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State Board · Class 11 · Mathematics Part 2

Lesson 8: Continuity

4 concepts · 32 practice questions with answers · Free to read

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Continuity at a point

From daily life

A line drawn without lifting the pencil is unbroken; if the pencil has to be lifted on the way there is a break. This unbroken nature of a graph is continuity. Building on classes 9–10 : Since classes 9–10 we have drawn graphs; here is the exact condition for them to be unbroken.

Simple definition

f is continuous at a if three things hold: (1) f(a) exists, (2) lim f(x) as x → a exists, (3) the limit equals f(a).

Rule 1: Checking the three conditions — left limit = right limit = f(a); if any one fails the function is discontinuous. For a function in pieces test both formulas at the joining point.

a1 f(a) exists2 limit exists3 limit = f(a)no lift of pencil

In the figure : An unbroken graph and the three conditions.

How is the formula obtained? : If the limit and f(a) differ, the point and the curve do not meet and there is a break. When all three conditions agree there is no break.

  • Easy : f(x) = x² + 1: at a = 2, f(2) = 5 and the limit is 5; continuous.
  • Medium : f(x) = (x² − 9)/(x − 3) (x ≠ 3), k (x = 3): the limit is 6; continuous when k = 6.
  • Exam-type : f(x) = kx + 1 (x ≤ 2), 3x − 2 (x > 2): left 2k + 1, right 4; 2k + 1 = 4, so k = 3/2.
Remember

All three conditions must agree; the existence of the limit alone is not enough.

Common mistake

Checking only the limit and forgetting that f(a) may not exist.

Shortcut

To find an unknown k set left limit = right limit = f(a) at the joining point. Memory clue : "Left, right, value: all three equal".

CET trap

Note which formula gives f(2): the one with x ≤ 2.

CET-style example : If f(x) = x + k (x < 1), 3 (x ≥ 1) is continuous at x = 1, then k = ? (A) 2 (B) 1 (C) 3 (D) 0

Answer : (A). 1 + k = 3. Time-saving method : The left limit is 1 + k and the value is 3.

Spot the difference

ConditionMeaning
1f(a) exists
2limit exists (left = right)
3limit = f(a)

Practice questions

Easy
  1. 1Is a polynomial continuous at every point?
  2. 2Is f(x) = 1/(x − 2) continuous at x = 2?
  3. 3Find k so that f(x) = (x² − 4)/(x − 2) (x ≠ 2), k (x = 2) is continuous.
Medium
  1. 4f(x) = 3x − 1 (x < 2), k (x = 2), x + 3 (x > 2): find k for continuity.
  2. 5f(x) = 2x + k (x ≤ 1), 5 (x > 1): find k for continuity.
  3. 6Is [x] continuous at x = 2?
Challenging
  1. 7f(x) = sin 3x/x (x ≠ 0), k (x = 0): find k for continuity.
  2. 8f(x) = (√(x + 1) − 1)/x (x ≠ 0), k (x = 0): find k for continuity.
Show answers (try them yourself first)
  1. 1होय.
  2. 2नाही ; f(2) अस्तित्वात नाही.
  3. 3k = 4.
  4. 4दोन्ही बाजूंची सीमा 5 ; k = 5.
  5. 52 + k = 5 ; k = 3.
  6. 6नाही ; डावी सीमा 1, उजवी 2.
  7. 7सीमा 3 ; k = 3.
  8. 8सीमा 1/2 ; k = 1/2.

Types of discontinuity

From daily life

A small pothole can be filled so the road is whole again; a broken bridge needs a jump to the other side; a gorge ends the road. Discontinuities are of three similar kinds. Building on classes 9–10 : Since classes 9–10 we have seen holes and jumps in graphs; here we name them.

Simple definition

Removable: the limit exists but f(a) is missing or different (a hole); redefining f(a) makes it continuous. Jump: left and right limits exist but differ. Infinite: a limit of ±∞ (a vertical line).

Rule 1: How to recognise — Find the left and right limits: equal means removable; different and finite means a jump; infinite means an infinite discontinuity. Size of the jump = right − left.

removablejumpinfinitelimit existslimits differlimit is ∞

In the figure : Three kinds: removable hole, jump, infinite.

How is the formula obtained? : (x² − 1)/(x − 1) = x + 1 (x ≠ 1); the limit at 1 is 2 but f(1) is missing. Defining f(1) = 2 removes the break, hence removable.

  • Easy : f(x) = (x² − 1)/(x − 1) at 1: limit 2, f(1) missing; removable. Take f(1) = 2.
  • Medium : f(x) = [x] at 2: left 1, right 2; the jump is of size 1.
  • Exam-type : f(x) = 1/(x − 3) at 3: limit ±∞; infinite discontinuity. f(x) = x² (x < 1), x + 2 (x ≥ 1): left 1, right 3; jump 2.
Remember

A jump cannot be removed by redefining; only a removable one can.

Common mistake

Mixing a hole with a jump: in a hole both limits are equal, in a jump they differ.

Shortcut

First write the left and right limits; then decide the name. Memory clue : "Equal hole, different jump, infinite gorge".

CET trap

The discontinuity of ∣x∣/x at 0 is a jump, not a hole.

CET-style example : The discontinuity of f(x) = ∣x∣/x at x = 0 is? (A) jump (B) removable (C) infinite (D) continuous

Answer : (A). Left −1, right 1. Time-saving method : Both finite and different: a jump.

Spot the difference

TypeLeft / right limitsExample
Removableequal(x² − 1)/(x − 1) at 1
Jumpdifferent, finite[x] at 2
Infiniteone is ∞1/(x − 3) at 3

Practice questions

Easy
  1. 1What kind of discontinuity does (x² − 9)/(x − 3) have at 3?
  2. 2What kind of discontinuity does [x] have at 5?
  3. 3What kind of discontinuity does 1/x have at 0?
Medium
  1. 4How can the discontinuity of (x² − 4)/(x + 2) at x = −2 be removed?
  2. 5What is the jump of [x] at 5?
  3. 6Is the discontinuity of 1/x removable?
Challenging
  1. 7f(x) = x + 1 (x < 3), x² (x ≥ 3): find the jump at 3.
  2. 8Why is a jump discontinuity not removable?
Show answers (try them yourself first)
  1. 1निवारणीय (सीमा 6).
  2. 2उडी.
  3. 3अनंत.
  4. 4f(−2) = −4 ठरवा.
  5. 55 − 4 = 1.
  6. 6नाही ; सीमा अनंत.
  7. 79 − 4 = 5.
  8. 8डावी आणि उजवी सीमा भिन्न असल्याने एक मूल्य ठरवून जुळत नाही.

Continuity on an interval and properties

From daily life

To check whether a road is unbroken from one village to another we check every point, and at the ends we can arrive only from one side. Continuity on an interval is measured the same way. Building on classes 9–10 : Since classes 9–10 we know intervals and graphs; here we study continuity over them.

Simple definition

f is continuous on (a, b) if it is continuous at every point. It is continuous on [a, b] if it is continuous on (a, b), right-continuous at a and left-continuous at b.

Rule 1: Properties — Sums, differences and products of continuous functions are continuous; a quotient is continuous where the denominator is not zero; a composite is continuous. Polynomials, sin, cos, eˣ and ∣x∣ are continuous everywhere; a rational function wherever its denominator is not zero.

abfrom rightfrom leftcontinuous on [a, b]

In the figure : Continuous on [a, b]: from the right at a, from the left at b.

How is the formula obtained? : The domain of f(x) = √(x − 1) is [1, ∞); there is no function to the left of 1, so only right-continuity is tested there, and it holds.

  • Easy : f(x) = √(x − 1) on [1, ∞): at 1 from the right √0 = 0; continuous.
  • Medium : (x + 1)/(x² − 4) is discontinuous at x = 2 and x = −2 (denominator zero).
  • Exam-type : (x² + 1)/(x² − 5x + 6) = (x² + 1)/((x − 2)(x − 3)) is discontinuous at x = 2 and 3.
Remember

At an end point a one-sided limit is enough.

Common mistake

Forgetting to find where the denominator is zero; those are the points of discontinuity.

Shortcut

For a rational function solve denominator = 0; it is continuous everywhere else. Memory clue : "Denominator zero, break only there".

CET trap

If both top and bottom vanish the break may be removable; cancel the common factor and check.

CET-style example : At how many points is f(x) = 1/(x² − 9) discontinuous? (A) 2 (B) 1 (C) 3 (D) 0

Answer : (A). x = 3 and −3. Time-saving method : Denominator (x − 3)(x + 3) = 0.

Spot the difference

IntervalCondition
(a, b)continuous at every point
[a, b]continuous on (a, b) + from the right at a + from the left at b

Practice questions

Easy
  1. 1Is 3x² + 1 continuous everywhere?
  2. 2How many points of discontinuity does 1/(x − 5) have?
  3. 3Write the points of discontinuity of x/(x² − 1).
Medium
  1. 4Write the points of discontinuity of 1/(x² − 5x + 6).
  2. 5Is √(4 − x²) continuous on [−2, 2]?
  3. 6Is ∣x − 3∣ continuous everywhere?
Challenging
  1. 7How many points of discontinuity does 1/(x³ − x) have?
  2. 8Is [x] continuous on (2, 3)?
Show answers (try them yourself first)
  1. 1होय.
  2. 21 (x = 5).
  3. 3x = 1 आणि −1.
  4. 4x = 2 आणि 3.
  5. 5होय ; टोकांवर एकतर्फी.
  6. 6होय.
  7. 73 (x = 0, 1, −1).
  8. 8होय ; तेथे ते स्थिर 2 आहे.

Intermediate value theorem

From daily life

If the temperature rises continuously from 18° in the morning to 30° at noon, it must have been 25° at some time in between. A graph that climbs without jumping takes every value on the way. Building on classes 9–10 : Since classes 9–10 we have read solutions of equations from graphs; here we decide in which interval a solution lies.

Simple definition

If f is continuous on [a, b] and f(a), f(b) have opposite signs, then f(c) = 0 for at least one c in (a, b). In general f takes every value between f(a) and f(b).

Rule 1: Interval of a root — First state that the function is continuous; then take two values and look at the signs. A sign change means a root in that interval. Halving repeatedly shrinks the interval.

f(a) < 0f(b) > 0cf(c) = 0 for some c in (a, b)

In the figure : f(a) < 0 and f(b) > 0; the graph crosses the axis at c.

How is the formula obtained? : A continuous curve going from below the axis to above it must cross the axis; so f(c) = 0 there.

  • Easy : x³ − x − 1 = 0: f(1) = −1, f(2) = 5; the sign changes, so a root lies in (1, 2).
  • Medium : cos x − x = 0: f(0) = 1 > 0 and f(1) = cos 1 − 1 < 0; a root lies in (0, 1).
  • Exam-type : x³ + x − 3: f(1) = −1, f(2) = 7; halving gives f(1.5) = 1.875 > 0; the root lies in (1, 1.5).
Remember

The theorem needs continuity: 1/x has f(−1) < 0 < f(1) but no root at 0.

Common mistake

Assuming a sign change means exactly one root. The theorem gives at least one.

Shortcut

Put the whole numbers 0, 1, 2, 3 in turn and find the first sign change. Memory clue : "Sign changed, root found".

CET trap

If both ends have the same sign there may still be a root; the theorem is silent.

CET-style example : In which interval does x³ − 2x − 5 = 0 have a root? (A) (2, 3) (B) (0, 1) (C) (1, 2) (D) (3, 4)

Answer : (A). f(2) = −1, f(3) = 16. Time-saving method : f(0), f(1), f(2) are negative and f(3) is positive; hence (2, 3).

Spot the difference

Functionf(a)f(b)Root
x³ − x − 1f(1) = −1f(2) = 5(1, 2)
cos x − xf(0) = 1f(1) < 0(0, 1)
x³ − 2x − 5f(2) = −1f(3) = 16(2, 3)

Practice questions

Easy
  1. 1x² − 4x + 1: write the signs of f(0) and f(1) and conclude.
  2. 2Between which integers does the root of x³ = 9 lie?
  3. 3Which property of the function does the intermediate value theorem need?
Medium
  1. 4In which interval does x³ + 2x − 4 = 0 have a root?
  2. 5Narrow the root above further (take 1.5).
  3. 6Between which integers does the root of eˣ = 3 lie?
Challenging
  1. 7In which two intervals do the two different roots of x⁴ − 3x − 1 = 0 lie?
  2. 8Find f(−1) and f(0) for this function.
Show answers (try them yourself first)
  1. 1f(0) = 1 > 0, f(1) = −2 < 0 ; मूळ (0, 1).
  2. 22 आणि 3 (8 < 9 < 27).
  3. 3संततता.
  4. 4(1, 2) ; f(1) = −1, f(2) = 8.
  5. 5f(1.5) = 2.375 > 0 ; (1, 1.5).
  6. 61 आणि 2 (e ≈ 2.718 < 3 < e²).
  7. 7(−1, 0) आणि (1, 2).
  8. 8f(−1) = 3, f(0) = −1.

Lesson 8: Continuity — quick revision

TopicFormula / point
Continuityf(a) exists; the limit exists; limit = f(a)
Removablelimit exists but f(a) missing or different; can be repaired
Jumpleft ≠ right (finite); size = right − left
Infinitelimit ±∞; vertical line
Intervalon [a, b]: from the right at a, from the left at b
Intermediate valuef(a) f(b) < 0 and f continuous gives f(c) = 0

🎯 Most important points for the exam

  • Test all three conditions.
  • For a function in pieces make left = right at the joining point.
  • A break occurs where the denominator is zero.
  • Look for a sign change when locating a root.
  • Without continuity the theorem does not apply.
Study mantra

"Left, right, value: all three equal".