NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4.

- Application of Derivatives Class 12 Ex 6.1
- Application of Derivatives Class 12 Ex 6.2
- Application of Derivatives Class 12 Ex 6.3
- Application of Derivatives Class 12 Ex 6.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Application of Derivatives |

Exercise |
Ex 6.4 |

Number of Questions Solved |
9 |

Category |
NCERT Solutions |

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

**Question 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)

**Solution:**

(i) y + ∆y =

=

=

∴ x = 25

∆x = 0.3

⇒ y = √x

**Question 2.**

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

**Solution:**

f(x+∆x) = f(2.01), f(x) = f (2) = 4.2² + 5.2 + 2 = 28,

f’ (x) = 8x + 5 Now, f(x + ∆x) = f(x) + ∆f(x)

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

= 28 + (16 + 5) x 0.01

= 28 + 21 x 0.01

= 28 + 0.21

Hence,f(2 x 01)

= 28 x 21.

**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) = 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.

**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 meters.

Increase in side = 1% = 0.01 × x = 0.01 x

Volume of cube V= x^{3}

∴ ∆v = × ∆x

= 3x² × 0.01 x

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

**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 cube = 6x² m² = S

∴ × ∆x = 12x × (- 0.01 x)

= – 0.12 x² m².

**Question 6.**

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

**Solution:**

Radius of the sphere = 7m : ∆r = 0.02 m.

Volume of the sphere V =

= 4π × 7² × 0.02

= 3.92 πm³

**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 = 0.03m

Surface area of sphere S = 4πr²

∆s = × ∆r

= 8πr × ∆r

= 8π × 9 × 0.03

= 2.16 πm².

**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=30, ∆x=.02,

∆f(x) = f(x + ∆x) – f(x)

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

Now f(x) = 3×2 + 15x + 5; f(3) = 77, f’ (x) = 6x + 15

f’ (3) = 33

∴ f (3.02) = 87 + 33 x 0 02 = 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

**Solution:**

(c) Side of a cube = x meters

Volume of cube = x³,

for ∆x. ⇒ 3% of x = 0.03 x

Let ∆v be the change in v0l. ∆v = x ∆x = 3x² × ∆x

But, ∆x = 0.03 x

⇒ ∆v = 3x² x 0.03 x

= 0.09 x³m³

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