# NCERT solution class 12 chapter 6 Application of Derivatives exercise 6.4 mathematics part 1

## EXERCISE 6.4

#### 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)

(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 = 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 = 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 = 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

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

f(2.01) = f(+ 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.

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%.

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%

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.

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.

Let 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 (3.02) is

A. 47.66 B. 57.66 C. 67.66 D. 77.66

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

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

#### 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