>

Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.1

Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.1

Continuity and Differentiability Exercise 5.1 Solutions

1. Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

Solution

The given function is f(x) = 5x - 3 
At x = 0, f(0) = 5× 0 -3 = -3 

Therefore, f is continuous at x = 5 


2. Examine the continuity of the function f (x) = 2x2 – 1 at x = 3.

Solution

The given function is f(x) = 2x2 - 1 
At x = 3, f(x) = f(3) = 2× 32 - 1 = 17 

Thus, f is continuous at x = 3


3. 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 

Solution

(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. 

(iv) The given function is 
This function f is defined at all points of the real line. 
Let c be a point on a real line. Then, c < 5 or c = 5 or c > 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, f(c) = f(5) = (5 - 5) = 0 

Therefore, f is continuous at x = 5 
Case III : c > 5 
Then, f(c) = f(5) = c - 5 

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 


4. Prove that the function f(x) = xn is continuous at x = n, where n is a positive integer. 
Solution
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 . 

Therefore, f is continuous at n, where n is a positive integer

5. Is the function f defined by f(x)=  continuous at x = 0? At x = 1? At x = 2?
Solution
The given function f is f(x) = 
At x = 0, 
It is evident that f is defined at 0 and its value at 0 is 0 .

Therefore, f is continuous at x = 0 
At x = 1, 
f is defined at 1 and its value at 1 is 1. 
The left hand limit of f at x = 1 is, 

Therefore, f is not continuous at x = 1 
At x = 2, 
f is defined at 2 and its value at 2 is 5. 

Therefore, f is continuous at x = 2 

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

Therefore, f is continuous at all points x, such that x < 2. 
Case II : c > 2 
Then, 
f(c) = 2c - 3 

Therefore, f is continuous at all points x, such that x > 2
Case III : c = 2 
Then, the left hand limit of f at x = 2 is, 
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. 

7. Find all points of discontinuity of f, where f is defined by 
Solution
The given function f is 
The given function f is defined at all the points of the real line. 
Let c be a  point on the real line. 
Case I:
If c < -3, then f(c) = -c + 3 

Therefore, f is continuous at all points x, such that x < -3 
Case II: 
If c = -3, then f(-3) = -(-3) + 3 = 6 

Therefore, f is continuous at x = -3
Case III : 
If -3 < c <3, then f(c) 
Therefore, f is continuous in (-3, 3). 
Case IV : 
If c = 3, then the left hand limit of f at x = 3 is  

It is observed that the left and right hand limit of f at x = 3 do not coincide. 
Therefore, f is not continuous at x = 3 
Case V : 

Therefore, f is continuous at all points x, such that x > 3
Hence, x = 3 is the only point of discontinuity of f. 

8. Find all points of discontinuity of f, where f is defined by  
Solution
The given function f is 
It is known that, x < 0 ⇒ |x| = -x and x > 0 ⇒ |x| = x 
Therefore, the given function can be rewritten as 

The given function f is defined at all the points of the real line. 
Let c be a point on the real line .  
Case I:
If c < 0, then f(c) = -1 

Therefore, f is continuous at all points x < 0
Case II : 
If c = 0, Then the left hand limit of f at x = 0 is 

It is observed that the left and right hand limit of f at x = 0 do not coincide. 
Therefore, f is not continuous at x = 0 
Case III : 
If c > 0, then f(c) = 1 

Therefore, f is continuous at all points x, such that x > 0 
Hence, x = 0 is the only point of discontinuity of f. 

9. Find all points of discontinuity of f, where f is defined by 
Solution
The given function f is 
It is known that, x < 0 ⇒ |x| = -x 
Therefore, the given function can be rewritten as 
Therefore, the given function is a continuous function. 
Hence, the given function has no point of discontinuity. 

10. Find all points of discontinuity of f, where f is defined by
Solution
The given function f is 
The given function f is defined at all the points of the real line. 
Let C be a point on the real line. 
Case I : 

Therefore, f is continuous at all points x, such that x < 1 
Case II : 
If c = 1, then f(c) = f(1) = 1 + 1 = 2 
The left hand limit of f at x  = 1 is, 
Therefore, f is continuous at all points x, such that x > 1. 
Hence, the given function f has no point of discontinuity. 

11. Find all points of discontinuity of f, where f is defined 
Solution
The given function f is 
The given function f is defined at all the points of the points of the real line .
Let c be a point on the real line. 
Case I:
Therefore, f is continuous at all points x, such that x > 2
Thus, the given function f is continuous at every point on the real line. 
Hence, f has no point of discontinuity.

12.Find all points of discontinuity of f, where f is defined by f(x)=
Solution

The give function f is 
The given function f is defined at all the points of the real line. 
Let c be a point on the real line. 
Case I: 

Therefore, f is continuous at all points x, such that x < 1 
Case II : 
If c = 1, then the left hand limit of f at x = 1 is, 
It is observed that the left and right hand limit of f at x = 1 do not coincide. 
Therefore, f is not continuous at x = 1 .
Case III : 
If c > 1, then f(c) = c2 

Therefore, f is continuous at all points x, such that x > 1 
Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity. 

13. Is the function defined by   a continuous function ? 
Solution
The given function is 
The given function f is defined at all the points of the real line. 
Let c be a point on the real line. 
Case I: 
Therefore, f is continuous at all points x, such that x < 1 
Case II : 
If c = 1, then f(1) = 1 + 5 = 6 
The left hand limit of f at x = 1 is,  
It is observed that the left and right hand limit of f at x = 1 do not coincide. 
Therefore, f is not continuous at x = 1 
Case III : 
Therefore, f is continuous at all points x, such that x > 1
Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

14. Discuss the continuity of the function f, where f is defined by 
Solution
The given function is 
The given function is defined at all points of the interval [0, 10]. 
Let c be a point in the interval [0, 10]. 
Case I : 
Therefore, f is continuous in the interval [0, 1]. 
Case II : 
If c = 1, then f(3) = 3 
The left hand limit of f at x = 1 is, 
It is observed that the left and right hand limits of f at x = 1 do not coincide. 
Therefore, f is not continuous at x = 1 
Case III : 

Therefore, f is continuous at all points of the interval (1, 3). 
Case IV : 
If c = 3,  then f(c) = 5 
The left hand limit of f at x = 3 is, 
It is observed that the left and right hand limits of f a x = 3 do not coincide. 
Therefore, f is not continuous at x = 3
Case V : 

Therefore, f is continuous at all points of the interval (3, 10). 
Hence, f is not continuous at x = 1 and x = 3

15. Discuss the continuity of the function f, where f is defined by 
Solution
The given function is 
The given function is defined at all points of the real line. 
Let c be a point on the real line. 
Case I : 
If c < 0 , then f(c) = 2c 

Therefore, f is continuous at all points x, such that x < 0
Case II : 
If c = 0, then f(c) = f(0) = 0 
The left hand limit of f at x = 0 is 

Therefore, f is continuous at all points of the interval (0, 1)
Case IV : 
If c = 1, then f(c) = f(1) = 0 
The left hand limit  of f at x = 1 is, 
It is observed that the left and right hand limits of f at x = 1 do not coincide. 
Therefore, f is not continuous at x = 1 
Case V : 

Therefore, f is continuous at all points x, such that x > 1 
Hence, f is not continuous only at x = 1 

16. Discuss the continuity of the function f, where f is defined by 
Solution
The given function f is 
The given function is defined at all points of the real line. 
Let c be a point on the real line. 
Case I : 

Therefore, f is continuous at all points x, such that x < -1
Case II : 
If c = -1, then f(c) = f(-1) = -2 
The left hand limit of f at x = -1 is , 

Therefore, f is continuous at x = -1
Case III :
If - < c < 1, then f(c) = 2c 

Therefore, f is continuous at all points of the interval (-1, 1). 
Case IV :
If c = 1, then f(c) = f(1) = 2 × 1 = 2
The left hand limit of f at x = 1 is, 

Therefore, f is continuous at all points x, such that x > 1 
Thus, from the above observations, it can be concluded that f is continuous at all points of the real line. 

17. Find the relationship between a and b so that the function f defined by  is continuous at x = 3. 
Solution

Therefore, from (1), we obtain 
3a + 1 = 3b + 3 
⇒ 3a + 1 = 3b + 3 
⇒ 3a = 3b = 2 
⇒a = b + 2/3 
Therefore, the required relationship is given by , a = b + 2/3 

18. For what value of λ is the function defined by  continuous at x = 0 ? What about continuity at x = 1 ? 
Solution
The given function f is 
If f is continuous at x = 0, then  

⇒ λ(02 – 2 × 0) = 4 × 0 + 1 = 0
⇒ 0 = 1 = 0, which is not possible 
Therefore, there is no value of λ for which f is continuous at x = 0 
At x = 1, 
f(1) = 4x + 1 = 4 × 1 + 1 = 5 

Therefore, for any values of λ, f is continuous at x = 1 

19. Show that the function defined by g(x) = x = [x] is discontinuous at all integral point. Here [x] denotes the greatest integer less than or equal to x.
Solution
The given function is g(x) = x - [x] 
It is evident that g is defined at all integral points.  
Let n be an integer. 
Then,  
g(n) = n - [n] = n - n = 0 
The left hand limit of f at x = n is, 

It is observed that the left and right hand limits of f at x = n do not coincide. 
Therefore, f is not continuous at x = n 
Hence, g is discontinuous at all integral points. 

20. Is the function defined by f(x) = x2 - sinx + 5 continuous at x = π ? 
Solution
The given function is f(x) = x2 - sinx + 5 
It is evident that f is defined at x = π 
At x = π, f(x) = f(π) = π2  - sinπ + 5 = π2  - 0 + 5 = π2  + 5 

Therefore, the given function f is continuous at x = π 

21. Discuss the continuity of the following functions.
(a) f (x) = sin x + cos x
(b) f (x) = sin x − cos x
(c) f (x) = sin x × cos x
Solution
It is known that if g and h are two continuous functions, then (g + h), (g - h), and (g. h) are also continuous. 
It has to proved first that g(x) = sin x and h (x) = cos x are continuous functions. 
Let g(x) = sinx  
It is evident that g(x) = sin x is defined for every real number. 
Let c be a real number. Put x = c + h 
If x → c, then h → 0 
g(c) = sinc 
Therefore, g is a continuous function. 
Let h (x) = cos x 
It is evident that h(x) = cos x is defined for every real number.  
Let C be a real number. Put x = c + h 
If x → c, then h → 0 
h(c)  = cos c 
Therefore, h is a continuous function . 
Therefore, it can be concluded that 
(a) f(x) = g(x) + h(x) = sin x  + cos x is a continuous function 
(b) f(x) = g(x) - h(x) = sin x  - cos x is a continuous function 
(c) f(x) = g(x) × h(x) = sin x × cos x is a continuous function 

22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions,
Solution
It is known that if g and h are two continuous functions, then 
(i)  h(x)/g(x) , g(x) ≠ 0 is continuous  
(ii) 1/g(x), g(x) ≠ 0 is continuous  
(iii) 1/h(x), h(x) ≠ 0 is continuous 
It has to be proved first that g(x) = sin x and h(x) = cos x are continuous functions . 
Let g(x) = sin x 
It is evident that g(x) = sin x is defined for every real number. 
Let c be a real number. Put x = (c + h)
if x → c, then h → 0 
g(c) = sin c 
Therefore, g is a continuous function. 
Let h(x) = cos x 
It is evident that h(x) = cos x is defined for every real number. 
Let c be a real number. Put x = c + h 
If x → c, then h → 0 
h(c) = cos c 
Therefore, h(x) = cos x is continuous function 
It can be concluded that, 
cosec x = 1/sin x, sin x ≠ 0 is continuous 
⇒ cosec x, x ≠ nπ (n ∊ Z) is continuous 
Therefore, cosecant if continuous except at x = np, n ∊ Z 
sec x = 1/cos x, cos x ≠ 0 is continuous 
⇒ sec x, x ≠ (2n + 1)(π/2)  (n ∊ Z) is continuous 
Therefore, secant is continuous except at x = (2n + 1)(π/2)  (n ∊ Z) 
cot x = cos x/sin x , sin x ≠ 0 is continuous 
⇒ cot x, x ≠ nπ (n ∈ Z) is continuous 
Therefore, cotangent is continuous except at x = np, n ∈ Z 

23. Find the points of discontinuity of f, where 
Solution
The given function f is 
It is evident that f is defined at all points of the real line. 
Let c be a real number. 
Case I : 
Therefore, f is continuous at all points x, such that x > 0 
Case III : 
If c = 0, then f(c) = f(0) = 0 + 1 = 1 
The left hand limit of f at x = 0 is, 
Therefore, f is continuous at x = 0 
From the above observations, it can be concluded that f is continuous at all points of the real line. 
Thus, f has no point of discontinuity. 

24.  Determine if f defined by  is a continuous function ? 
Solution
He given function f is 
It is evident that f is defined at all points of the real line. 
Let C be a real number. 
Case I : 
if c ≠ 0, then f(c) = c2 sin 1/c
Therefore, f is continuous at x = 0 
From the above observations, it can be concluded that f is continuous at every point of the real line. 
Thus, f is a continuous function . 

25. Examine the continuity of f, where f is defined b
Solution
The given function f is 
It is evident that f is defined at all points of the real line. 
Let c be a real number. 
Case I : 
If c ≠ 0, then f(c) = sin c - cos c 

Therefore, f is continuous at all points x, such that x  ≠ 0 
Case II : 
If c = 0, then f(0) = -1
Therefore, f is continuous at x = 0 
From the above observations, it can be concluded that f is continuous at every point of the real line. 
Thus, f is a continuous function. 

26. Find the values of so that the function f is continuous at the indicated point.
 at x = π/2 
Solution
The given function f is 
The given function f is continuous at x = π/2, if f is defined at x =π/2 and if the value of the f at x = π/2 equals the limit of f at x = π/2. 
It is evident that f is defined at x = π/2 and f(π/2) = 3 
⇒ k/2 = 3 
⇒ k = 6 
Therefore, the required value of k is 6.

27. Find the values of k so that the function f is continuous at the indicated point. 
 at x = 2 
Solution
The given function is 
The given function f is continuous at x = 2, if f is defined at x = 2 and if the value of f at x = 2 equals the limit of f at x = 2
It is evident that f is defined at x = 2 and f(2) = k(2)2  = 4k 

⇒ k × 22  = 3 = 4k 
⇒ 4k = 3 
⇒ k = 3/4 
Therefore, the required value of k is 3/4. 

28. Find the values of k so that the function f is continuous at the indicated point. 
 at x = π
Solution
The given function is 
The given function f is continuous at x = π , if f is defined at x = π and if the value of f at x = π equals the limit of f at x = π
⇒ kπ + 1 = cos π = kπ + 1 
⇒ kπ + 1 = -1 = kπ + 1 
⇒ k = -2/π
Therefore, the required value of k is -2/π. 

29. Find the values of k so that the function f is continuous at the indicated point. 
 at x = 5 
Solution
The given function f is 
The given function f is continuous at x = 5, if f is defined at x = 5 and if the value of f at x = 5 equals the limit of f at x = 5 
It is evident that f is defined at x = 5 and  f(5) = kx + 1 = 5k + 1 

⇒ 5k + 1 = 15 - 5 = 5k + 1 
⇒ 5k + 1 = 10 
⇒ 5k = 9 
⇒ k = 9/5 
Therefore, the required value of k is 9/5. 

30. Find the values of a and b such that the function defined by  is a continuous function. 
Solution
The given function f is 
It is evident that the given function f is defined at all points of the real line. 
If f is a continuous function, then f is continuous at all real numbers. 
In particular, f is continuous at x = 2 and x = 10 
Since f is continuous at x = 2, we obtain 

On subtracting equation (1) from equation (2), we obtain 
8a = 16 
⇒ a = 2 
By putting a = 2 in equation (1), we obtain 
2×2 + b = 5 
⇒ 4 + b = 5 
⇒ b = 1 
Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectiely. 

31. Show that the function defined by f(x) = cos (x2 ) is a continuous function . 
Solution
The given function is f(x) = cos (x2
This function f is defined for every real number and f can be written as the composition of two functions as, 
f = g o h, were g(x) = cos x and h(x) = x2 
[∵ (goh)(x)  = g[h(x)] = g(x2 ) = cos(x2 ) = f(x) ] 
It has to be first proved that g(x) = cos x and h(x) = x2  are continuous functions. 
It is evident that g is defined for every real number. 
Let c be a real number. 
Then, g(c) = cos c
Put x = c + h 
If x ⟶ c, then h ⟶ 0 
Therefore, g(x) = cos x is continuous function. 
h(x) = x2  
Clearly, h is defined for every real number. 
Let k be a real number, then h (k) = k2  

Therefore, h is a continuous function. 
It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g(c), then (f o g ) is continuous at c. 
Therefore, f(x) = (goh)(x) = cos(x2 ) is a continuous function. 

32. Show that the function defined by f(x) = |cos x| is a continuous function.
Solution
The given function is f(x) = |cos x|
This function f is defined for every real number and f can be written as the composition of two functions as, 
f = g o h, where g(x) = |x| and h(x) = cos x 
[∵ (goh) (x) = g{h (x)} = g(cos x) = |cos x| = f(x)]
It has to be first proved that g(x) = |x| and h(x) = cos x are continuous functions. 
g(x) = |x| can be written as 

Clearly, g is defined d for all real numbers. 
Let c be a real number. 
Case I : 
Therefore, g is continuous at all points x, such that x < 0 
Case II : 

Therefore, g is continuous at all points x, such that x > 0 
Case III : 
If  c = 0, then g(c) = g(0) = 0 
Therefore, g is continuous at x = 0 
From the above three observations, it can be concluded that g is continuous at all points. 
h(x) = cos x 
It is evident that h(x) = cos x is defined for every real number. 
Let c be a real number. Put x = c + h 
If x → c, then h → 0 
h(c) = cos c 
Therefore, h(x) = cos x is a continuous function. 
It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c. 
Therefore, f(x) = (goh)(x) = g{h(x)} = g(cos x) = |cos x| is a continuous function . 

33. Examine sin |x| is a continuous function.
Solution
Let f(x) = sin |x| 
This function f is defined for every real number and f can be written as the composition of  two functions as, 
f = g o  h, where g(x) = |x| and h(x) = sin x 
[ ∵ (goh)(x) = {h(x)} = g(sin x) = |sin x| = f(x)]
It has to be proved first that g(x) = |x| and h(x) = sin x  are continuous functions. 
g(x) = |x| can be written as 

Clearly, g is defined for all real numbers. 
Let c be a real number. 
Case I : 
Therefore, g is continuous at all points x, such that x <  0 
Case II : 

Therefore, g is continuous at all points x, such that x > 0 
Case III : 
If c = 0, then g(c) = g(0) = 0 

Therefore, g is continuous at x = 0 
From the above three observations, it can be concluded that g is continuous at all points. 
h(x) = sin x 
It is evident that h(x) = sin x is defined for every real number. 
Let c be real number. Put x = c + k 
If x → c, then k→ 0 
h (c) = sin c 

Therefore, h is a continuous function. 
It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g(c), then (f  o  g) is continuous at c. 
Therefore, f(x) = (goh)(x) = g[h(x)] = g(sin x) = |sin x| is a continuous function . 

34. Find all the points of discontinuity of f defined by f(x) = |x| - |x + 1|. 
Solution
The given function is f(x) = |x| - |x + 1| 
The two functions, g and h, are defined as 
g(x) = |x| and h(x) = |x + 1| 
Then, f = g - h 
The continuity of g and h is examined first. 
g(x) = |x| can be written as 

Clearly, g is defined for all real numbers. 
Let c be a real number. 
Case I : 
Therefore, g is continuous at all points x, such that x < 0 
Case II : 

Therefore, g is continuous at all points x, such that x > 0 
Case III : 
If c = 0, then g(c) = g(0) = 0 
Therefore, g is continuous at x = 0 
From the above three observations, it can be concluded that g is continuous at all points. 
h(x) = |x + 1| can be written as 
Clearly, h is defined for every real number. 
Let c be a real number. 
Case I : 
Therefore, h is continuous at all points x, such that x < -1 
Case II : 
Therefore, h is continuous at all points x, such that x > - 1 
Case III : 
If c = -1, then h(c) = h(-1) = -1 + 1 = 0 

Therefore, h is continuous at x = -1 
From the above three observations, it can be concluded that h is continuous at all points of the real line. 
g and h are continuous functions. Therefore, f  = g - h is also a continuous function. 
Therefore, f has no point of discontinuity.
Previous Post Next Post