NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8.

- Continuity and Differentiability Class 12 Ex 5.1
- Continuity and Differentiability Class 12 Ex 5.2
- Continuity and Differentiability Class 12 Ex 5.3
- Continuity and Differentiability Class 12 Ex 5.4
- Continuity and Differentiability Class 12 Ex 5.5
- Continuity and Differentiability Class 12 Ex 5.6
- Continuity and Differentiability Class 12 Ex 5.7

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Continuity and Differentiability |

Exercise |
Ex 5.8 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc 5.8

**Question 1.**

Verify Rolle’s theorem for the function

f(x) = x² + 2x – 8,x∈ [-4,2]

**Solution:**

Now f(x) = x² + 2x – 8 is a polynomial

∴ it is continuous and derivable in its domain x∈R.

Hence it is continuous in the interval [-4,2] and derivable in the interval (- 4,2)

f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0,

f(2) = 2² + 4 – 8 = 8 – 8 = 0

Conditions of Rolle’s theorem are satisfied.

f'(x) = 2x + 2

∴ f’ (c) = 2c + 2 = 0

or c = – 1, c = – 1 ∈ [-4,2]

Thus f’ (c) = 0 at c = – 1.

**Question 2.**

Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?

(i) f(x) = [x] for x ∈ [5,9]

(ii) f (x) = [x] for x ∈ [-2,2]

(iii) f (x) = x² – 1 for x ∈ [1,2]

**Solution:**

(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6,7,8 Hence Rolle’s theorem is not applicable

(ii) f (x) = [x] is not continuous and derivable at -1, 0, 1. Hence Rolle’s theorem is not applicable.

(iii) f(x) = (x² – 1),f(1) = 1 – 1 = 0,

f(2) = 22 – 1 = 3

f(a)≠f(b)

Though it is continous and derivable in the interval [1,2].

Rolle’s theorem is not applicable.

In case of converse if f (c)=0, c ∈ [a, b] then conditions of rolle’s theorem are not true.

(i) f (x) = [x] is the greatest integer less than or equal to x.

∴f(x) = 0, But fis neither continuous nor differentiable in the interval [5,9].

(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [-2,2].

(iii) f (x)=x² – 1, f'(x)=2x. Here f'(x) is not zero in the [1,2], So f (2) ≠ f’ (2).

**Question 3.**

If f: [-5,5] –>R is a differentiable function and if f (x) does not vanish anywhere then prove that f (- 5) ≠ f (5).

**Solution:**

For Rolle’s theorem

If (i) f is continuous in [a, b]

(ii) f is derivable in [a, b]

(iii) f (a) = f (b)

then f’ (c)=0, c e (a, b)

∴ f is continuous and derivable

but f (c) ≠ 0 =>f(a) ≠ f(b) i.e., f(-5)≠f(5)

**Question 4.**

Verify Mean Value Theorem, if

f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4.

**Solution:**

f (x) = x² – 4x – 3. It being a polynomial it is continuous in the interval [1,4] and derivable in (1,4), So all the condition of mean value theorem hold.

then f’ (x) = 2x – 4,

f’ (c) = 2c – 4

f(4)= 16 – 16 – 3 = – 3,

f(1)= 1 – 4 – 3 = – 6

Then there exist a value c such that

**Question 5.**

Verify Mean Value Theorem, if f (x)=x^{3} – 5x^{2} – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1,3) for which f’ (c) = 0.

**Solution:**

f (x)=x^{3} – 5x^{2} – 3x,

It is a polynomial. Therefore it is continuous in the interval [1,3] and derivable in the interval (1,3)

Also, f'(x)=3x²-10x-3

**Question 6.**

Examine the applicability of Mean Value theroem for all three functions given in the above Question 2.

**Solution:**

(i) F (x)= [x] for x ∈ [5,9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.

(ii) f (x) = [x], for x ∈ [-2,2],

Again f (x) = [x] in the interval [-2,2] is neither continous, nor differentiable.

(iii) f(x) = x²-1 for x ∈ [1,2], It is a polynomial. Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)

f (x) = 2x, f(1) = 1 – 1 = 0 ,

f(2) = 4 – 1 = 3, f'(c) = 2c

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