NCERT Solutions Class 12 Mathematics Chapter 5 Continuity and Differentiability

Exercise 5.1

Question. Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
Answer :
The given function is f(x) = 5x - 3 
At x = 0, f(0) = 5× 0 -3 = -3 

Question. Examine the continuity of the function f (x) = 2x² – 1 at x = 3.
Answer :
The given function is f(x) = 2x² - 1 

Question. Examine the following functions for continuity. 
(i) f(x) = x - 5 
(ii) f(x) = [1/(x- 5)] , x ≠ 5 
(iii) f(x) = (x² - 25)/(x + 5), x ≠ - 5 
(iv) f(x) = |x - 5|, x ≠ 5 
Answer :
(i) The given function is f(x) = x - 5
It is evident that f is defined at every real number k and its value at k is k - 5 .
It is also observed that

Hence, f is continuous at every real number and therefore, it is a continuous function. 

(ii) The given function is f(x) =  [1/(x- 5)] , x ≠ 5 
For any real number k ≠ 5, we obtain

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function. 

(iii) The given function is f(x) = (x² - 25)/(x + 5), x ≠ - 5
For any real number c ≠ - 5 , we obtain 

Hence, f is continuous at every point in the domain of f and therefore, it is a continuous function.

Therefore, f is continuous at all real numbers greater than 5.
Hence, f is continuous at every real number and therefore, it is a continuous function 

Question. Prove that the function f(x) = xⁿ is continuous at x = n, where n is a positive integer. 
Answer :
The given function is f(x) = xⁿ 
It is evident that f is defined at all positive integers, n, and its value at n is nⁿ .

Question. Is the function f defined by f(x) =

continuous at x = 0? At x = 1? At x = 2?
Answer :

Question. Find all points of discontinuity of f,  where f is defined by 

Answer :
It is evident that the given function f is defined at all the points of the real line. 
Let c be a point on the real line. Then, three cases arise. 
c < 2
c > 2 
c = 2 
Case I : c < 2 
f(c) = 2c + 3 
Then, 

It is observed that the left and right hand limit of f at x = 2 do not coincide. 
Therefore, f is not continuous at x = 2 .
Hence, x = 2 is the only point of discontinuity of f.