NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives Ex 6.4

NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives Ex 6.4

Page No 216:

Question 1:

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

(i)  (ii)  (iii) 

(iv)  (v)  (vi) 

(vii)  (viii)  (ix) 

(x)  (xi)  (xii) 

(xiii)  (xiv)  (xv) 

Answer:

(i) 

Consider. Let x = 25 and Δx = 0.3.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 0.03 + 5 = 5.03.

(ii) 

Consider. Let x = 49 and Δx = 0.5.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 7 + 0.035 = 7.035.

(iii) 

Consider. Let x = 1 and Δx = − 0.4.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 1 + (−0.2) = 1 − 0.2 = 0.8.

(iv) 

Consider. Let x = 0.008 and Δx = 0.001.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 0.2 + 0.008 = 0.208.

(v) 

Consider. Let x = 1 and Δx = −0.001.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 1 + (−0.0001) = 0.9999.

(vi) 

Consider. Let x = 16 and Δx = −1.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 2 + (−0.03125) = 1.96875.

(vii) 

Consider. Let x = 27 and Δx = −1.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 3 + (−0.0370) = 2.9629.

(viii) 

Consider. Let x = 256 and Δx = −1.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 4 + (−0.0039) = 3.9961.

(ix) 

Consider. Let x = 81 and Δx = 1.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 3 + 0.009 = 3.009.

(x) 

Consider. Let x = 400 and Δx = 1.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 20 + 0.025 = 20.025.

(xi) 

Consider. Let x = 0.0036 and Δx = 0.0001.

Then,

Now, dy is approximately equal to Δy and is given by,

Thus, the approximate value ofis 0.06 + 0.00083 = 0.06083.

(xii) 

Consider. Let x = 27 and Δx = −0.43.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 3 + (−0.015) = 2.984.

(xiii) 

Consider. Let x = 81 and Δx = 0.5.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 3 + 0.0046 = 3.0046.

(xiv) 

Consider. Let x = 4 and Δx = − 0.032.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 8 + (−0.096) = 7.904.

(xv) 

Consider. Let x = 32 and Δx = 0.15.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 2 + 0.00187 = 2.00187.

Question 2:

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

Answer:

Let x = 2 and Δx = 0.01. Then, we have:

f(2.01) = f(x + Δx) = 4(x + Δx)² + 5(x + Δx) + 2

Now, Δy = f(x + Δx) − f(x)

∴ f(x + Δx) = f(x) + Δy

≈

Hence, the approximate value of f (2.01) is 28.21.

Question 3:

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

Answer:

Let x = 5 and Δx = 0.001. Then, we have:

Hence, the approximate value of f (5.001) is −34.995.

Question 4:

Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.

Answer:

The volume of a cube (V) of side x is given by V = x³.

Hence, the approximate change in the volume of the cube is 0.03x³ m³.

Question 5:

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

Answer:

The surface area of a cube (S) of side x is given by S = 6x².

Hence, the approximate change in the surface area of the cube is 0.12x² m².

Question 6:

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

Answer:

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Then,

r = 7 m and Δr = 0.02 m

Now, the volume V of the sphere is given by,

Hence, the approximate error in calculating the volume is 3.92 π m³.

Question 7:

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 in surface area.

Answer:

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Then,

r = 9 m and Δr = 0.03 m

Now, the surface area of the sphere (S) is given by,

S = 4πr²

Hence, the approximate error in calculating the surface area is 2.16π m².

Question 8:

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

Answer:

Let x = 3 and Δx = 0.02. Then, we have:

Hence, the approximate value of f(3.02) is 77.66.

The correct answer is D.

Question 9:

The approximate change in the volume of a cube of side x metres 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

Answer:

The volume of a cube (V) of side x is given by V = x³.

Hence, the approximate change in the volume of the cube is 0.09x³ m³.

The correct answer is C.