** NCERT Solutions for Maths Class 12 Exercise 6.4**

Hello Students! Welcome to **maths-formula.com**. In this post, you will find the complete** ****NCERT Solutions for Maths Class 12 Exercise 6.4**.

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Class 12th is a very crucial stage of your student’s life, since you take all important decisions about your career on this stage. Mathematics plays a vital role to take decision for your career because if you are good in mathematics, you can choose engineering and technology field as your career.

**NCERT Solutions for Maths Class 12 Exercise 6.4** helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

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**NCERT Solutions for Maths Class 12 Exercise 6.4** are prepared by the experienced teachers of CBSE board. If you are preparing for JEE Mains and NEET level exams, then it will definitely make your foundation strong.

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**NCERT Solutions for Maths Class 12 Exercise 6.1**

**NCERT Solutions for Maths Class 12 Exercise 6.2**

**NCERT Solutions for Maths Class 12 Exercise 6.3**

**NCERT Solutions for Maths Class 12 Exercise 6.5**

**NCERT Solutions for Maths Class 12 Exercise 6.4**

**Maths Class
12 Ex 6.4 Question 1.**

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

**Solution:**

**Maths Class
12 Ex 6.4 Question 2.**

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

**Solution:**

f(x + ∆x) =
f(2.01) = f(2 + 0.01), f(x) = f(2) = 4 × 2² + 5 × 2 + 2 = 28

f’(x) = 8x + 5

We have, x = 2
and ∆x = 0.01

Now, f(x + ∆x) =
f(x) + ∆f(x)

= f(x) + f’(x) × ∆x = 28 + (8x + 5) ∆x

= 28 + (16 + 5) × 0.01

= 28 + 21 × 0.01

= 28 + 0.21

Hence, f(2.01) = 28.21

**Maths Class
12 Ex 6.4 Question 3.**

Find the
approximate value of f(5.001), where f(x) = x^{3} – 7x^{2} +15.

**Solution:**

Let x + ∆x =
5.001, x = 5 and ∆x = 0.001,

f(x) = x^{3} – 7x^{2} +15

f(5) = –35

f(x + ∆x) = f(x) + ∆f(x) = f(x) + f'(x).∆x

= (x^{3} –
7x² + 15) + (3x² – 14x) × ∆x

f(5.001) = –35 + (3 × 5² – 14 × 5) × 0.001

⇒ f(5.001) = –35 + 0.005

= –34.995

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

**Maths Class
12 Ex 6.4 Question 4.**

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

**Solution:**

The side of the
cube = x metres

Increase in side (∆x) = 1% = 0.01 × x = 0.01 x

Volume of cube V = x^{3}

⇒ dV/dx = 3x^{2 }

∴ change in volume (∆V) = dV/dx ×
∆x

= 3x² ×
0.01 x

= 0.03 x^{3} m^{3}

Hence, the
approximate change in volume is 0.03 x^{3} m^{3}.

**Maths Class
12 Ex 6.4 Question 5.**

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

**Solution:**

The side of the
cube = x m

Decrease in side = 1% = 0.01 x

Increase in side = ∆x = –0.01 x

Surface area of a cube (S) = 6x² m²

⇒ dS/dx = 12x^{ }

∴ change in the surface area (∆S) =
dS/dx × ∆x = 12x × (–0.01 x)

= –0.12
x² m²

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

**Maths Class
12 Ex 6.4 Question 6.**

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.

**Solution:**

Radius of the
sphere (r) = 7 m; ∆r = 0.02 m

Volume of the sphere (V) = 4/3 × Ï€r^{3}

⇒ dV/dr = 4/3 × Ï€ × 3r^{2}

∆V = dV/dr × ∆r = 4/3 × Ï€ × 3r^{2 }× ∆r

= 4Ï€ × 7² × 0.02

= 3.92Ï€ m³

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

**Maths Class
12 Ex 6.4 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 its surface area.

**Solution:**

Radius of the
sphere = 9 m; ∆r = 0.03 m

Surface area of sphere (S) = 4Ï€r²

⇒ dS/dr = 4Ï€ × 2r = 8Ï€r

∆S = dS/dr ×
∆r

= 8Ï€r × ∆r

= 8Ï€ × 9 × 0.03

= 2.16 Ï€ m²

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

**Maths Class
12 Ex 6.4 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

**Solution:**

(D) x + ∆x =
3.02, where x = 3, ∆x = 0.02,

∆f(x) = f(x + ∆x) – f(x)

⇒ f(x + ∆x) = f(x) + ∆f(x) = f(x) + f’(x).∆x

Now, f(x) = 3x² + 15x + 5; f(3) = 77, f’(x) = 6x + 15

f’(3) = 33

∴ f(3.02) = 77 + 33 × 0.02 = 77.66

Hence, the correct answer is option (D).

**Maths Class
12 Ex 6.4 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

**Solution:**

(C) Side of a
cube = x metres

Volume of a cube (V) = x³,

⇒ dV/dx = 3x²

Change in side (∆x)
= 3% of x = 0.03 x

Let ∆V be the change in volume.

Then, ∆V = dV/dx ×
∆x = 3x² × ∆x

But, ∆x = 0.03 x

⇒ ∆V = 3x² × 0.03 x

= 0.09 x³ m³

Hence, the correct answer is option (C).

**NCERT Solutions for Maths Class 12 Exercise 6.1**

**NCERT Solutions for Maths Class 12 Exercise 6.2**

**NCERT Solutions for Maths Class 12 Exercise 6.3**