NCERT Solutions for Maths Class 12 Exercise 5.8

NCERT Solutions for Maths Class 12 Exercise 5.8

                                                                      

       NCERT Solutions for Maths Class 12 Exercise 5.8


Hello Students! Welcome to maths-formula.com. In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 5.8.

 

You can download the PDF of NCERT Books Maths Chapter 5 for your easy reference while studying NCERT Solutions for Maths Class 12 Exercise 5.8.

 

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 5.8 helps you to solve each and every problem with step by step explanation which makes you strong in mathematics.

 

All the schools affiliated with CBSE, follow the NCERT books for all subjects. You can check your syllabus from NCERT Syllabus for Mathematics Class 12.

 

NCERT Solutions for Maths Class 12 Exercise 5.8 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.

 

If you want to recall All Maths Formulas for Class 12, you can find it by clicking this link.

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NCERT Solutions for Maths Class 12 Exercise 5.1

NCERT Solutions for Maths Class 12 Exercise 5.2

NCERT Solutions for Maths Class 12 Exercise 5.3

NCERT Solutions for Maths Class 12 Exercise 5.4

NCERT Solutions for Maths Class 12 Exercise 5.5

NCERT Solutions for Maths Class 12 Exercise 5.6

NCERT Solutions for Maths Class 12 Exercise 5.7

 

NCERT Solutions for Maths Class 12 Exercise 5.8

 

Maths Class 12 Ex 5.8 Question 1.

Verify Rolle’s theorem for the function f(x) = x² + 2x – 8, x [-4, 2].

Solution:
Here, f(x) = x² + 2x – 8 is a polynomial.
Therefore, it is continuous and differentiable in its domain x
R.
Hence, it is continuous in the interval [–4, 2] and differentiable in the interval (–4, 2).
f(–4) = (–4)² + 2(–4) – 8 = 16 – 8 – 8 = 0 and f(2) = 2² + 4 – 8 = 8 – 8 = 0
Conditions of Rolle’s theorem are satisfied because f(–4) = f(2) = 0.
f'(x) = 2x + 2
f’(c) = 2c + 2 = 0 or c = –1

Hence, Rolle’s theorem is verified at c = –1 [–4, 2].

Maths Class 12 Ex 5.8 Question 2.

Examine if Rolle’s theorem is applicable to any of the following functions. Can you say something 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:

By Rolle’s theorem, f : [a, b] → R, if         

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

(c) f(a) = f(b)

Then, there exists some c (a, b) such that f′(c) = 0.

Thus, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions given above.


(i) We have, f(x) = [x] for x [5, 9]

     Since, the given function f(x) is not continuous at every integral point.

     In general, f(x) is not continuous at x = 5 and x = 9.

     Therefore, f(x) is not continuous in [5, 9]

     Also, f(5) = [5] = 5 and f(9) = [9] = 9

     Thus, f(5) ≠ f(9)

Since the LHD and RHD of the function f(x) at n (5, 9) are not equal, the function f(x) is not differentiable in (5, 9).

Hence, Rolle’s theorem is not applicable to the function f(x) = [x] for x [5, 9].


(ii) We have, f(x) = [x] for x
[2, 2]

     Since, the given function f(x) is not continuous at every integral point.

     In general, f(x) is not continuous at x = 2 and x = 2.

     Therefore, f(x) is not continuous in [2, 2]

     Also, f(2) = [2] = 2 and f(2) = [2] = 2

     Thus, f(2) ≠ f(2)

Since the LHD and RHD of the function f(x) at n (–2, 2) are not equal, the function f(x) is not differentiable in (–2, 2).  

Hence, Rolle’s theorem is not applicable to the function f(x) = [x] for x [–2, 2].

 

(iii) We have, f(x) = x² – 1 for x [1, 2]

Since f is a polynomial, it is continuous in [1, 2] and differentiable in (1, 2).

Now, f(1) = 1 – 1 = 0 and f(2) = 4 – 1 = 3
Here, f(a) ≠ f(b)
Since the function does not satisfy the conditions of Rolle’s theorem

Hence, Rolle’s theorem is not applicable to the given function f(x) = x² – 1 for x [1, 2].

In case of converse of Rolle’s theorem, if f(c) = 0, c [a, b], then the 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 f is neither continuous nor differentiable in the interval [5, 9].
(ii) Here also, 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 interval [1, 2]. So, f(2) ≠ f’(2).

Maths Class 12 Ex 5.8 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:
We have, f : [−5, 5→ R is a differentiable function.

Since every differentiable function is a continuous function, we obtain

f is continuous on [−5, 5]

f is continuous on (−5, 5)

Thus, by the Mean Value Theorem, there exists c (5, 5) such that 

f’(c) = [f(5) – f(–5)]/[5 – (–5)]  

10f’(c) = [f(5) – f(–5)]

It is also given that f′(x) does not vanish anywhere.

Therefore, f′(c) ≠ 0

Thus, 10f′(c) ≠ 0

f(5) − f(−5) ≠ 0

f(5) ≠ f(−5)

10f(c) 0

f(5) f(5) 0

f(5) f(5)

Hence proved.

Maths Class 12 Ex 5.8 Question 4.

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

Solution:
We have, f(x) = x² – 4x – 3. f being a polynomial, it is continuous in the interval [1, 4] and differentiable in (1, 4). So, all the conditions of mean value theorem hold.
Now, f’(x) = 2x – 4,
Then, f’(c) = 2c – 4
f(4) = 16 – 16 – 3 = –3,
f(1) = 1 – 4 – 3 = –6
Then there exist a value c such that

Hence, the mean value theorem is verified for the given function.

Maths Class 12 Ex 5.8 Question 5.

Verify Mean Value Theorem, if f(x) = x3 – 5x2 – 3x in the interval [a, b], where a = 1 and b = 3. Find all c (1, 3) for which f’(c) = 0.

Solution:
We have, f(x) = x3 – 5x2 – 3x,
It is a polynomial. Therefore, it is continuous in the interval [1, 3] and differentiable in the interval (1, 3).
Also, f'(x) = 3x² – 10x – 3, then  f'(c) = 3c² – 10c – 3

f(1) = 1 – 5 – 3 = –7 and f(3) = 27 – 45 – 9 = –27

f'(c10

⇒ 3c2 − 10– 3 10

⇒ 3c− 100

⇒ 3c− 3− 70

⇒ 3c(− 1− 7(− 1

⇒ (− 1(3− 70

⇒ 17/3   [where 7/3 ∈ (13)]

Thus, Mean Value Theorem is verified for the given function and c = 7/3 ∈ (1, 3) is the only point for which f'(c) = 0.

Maths Class 12 Ex 5.8 Question 6.

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

Solution:
Mean Value Theorem states that for a function f : [a, b] → R, if

a.      f is continuous on [a, b]

b.     f is continuous on (a, b)

Then there exists some c (a, b) such that 

Thus, Mean Value Theorem is not applicable to those functions that do not satisfy any of three conditions given above.

(i) We have f(x) = [x] for x [5, 9]

Since, the given function f(x) is not continuous at every integral point.

In general, f(x) is not continuous at x = 5 and x = 9

Therefore, f(x) is not continuous in [5, 9]

The differentiability of f in (5, 9) is checked as follows.

Let n be an integer such that n (5, 9)

Since LHD and RHD of f at x = n are not equal, f is not differentiable at x = n.

Therefore, f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the Mean Value Theorem.

Thus, Mean Value Theorem is not applicable for f(x) = [x] for x [5, 9]

 

(ii) We have, f(x) = [x] for x [−2, 2]

Since, the given function f(x) is not continuous at every integral point.

In general, f(x) is not continuous at x = −2 and x = 2

Therefore, f(x) is not continuous in [−2, 2]

The differentiability of f in (−2, 2) is checked as follows.

Let n be an integer such that n (−2, 2)

Since LHD and RHD of f at x = n are not equal, f is not differentiable at x = n

Therefore, f is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Thus, Mean Value Theorem is not applicable for f(x) = [x] for x [−2, 2].

 

(iii) We have, f(x) = x2 − 1 for x [1, 2]

Since, f being a polynomial function is continuous in [1, 2] and is differentiable in (1, 2)

It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for f(x) = x2 − 1 for x [1, 2].

It can be proved as follows.

We have, f(x) = x2 − 1

Then, f(1) = (1)2 – 1 = 0 and f(2) = (2)2 – 1 = 3

Hence, f′(x) = 2x

Thus, f′(c) = 3

2c = 3

c = 3/2

c = 1.5 [where 1.5 [1, 2]]


NCERT Solutions for Maths Class 12 Exercise 5.1

NCERT Solutions for Maths Class 12 Exercise 5.2

NCERT Solutions for Maths Class 12 Exercise 5.3

NCERT Solutions for Maths Class 12 Exercise 5.4

NCERT Solutions for Maths Class 12 Exercise 5.5

NCERT Solutions for Maths Class 12 Exercise 5.6

NCERT Solutions for Maths Class 12 Exercise 5.7

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