Easy Method for Continuity 12th Ncert Pdf
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NCERT Class 12 Maths Textbook Chapter 5 With Answer PDF Free Download
Chapter 5: Continuity and Differentiability
5.1 Introduction
This chapter is essentially a continuation of our study of differentiation of functions in Class XI.
We had learned to differentiate certain functions like polynomial functions and trigonometric functions.
In this chapter, we introduce the very important concepts of continuity, differentiability, and relations between them. We will also learn the differentiation of inverse trigonometric functions.
Further, we introduce a new class of functions called exponential and logarithmic functions. These functions lead to powerful techniques of differentiation.
We illustrate certain geometrically obvious conditions through differential calculus. In the process, we will learn some fundamental theorems in this area.
5.3.2 Derivatives of implicit functions
Until now we have been differentiating various functions given in the form y = f (x). But it is not necessary that functions are always expressed in this form.
For example, consider one of the following relationships between x and y:
x – y – π = 0
x + sin XY – y = 0
In the first case, we can solve for y and rewrite the relationship as y = x – π. In the second case, it does not seem that there is an easy way to solve for y.
Nevertheless, there is no doubt about the dependence of y on x in either of the cases.
When a relationship between x and y is expressed in a way that it is easy to solve for y and write y = f (x), we say that y is given as an explicit function of x.
In the latter case, it is implicit that y is a function of x and we say that the relationship of the second type, above, gives function implicitly. In this subsection, we learn to differentiate implicitly functions.
| Author | NCERT |
| Language | English |
| No. of Pages | 47 |
| PDF Size | 596 KB |
| Category | Mathematics |
| Source/Credits | ncert.nic.in |
NCERT Solutions Class 12 Maths Chapter 5 Continuity and Differentiability
Question 1:
Prove that the function is continuous at
ANSWER:
The given function is f(x)=5x−3At x=0, f(0)=5×0−3=−3limx→0f(x)=limx→0(5x−3)=5×0−3=−3∴limx→0f(x)=f(0)The given function is fx=5x-3At x=0, f0=5×0-3=-3limx→0fx=limx→05x-3=5×0-3=-3∴limx→0fx=f0
Therefore,f is continuous atx = 0
Therefore,fis continuous atx = −3
Therefore,f is continuous atx = 5
Question 2:
Examine the continuity of the function
.
ANSWER:
Thus,f is continuous atx = 3
Question 3:
Examine the following functions for continuity.
(a)
(b)
(c)
(d)
ANSWER:
(a) The given function is
It is evident thatf is defined at every real numberk and its value atk isk − 5.
It is also observed that,
Hence, fis continuous at every real number, and therefore, it is a continuous function.
(b) The given function is
For any real numberk ≠ 5, we obtain
Hence,f is continuous at every point in the domain off, and therefore, it is a continuous function.
(c) The given function is
For any real numberc ≠ −5, we obtain
Hence,f is continuous at every point in the domain off, and therefore, it is a continuous function.
(d) The given function is
This functionf is defined at all points of the real line.
Letc be a point on a real line. Then,c < 5 orc = 5 orc > 5
Case I:c < 5
Then,f(c) = 5 −c
Therefore,f is continuous at all real numbers less than 5.
Case II :c = 5
Then,
Therefore,fis continuous atx = 5
Case III:c > 5
Therefore,f is continuous at all real numbers greater than 5.
Hence, fis continuous at every real number, and therefore, it is a continuous function.
Question 4:
Prove that the function
is continuous atx =n, wheren is a positive integer.
ANSWER:
The given function isf (x) =x n
It is evident thatf is defined at all positive integers,n, and its value atn isn n .
Therefore,fis continuous atn, wheren is a positive integer.
Question 5:
Is the functionf defined by
continuous atx = 0? Atx = 1? Atx = 2?
ANSWER:
The given functionf is
Atx = 0,
It is evident that fis defined at 0 and its value at 0 is 0.
Therefore,f is continuous atx = 0
Atx = 1,
fis defined at 1 and its value at 1 is 1.
The left-hand limit of fatx = 1 is,
The right-hand limit offatx = 1 is,
Therefore, fis not continuous atx = 1
Atx= 2,
fis defined at 2 and its value at 2 is 5.
Therefore,f is continuous atx= 2
NCERT Class 12 Maths Textbook Chapter 5 With Answer PDF Free Download
Source: https://panotbook.com/continuity-and-differentiability-class-12/
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