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State Board · Class 11 · Mathematics Part 2

Lesson 6: Functions

6 concepts · 48 practice questions with answers · Free to read

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Function, domain and range

From daily life

In a shop the price is fixed by the weight: one weight, one price. A rule that gives exactly one output for every input is a function. Building on classes 9–10 : Since classes 9–10 we know equations such as y = 2x + 1 and their graphs; here we treat them as functions.

Simple definition

A rule f : A → B that joins every x of A with exactly one y of B is a function. A is the domain, B the co-domain, and the set of values y actually obtained is the range.

Rule 1: Finding the domain — No zero in a denominator: for 1/(x − a), x ≠ a. No negative under a square root: for √(x − a), x ≥ a. If both occur obey both. If a vertical line cuts the graph twice, it is not a function.

AB1232468domainco-domainrange = {2, 4, 6} (8 has no pre-image)

In the figure : f(x) = 2x: A = {1, 2, 3}, B = {2, 4, 6, 8}; range {2, 4, 6}.

How is the formula obtained? : In a function one x cannot have two y values, but two x may share one y. So {(1, 2), (1, 3)} is not a function, but {(1, 2), (2, 2)} is.

  • Easy : f(x) = 3x − 2: f(4) = 3 × 4 − 2 = 10.
  • Medium : f(x) = √(x − 3)/(x − 5): need x − 3 ≥ 0 and x ≠ 5; domain [3, ∞) with 5 removed.
  • Exam-type : f(x) = √(9 − x²): 9 − x² ≥ 0, that is −3 ≤ x ≤ 3; domain [−3, 3], range [0, 3].
Remember

The range is a subset of the co-domain; they need not be equal.

Common mistake

Forgetting to apply both the denominator and the square-root conditions together.

Shortcut

For the domain first write "what is not allowed": division by zero and the square root of a negative. Memory clue : "No zero below, no negative under a root".

CET trap

For 1/√(x − 2) x = 2 is not allowed because the denominator is 0; the domain is (2, ∞).

CET-style example : The domain of f(x) = 1/√(x − 2) is? (A) (2, ∞) (B) [2, ∞) (C) (−∞, 2) (D) R − {2}

Answer : (A). x − 2 > 0. Time-saving method : Under the root positive and denominator not zero, so x − 2 > 0 (not equal).

Spot the difference

SetMeaningf(x) = 2x, A = {1, 2, 3}
Domaininputs x{1, 2, 3}
Co-domaingiven B{2, 4, 6, 8}
Rangevalues obtained{2, 4, 6}

Practice questions

Easy
  1. 1f(x) = x² + 1: find f(3).
  2. 2Is {(1, 2), (1, 3)} a function?
  3. 3f(x) = 5 − x: find f(−2).
Medium
  1. 4Write the domain of 1/(x − 4).
  2. 5Write the domain of √(x + 5).
  3. 6Write the range of f(x) = x² on R.
Challenging
  1. 7Write the domain of √(x − 2) + √(6 − x).
  2. 8Write the range of f(x) = √(16 − x²).
Show answers (try them yourself first)
  1. 110.
  2. 2नाही ; 1 ला दोन y.
  3. 37.
  4. 4R − {4}.
  5. 5[−5, ∞).
  6. 6[0, ∞).
  7. 7[2, 6].
  8. 8[0, 4].

Types of functions

From daily life

On a bus each passenger has a separate seat and no seat is empty, so passengers and seats are perfectly paired. In functions we look at the same kinds of properties. Building on classes 9–10 : Since classes 9–10 we have found y for each x; here we study properties of the pairing.

Simple definition

One-one: different x give different y. Onto: every element of the co-domain is the y of some x (range = co-domain). If both hold the function is bijective.

Rule 1: Counting functions — m elements → n elements: all functions nᵐ; one-one ⁿPₘ (n ≥ m); when m = n bijective n!.

one-oneontobothnot ontonot one-onebijectiveone-one: no two share an imageonto: every B element is hit

In the figure : Three diagrams: one-one, onto, bijective.

How is the formula obtained? : Each x has n choices of y, m times: nᵐ. For one-one no y can be reused: n × (n − 1) × … = ⁿPₘ.

  • Easy : f(x) = 2x + 3 (R → R) is one-one and onto, because from y = 2x + 3 we always get x = (y − 3)/2.
  • Medium : f(x) = x² (R → R): f(2) = f(−2) = 4, so not one-one; −1 is nobody’s square, so not onto.
  • Exam-type : f(x) = x² from [0, ∞) to [0, ∞) is bijective. From 3 elements to 5: one-one functions ⁵P₃ = 60 and all functions 5³ = 125.
Remember

To test one-one, take f(a) = f(b) and see whether a = b follows.

Common mistake

Changing the domain or co-domain can change the type; x² is not one-one on R but is on [0, ∞).

Shortcut

If a horizontal line cuts the graph twice, it is not one-one. Memory clue : "Horizontal line twice means not one-one".

CET trap

When counting one-one functions with m > n the answer is 0; ⁿPₘ does not apply.

CET-style example : How many one-one functions are there from {1, 2, 3} to {1, 2, 3, 4, 5}? (A) 60 (B) 125 (C) 15 (D) 243

Answer : (A). ⁵P₃ = 5 × 4 × 3 = 60. Time-saving method : 125 counts all functions; one-one means no repeats.

Spot the difference

TypeConditionExample
One-onedifferent x → different y2x + 3
Ontorange = co-domain2x + 3 (R → R)
Bijectivebothx² ([0, ∞) → [0, ∞))

Practice questions

Easy
  1. 1How many functions are there from a 2-element set to a 3-element set?
  2. 2Of these how many are one-one?
  3. 3Is f(x) = 3x − 1 one-one?
Medium
  1. 4Is f(x) = x² from {−1, 0, 1} to {0, 1} one-one?
  2. 5How many bijections are there from a 4-element set onto itself?
  3. 6How many onto functions are there from 3 elements to 2 elements?
Challenging
  1. 7How many onto functions are there from 4 elements to 2 elements?
  2. 8How many one-one functions are there from 3 elements to 4 elements?
Show answers (try them yourself first)
  1. 13² = 9.
  2. 2³P₂ = 6.
  3. 3हो ; 3a − 1 = 3b − 1 वरून a = b.
  4. 4नाही ; f(−1) = f(1) = 1.
  5. 54! = 24.
  6. 62³ − 2 = 6.
  7. 72⁴ − 2 = 14.
  8. 8⁴P₃ = 24.

Standard functions

From daily life

To tell how far the temperature is from zero we drop the sign and give the size only. A bus fare that rises in steps with each stage is a staircase. Building on classes 9–10 : Since classes 9–10 we know absolute value and rounding down; here they appear as functions.

Simple definition

∣x∣ = x (x ≥ 0), −x (x < 0). [x] = the greatest integer less than or equal to x. {x} = x − [x] (fractional part). sgn x = 1 (x > 0), 0 (x = 0), −1 (x < 0).

Rule 1: Useful rules — ∣x∣ < a means −a < x < a; ∣x∣ = a means x = ±a; [x] = n means n ≤ x < n + 1; 0 ≤ {x} < 1.

∣x∣[x]sgn x∣x∣ ≥ 0 always; [x] makes steps; sgn x is −1, 0 or 1

In the figure : Three graphs: ∣x∣, [x], sgn x.

How is the formula obtained? : For [x] we go down even when x is negative: [−2.3] = −3, not −2. So {−2.3} = −2.3 − (−3) = 0.7.

  • Easy : ∣−5∣ = 5; [3.7] = 3; [−2.3] = −3; {−2.3} = 0.7.
  • Medium : ∣x − 2∣ = 5: x − 2 = 5 or −5; x = 7 or −3. [x] = 3: 3 ≤ x < 4.
  • Exam-type : ∣x − 1∣ < 3: −3 < x − 1 < 3, that is −2 < x < 4. Also [x] + {x} = x always.
Remember

∣x∣ is never negative; {x} always lies from 0 up to but not including 1.

Common mistake

Taking [−2.3] = −2. The correct value is −3, the greatest integer not above −2.3.

Shortcut

For [x] look to the left on the number line and take the nearest integer. Memory clue : "The step below".

CET trap

∣x∣ > a splits into two parts: x > a or x < −a; not the middle part.

CET-style example : [−3.2] + [3.2] = ? (A) −1 (B) 0 (C) −2 (D) 1

Answer : (A). −4 + 3 = −1. Time-saving method : For the negative number go one step down: −3.2 → −4.

Spot the difference

Functionfor x = −2.3for x = 3.7
∣x∣2.33.7
[x]−33
{x}0.70.7
sgn x−11

Practice questions

Easy
  1. 1Find ∣−8∣.
  2. 2Find [5.9].
  3. 3Find [−0.5].
Medium
  1. 4Solve ∣x + 4∣ = 6.
  2. 5Find {−1.6}.
  3. 6Write the x with [x] = −2.
Challenging
  1. 7Solve ∣2x − 3∣ ≤ 5 (integer x).
  2. 8Find [−1.5] × [1.5].
Show answers (try them yourself first)
  1. 18.
  2. 25.
  3. 3−1.
  4. 4x = 2 किंवा −10.
  5. 5−1.6 − (−2) = 0.4.
  6. 6−2 ≤ x < −1.
  7. 7−5 ≤ 2x − 3 ≤ 5 ; −1 ≤ x ≤ 4 : −1, 0, 1, 2, 3, 4.
  8. 8(−2) × 1 = −2.

Algebra of functions and composite function

From daily life

Grain is milled to flour and flour is made into bread: the first machine takes the input and hands its output to the second. Such a chain is a composite function. Building on classes 9–10 : Since classes 9–10 we have added and multiplied algebraic expressions; here we do it for functions.

Simple definition

(f + g)(x) = f(x) + g(x); (fg)(x) = f(x) · g(x); (f/g)(x) = f(x)/g(x), g(x) ≠ 0. The composite function (f ∘ g)(x) = f(g(x)): first g, then f.

Rule 1: Domains — For sum and product take the common part of the two domains; for a quotient remove g(x) = 0; for f ∘ g the range of g must lie in the domain of f.

xgg(x)fyf ∘ g (x) = f(g(x))g works first, f works nextf ∘ g and g ∘ f are usually different

In the figure : x → g → f: f(g(x)).

How is the formula obtained? : f(x) = 2x + 1, g(x) = x²: f(g(x)) = 2x² + 1 and g(f(x)) = (2x + 1)²; they differ, so order matters.

  • Easy : f(x) = 2x + 1, g(x) = x²: (f ∘ g)(2) = f(4) = 9; (g ∘ f)(2) = g(5) = 25.
  • Medium : f(x) = x + 2, g(x) = 3x: (f ∘ g)(x) = 3x + 2; (g ∘ f)(x) = 3x + 6.
  • Exam-type : f(x) = √x, g(x) = x − 4: f ∘ g = √(x − 4), domain x ≥ 4; the domain of f/g is x ≥ 0, x ≠ 4.
Remember

In (f ∘ g)(x) the inner function g acts first.

Common mistake

Taking f ∘ g and g ∘ f as the same. They usually differ.

Shortcut

Find the inner value first and put it into the outer function. Memory clue : "Inside first, outside next".

CET trap

(f ∘ g)(2) means f(g(2)), not g(f(2)).

CET-style example : If f(x) = x + 2 and g(x) = 3x then (f ∘ g)(2) = ? (A) 8 (B) 12 (C) 10 (D) 14

Answer : (A). g(2) = 6, f(6) = 8. Time-saving method : 12 is g(f(2)); watch the order.

Spot the difference

Functionf(x) = x + 2, g(x) = 3xfor x = 2
f ∘ g3x + 28
g ∘ f3x + 612
f + g4x + 210
fg3x² + 6x24

Practice questions

Easy
  1. 1f(x) = x + 3, g(x) = 2x: find (f + g)(2).
  2. 2Find (fg)(1).
  3. 3Find f(g(1)).
Medium
  1. 4f(x) = x², g(x) = x − 1: find (f ∘ g)(3).
  2. 5Find (g ∘ f)(3).
  3. 6Find (f − g)(2).
Challenging
  1. 7f(x) = 2x + 3, g(x) = (x − 3)/2: find f(g(7)) and state the conclusion.
  2. 8f(x) = x² + 1, g(x) = √(x − 1): find g(f(−2)).
Show answers (try them yourself first)
  1. 1(f + g)(2) = 5 + 4 = 9.
  2. 2(fg)(1) = 4 × 2 = 8.
  3. 3g(1) = 2, f(2) = 5.
  4. 4g(3) = 2, f(2) = 4.
  5. 5f(3) = 9, g(9) = 8.
  6. 64 − 1 = 3.
  7. 7g(7) = 2, f(2) = 7 ; f(g(x)) = x, म्हणजे दोन्ही एकमेकांचे व्यस्त.
  8. 8f(−2) = 5, g(5) = 2.

Inverse function

From daily life

Reversing the rule that turns Celsius into Fahrenheit gives Fahrenheit back to Celsius. A rule that can be reversed completely gives an inverse function. Building on classes 9–10 : Since classes 9–10 we have solved an equation for x; here we use the same method to find an inverse.

Simple definition

If f : A → B is one-one and onto, then f⁻¹ : B → A satisfies f⁻¹(f(x)) = x and f(f⁻¹(y)) = y. Only a bijective function has an inverse.

Rule 1: Method to find the inverse — Write y = f(x); solve for x; swap x and y. The graph of f⁻¹ is the mirror image in the line y = x.

y = xff⁻¹(1, 3)(3, 1)the graphs are mirror images in y = x

In the figure : f(x) = 2x + 1 and f⁻¹: (1, 3) becomes (3, 1).

How is the formula obtained? : f(1) = 3 means f⁻¹(3) = 1; every point (a, b) becomes (b, a), and this swap is exactly the mirror image in the line y = x.

  • Easy : f(x) = 3x − 4: y = 3x − 4, x = (y + 4)/3; f⁻¹(x) = (x + 4)/3.
  • Medium : f(x) = x³ + 1: x = (y − 1)¹ᐟ³; f⁻¹(x) = (x − 1)¹ᐟ³.
  • Exam-type : f(x) = (x + 1)/(x − 2): y(x − 2) = x + 1; x(y − 1) = 2y + 1; f⁻¹(x) = (2x + 1)/(x − 1).
Remember

f⁻¹ does not mean 1/f; it is the reversing function.

Common mistake

Forgetting to swap x and y. In the end write the function in terms of x.

Shortcut

To find f⁻¹(k) just solve f(x) = k; no need for the inverse formula. Memory clue : "Which x gave this?".

CET trap

x² has no inverse on R because it is not one-one; the domain must be restricted to [0, ∞).

CET-style example : If f(x) = 2x + 5 then f⁻¹(9) = ? (A) 2 (B) 7 (C) 4 (D) 14

Answer : (A). 2x + 5 = 9, x = 2. Time-saving method : Solve f(x) = 9 without finding the formula.

Spot the difference

f(x)f⁻¹(x)Check
3x − 4(x + 4)/3f(2) = 2, f⁻¹(2) = 2
x³ + 1(x − 1)¹ᐟ³f(2) = 9, f⁻¹(9) = 2
2x + 5(x − 5)/2f(2) = 9, f⁻¹(9) = 2

Practice questions

Easy
  1. 1f(x) = x + 1: find f⁻¹(5).
  2. 2Write the inverse of f(x) = x + 7.
  3. 3If f(x) = x + 7 find f⁻¹(10).
Medium
  1. 4f(x) = 5x − 3: find f⁻¹(12).
  2. 5g(x) = 4x − 3: find g(3).
  3. 6Write the inverse f⁻¹(x) of f(x) = 2x.
Challenging
  1. 7f(x) = (x + 1)/(x − 2): find f⁻¹(3).
  2. 8In which line are the graphs of f(x) = 2x + 1 and f⁻¹ mirror images?
Show answers (try them yourself first)
  1. 1x + 1 = 5, x = 4.
  2. 2f⁻¹(x) = x − 7.
  3. 310 − 7 = 3.
  4. 45x − 3 = 12, x = 3.
  5. 59.
  6. 6f⁻¹(x) = x/2.
  7. 7f⁻¹(3) = (2 × 3 + 1)/(3 − 1) = 7/2.
  8. 8y = x.

Exponential and logarithmic functions

From daily life

A bank balance grows by a fixed factor each year: 1000, 1100, 1210 …. This is exponential growth. "In how many years will it double?" is the reverse question, answered by a logarithm. Building on classes 9–10 : Since classes 9–10 we know the laws of indices; here they become a function and its inverse, the logarithm.

Simple definition

The function aˣ (a > 0, a ≠ 1) has domain R and range (0, ∞). log_a x = y means aʸ = x; its domain is (0, ∞) and its range is R. They are inverses of each other.

Rule 1: Laws of logarithms — log(mn) = log m + log n; log(m/n) = log m − log n; log mⁿ = n log m; log_a a = 1; log_a 1 = 0.

(0, 1)(1, 0)2ˣlog₂ x

In the figure : 2ˣ and log₂ x: (0, 1) and (1, 0); mirror images in y = x.

How is the formula obtained? : With m = aᵖ and n = aᑫ we get mn = aᵖ⁺ᑫ. So log(mn) = p + q = log m + log n.

  • Easy : log₂ 8 = 3 because 2³ = 8; log₁₀ 0.001 = −3.
  • Medium : log₂ 6 + log₂ (4/3) = log₂ 8 = 3; 3 log₂ 5 = log₂ 125.
  • Exam-type : Given log₁₀ 2 = 0.3010: log₁₀ 8 = 3 × 0.3010 = 0.9030; log₁₀ 5 = log₁₀ (10/2) = 1 − 0.3010 = 0.6990.
Remember

log (a + b) is not log a + log b; the laws are only for product, quotient and power.

Common mistake

Forgetting that the domain of log is (0, ∞) and taking log of a negative or zero.

Shortcut

Write log_a x = y in the exponential form aʸ = x; the equation becomes easy. Memory clue : "Log means exponent".

CET trap

In a log equation keep only the value for which every log argument is positive; reject the others.

CET-style example : log₃ 81 − log₂ 8 = ? (A) 1 (B) 7 (C) 4 (D) 3

Answer : (A). 4 − 3 = 1. Time-saving method : Recognise 3⁴ = 81 and 2³ = 8.

Spot the difference

Pointaˣlog_a x
DomainR(0, ∞)
Range(0, ∞)R
Key point(0, 1)(1, 0)

Practice questions

Easy
  1. 1Find log₅ 25.
  2. 2Find log₂ 32.
  3. 3Find log₁₀ 1.
Medium
  1. 4Find log₃ (1/9).
  2. 5Find log₂ 3 + log₂ (8/3).
  3. 6Solve 2ˣ = 64.
Challenging
  1. 7Solve log₂ (x + 1) = 4.
  2. 8Solve log₂ x + log₂ (x − 2) = 3.
Show answers (try them yourself first)
  1. 12.
  2. 25.
  3. 30.
  4. 4−2.
  5. 5log₂ 8 = 3.
  6. 6x = 6.
  7. 7x + 1 = 16 ; x = 15.
  8. 8x(x − 2) = 8 ; x = 4 किंवा −2 ; −2 वगळा ; x = 4.

Lesson 6: Functions — quick revision

TopicFormula / point
Functionexactly one y for each x; domain, co-domain, range
Domaindenominator ≠ 0; under a root ≥ 0
Typesone-one, onto, bijective; nᵐ, ⁿPₘ, n!
Standard∣x∣, [x], {x} = x − [x], sgn x
Composite(f ∘ g)(x) = f(g(x)): g first
Inversesolve y = f(x), swap x and y; graph mirrors in y = x
Logarithmlog(mn) = log m + log n; log mⁿ = n log m; inverse of aˣ

🎯 Most important points for the exam

  • First write what is not allowed for the domain.
  • [−2.3] = −3.
  • In f ∘ g, g acts first.
  • f⁻¹ does not mean 1/f.
  • The domain of log is (0, ∞).
Study mantra

"Inside first, outside next".