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
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?
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
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,
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.
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)
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.
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.
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
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.
Hence, the given function has no point of discontinuity.
10. Find all points of discontinuity of f, where f is defined by
Hence, the given function f has no point of discontinuity.
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. 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,
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:
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.
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)=
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.
Therefore, f is continuous at all points x, such that x ≠ 0
Case II :
at x = π/2 
at x = 2
at x = π
at x = 5
On subtracting equation (1) from equation (2), we obtain
Therefore, h is a continuous function.
Clearly, g is defined d for all real numbers.
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.
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,
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 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 ?
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:
From the above observations, it can be concluded that f is continuous at every point of the real line.
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,
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 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.
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 :
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,
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,
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
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,
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
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 
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
Therefore, f is continuous at all points x, such that x > 0
Case III :
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.
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.
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 ?
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
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.
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
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
(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
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
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
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
(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
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.
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
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 ≠ nÏ€ (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)(Ï€/2) (n ∊ Z) is continuous
cosec x = 1/sin x, sin x ≠ 0 is continuous
⇒ cosec x, x ≠ nÏ€ (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)(Ï€/2) (n ∊ Z) is continuous
Therefore, secant is continuous except at x = (2n + 1)(Ï€/2) (n ∊ Z)
cot x = cos x/sin x , sin x ≠ 0 is continuous
cot x = cos x/sin x , sin x ≠ 0 is continuous
⇒ cot x, x ≠ nÏ€ (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.
It is evident that f is defined at all points of the real line.
Let c be a real number.
Case I :
Case III :
If c = 0, then f(c) = f(0) = 0 + 1 = 1
The left hand limit of f at x = 0 is,
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.
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 ?
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.
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.
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 k 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.
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
⇒ 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
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.
⇒ 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 = πSolution
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
⇒ kÏ€ + 1 = cos Ï€ = kÏ€ + 1
⇒ kÏ€ + 1 = -1 = kÏ€ + 1
⇒ k = -2/Ï€
Therefore, the required value of k is -2/Ï€.
⇒ kÏ€ + 1 = -1 = kÏ€ + 1
⇒ k = -2/Ï€
Therefore, the required value of k is -2/Ï€.
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
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.
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.
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
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
⇒ 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.
⇒ 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
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.
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,
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.
[∵ (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 :
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.
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
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,
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.
Therefore, g is continuous at all points x, such that x > 0
Case III :
Therefore, h is a continuous function.
Therefore, g is continuous at all points x, such that x > 0
Case III :
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 :
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.
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
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
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.
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.
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 :
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
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 :
Case II :
Therefore, h is continuous at all points x, such that x > - 1
Case III :
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.
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.