Continuity | Mathematics (Maths) for JEE Main & Advanced PDF Download

Definition 

Continuity at a Point: A function f is continuous at c if the following three conditions are met.

  •   f(x) is defined.
  • Continuity | Mathematics (Maths) for JEE Main & Advanced
  • Continuity | Mathematics (Maths) for JEE Main & Advanced

In other words function f(x) is said to be continuous at x = c , if Continuity | Mathematics (Maths) for JEE Main & Advanced 

Symbolically f is continuous at  x = c

Continuity | Mathematics (Maths) for JEE Main & Advanced

One-sided Continuity

  • A function f defined in some neighbourhood of a point c for c ⇒ c is said to be continuous at c from the left if
    Continuity | Mathematics (Maths) for JEE Main & Advanced
  • A function f defined in some neighbourhood of a point c for x ³ c is said to be continuous at c from the right if
    Continuity | Mathematics (Maths) for JEE Main & Advanced
  • One-sided continuity is a collective term for functions continuous from the left or from the right.
  • If the function f is continuous at c, then it is continuous at c from the left and from the right . Conversely, if the function f is continuous at c from the left and from the right, then
    Continuity | Mathematics (Maths) for JEE Main & Advancedexists & Continuity | Mathematics (Maths) for JEE Main & Advanced
  • The last equality means that f is continuous at c.
  • If one of the one-sided limits does not exist, thenContinuity | Mathematics (Maths) for JEE Main & Advanced does not exist either. In this case, the point c is a discontinuity in the function, since the continuity condition is not met.

Question for Continuity
Try yourself:
Which of the following conditions must be met for a function to be continuous at a point?
View Solution

Continuity In An Interval

  • A function f is said to be continuous in an open interval (a , b) if f is continuous at each & every point ∈(a, b).
  • A function f is said to be continuous in a closed interval [a,b] if:
    (i) f is continuous in the open interval (a , b) &
    (ii) f is right continuous at `a' i.e.
    Continuity | Mathematics (Maths) for JEE Main & Advanced   
    (iii) f is left continuous at `b' i.e.
     Continuity | Mathematics (Maths) for JEE Main & Advanced

A function f can be discontinuous due to any of the following three reasons:

  • Continuity | Mathematics (Maths) for JEE Main & Advanced does not exist i.e.  
    Continuity | Mathematics (Maths) for JEE Main & Advanced
  • f(x) is not defined at x= c
  • Continuity | Mathematics (Maths) for JEE Main & Advanced 
  • Geometrically, the graph of the function will exhibit a break at x= c.

Example 1. Test the following functions for continuity
(a) 2x5 - 8x2 + 11 / x4 + 4x3 + 8x2 + 8x +4
(b) f(x) = 3sin3x + cos2x + 1 / 4cos x - 2
Solution
(a) A function representing a ratio of two continuous functions will be (polynomials in this case) discontinuous only at points for which the denominator zero. But in this case (x4 + 4x3 + 8x2 + 8x + 4) = (x2 + 2x + 2)2 = [(x + 1)2 + 1]2 > 0 (always greater than zero)
Hence f(x) is continuous throughout the entire real line.
(b) The function f(x) suffers discontinuities only at points for which the denominator is equal to zero i.e. 4 cos x - 2 = 0 or cos x = 1/2 ⇒ x = xn = ± π/3 + 2nπ(n=0, ±1, ±2...) Thus the function f(x) is continuous everywhere, except at the point xn.

Example 2.

Continuity | Mathematics (Maths) for JEE Main & Advanced

Find A and B so as to make the function continuous.
SolutionAt x =  - π/2

Continuity | Mathematics (Maths) for JEE Main & Advanced
- π/2 - h
where h→ 0
Replace x by - π/2+h
where h → 0

Continuity | Mathematics (Maths) for JEE Main & Advanced

So B - A = 2 ...(i)
At x = π/2

Continuity | Mathematics (Maths) for JEE Main & Advanced
Replace x by π/2 - h
Replace x by π/2+h
where h→ 0
Continuity | Mathematics (Maths) for JEE Main & Advanced
So    A+B = 0      ...(ii)
Solving (i) & (ii),    B= 1, A = -1

Example 3. Test the continuity of f(x) at x = 0 if 

Continuity | Mathematics (Maths) for JEE Main & Advanced
SolutionFor x < 0,
Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced
L.H.L. = R.H.L. ≠ f(0) Hence f(x) is discontinuous at x = 0.

Example 4. If f(x) be continuous function for all real values of x and satisfies;
x2 + {f(x) – 2} x + 2√3 – 3 – √3 . f(x) = 0, for  x ∈ R. Then find the value of f(√3 ).
Solution. As f(x) is continuous for all x ∈ R.
Thus,
Continuity | Mathematics (Maths) for JEE Main & Advanced
where
f(x) = x2 - 2x + 2√3 - 3 / √3 - x, x ≠ √3

Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced

f(√3) = 2(1-√3).

Example 5. 
Continuity | Mathematics (Maths) for JEE Main & Advanced
If f (x) is continuous at x = 0, then find the value  of (b+c)3-3a.
Solution.   

Continuity | Mathematics (Maths) for JEE Main & Advanced

N→ 1 + a + b D→ 0
for existence of limit a + b + 1 = 0

Continuity | Mathematics (Maths) for JEE Main & Advanced
Continuity | Mathematics (Maths) for JEE Main & Advanced
limit of Nr ⇒ 2a+8b = 0 ⇒ a = -4b
hence
-4b+b = -1  
⇒ b = 1/3 and a = -4/3

Continuity | Mathematics (Maths) for JEE Main & Advanced
= 8 sin2x - 2 sin22x / 3x4 = 8 sin2x - 8sin2xcos2x / 3x4

= 8 / 3 . sin2x / x2 . sin2x / x2 = 8 / 3

⇒ eA = 1 / 2 (e2x A / x + B / x) ⇒ x . eA = 1 / 2 (e2x . A + B)

Example 6. 

Continuity | Mathematics (Maths) for JEE Main & Advanced

If f is continuous at x = 0, then find the values of a, b, c & d.
Solution. 

Continuity | Mathematics (Maths) for JEE Main & Advanced  ,
for existence of limit  a + b + 5 = 0

Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced
= a / 2 + 5 / 2 - a = 3
⇒ a = - 1 ⇒ b = - 4

Continuity | Mathematics (Maths) for JEE Main & Advanced 
for existence of limit  c = 0

Continuity | Mathematics (Maths) for JEE Main & Advanced

= ed = 3 ⇒ d = ln 3

Example 7. Let f(x) = x3 = 3x2 + 6 ∀ x ∈ R and
Continuity | Mathematics (Maths) for JEE Main & Advanced
Test continuity of g (x) for  x ∈ [-3, 1].
Solution. Since f(x) = x3 - 3x2 + 6  ⇒ f'(x) = 3x2 - 6x = 3x (x - 2) for maxima and minima f'(x) = 0

Continuity | Mathematics (Maths) for JEE Main & Advanced

x = 0, 2
f"(x) = 6x – 6
f" (0) = –6 < 0 (local maxima at x = 0)
f" (2) = 6 > 0 (local minima at x = 2)
x3 – 3x2 + 6 = 0 has maximum 2 positive and 1 negative real roots. f(0) = 6.
Now graph of f(x) is :
Clearly f(x) is increasing in (– ∝, 0) U (2, ∝) and decreasing in (0, 2)
⇒ x + 2 < 0 ⇒ x < – 2 ⇒ –3 ≤ x < – 2
⇒ –2 ≤ x + 1 < –1 and –1 ≤ x + 2 < 0
in both cases f(x) increases (maximum) of g(x) = f(x + 2)
g(x) = f(x + 2); –3 ≤ x < – 2 ...(1)
and if x + 1 < 0 and 0 ≤ x + 2 < 2
– 2 ≤ x < –1 then g(x) = f(0)
Now for x + 1 ≥ 0 and x + 2 < 2 ⇒ –1 ≤ x < 0, g(x) = f(x + 1)

Continuity | Mathematics (Maths) for JEE Main & Advanced
Hence g(x) is continuous in the interval [–3, 1].

Example 8. Given the function, 
f(x) = x [ 1 / x(1 + x) + 1 / (1 + x)(1 + 2x) + 1 / (1 + 2x)(1 + 3x) + ....upto ∞

Find f (0) if f (x) is continuous at x = 0.
Solution. 

Continuity | Mathematics (Maths) for JEE Main & Advanced
f(x) = 2 / 1 + x - 1 / 1 + nx upto n terms when x ≠0.
Hence 
Continuity | Mathematics (Maths) for JEE Main & Advanced

Example 9.  Let f: R →R be a function which satisfies f(x+y3) = f(x) + (f(y))3 ∀ x, y ∈ R. If f is continuous at x = 0, prove that f is continuous every where.
Solution. 
To prove
Continuity | Mathematics (Maths) for JEE Main & Advanced

Put x = y = 0 in the given relation f(0) = f(0) + (f(0))3 ⇒ f(0) = 0
Since f is continuous at x = 0
To prove
Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced
= f(x) + 0 = f(x).

Hence f is continuous for all x ∈ R.

Question for Continuity
Try yourself:
Which of the following statements is true about continuity of a function in a closed interval?
View Solution

Theorems of Continuity

Theorem 1. If f & g are two functions that are continuous at x= c then the functions defined by F1(x) = f(x) ± g(x) ; F2(x) = K f(x)  K any real number ; F3(x) = f(x).g(x) are also continuous at x= c.
Further, if g (c) is not zero, then  Continuity | Mathematics (Maths) for JEE Main & Advanced  is also continuous at x = c.

Theorem 2. If  f(x) is continuous & g(x) is discontinuous at x = a  then the product function ø(x) = f(x) . g(x) is not necessarily discontinuous at x = a .

Continuity | Mathematics (Maths) for JEE Main & Advanced

Theorem 3. If f(x) and g(x)  both  are  discontinuous  at  x = a  then  the  product  function ø(x) = f(x) . g(x) is not necessarily discontinuous at  x = a .

Continuity | Mathematics (Maths) for JEE Main & Advanced

Theorem 4: Intermediate Value Theorem 

  • If f is continuous on the closed interval [a, b] and k is any number between f(a) and f(b), then there is at least one number c in [a, b] such that f(c) = k.

Note:

  • The Intermediate Value Theorem tells that at least one c exists, but it does not give a method for finding c. Such theorems are called existence theorems.
  • As a simple example of this theorem, consider a person's height. Suppose that a girl is 5 feet tall on her thirteenth birthday and 5 feet 7 inches tall on her fourteenth birthday. Then, for any height h between 5 feet and 7 inches, there must have been a time t when her height was exactly h. This seems reasonable because human growth is continuous and a person's height does not abruptly change from one value to another.
  • The Intermediate Value Theorem guarantees the existence of at least one number c in the closed interval [a, b]. There may, of course, be more than one number c such that f(c) = k, as shown in Figure 1. A function that is not continuous does not necessarily possess the intermediate value property. For example, the graph of the function shown in Figure 2 jumps over the horizontal line given by y = k and for this function there is no value of c in [a, b] such that f(c) = k.
  • The Intermediate Value Theorem often can be used to locate the zeroes of a function that is continuous on a closed interval. Specifically, if f is continuous on [a, b] and f(a) and f(b) differ in sign, then the intermediate Value Theorem guarantees the existence of at least one zero of f in the closed interval [a, b].
    Continuity | Mathematics (Maths) for JEE Main & Advanced   Continuity | Mathematics (Maths) for JEE Main & Advanced   Continuity | Mathematics (Maths) for JEE Main & Advanced

 

Example 10. Use the Intermediate Value Theorem to show that the polynomial function f(x) = x3 + 2x - 1 has a zero in the interval [0, 1]

Sol. Note that f is continuous on the closed interval [0, 1]. Because

f(0) = 03 + 2(0) - 1 = -1         and                f(1) = 13 + 2(1) - 1 = 2

it follows that f(0) < 0 and f(1) > 0. You can therefore apply the Intermediate Value Theorem to conclude that there must be some c in [0, 1] such that f(c) = 0, as shown in Figure 3.


Example 11. State intermediate value theorem and use it to prove that the equation Continuity | Mathematics (Maths) for JEE Main & Advanced  has at least one real root.

Sol. Let f (x) = Continuity | Mathematics (Maths) for JEE Main & Advanced   first, f (x) is continuous on [5, 6]

Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced

Continuity | Mathematics (Maths) for JEE Main & Advanced

Hence by intermediate value theorem É at least one value of c ∈ (5, 6) for which f (c) = 0

Continuity | Mathematics (Maths) for JEE Main & Advanced

 c is root of the equation  Continuity | Mathematics (Maths) for JEE Main & Advanced

 

Example 12. If f(x) be a continuous function in Continuity | Mathematics (Maths) for JEE Main & Advanced then prove that there exists point Continuity | Mathematics (Maths) for JEE Main & Advanced

Sol. 

Let g(x) = f(x) – f(x + π) ....(i)

at x = π; g(π) = f(π) – f(2π) ....(ii)

at x = 0, g(0) = f(0) – f(π) ...(iii)

adding (ii) and (iii), g(0) + g(π) = f(0) – f(2π)

⇒ g(0) + g(π) = 0 [Given f(0) = f(2π) ⇒ g(0) = –g(π)

⇒ g(0) and g(π) are opposite in sign.

⇒ There exists a point c between 0 and p such g(c) = 0 as shown in graph;

From (i) putting x = c g(c) = f(c) – f(c +π) = 0 Hence, f(c) = f(c + π)

The document Continuity | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on Continuity - Mathematics (Maths) for JEE Main & Advanced

1. What is the definition of one-sided continuity?
Ans. One-sided continuity refers to the continuity of a function at a given point from either the left or the right side of that point.
2. What does continuity in an interval mean?
Ans. Continuity in an interval refers to a function being continuous for all values within that interval, meaning there are no abrupt changes or discontinuities.
3. What are some important theorems of continuity?
Ans. Some important theorems of continuity include the Intermediate Value Theorem, the Extreme Value Theorem, and the Bolzano's Intermediate Value Theorem.
4. How can one determine if a function is continuous at a specific point?
Ans. To determine if a function is continuous at a specific point, you need to check if the limit of the function exists at that point and if the function value at that point is equal to the limit value.
5. How can one determine if a function is continuous on a closed interval?
Ans. To determine if a function is continuous on a closed interval, you need to check if the function is continuous at every point within the interval and at the endpoints of the interval.
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