Continuity & Differentiability(Ex. 5.1)

 Exercise 5.1

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

Solun:- Given function is f(x) = 5x-3

We know that for continuous function:-

Put x=0

= -3

Put x=0 in given function:-

⇒ f(0) = -3

Put x = -3

= -18

Put x = -3 in given function:-

⇒ f(-3) = -18

Put x = 5

= 22

Put x = 5 in given function:-

⇒ f(5) = 22

Hence f(x) is continuous at x=0,at x=-3 and at x=5.

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

Solun:- Given function is f(x) = 2x2-1

We know that for continuous function:-

Put x=3

= 17

Put x=3 in given function:-

⇒ f(3) = 17

Hence f is continuous at x=3.

3. Examine the following functions for continuity. 

(a) f(x) = x-5

Solun:- Given function is f(x) = x-5

We know that for continuous function:-

Put x=k

= k-5

Put x=k in given function:-

⇒ f(k) = k-5

Hence f is continuous at all real numbers.

(b) f(x) = 1/(x-5),  x ≠ 5

Solun:- Given function is f(x) = 1/(x-5)

We know that for continuous function:-

Put x=k

= 1/(k-5)

Put x=k in given function:-

⇒ f(k) = 1/(k-5)

Hence f is continuous at all real numbers except x=5.

Solun:- Given function is f(x) = x2-25 / (x+5)

We know that for continuous function:-

Put x=k

= (k-5)

Put x=k in given function:-

Hence f is continuous at all real numbers except x = -5.

(d) f(x) = |x-5|

Solun:- Given function is f(x) = |x-5|

We know that for continuous function:-

Case 1: k>5

⇒ f(x) = x-5

Put x=k

= (k-5)

Put x=k in given function:-

⇒ f(k) = (k-5)

Case 2: k=5

⇒ f(x) = 0

Put x=k

= 0

Put x=k in given function:-

⇒ f(k) = 0

Case 3: k<5

⇒ f(x) = 5-x

Put x=k

= 5-k

Put x=k in given function:-

⇒ f(k) = 5-k

Hence f is continuous at all real numbers.

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

Solun:- Given function is f(x) = xn

We know that for continuous function:-

Put x=n

= nn

Put x=n in given function:-

⇒ f(n) = nn

Hence f is continuous at all positive integers.

5. Is the function f defined by

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

Solun:- Given function is

At x=0

⇒ f(x) = x

We know that for continuous function:-

Put x=0

= 0

Put x=0 in given function:-

⇒ f(0) = 0

Hence f is continuous at x=0.

At x = 1

Put x = 1

= 1

Put x = 1

= 5

Because, R.H.L ≠ L.H.L

So, f is not continuous at k=1.

Hence, f is discontinuous at x=1.

At x = 2

⇒ f(x) = 5

Put x = 2

= 5

Put x=2 in given function:-

⇒ f(2) = 5

f is continuous at x=2.

Hence f is continuous at x=0 and x=2 but not continuous at x=1.

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

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k<2

⇒ f(x) = 2x+3

Put x=k

= 2k+3

Put x=k in given function:-

⇒ f(k) = 2k+3

Hence f is continuous at all points k, such that k<2.

Case 2: k > 2

Put x=k

= 2k-3

Put x=k in given function:-

⇒ f(k) = 2k-3

Hence f is continuous at all points k, such that k>2.

Case 3: k = 2

Put x=2

= 7

Put x=2

= 1

Because, R.H.L ≠ L.H.L

So, f is not continuous at k=2.

Hence, f is discontinuous at x=2.

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k < - 3

Put x=k

= - k+3

Put x=k in given function:-

⇒ f(k) = - k+3

Hence f is continuous at all points k, such that k < - 3.

Case 2: - 3 < k < 3

Put x=k

= - 2k

Put x=k in given function:-

⇒ f(k) = - 2k

Hence f is continuous at all points k, such that - 3 < k < 3.

Case 3: k > 3

Put x=k

= 6k+2

Put x=k in given function:-

⇒ f(k) = 6k+2

Hence f is continuous at all points k, such that k > 3.

Case 4: k = - 3

Put x = - 3

= 6

Put x = - 3

= 6

Put the value of x in given Eq.

f(-3) = 6

Hence f is continuous at k = - 3.

Case 5: k = 3

Put x=3

= 20

Put x=3

= - 6

Because R.H.L ≠ L.H.L

So, f is not continuous at k=3.

Hence, f is discontinuous at x=3.

Solun:- Given function is

We know that

We know that for continuous function:-

Let k be a real number then

Case 1: k > 0

⇒ f(x) = 1

Put x=k

= 1

Put x=k in given function:-

⇒ f(k) = 1

Hence f is continuous at all points k, such that k > 0.

Case 2: k < 0

⇒ f(x) = - 1

Put x=k

= - 1

Put x=k in given function:-

⇒ f(k) = - 1

Hence f is continuous at all points k, such that k < 0.

Case 3: k = 0

Put x=0

= 1

Put x=0

= - 1

Because R.H.L ≠ L.H.L

So, f is not continuous at k=0.

Hence, f is discontinuous at x=0.

Solun:- Given function is

We know that

We know that for continuous function:-

Let k be a real number then

Case 1: k < 0

⇒ f(x) = - 1

Put x=k

= - 1

Put x=k in given function:-

⇒ f(k) = - 1

Hence f is continuous at all points k, such that k < 0.

Case 2: k > 0

⇒ f(x) = - 1

Put x=k

= - 1

Put x=k in given function:-

⇒ f(k) = - 1

Hence f is continuous at all points k, such that k > 0.

Case 3: k = 0

Put x=0

= - 1

Put x=0

= - 1

Put x=0 in given function:-

⇒ f(0) = - 1

Because R.H.L = L.H.L = f(0)

So, f is continuous at x=0.

Hence there is no point of discontinuity.

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k < 1

⇒ f(x) = x2+1

Put x=k

= k2+1

Put x=k in given function:-

⇒ f(k) = k2+1

Hence f is continuous at all points k, such that k < 1.

Case 2: k > 1

⇒ f(x) = x+1

Put x=k

= k+1

Put x=k in given function:-

⇒ f(k) = k+1

Hence f is continuous at all points k, such that k > 1.

Case 3: k = 1

Put x=1

= 2

Put x=1

= 2

Put x=1 in given function:-

⇒ f(1) = 2

Because R.H.L = L.H.L = f(0)

So, f is continuous at x=1.

Hence there is no point of discontinuity.

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k < 2   ⇒ f(x) = x3-3

Put x=k

= k3 - 3

Put x=k in given function:-

⇒ f(k) = k3 - 3

Hence f is continuous at all points k, such that k < 2.

Case 2: k > 2

⇒ f(x) = x2+1

Put x=k

= k2+1

Put x=k in given function:-

⇒ f(k) = k2+1

Hence f is continuous at all points k, such that k > 2.

Case 3: k = 2

Put x=2

= 5

Put x=2

= 5

Put x=2 in given function:-

⇒ f(2) = 5

Because R.H.L = L.H.L = f(0)

So, f is continuous at x=2.

Hence there is no point of discontinuity.

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k < 1   ⇒ f(x) = x10 - 1

Put x=k

= k10 - 1

Put x=k in given function:-

⇒ f(k) = k10 - 1

Hence f is continuous at all points k, such that k < 1.

Case 2: k > 1

⇒ f(x) = x2

Put x=k

= k2

Put x=k in given function:-

⇒ f(k) = k2

Hence f is continuous at all points k, such that k > 1.

Case 3: k = 1

Put x=1

= 1

Put x=1

= 0

Because R.H.L ≠ L.H.L

So, f is not continuous at x=1.

Hence f is discontinuous at x=1.

13. Is the function defined by

a continuous function?

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k < 1   

⇒ f(x) = x + 5

Put x=k

= k+5

Put x=k in given function:-

⇒ f(k) = k+5

Hence f is continuous at all points k, such that k < 1.

Case 2: k > 1

⇒ f(x) = x-5

Put x=k

= k - 5

Put x=k in given function:-

⇒ f(k) = k - 5

Hence f is continuous at all points k, such that k > 1.

Case 3: k = 1

Put x=1

= -4

Put x=1

= 6

Because R.H.L ≠ L.H.L

So, f is not continuous at x=1.

Hence f is discontinuous at x=1.

Discuss the continuity of the function f, where f is defined by

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: 0 ≤ k < 1   ⇒ f(x) = 3

Put x=k

= 3

Put x=k in given function:-

⇒ f(k) = 3

Hence f is continuous at all points k, such that 0 ≤ k < 1.

Case 2: k = 1

Put x=1

= 3

Put x=1

= 4

Because R.H.L ≠ L.H.L

So, f is not continuous at x=1.

Hence f is discontinuous at x=1.

Case 3: 1 < k < 3

⇒ f(x) = 4

Put x=k

= 4

Put x=k in given function:-

⇒ f(k) = 4

Hence f is continuous at all points k, such that 1 < k < 3.

Case 4: k = 3

Put x=3

= 4

Put x=3

= 5

Because R.H.L ≠ L.H.L

So, f is not continuous at x=3.

Hence f is discontinuous at x=3.

Case 5: 3 < k ≤ 10

⇒ f(x) = 5

Put x=k

= 5

Put x=k in given function:-

⇒ f(k) = 5

So, f is continuous at all points k, such that 3 < k ≤ 10.

Hence f is discontinuous at x=1 and x=3.

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k<0   ⇒ f(x) = 2x

Put x=k

= 2k

Put x=k in given function:-

⇒ f(k) = 2k

Hence f is continuous at all points k, such that k < 0.

Case 2: k = 0

Put x=0

= 0

Put x=0

= 0

Put x=0 in given function:-

⇒ f(k) = 0

Because R.H.L = L.H.L = f(0)

Hence f is continuous at x=0.

Case 3: 0 < k < 1

⇒ f(x) = 0

Put x=k

= 0

Put x=k in given function:-

⇒ f(k) = 0

Hence f is continuous at all points k, such that 0 < k < 1.

Case 4: k = 1

Put x=1

= 0

Put x=1

= 4

Because R.H.L ≠ L.H.L

So, f is not continuous at x=1.

Hence f is discontinuous at x=1.

Case 5: k > 1

⇒ f(x) = 4x

Put x=k

= 4k

Put x=k in given function:-

⇒ f(k) = 4k

Hence f is continuous at all points k, such that k > 1.

Hence f is only discontinuous at x=1.

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k < - 1   ⇒ f(x) = - 2

Put x=k

= - 2

Put x=k in given function:-

⇒ f(k) = - 2

Hence f is continuous at all points k, such that k < - 1.

Case 2: k = - 1

Put x = - 1

= - 2

Put x = - 1

= - 2

Put x = - 1 in given function:-

⇒ f(-1) = - 2

Because R.H.L = L.H.L = f(-1)

Hence f is continuous at x = - 1.

Case 3: -1 < x < 1

⇒ f(x) = 2x

Put x=k

= 2k

Put x=k in given function:-

⇒ f(k) = 2k

Hence f is continuous at all points k, such that -1 < x < 1.

Case 4: k = 1

Put x=1

= 2

Put x=1

= 2

Put x=1 in given function

⇒ f(1) = 2

Because R.H.L = L.H.L = f(1)

So, f is continuous at x=1.

Case 5: k > 1

⇒ f(x) = 2

Put x=k

= 2

Put x=k in given function:-

⇒ f(k) = 2

Hence f is continuous at all points k, such that k > 1.

Hence f is continuous at all points.

17. Find the relationship between a and b so that the function f defined by

is continuous at x = 3.

Solun:- Given function is

continuous function:-

At x=3:-

Put x=3

= 3a+1

Put x=3

= 3b+3

Given functions is continuous then

R.H.L = L.H.L = f(3)

⇒ 3a+1 = 3b+3 = 3a+1

⇒ 3a-3b = 2

⇒ a-b = 2/3

⇒ 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?

Solun:- Given function is

continuous function:-

At x=0:-

Put x=0

= 0

Put x=0

= 1

Hence R.H.L ≠ L.H.L

So there is no such λ exists.

At x=1

Let k is a real number

f(x) = 4x+1

Put x=k

= 4k+1

Put x=k in given function:-

⇒ f(k) = 4k+1

Hence f is continuous at all points k, such that k=1 at any value of λ.

19. Show that the function defined by g(x) = x - [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Solun:- Given function is g(x) = x - [x]

Let n be an integer.

g(k) = k - [k] = k - k = 0

Because R.H.L ≠ L.H.L

Hence given function is discontinuous at all integers.

20. Is the function defined by f(x) = x2 - sin x + 5 continuous at x = π?

Solun:- Given function is f(x) = x2 - sin x + 5

f(π) = π2 - sin π + 5 = π2 + 5     (sin π = 0)

Put x = π

= π2 - sin π + 5

= π2 + 5

Hence 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

Solun:- Let g(x) = sin x and h(x) = cos x

Let c be a real number then

g(c) = sin c

Put x=c

= sin c

Hence g is continuous at x = c.

h(c) = cos c

Put x=c

= cos c

Hence h is continuous at x = c.

We know that if g and h are two continuous functions then

g+h, g-h and g.h are also continuous.

(a) f(x) = sin x + cos x

⇒  f(x) = g(x) + h(x)

Because g(x) and h(x) are continuous functions then g(x)+h(x) is continuous.

Hence f(x) is also a continuous function.

(b) f(x) = sin x - cos x

⇒  f(x) = g(x) - h(x)

Because g(x) and h(x) are continuous functions then g(x)-h(x) is continuous.

Hence f(x) is also a continuous function.

(a) f(x) = sin x . cos x

⇒  f(x) = g(x) . h(x)

Because g(x) and h(x) are continuous functions then g(x).h(x) is continuous.

Hence f(x) is also a continuous function.

22. Discuss the continuity of the cosine, cosecant, secant, and cotangent functions.

Solun:- Let g(x) = sin x and h(x) = cos x

Let c be a real number then

g(c) = sin c

Put x=c

= sin c

Hence g is continuous at x = c.

h(c) = cos c

Put x=c

= cos c

h is continuous at x = c.

Hence cosine is a continuous function.

We know that cosec x=1/sin x

cosec x = 1/g(x)  given sin x ≠ 0

We know that if g(x) is continuous then 1/g(x) is also continuous.

Hence cosecant is a continuous function except for x = nπ, n∈Z.

We know that sec x=1/cos x

sec x = 1/h(x)     given cos x ≠ 0

We know that if h(x) is continuous then 1/h(x) is also continuous.

Hence secant is a continuous function except for x = (2n+1)π/2, n∈Z.

We know that cot x=cos x/sin x

cot x = h(x)/g(x)

We know that if g(x) and h(x) are continuous functions then h(x)/g(x) is also a continuous function.

Hence cotangent is a continuous function except for x = nπ, n∈Z.

23. Find all points of discontinuity of f, where 

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k < 0   

⇒ f(x) = sin x / x

Put x=k

= sin k / k

Put x = k in given function:-

⇒ f(k) = sin k / k

Hence f is continuous at all points k, such that k < 0.

Case 2: k = 0

= 1

Put x=0

= 1

Put x=0 in given function:-

⇒ f(0) = 1

Because R.H.L = L.H.L = f(0)

Hence f is continuous at x=0.

Case 3: k > 0   

⇒ f(x) = x+1

Put x=k

= k+1

Put x=k in given function:-

⇒ f(k) = k+1

So, f is continuous at all points k, such that k > 0.

Hence there is no point of discontinuity.

24. Determine if f defined by 

is a continuous function?

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k ≠ 0   

⇒ f(x) = x2sin(1/x)

Hence f is continuous at all points k, such that k ≠ 0.

Case 2: k = 0

Put x=0

= 0

Put x=0

= 0

Put x=0 in given function:-

⇒ f(0) = 0

Because R.H.L = L.H.L = f(0)

Hence f is continuous at all x.

25. Examine the continuity of f, where f is defined by 

Solun:- Given function is

We know that for continuous function:-

Let k be a real number then

Case 1: k≠0   ⇒ f(x) = sin x - cos x

Put x=k

= sink - cos k

Put x=k in the given function

⇒ f(k) = sin k - cos k

Hence f is continuous at all points k, such that k ≠ 0.

Case 2: k = 0

Put x=0

= - 1

Put x=0

= - 1

Put x=0 in given function:-

⇒ f(0) = - 1

Because R.H.L = L.H.L = f(0)

Hence f is continuous at all x.

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

Solun:- Given function is

continuous function:-

At x = π/2:-

Put x=π/2

= k/2

Given functions is continuous then

L.H.L = f(π/2)

⇒ k/2 = 3

⇒ k = 6

Solun:- Given function is

continuous function:-

At x = 2:-

Put x=2

= 3

Given functions is continuous then

R.H.L = f(2)

⇒ 3 = k(2)2

⇒ 4k = 3

⇒ k = 3/4

Solun:- Given function is

continuous function:-

At x = π:-

Put x=π

= - 1

Given functions is continuous then

R.H.L = f(π)

⇒ - 1 = π(k)+1

⇒ π(k) = - 2

⇒ k = - 2/π

Solun:- Given function is

continuous function:-

At x = 5:-

Put x=5

= 10

Given functions is continuous then

R.H.L = f(5)

⇒ 10 = 5k+1

⇒ 5k = 9

⇒ k = 9/5

30. Find the values of a and b such that the function defined by

is continuous function.

Solun:- Given function is

continuous function:-

At x = 2:-

Put x=2

= 2a+b

Given functions is continuous then

R.H.L = f(2)

⇒ 2a+b = 5 ……………(1)

At x = 10:-

Put x=10

= 10a+b

Given functions is continuous then

L.H.L = f(10)

⇒ 10a+b = 21 ……………(2)

Solving Eq. 1 and 2:-

⇒ a = 2 and b = 1
31. Show that the function defined by f(x) = cos(x2) is a continuous function.

Solun:- Let g(x) = cos x  and  h(x) = x2

Let c be a real number then

g(x) = cos x

Put x=c

= cos c

Put value of x is given function

⇒ g(c) = cos c

Hence g is continuous at x = c.

Put x=c

= c2

Put value of x is given function

⇒ h(c) = c2

Hence h is continuous at x = c.

⇒ goh(x) = g(h(x))

⇒ goh(x) = cos(x2)

We know that if g is continuous at c and if f is continuous at g(c), then goh(x) is continuous at c provided g and h are real-valued functions such that goh is defined at c.

Hence cos (x2) is a continuous function.

32. Show that the function defined by f(x) = |cos x| is a continuous function.

Solun:- Let g(x) = |x|  and  h(x) = cos x

Let c be a real number then

⇒ g(x) = |x|

We know that

Let c be a real number then

Case 1: c > 0

Put x=c

= c

Put x=c in given function:-

⇒ g(c) = c

Hence g is continuous at all points c, such that c > 0.

Case 2: c < 0

Put x=c

= - c

Put x=c in given function:-

⇒ g(c) = - c

Hence g is continuous at all points c, such that c < 0.

Case 3: c = 0

Put x=0

= 0

Put x=0

= 0

Put x=0 in given function:-

⇒ g(0) = 0

Because R.H.L = L.H.L = f(0)

Hence g is continuous at x = c.

h(x) = cos x

Put x=c

= cos c

Put value of x is given function

⇒ h(c) = cos c

Hence h is continuous at x = c.

⇒ goh(x) = g(h(x))

⇒ goh(x) = |cos x|

We know that if g is continuous at c and if f is continuous at g(c), then goh(x) is continuous at c provided g and h are real-valued functions such that goh is defined at c.

Hence |cos x| is a continuous function.

33. Examine that sin |x| is a continuous function.

Solun:- Let g(x) = sin x  and  h(x) = |x|

Let c be a real number then

g(x) = sin x

Put x=c

= sin c

Put value of x is given function

⇒ g(c) = sin c

Hence g is continuous at x = c.

⇒ h(x) = |x|

We know that

Let c be a real number then

Case 1: c > 0

Put x=c

= c

Put x=c in given function:-

⇒ h(c) = c

Hence h is continuous at all points c, such that c > 0.

Case 2: c < 0

Put x=c

= - c

Put x=c in given function:-

⇒ h(k) = - c

Hence h is continuous at all points c, such that c < 0.

Case 3: c = 0

Put x=0

= 0

Put x=0

= 0

Put x=0 in given function:-

⇒ f(0) = 0

Because R.H.L = L.H.L = f(0)

So, h is continuous at x=0.

Hence h is continuous at x = c.

⇒ goh(x) = g(h(x))

⇒ goh(x) = sin |x|

We know that if g is continuous at c and if f is continuous at g(c), then goh(x) is continuous at c provided g and h are real-valued functions such that goh is defined at c.

Hence sin |x| is a continuous function.

34. Find all the points of discontinuity of function defined by f(x) = |x| - |x+1|

Solun:- Given f(x) = |x| - |x+1|

Let g(x) = |x|  and h(x) = |x+1|

Let c be a real number then

⇒ g(x) = |x|

We know that

Let c be a real number then

Case 1: c > 0

Put x=c

= c

Put x=c in given function:-

⇒ g(c) = c

Hence g is continuous at all points c, such that c > 0.

Case 2: c < 0

Put x=c

= - c

Put x=c in given function:-

⇒ g(c) = - c

Hence g is continuous at all points c, such that c < 0.

Case 3: c = 0

Put x=0

= 0

Put x=0

= 0

Put x=0 in given function:-

⇒ g(0) = 0

Because R.H.L = L.H.L = f(0)

Hence g is continuous at x = c.

⇒ h(x) = |x + 1|

We know that

Let c be a real number then

Case 1: c > -1

Put x=c

= c+1

Put x=c in given function:-

⇒ h(c) = c+1

Hence h is continuous at all points c, such that c > 0.

Case 2: c < -1

Put x=c

= - (c+1)

Put x=c in given function:-

⇒ h(k) = - (c+1)

Hence h is continuous at all points c, such that c < 0.

Case 3: c = 0

Put x = - 1

= 0

Put x = - 1

= 0

Put x = - 1 in given function:-

⇒ f(-1) = 0

Because R.H.L = L.H.L = f(0)

So, h is continuous at x=0.

Hence h is continuous at x = c.

We know that if g and h are two continuous functions then g-h is also a  continuous function.

Given f(x) = |x| - |x+1|

f(x) = g(x) - h(x)

Hence given function is continuous function.

Download PDF of Exercise 5.1

See Also:-

Notes of Continuity & Differentiability

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