Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7, Exercise 5.8

Exericse 5.7

Question 1.
x² + 3x + 2 = y(say)
Solution:

 Question 2.
x20 = y(say)
Solution:

Question 3.
x.cos x = y(say)
Solution:

 Question 4.
log x = y (say)
Solution:

 Question 5.
x3 log x = y (say)
Solution:
x3 log x = y

 Question 6.
ex sin5x = y
Solution:
ex sin5x = y

 Question 7.
e6x cos3x = y
Solution:
e6x cos3x = y

Question 8.
tan-1 x = y
Solution:

 Question 9.
log(logx) = y
Solution:
log(logx) = y

 Question 10.
sin(log x) = y
Solution:
sin(log x) = y

 Question 11.
If y = 5 cosx – 3 sin x, prove that 
Solution:



Hence proved

 Question 12.
If y = cos-1 x, Find  in terms of y alone.
Solution:

Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that

Solution:
Given that
y = 3 cos (log x) + 4 sin (log x)

 Question 14.

Solution:
Given that

Question 15.
If y = 500e7x + 600e-7x, show that .
Solution:
we have
y = 500e7x + 600e-7x

 Question 16.
If ey(x+1) = 1,show that 
Solution:

Question 17.
If y=(tan-1 x)² show that (x²+1)²y2+2x(x²+1)y1=2
Solution:
we have
y=(tan-1 x)²

 

Exercise 5.8

September 4, 2019 by Bhagya Leave a Comment

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]

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)=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:
f (x)=x3 – 5x2 – 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

Thus f’ (c) = 0 at c = – 1.