

x+â–łx = 2.01

â–łx = 2.01-2 = 0.01

y = 4x2+5x+2

= 4(2)2+5(2)+2 = 28

= 8x+5

= 8(2)+5 = 21

â–ły = dy =  dx

â–ły = (21) â–łx

â–ły = (21) (0.01) = 0.21

Hence,

f(2.01) = y+â–ły = 28 + 0.21 = 28.21

### Question 11: Find the appropriate value of f(5.001), where f(x) = x3-7x2+15

Solution:

Considering the function as

y = f(x) = x3-7x2+15

Taking x = 5, and

x+â–łx = 5.001

â–łx =5.001-5 = 0.001

y = x3-7x2+15

= (5)3-7(5)2+15 = -35

= 3x2-14x

= 3(5)2-14(5) = 5

â–ły = dy =  dx

â–ły = (5) â–łx

â–ły = (5) (0.001) = 0.005

Hence,

f(5.001) = y+â–ły = -35 + 0.005 = -34.995

### Question 12: Find the appropriate value of log10 1005, given that log10 e=0.4343

Solution:

Considering the function as

y = f(x) = log10 x

Taking x = 1000, and

x+â–łx = 1005

â–łx =1005-1000 = 5

y = log10 x =

= log10 1000 = 3

= 0.0004343

â–ły = dy =  dx

â–ły = (0.0004343) â–łx

â–ły = (0.0004343) (5) = 0.0021715

Hence, log10 1005 = y+â–ły = 3 + 0.0021715 = 3.0021715

### Question 13: If the radius of a sphere is measured as 9cm with an error of 0.03m, find the approximate error in calculating its surface area.

Solution:

According to the given condition,

As, Surface area = 4Ď€x2

Let â–łx be the change in the radius and â–ły be the change in the surface area

x = 9

â–łx =  0.03m = 3cm

x+â–łx = 9+3 = 12cm

= 4Ď€x2 = 4Ď€(9)2 = 324 Ď€

= 8Ď€x

= 8Ď€(9) = 72Ď€

â–ły = dy =  dx

â–ły = (72Ď€) â–łx

â–ły = (72Ď€) (3) = 216 Ď€

Hence, approximate error in surface area of the sphere is 216 Ď€ cm2

### Question 14: Find the approximate change in the surface area of a cube as side x meters caused by decreasing the side by 1%.

Solution:

According to the given condition,

As, Surface area = 6x2

Let â–łx be the change in the length and â–ły be the change in the surface area

â–łx/x Ă— 100 =  1

= 6(2x) = 12x

â–ły =  â–łx

â–ły = (12x) (x/100)

â–ły = 0.12 x2

Hence, the approximate change in the surface area of a cubical box is  0.12 x2 m2

### Question 15: If the radius of a sphere is measured as 7m with an error of 0.02m, find the approximate error in calculating its volume.

Solution:

According to the given condition,

As, Volume of sphere = Ď€x3

Let â–łx be the error in the radius and â–ły be the error in the volume

x = 7

â–łx =  0.02 cm

Ď€(3x2) = 4Ď€x2

= 4Ď€(7)2 = 196Ď€

â–ły = dy =  dx

â–ły = (196Ď€) â–łx

â–ły = (196Ď€) (0.02) = 3.92 Ď€

Hence, approximate error in volume of the sphere is 3.92 Ď€ cm2

### Question 16: Find the approximate change in the volume of a cube as side x meters caused by increasing the side by 1%.

Solution:

According to the given condition,

As, Volume of cube = x3

Let â–łx be the change in the length and â–ły be the change in the volume

â–łx/x Ă— 100 =  1

= 3x2

â–ły =  â–łx

â–ły = (3x2) (x/100)

â–ły = 0.03 x3

Hence, the approximate change in the volume of a cubical box is 0.03 x3 m3

My Personal Notes arrow_drop_up