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Mathematics (Maths) Class 12

JEE : Doc: Continuity of a Function JEE Notes | EduRev

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A. Definition of Continuity 

Continuity at a Point: A function f is continuous at c if the following three conditions are met.
(i) f(x) is defined.
(ii)Doc: Continuity of a Function JEE Notes | EduRev

(iii)Doc: Continuity of a Function JEE Notes | EduRev

In other words function f(x) is said to be continuous at x = c , if Doc: Continuity of a Function JEE Notes | EduRev 

Symbolically f is continuous at  x = c

Doc: Continuity of a Function JEE Notes | EduRev

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
    Doc: Continuity of a Function JEE Notes | EduRev
  • A function f defined in some neigbourhood of a point c for x ³ c is said to be continuous at c from the right if
    Doc: Continuity of a Function JEE Notes | EduRev
  • 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
    Doc: Continuity of a Function JEE Notes | EduRevexists & Doc: Continuity of a Function JEE Notes | EduRev
  • The last equality means that f is continuous at c.
  • If one of the one-sided limits does not exist, thenDoc: Continuity of a Function JEE Notes | EduRev does not exist either. In this case, the point c is a discontinuity in the function, since the continuity condition is not met.

Continuity In An Interval
(a) A function f is said to be continuous in an open interval (a , b) if f is continuous at each & every point ∈(a, b).
(b) A function f is said to be continuous in a closed interval [a,b] if:

  • f is continuous in the open interval (a , b) &
  • f is right continuous at `a' i.e.
    Doc: Continuity of a Function JEE Notes | EduRev    
  • f is left continuous at `b' i.e.
     Doc: Continuity of a Function JEE Notes | EduRev

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

  • Doc: Continuity of a Function JEE Notes | EduRev does not exist i.e.  
    Doc: Continuity of a Function JEE Notes | EduRev
  • f(x) is not defined at x= c
  • Doc: Continuity of a Function JEE Notes | EduRev 
  • 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.

Doc: Continuity of a Function JEE Notes | EduRev

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

Doc: Continuity of a Function JEE Notes | EduRev
- π/2 - h
where h→ 0
Replace x by - π/2+h
where h → 0

Doc: Continuity of a Function JEE Notes | EduRev

So B - A = 2 ...(i)
At x = π/2

Doc: Continuity of a Function JEE Notes | EduRev
Replace x by π/2 - h
Replace x by π/2+h
where h→ 0
Doc: Continuity of a Function JEE Notes | EduRev
So    A+B = 0      ...(ii)
Solving (i) & (ii),    B= 1, A = -1

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

Doc: Continuity of a Function JEE Notes | EduRev
SolutionFor x < 0,
Doc: Continuity of a Function JEE Notes | EduRev

Doc: Continuity of a Function JEE Notes | EduRev

Doc: Continuity of a Function JEE Notes | EduRev

Doc: Continuity of a Function JEE Notes | EduRev
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,
Doc: Continuity of a Function JEE Notes | EduRev
where
f(x) = x2 - 2x + 2√3 - 3 / √3 - x, x ≠ √3

Doc: Continuity of a Function JEE Notes | EduRev

Doc: Continuity of a Function JEE Notes | EduRev

f(√3) = 2(1-√3).

Example 5. 
Doc: Continuity of a Function JEE Notes | EduRev
If f (x) is continuous at x = 0, then find the value  of (b+c)3-3a.
Solution.   

Doc: Continuity of a Function JEE Notes | EduRev

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

Doc: Continuity of a Function JEE Notes | EduRev
Doc: Continuity of a Function JEE Notes | EduRev
limit of Nr ⇒ 2a+8b = 0 ⇒ a = -4b
hence
-4b+b = -1  
⇒ b = 1/3 and a = -4/3

Doc: Continuity of a Function JEE Notes | EduRev
= 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. 

Doc: Continuity of a Function JEE Notes | EduRev

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

Doc: Continuity of a Function JEE Notes | EduRev  ,
for existence of limit  a + b + 5 = 0

Doc: Continuity of a Function JEE Notes | EduRev

Doc: Continuity of a Function JEE Notes | EduRev
= a / 2 + 5 / 2 - a = 3
⇒ a = - 1 ⇒ b = - 4

Doc: Continuity of a Function JEE Notes | EduRev 
for existence of limit  c = 0

Doc: Continuity of a Function JEE Notes | EduRev

= ed = 3 ⇒ d = ln 3

Example 7. Let f(x) = x3 = 3x2 + 6 ∀ x ∈ R and
Doc: Continuity of a Function JEE Notes | EduRev
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

Doc: Continuity of a Function JEE Notes | EduRev

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)

Doc: Continuity of a Function JEE Notes | EduRev
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. 

Doc: Continuity of a Function JEE Notes | EduRev
f(x) = 2 / 1 + x - 1 / 1 + nx upto n terms when x ≠0.
Hence 
Doc: Continuity of a Function JEE Notes | EduRev

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
Doc: Continuity of a Function JEE Notes | EduRev

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
Doc: Continuity of a Function JEE Notes | EduRev

Doc: Continuity of a Function JEE Notes | EduRev

Doc: Continuity of a Function JEE Notes | EduRev
= f(x) + 0 = f(x).

Hence f is continuous for all x ∈ R.

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