20_JEE_IL-Achivers_Maths_Functions_V2

FUNCTIONS CHAPTER 20

Chapter Outline

20.1 Relation as a Function

20.2 Real Valued Functions

20.3 Methods of Finding Domain and Range of a Function

20.4 Algebra of Functions

20.5 Graphical Transformations

20.6 Kinds of Functions

20.7 Composition of a Function

20.8 Inverse of a Function

■ A function is a rule that assigns each input (from a set called the domain) to exactly one output (in the range).

■ It is commonly written as f(x), where x is the input, and f(x)is the output.

■ Functions help describe relationships between variables in mathematics, science, and real-world applications.

■ Types of functions include polynomial, rational, trigonometric, exponential, and logarithmic.

■ They can be represented using equations, graphs, tables, or mappings.

20.1 RELATION AS A FUNCTION

A function f from a set A to a set B associates each element of A to a unique element of B.

■ Denoted as: f : A → B

(Read as: f is mapping from A to B)

■ Every function is a relation, but not every relation a function.

A relation R = {(a,b): a∈A, b∈B} is a function if:

■ Each element of A has a unique image in B (a,b) ∈ R for all a ∈ A.

■ No element in A has multiple images if (a,b) ∈ R and (a,c) ∈ R, then b=c.

20.1.1 Image, Pre-image, Value of a Function

■ Image: y is the image of x under f, where (x,y) ∈ f.

■ Pre-image: x is the pre-image of y under f.

Function Components:

■ Domain: A (set of all inputs)

■ Co-domain: B (set where outputs belong)

■ Range: {()} :,faaA ∈ subset of B.

Function Value at x=a y = f(x) ⇒ f(a)= y

Solved example

1. If a function f : A → B is defined as f(x) = x2 , then find the range of f. Sol. Since, the given relation is square function, it gives only positive numbers. So, the range of f(x) = x2 is set of all non-negative real numbers.

Try yourself:

1. If f(x) = x2 , find the value of ()() 1.11 1.11 ff . Ans: 2.1

20.1.2 Check Whether the Given Relation Is a Function

To verify from ordered pairs:

■ Domain check: First elements in ordered pairs must cover all of A.

■ Uniqueness: No two ordered pairs should have the same first element.

From a Venn diagram:

■ Arrows must originate from every element in A.

■ No element in A should have multiple outgoing arrows.

From a Graph:

■ Vertical Line Test: If a vertical line intersects the graph at more than one point, it is not a function.

Solved example

2. I f X = {1,2,3,4} and Y = {1,5,9,11,15,16}, determine which of the following are functions from X to Y.

ƒ ()()()() {} 1 1,1,2,11,3,1,4,15 f =

ƒ ()()() {} 2 1,1,2,9,3,5 f =

ƒ ()()()()() {} 3 1,5,2,9,3,3,4,5,2,11 f =

Sol. The re lation f 1 is a function because all the elements of X have images and each element of X has a unique image.

The relation f2 is not a function because 4∈X has no image.

The relation f3 is not a function because 2∈X has two images.

■ When relation is given in Venn diagram, check the following to decide whether or not it is a function.

‰ The arrows initiate from every element of first set and terminate at any element of second set.

‰ Only one arrow must initiate from each element of the first set.

Solved example

3. Check whether the following relations are functions or not and give reason.

Sol. Observe the first arrow diagram, 3 ∈ A has no image in B. So, the relation f1 is not a function.

Observe the second arrow diagram, all the elements of A have unique images, but not 4. So, the relation f2 is not a function.

■ If any vertical line cuts the continuous graph of relation at only one point, then the relation is a function.

Solved example

4. Check whether or not the relation defined as f(x) = x2 on the set of real numbers is a function.

Sol. Draw the graph of f(x) = x2 .

Any vertical line intersects the above graph at most at one point. So, the relation is a function.

Try yourself:

2. Check whether the relation defined as f(x) = x3 on the set of real numbers is a function or not.

Ans: Yes

20.1.3 Number of Functions

For two finite sets Aand B

If n(A) = p and n(B) = q, then the total number of functions from A to B is: qp .

(Each element in A has q choices in B, so total possibilities =qp.)

Solved example

5. What is the number of functions defined on the set A = {1,2,3,4,5}?

Sol. Given, set A = {1,2,3,4,5} Here, n(A) = 5, so the number of functions defined on the set A is 55 = 3125

Try yourself:

3. What is the number of functions defined on the set A = {1,2,3,4}?

Ans: 44 = 256

20.1.4 Equality of Two Functions

Two functions f, and g are equal if:

■ Same domain: Dom (f)=Dom(g) for all x in the domain

■ For every element x in the domain ,f(x) =g(x)

Solved example

6. Find the domain for which the function f(x) = 2x2–1 and g(x) = 1 + 3x are equal.

Sol. Given, f(x) = 2x2–1 and g(x) = 1+3x Suppose that f(x) = g(x)

Try yourself:

4. What are the restrictions for which the functions f ( x ) = sec2x – tan2x and g ( x ) = 1 are equal?

Ans: Domain must restrict.

TEST YOURSELF

1. Let A be the set of n elements. Then the number of functions defined from the set A to the set A is

(1) n2 (2) nn (3) 2n (4) n!

2. What is the main difference between a relation and a function?

(1) A relation may have multiple outputs for a given input, while a function has only one output for each input.

(2) A function may have multiple outputs for a given input, while a relation has only one output for each input.

(3) A relation is a set of ordered pairs, while a function is a set of numbers.

(4) A relation is a one-to-one mapping, while a function is a many -to-one mapping.

3. Which of following relations is a function?

(1) {(–2, 3), (1, 5), (–2, 7)}

(2) {(3, 5), (2, 4), (3, 3)}

(3) {(1, 2), (2, 4), (3, 6)}

(4) {(1, 2), (2, 2), (2, 3)}

4. If 2,2, 88 yy fxxxy  +−=  then f(m, n) + f(n, m) = 0 (1) only when m = n (2) only when m ≠ n (3) only when m = –n (4) for all m and n

5. If g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by the formula g(x) = ax + b, then (a, b)= (1) (2, 1) (2) (–2, 1)

CHAPTER 20: Functions

(3) (–2, –1) (4) (2, –1)

6. If f(x) = x2 – 2x + 4, then the set of values of x satisfying f(x –1) = f(x +1) is (1) {–1} (2) {–1, 1} (3) {1} (4) {1, 2}

Answer Key (1) 2 (2) 1 (3) 3 (4) 4 (5) 4 (6) 3

20.2 REAL VALUED FUNCTIONS

■ A function f : A →B is a real-valued function if B is subset of R. If both A and R are subsets of R, then f is a real function.

Types of real functions:

■ Algebraic functions: Functions obtained using a finite number of fundamental operations (addition, subtraction, multiplication, division, root extraction) on polynomial functions.

■ Transcendental functions: Functions that are not algebraic, e.g., exponential, logarithmic, trigonometric, and inverse trigonometric functions.

20.2.1 Polynomial Functions

■ A function f : R → R is a polynomial function of degree

■ () 12 012 ... nnn nfxaxaxaxa =++++ ,where a0 ≠ 0 and and n is a non-negative integer.

■ Domain: R

■ Example:

‰ Linear: f(x) = ax + b

‰ Quadratic: f(x) = ax2 + bx + c

■ The domain of polynomial function is R , the range of odd degree polynomial functions is R, and the range of even degree polynomial function is set of all non-negative real numbers.

Try yourself:

5. Let f = {(–1,–3),(0,–1),(1,1),(2,3)} be a linear function defined on the set of all real numbers. Then, find f(x).

Ans: 2x – 1

20.2.2 Constant Function

■ Defined as f(x) = k for all x ∈ R.

■ Graph: Horizontal line at y = k

■ Range: {k} (singleton set) 4 x y (0,k)

f(x) = k 2 3 1 -2 -3 -4 -1

Solved example

7.Describe the graph of the function f(x) = 10.

Sol. The graph of f(x) = 10 is a straight line parallel to x-axis and above the x- axis at a distance of 10 units from it.

Try yourself:

6. What is the domain and range of a function f(x) = sin2 x + cos2 x for all x ∈ R?

Ans: Domain is R and range is {1}

20.2.3 Identity Function

■ Defined as f(x) = x for all x ∈ R.

■ Graph: Line y = x bisecting the first and third quadrants.

■ Domain and Range: R

Solved example

8. If the graph of a function is a straight line passing through the origin and making an angle of 45° with the x–axis, then what is the type of the function?

Sol. It is an identity function.

Try yourself:

7. What is the graph of f ( x ) = – x ? Find its domain and range.

Properties of modulus function:

For any real number x,

■ xaaxa <⇔−<<

■ xaaxa ≤⇔−≤≤

■ or xaxaxa >⇔<−>

■ or xaxaxa ≥⇔≤−≥

■ ()() ,, axbxbaab <<⇔∈−−∪

■ ,, axbxbaab ≤≤⇔∈−−∪

For any two real numbers x,y,

■ xyxy +=+ ; here, both x,y are of same sign

■ xyxy −≤+

■ xyxy +≤+

■ xyxy +≥−

Solved example

20.2.4 Modulus Function

■ Defined as f(x) = |x|: ()≥

if 0 if 0 xx fx xx

■ Graph: y = x for x > 0, y = –x for x < 0.

■ Domain: R

■ Range: [0, ∞ )

9. Write the function g(x) = x–|x|, as a piecewise function.

Sol. The given function is g(x) = x–|x|

When x > 0, g(x) = x–x = 0

When x < 0, g(x) = x+x = 2x

When x = 0, g(x) = 0

Therefore, () 00 20 ifx gxxifx ≥  =<  

Try yourself:

8. What is 2 cos x for x ∈ (0,π)?

Ans: cos,0.cos, 22 xxxx ππ π<<−<<

20.2.5 Greatest Integer Function

■ Defined as f ( x ) = [ x ], which gives the greatest intefer < x.

■ Graph: Step function

■ Domain: R

■ Range: Z

Properties of greatest integer function:

■ If n is any integer, then [– n] = –n.

■ If x is any real number, [ x+k ] = [ x ]+ k , where k is integer.

■ If x ∉ Z, then [–x] = –[x]–1 and

‰ [x] + [–x] = –1

‰ [x] – [–x] = 2[x]+1

■ If x ∈ Z, and k ∈ Z then,

‰ [x] ≥ k ⇔ x ≥ k

‰ [x] ≤ k ⇔ x < k+1

‰ [x] < k ⇔ x < k

CHAPTER 20: Functions

Solved example

10. If x lies in the first quadrant, then what is the value of [sin x]?

Sol. The value of sin x is between 0 and 1, when x lies in the first quadrant. Hence, [sin x] = 0.

11. What are the values of [3.4],[2],[–2.5],[–5]?

Sol. The value of [3.4] is greatest integer, which is less than 3.4, so [3.4] = 3 and [2] = 2.

The value of [–2.5] is greatest integer, which is less than –2.5, so [–2.5] = –3 and [–5] = –5.

Try yourself:

9. For a real number x = 0.5, what is the value of 12 2023 ... ? 202420242024 xxxx

Ans: 1012

20.2.6 Least Integer Function

■ Defined as f(x) = (x), which gives the least interger > x.

■ Graph: Step function

■ Domain: R

■ Range: Z

Properties of least integer function:

■ If x is an integer, then

‰ (x) = x

‰ (x) + (–x) = 0

‰ (x) – (–x) = 2(x)

■ If x is not an integer, then

‰ (x) = (x)+1

‰ (x) + (–x) = 1

‰ (x) – (–x) = 2(x)–1

‰ ()

Solved example

12. Find the value of k, if (–2.3)k – 2k + (8.34) k = (4.2.3) + (2.23), where (.) represents the least integer function.

Sol. Given:

(–2.3)k – 2k + (8.34)k = (4.23) + (2.23)

⇒ – 2k – 2k + 9k = 5 + 3

⇒ 5k = 8

⇒ k = 8/5

Try yourself:

10. Prove that

20.2.7 Fractional Part Function

■ Defined as f(x) = x–[x].

■ Graph: Periodic between 0 and 1.

■ Domain: R

■ Range: [0, 1)

Properties of fractional part function:

■ If x is an integer, then ‰ {x} + {–x} = 0

‰ {x} = 0

■ If x is not an integer, then

‰ {[x]} = 0

‰ [{x}] = 0

‰ {x} = x, when x ∈ (0,1]

Solved example

13. What is the value of {3.45}?

Sol. We know that {x} = x – [x]. {}3.453.453.45 3.453 0.45 =−  

Try yourself:

11. What is the value of {[7]}?

20.2.8 Signum Function

■ Defined as: ()  −<  ==

10 sgn00 10 ifx xifx ifx

>

Ans: 0

■ Graph: Three distinct horizontal segments.

■ Domain: R

■ Range: {–1, 0, 1}

Solved example

14. What is the value of f(–3.002), where f(x) is a signum function?

Sol. We know that if x is negative, the signum function gives –1.

Hence, f(–3.002) = –1

Try yourself:

12. What is the value of f ({–3.45}), where f is signum function and { } is fractional part function?

Ans: 1

20.2.9 Exponential Function

■ Defined as f(x) = ax, where a > 0, a ≠ 1.

■ Domain: R

■ Range: (0, ∞ )

■ Graph: Increasing if a > 1, decreasing if 0 < a < 1.

20.2.10

Logarithmic Function

■ Defined as f(x) = loga x, where x > 0, a > 1.

■ Domain: (0, ∞ )

■ Range: R

■ Graph: Mirroe image of ax w.r.t. y=x

Properties:

20.2.11 Reciprocal Function

■ Defined as () 1 , fx x =

■ Domain: R–{0}.

■ Range: R–{0}.

■ Graph: Hyperbola in first and third quadrants.

20.2.13 Cube Function and Cube Root Function

■ Cube function: f(x) = x3

■ Domain and Range: R.

■ Graph: Similar sign as x.

20.2.12 Square Function and Square Root Function

■ Square function: f(x) = x2

■ Domain: R

■ Range: [0, ∞ ).

■ Graph: Parabola.

Cube root function: () 3 fxx =

■ Domain and Range: R

■ Graph: Increasing function .

Square root function: () fxx =

■ Domain: [0, ∞ ).

■ Range: [0, ∞ ).

■ Graph: Increasing curve.

20.2.14 Explicit and Implicit Functions

■ A function in which a dependent variable can be expressed in terms of independent

variables is called explicit function . The explicit function can be represented as y = f(x).

■ If the function cannot be expressed directly in terms of independent variable, then the function is said to be implicit function . The implicit function can be represented as f(x,y) = 0.

■ Example: y = sin x + log x is an explicit function, whereas ax2 + 2hxy + by2 = 0 is an implicit function.

20.2.15 Piece-wise Defined Functions

■ A function f ( x ) is defined differently on disjoint subsets of its domain.

‰ The piece-wise definition of modulus function f(x) = |x| is () 0 0 fxxifx xifx

Try yourself:

13. Draw the graph of () 42 22 2 xx fxxx xx

20.2.16 Rational Function

A function of the form () () fx gx , where g(x) ≠ 0.

Example: 22 32 24734 , 32 257 xxxx xxx ++++ +++

20.2.17 Periodic Function

A function f is periodic if there exists p such that f(x+p) = f(x) for all x.

■ Example: sin x, cos x

■ Properties:

‰ f(x+kp) = f(x) for integer k.

‰ kf(x) is periodic with period p.

CHAPTER 20: Functions

‰ f(ax+b) is periodic with period p a

Solved example

15. What is the period of the function f(x) = x–[x–b], b ∈ R where [ ] represents the greatest integer function?

Sol. Given: f(x) = x –[x–b], b ∈ R can be written as f(x) = x – (x – b – {x–b}), b ∈ R=b+ {x–b}

∴ the period of the function f(x) is 1.

Try yourself:

14. What is the period of the function f(x) = sinx + cosx?

Ans: 2π

20.2.18 Even and Odd Functions

■ A function : fAR → is

Even Function

■ A function is even if: f(–x) = f(x), ∀ x ∈ A

Properties:

■ The graph of an even function is symmetric about y-axis. y Ox

■ If ( x , y ) is a point on the graph of even function, then (–x, y) is also on the graph.

■ Every constant function is an even function.

Odd Function

■ A function is odd if ()() fxfx −=− xA∀∈

Properties:

■ The graph of an odd function is symmetric about the origin, and it is symmetrical in opposite quadrants. y Ox

Try yourself:

15. The function () 1 2 1 x fxxx e =++ on R – {0} is ______ function.

Ans: vene

TEST YOURSELF

1. Given, the function () 2 xx fxaa + = (where

a > 2). Then, f(x + y)+f(x − y) =

(1) 2f(x). f(y)

■ If ( x , y ) is a point on the graph of odd function, then (–x, –y) is also a point on the graph.

Additional Properties

■ The even power of an even or odd function is always an even function.

■ Every function can be uniquely expressed as a sum of an even function and an odd function:

f(x) = g(x) + h(x) where g(x) is even and h(x) is odd.

Solved example

16. I f f ( x )and g ( x ) are two functions, with all real numbers as their domains, then show that

h(x) = { f(x) + f(–x)} { g(x) – g(–x)} is an odd function.

Sol: Given:

h(x) = { f(x) + f(–x)} { g(x) – g(–x)}

Consider, ()()(){}()() {} {()()}()() {} () hxfxfxgxgx fxfxgxgx hx −=−+−− =−+−−− =−

Therefore, the function h(x) is odd function

(2) f(x). f(y) (3) () () fx fy

(4) f(x) – f(y)

2. If a function f(x) is given by 5 () 55 x fxx = + , then, the sum of the series 12339 20202020 ffff

is equal to (1) 29 2 (2) 49 2 (3) 19 2 (4) 39 2

3. Suppose f :[−2,2]→ is defined by () 1for 20 1for 02 x fx xx −−≤≤   =−≤≤

Then, {x ∈ [−2, 2] : x ≤ 0 and f(|x|)= x} =

(1) {–1} (2) {0} (3) 1 2 

(4) ϕ

4. If () 2 log 2 faa a + = for 0 < a < 2, then 2 18 2 4 fa a

+

(1) f(a)

(2) 2f(a)

(3) 1 () 2 fa (4) – f(a)

5. If () 9 93 x fxx = + , then

6. If ()() 10 ,10,10 10 fxx ex x + =∈− and

20.3 METHODS OF FINDING DOMAIN AND RANGE OF A FUNCTION

A function’s domain comprises all valid input values, while its range encompasses all possible output values.

20.3.1 Finding Domain of a Function

Definition:

The domain of a function is the set of all elements where the function is well-defined.

Rules for finding the domain

■ Algebraic functions:

‰ Denominator ≠ 0.

‰ Expression under an even root must be > 0.

‰ Logarithmic expressions must be > 0.

■ Trigonometric functions:

‰ sin x, cos x: Defined for all real x.

‰ tan x, sec x: Defined for x ≠ (2n+1)π/2.

7. Number of solutions of the equation 2[x]= x + {x} is_____. (where([⋅] is greatest integer function, { } is fractional part function).

8. If f ( x ) is an even function and g ( x ) is an odd function, then f(x) ⋅ g(x) is____ function.

(1) even (2) odd (3) neither even nor odd (4) either even or odd

9. If f ( x ) = ax 5 + bx 3 + cx + d is an odd function, then d =

(1) 0 (2) 1 (3) –1 (4) Any real number

Answer Key

(1) 1 (2) 4 (3) 3 (4) 1 (5) 997.5 (6) 0.5 (7) 2 (8) 2 (9) 1

‰ csc x, cot x: Defined for x ≠ nπ.

■ Logarithmic Functions:

‰ loga(f(x)): Defined for f(x) > 0.

‰ log a( g ( x )/ h ( x )): Defined for g ( x ) > 0, h(x) > 0.

■ Wavy curve method (for inequalities):

‰ Solve f(x) < 0, f(x) > 0, f(x) ≤ 0, or f(x) ≥ 0.

‰ Mark roots (where f(x) = 0) with black dots.

‰ Mark discontinuities with circles.

‰ Draw a curve, alternating sign at odd powers, retaining the same sign at even powers.

‰ Include only intervals satisfying the given inequality.

Solved example

17. Let

Then, find the domain of ()fx .

Sol. Use wavy curve method or interval method.

First, keep the numbers –6,–2,0,1,3,7 on the number line, in which 0,7 are in hollow circle, as shown in the following figure.

Substitute x = –10, which is left to the least number.

We get,

Start the curve above the number line before –6 and cross at –6, touch at –2. Since it has even power, draw the curve below the number line and touch zero. Since it has even power, draw the curve below the number line only and touch 1, cross the curve at 1, cross the curve at 3, and cross the curve at 7, as shown below .

Methods to Find Range

1. Range of Basic Functions

■ Sum Functions: Use AM ≥ GM ≥ HM for positive quantities.

■ Quadratic Functions:

‰ If a > 0, minimum value is 4ac - b² / 4a. ‰ If a < 0, maximum value is 4ac - b² / 4a.

■ Trigonometric Functions:

‰ a cos x + b sin x: 2222 , abab

‰ a2 sec² x + b2 csc² x: ()) 2 , ab +∞ 

‰ a2 tan² x + b2 cot² x: ) 2, ab ∞ 

2. Range of Rational Functions

■ For y = P(x)/Q(x):

■ Find critical points from P ( x ) = 0 and Q(x) = 0.

■ Analyze limits as x → ±∞.

3. Strictly Increasing/Decreasing Functions

■ If f(x) is increasing in [a, b], Range = [f(a), f(b)].

Domain of ()fx is (– ∞ , –6] ∪ [1, 3] ∪ (7, ∞ ) ∪ {–2}.

Try yourself:

16. What is the solution of the inequality (3)(2)(5) 0 (1)(7) xxx xx −++ > +− ? Ans: (–5, –2) ∪ (–1, 3) ∪ (7, ∞ )

20.3.2 Finding Range of a Function

Definition:

■ The range is the set of all possible values of f(x).

■ If f(x) is decreasing Range = [f(b), f(a)].

4. Using Derivatives

■ Find f’(x) and identify critical points.

■ Determine max/min values to get the range.

Solved example

18. Find the range of the function

Let

1.

Now, D ≥ 0 ∀ x ∈ R

⇒ (y + 1)2 – 4(y – 1)2 ≥ 0

⇒ 3y2 – 10y + 3 ≤ 0

⇒ 3y2 – 9y – y + 3 ≤ 0

⇒ 3y(y – 3) – 1(y – 3) ≤ 0

⇒ (3y – 1)(y – 3) ≤ 0

⇒ y ∈ [1/3,3]

∴ Range is [1/3, 3]

CHAPTER 20: Functions

Try yourself:

17. Find the range of the function () 2 23fxxx x ++ = Ans: )( ,223223,  −∞−∪+∞

20.3.3 Domain and Range of Some Functions

■ The following table shows the domain and range of some standard functions.

TEST YOURSELF

1. The range of the function () 32 fxxx =−++ is

2. The domain of the function () 1 fx xx = is

(1) (0, ∞) (2) (−∞, 0)

(3) (−∞, ∞)−{0} (4) (−∞, ∞)

3. Domain of the real valued function

22 259 1 xxx −+++ is

(1)

(2)

55 , 33 

55 , 33

(3) 55 ,, 33

(4) 55 ,, 33

4. The domain of 32 () 32 fxxx xx =+ ++ is

(1) (−∞, 2) ∪ (3, ∞) (2) (2, 3)

(3) R (4) ϕ

5. Domain of ()() 2 10 1 log5 6 fxxx = is

(1) (0, 5) (2) (1, 4)

(3) (1, 3)

(4) 521521 , 22  −+   

6. The domain of the definition of the function ()() 3 10 2 3 log 4 fxxx x =+− is

(1) (1, 2)

(2) (–1, 0) ∪ (1, 2)

(3) (1, 2) ∪ (2, ∞) (4) (−1, 0) ∪ (1, 2) ∪ (2, ∞)

7. The range of the function f(x) = |x−1| + |x−2|, −1 ≤ x ≤ 3, is (1) [1, 3] (2) [1, 5] (3) [3, 5] (4) [2, 5]

8. Let A ={1, 2}, B ={3, 6} and f : A → B given by f ( x ) = x 2 + 2 and g : A → B given by g(x) = 3x. Then, the functions f(x) and g(x) are (1) equal (2) not equal (3) one-one (4) none of these

9. If a2 + b2 + c2 = 4, then the range of ab+bc +ca is [–2, k], then, k is _____.

10. If the range of f ( x ) = | x − 2|+ | x –12| is [m, ∞), then m is _____.

Answer Key

(1) 1 (2) 2 (3) 2 (4) 4 (5) 4 (6) 4 (7) 2 (8) 1 (9) 4 (10) 10

20.4 ALGEBRA OF FUNCTIONS

■ Functions can be combined using addition, subtraction, multiplication, and division to form new functions. These operations help in modeling and solving mathematical problems across various disciplines.

Basic Operations on Functions:

■ Let f : A → R and g: B → R

20.4.1 Adding a Scalar to a Function

■ Definition: (f+c)(x)=f(x)+c

■ Domain: Same as f(x).

Solved example

19. Find the value of (f + 3)(2), where f(x) = x2 + 3x–3.

Sol. (f + 3)(2) = 3+f(2) = 3+22 + 3(2)–3 = 10

Try yourself:

18. If f = {(1, 2), (2, –3), (3, –1)} then, find 2 + f .

Ans: {(1, 4), (2, –1), (3, 1)}

20.4.2 Scalar Multiplication of a Function

■ Definition: (cf)(x)=cf(x)

■ Domain: Same as f(x).

Solved example

20. Let f is a function defined by () 1 4 fx x = +

Then, find the function 2 f and its domain.

Sol. The given function is () 1 4 fx x = +

⇒()() 2 22 4 fxfx x =⋅= +

The domain of the function f is same as the domain of 2f, and it is set of all real numbers except –4.

Try yourself:

19. Let () 1 fxx=+ . Then, find the value of 2f(1).

Ans: 22

20.4.3 Reciprocal of a Function

■ Definition: () 1 fx is defined where f(x)≠0.

■ Domain: {()}=∈≠|0DxAfx

2. Operations on Two Functions

Let f: D₁ → R and g: D₂ → R be two functions.

Solved example

21. Find the reciprocal function for the function () 1 4 fx x = + and also find its domain.

Sol. Given () 1 4 fx x = +

f(x) – {x : f(x) = 0}

∴ the domain of () 1 x f is R–{–4}.

Try yourself:

20. Find the reciprocal function for the function f(x) = sinx and also find its domain.

Ans: Reciprocsl function of f(x) is cosec x and its domain is R–nπ, ∀ n ∈ Z

20.4.4 Addition and

Subtraction

of Functions

■ Definition: (f+g)(x) = f(x) + g(x)

■ Domain: D = D1∩ D2 (common domain).

Solved example

22. Let () 2 fxx=+ and () 2 4 gxx =− be any two functions. Then, find (f+g)(x) and also find its domain.

Sol. The function () 2 fxx=+ is defined for x + 2 ≥ 0 ⇒ x ≥- 2

Hence, the domain of the function () 2 fxx=+ is ) 1 2, D =−∞  

The function () 2 4 gxx =− is defined for 22 404 xx −≥⇒≤ . ⇒ x ∈ [–2, 2].

Hence, the domain of the function () 2 4 gxx =− is [–2, 2].

The sum of the above two functions is (f + g)(x). ⇒() ()()() 2 24 fgxfxgxxx +=+=++−

() ()()() 2 24 fgxfxgxxx +=+=++−

Reciprocal function to the above function is defined as ()()11 x ffx =

Hence, () 1 4 xx f =+

The domain of () 1 x f is domain of

Domain of (f+g)(x) is ) 2,2,22,2 −∞∩−=−

Try yourself:

21. If () 2 fxx=+ and () 2 4 gxx =− then, find (f–g)(x) and also find its domain.

Ans: [–2, 2]

IL ACHIEVER SERIES FOR JEE MATHEMATICS

20.4.5 Multiplication of Functions

■ Definition: (f ⋅ g)(x) = f(x) ⋅ g(x)

■ Domain: D = D1∩ D2.

Solved example

23. Let () 1 4 fx x = + and g(x) = x + 4. Then, find (f ⋅ g)(x) and find its domain.

Sol. The function () 1 4 fx x = + is defined for 404xx+≠⇒≠−

Hence, the domain of the function () 1 4 fx x = + is D1 = R–{–4}.

The function g(x) = x + 4 is defined for all real values of x

The product of two functions () ()()()() 1 41 4 fgxfxgxx x ⋅=⋅=+= +

The domain of the function (f ⋅ g)(x) is domain of f(x) ∩ domain of g(x)

⇒ Domain of (f ⋅ g)(x) is R–{–4}

Try yourself:

22. If () 2 fxx=+ and () 2 4 gxx =− then find (f ⋅ g)(x) and find its domain.

Ans: (f . g)(x) = 23 482xxx +−− and its domain is[–2, 2]

20.4.6 Division of Functions

■ Definition: ()() ()  =  ffx x ggx where g(x)≠0.

■ Domain: {()} 12 :0DDDxgx=∩−≠ .

Solved example

24. Let () 1 fxx=+ and () 2 9 gxx =−

Then, find ()fx g    and find its domain.

Sol. The function () 1 fxx=+ is defined for 101xx+≥⇒≥−

Hence, the domain of the function () 1 fxx=+ is ) 1 1, D =−∞  

The function () 2 9 gxx =− is defined for 22 909 xx −≥⇒≤

⇒ 3,3 x ∈− 

Hence, the domain of the function () 2 9 gxx =− is [–3, 3].

The quotient of the above two functions is ()fx g   

Now, ()() () 2 1 9 ffxx x ggxx + == 

Domain of ()fx g    is )(){} 1,3,3:0 xgx −∞∩−−=  

This is equal to {})() 1,33.31,33,3 −−−=−−∪−  

Try yourself:

23. If () 2 fxx=+ and () 2 4 gxx =− then find ()fx g    and find its domain.

Ans: 2 2 () 4 xf x xg + =  and its domain is (–2, 2)

20.4.7 Functional Equations

For a function f(x) satisfying:

■ Multiplicative Form:

‰ f(x+y) = f(x) f(y)

‰ f(x) =ax

■ Additive Form:

‰ f(x+y) = f(x) + f(y)

‰ f(x) = ax

■ Logarithmic Form:

‰ f(xy) = f(x) + f(y)

‰ f(x) = klogbx

■ Inverse Property:

‰ f(x+y) = f(x)

‰ f(x) = k

■ Alternative Multiplicative Form:

‰ f(x+y) + f(x–y) = 2f(x) f(y)

‰ f(x) = cosx

■ Power Form:

‰ f(x+y) = f(x) + f(y) + 2xy

‰ f(x) = x2

Solved example

25. If ()()() fxyfxfy +=+ and f(2) = 8, then find f(8).

Sol. We know that, if ()()() fxyfxfy +=+ , then f(x) = ax.

Since, ()2284faa==⇒= . Therefore, ()() 84832 f ==

Try yourself:

24. If ()()11fxffxf xx  +=  and f(2) = 9, then find the value of f(1).

Ans: 2

TEST YOURSELF

1. If ()() ()xaxb fx x = and () () () () () () fxfy xyxzyzyx + () () () fzk zxzyxyz += , then k is (1) a (2) b (3) ab (4) 3ab

CHAPTER 20: Functions

2. If () 2 1 23 0 , fxfxx x  −=≠   ,then f(2)= (1) 7 4 (2) 5 2 (3) –1 (4) 2

3. If f (x ) = ax 2+ bx + c for some a , b , c ∈ R, with a+b+c = 3 and f(x+y)= f(x) + f(y) + xy; ∀ x, y ∈ R, then () 10 1 n fn = ∑= (1) 330 (2) 225 (3) 165 (4) 190

4. The value of natural number a, for which f(a + k ) = 16(2 n −1), where the function satisfies the relation f ( x + y ) = f ( x ). f ( y ) for all natural numbers x and y and further  f(1) = 2 is (1) 16 (2) 2 (3) 3 (4) 4

5. If f : R−{0} → R is defined by () 1 fxx x =+ and if fk(x) = [(f(x)]k for K ≥ 1, then f 4(x) – f(x4) – 4f 2 (x)= (1) –2 (2) 2 (3) 1 (4) –1

6. If f ( x )+2 f (1− x ) = x 2 + 2 ∀ x ∈ R , then f ( x ) is given by (1) ()2 2 3 x (2) x2 − 2 (3) 1 (4) x2 + 2

7. If f(x) + 2f(1−x) = x2+1 ∀x∈R, then 3f(5) is equal to ______.

8. If f 3(x) − 3f 2(x) + 3f(x) − 1 = x6, then the value of f(0) is ______.

Answer Key

(1) 3 (2) 1 (3) 1 (4) 3 (5) 1 (6) 1 (7) 8 (9) 1

20.5 GRAPHICAL TRANSFORMATIONS

■ Graphical transformations involve modifying the graph of a function y = f(x)

by changing the independent variable x or dependent variable y These transformations help in sketching complex graphs using basic functions.

20.5.1 The Graph of y = f(x) + k, where k∈R

Vertical Shifts:

■ If k > 0 → Shift the graph upward by k units.

■ If k < 0 → Shift the graph downward by |k| units.

Solved example

26. Draw the graphs of y = sinx, y = 2 + sinx, and y = –2 + sinx

Sol. The graph of y = 2 +sin x can be obtained by shifting the graph of y = sin x in upward direction by a distance of 2 units, as shown below.

The graph of y = –2 + sinx can be obtained by shifting the graph of y = sin x in downward direction by a distance of 2 units, as shown below.

2

Try yourself:

25. Draw the graph of y = 2 + x2 , y = x2 and y = x2–2.

20.5.2 The Graph of y = f(x + k), where k∈R

Horizontal Shifts:

■ If k > 0 → Shift the graph left by |k| units.

■ If k < 0 → Shift the graph right by |k| units.

Solved example

27. Draw the graphs of y = x2 , y = (x + 2)2 and y = (x – 2)2.

Sol. The graph of y = (x + 2)2 can be obtained by shifting the graph of y = x2 to left by a distance of 2 units, as shown below.

The graph of y = (x – 2)2 can be obtained by shifting the graph of y = x2 to right by a distance 2 units, as shown below.

= (x + 2)2

Try yourself:

26. Draw the graph of y = x3 , y = (x + 3)2 and y = (x – 3)3

O y = x3 y = (x3+3) y = (x3–3)

Ans:

20.5.3 The Graph of y = kf (x), where k∈R

Vertical Scaling:

■ If k > 1 → Stretch the graph vertically by k times.

■ If 0 < k < 1 → Compress the graph vertically by k times

Try yourself:

27. Draw the graph of y = x3, y = 3x3 and 3 1 3 yx =

Solved example

28. Draw the graphs of y = sin(2 x ), y = sin x , () sin 2 x y = .

Sol. Graphs of y=sin(2x), y = sin

are as shown below

and

20.5.4 The Graph of y = f (kx), k∈R+ Horizontal Scaling:

■ If k > 1 → Compress the graph horizontally by k times.

■ If 0 < k < 1 → Stretch the graph horizontally by 1 k times.

Try yourself:

28. Draw the graph of y = cos(2 x ), y = cos x , cos 2 x y

20.5.5 The Graph of y = f(–x) Reflection about the Y-axis:

■ The graph is mirror-reflected about the y-axis.

Solved example

29. Draw the graph of y = (–x)3 is obtained from the graph of the function y = x3 .

Sol. Graph of y=(-x)3 is as follows.

Try yourself:

29. Draw the graph of y = e–x with the help of the graph of y = ex .

Try yourself:

30. Draw the graph of y = –{x} with the help of graph of y = {x}

1)

20.5.6 The Graph of y = –f(x)

Reflection about the X-axis:

■ The graph is mirror-reflected about the x-axis.

Solved example

30. Draw the graph of y = –ex is obtained from the graph of the function y = ex .

Sol. Graph of y = –ex is as follows.

20.5.7 The Graph of y = –(f(–x))

Reflection about Both Axes:

■ To obtain the graph of y = –(f(–x))

■ Start with y = f(x).

■ Reflect the graph about the y-axis (gives y = f(x)

■ Reflect the resulting graph about the x-axis (gives y = –(f(–x))

Solved example

31. Draw the graph of y = –e -x with the help of y = ex .

Sol. Graph of y = –e -x is as follows:

Try yourself:

31. Draw the graph of y = sinx – cosx with the help of the graph of y = sinx + cosx.

Ans:

20.5.8 The Graph of y = |f(x)|

Absolute Value Transformation:

■ To obtain the graph of y = |f(x)|:

■ Start with y = f(x).

■ Keep the portion where f(x) > 0 unchanged

■ Reflect the negative part (f(x) < 0) above the x-axis.

Solved example

32. Draw the graph of y = |sinx| from the graph of y = sinx.

Sol. Graph ofy = |sinx|as follows:

Ans: y = |logx| (1,0)

20.5.9 The Graph of y = f(|x|)

Reflection for Absolute Input:

■ To obtain the graph of y = f(|x|):

■ Start with y = f(x).

■ Keep the right half ( x>0) uncharged.

■ Mirror this part on the left side of the y-axis.

Solved example

33. Draw the graph of y = sin |x| with the help of the graph of y = sinx.

Sol. Steps to draw the graph of y = sin |x|:

Step I: Ignore the graph of y = sin x which is on the left side of y-axis.

Step II: Take the graph of y = sin x, which is on the right side of y-axis, as it is.

Step III: Take the image of portion in step 2 with respect to y-axis.

Finally, the graph of y = sin(| x |) is as shown below.

|x|

Try yourself:

32. Draw the graph of y = |log x| using the graph of y = logx.

Try yourself:

33. Draw the graph of y = e|x| with the help of the graph of y = ex .

e-x ex

20.5.10 The Graph of y =|f(|x|)|

Double Absolute Transform:

■ To obtain the graph of y = |f(|x|)|:

■ Start with y = f(x).

■ Apply y = f(x) transformation first (retain right-side graph and mirror it to the left)..

■ Apply y = f ( x ) transformation (reflect negative values above the x-axis)..

■ These steps will give you the graph of y = |f(|x|)|.

Solved example

34. Draw t he graph of y = |sin(| x |)| using the graph of y = sin(x).

Sol. Step I: Ignore the graph of y = sin x which is on the left side of y-axis.

Step II: Take the graph of y = sin x, which is on the right side of y-axis, as it is.

Step III: Take the image of portion in step 2 with respect to y-axis.

Step IV: The portion of the graph below x-axis will be shifted to above x-axis, as shown below. The graph of y = |sin(|x|)| is as shown in graph.

Try yourself:

34. Draw the graph of y = |cos(|x|)| using the graph of y = cos(x).

20.5.11 The Graph of |y| = f(x)

Absolute Value:

■ To obtain the graph of | y| = f(x):

■ Start with y = f(x).

■ Ignore the portion where f(x) < 0 (below the x-axis).

■ Reflect the upper part of the graph above x-axis to the lower half.

Example: Graph of | y| = logx:

■ Draw y = logx.

■ Ignore the negative part.

■ Reflect the upper part below the x-axis. y x

Try yourself:

35. Draw the graph of yx =

20.5.12 The Graph of |y| = |f(x)|

Double Absolute Transformation:

To obtain the graph of | y| = |f(x)|:

■ Start with y = f(x).

■ Reflect the graph below the x-axis to above the x-axis.

■ Reflect the entire graph again below the x-axis to complete the transformation.

Try yourself:

36. Draw the graph of |y| = |sin x|.

Ans:

TEST YOURSELF

1. Draw the graph of |y| = |log|x||.

2. Draw the graph of y = e–|x| .

3. Draw the graph of |y| = |x – 1| + |x + 1|.

4. Draw the graph of y = |sin x| + |cos x|.

Answer Key

1. Graph of | y | = |log| x || is as follows:

2. Graph of y =

3. Graph of |y| = |x – 1| + |x + 1| is as follows:

+1|

4. Graph of y = |sin x| + |cos x| is as follows:

20.6 KINDS OF FUNCTIONS

Functions can be categorized based on their mappings:

■ One-to- One (Injective) Function: Each element of the domain maps to a unique element in the codomain, ensuring no duplication.

■ Many-to-One Function: Multiple elements of the domain map to the same codomain element.

■ Onto (Surjective) Function: Every element in the codomain has at least one pre-image in the domain.

■ Into Function: Some elements in the codomain remain unmapped.

■ Bijective Function: A function that is both injective and surjective, creating a one-toone correspondence between domain and codomain.

20.6.1 One-to-One Function

■ A function : fAB → s one-to-one (injective) if different elements of A have different images in B.

■ Mathematically,

()() abfafb ≠⇒≠∀ a,b ∈ A

■ A function : fAB → s one-to-one (injective) if:

()() fafbab=⇔=∀ , abA ∈

Conditions for Injectivity:

■ If a function is given in the form of ordered pairs, and no two ordered pairs have the same second element, the function is oneto-one.

■ If f ( x ) is strictly increasing or strictly decreasing in its domain, then the function is one-to-one.

■ If f(x) (the derivative of f(x)) satisfies 0or0 ) , (() fxfxxA ′>′<∀∈ then the function is one-to-one.

Horizontal Line Test:

■ If the continuous graph of a function intersects every horizontal line (parallel to the x-axis) at most once, then the function is one-to-one.

■ This ensures that each input value corresponds to a unique output value.

To determine whether a given function is one-to-one, follow these methods:

■ Algebraic Method

‰ Assume ()() fxfy = .

‰ Show that x=y . using algebraic properties.

‰ If ()() fxfyxy =⇒= , then the function is one-to-one, otherwise, it is not.

■ Monotonicity Test

‰ Prove that the function is strictly increasing or strictly decreasing.

‰ A strictly increasing or decreasing function is always one-to-one.

■ Horizontal Line Test

‰ A function is one-to-one if no horizontal line intersects its graph more than once.

Number of One-to-One Functions

■ If n(A) = m amd n(B) =n:

‰ When mn ≤ , the number of one-to-one functions from A to B is ,nP m

‰ When mn > , one-to-one functions is not possible.

Solved example

35. Check if the function : fNN → , defined by () 2 1 faaa=++ , is one-to-one or not.

Sol: Given: () 2 1 faaa=++

Let , abN ∈ , such that ()() fafb = It implies that ()() () () 22 22 11 0 10 0 aabb abab abab ab ++=++ −+−= −++= −=

Hence, ()() fafbab =⇒=

Therefore, f(x)is one-to-one function.

Try yourself:

37. If A and B are two sets such that n(A) = 15, and n(B) = 20, then what is the number of injections from the set A to the set B?

Ans: 20 15P

20.6.2

Many-to-One Function

A function : fAB → i many-to-one if two or more elements of set A have the same image in B.

■ Mathematically, : fAB → is a many-toone, if there exists two distinct , xyA ∈ , such that ()() fxfy = .

■ If the function : fAB → is not one-to-one, then it is many-to-one function.

AB f x1 y1

x2

Key Properties:

■ A constant function (defined over multiple domain values) is always many-to-one.

■ Even functions and all polynomial functions of even degree are many-to-one.

■ Polynomial functions of odd degree may be either one-to-one or many-to-one.

■ If a function is given in set notation and at least two ordered pairs share the same second element, it is many-to-one.

■ A function is many-to-one if its graph is intersected more than once by any horizontal line (x-axis parallel test).

■ Periodic functions are always many-to-one.

Solved example

36. Let () 4 nA = and () 5 nB = . Find the number of all possible many-to-one functions from A to B.

Sol: Let () 4 nA = and () 5 nB = Number of all possible many-to-one functions from A to B.

()()()() nAnB nAnBP=− 45 4 5 P =− = 505

Try yourself:

38. Is the function 11 :, 22 fR  →  defined as () 2 1 fxx x = + , injective?

Ans: No

20.6.3 Onto Function or Surjection

■ A function f : A → B is onto if each element of B has at least one pre-image in A.

■ Mathematically, f is onto if for every b ∈ B, there exists at least one a ∈ A such that f(a) = b.

■ If the range of f ( x ) is equal to the codomain, the function is onto.

■ If any element of B has no pre-image in A, the function is not onto (also called into function) x

■ If ()() nAnB < , then the number of onto functions is zero.

■ If n ( A ) = n ( B ) = n, the number of onto functions is n!

■ If () nAm = and () 2 nB = , then the number of onto functions from A to B is 22 m .

■ If n(A) = m, n(B) = n, and 1 nm≤< , then the number of into functions from A to B is ()()() 123 123... mmm CCC nnnnnn −−−+−

Solved example

37. Is ) :0,fR →∞   defined by () 2 fxx = onto?

Sol: Yes, the function is onto because for every positive real number y , there exists a real number x such that () fxy = , where 2 yx = .

Try yourself:

39. If () 4 nA = and () 3 nB = , then find the number of onto functions from the set A to the set B

Ans: 36

20.6.4 Bijection

■ A function f : A → B is a bijection if it is one-to-one as well as onto (surjective)..

■ Injective Condition: ()() fxfyxy =⇒= for all , xyA ∈

■ Surjective Condition: For every yB ∈ ,there exists at least one xA ∈ such that () fxy = .

Number of onto functions:

■ If n(A) = m, n(B) = n, and ()() 1 nBnA≤≤ , then the number of onto functions from A to B is ()() 12 12...mmm CC nnnnn−−+−−

Number of Bijective Functions:

■ If A and B are two finite sets and f : A → B is bijective function, then n(A) = n (B).

■ The number of bijective functions from A to B is n(A)!

■ If n(A) ≠ n(B) then number of bijective functions = 0

Solved example

38. Is the function, f:R → R defined as () 3 51fxxx=++ , bijective?

Sol: Given: () 3 51fxxx=++

Now, () '2350, fxxxR =+>∀∈

()fx is strictly increasing function.

It is one-to-one.

Clearly, ()fx is continuous function and also in creasing on R.

() x Limfx →−∞ =−∞ and () x Limfx →∞ =∞

So, ()fx takes every value between −∞ and ∞

()fx is onto function.

∴ f is bijection

Try yourself:

40. Is the function, : fRR → , defined as () 3 fxx = , bijective?

Ans: Yes

TEST YOURSELF

1. The number of one one functions that can be defined from A = {1, 2, 3, 4, 5} into B = {a,b,c,d} is (1) 120 (2) 24 (3) 20 (4) 0

2. If f : R → R defined by () 25,if0 , 32,if0 xx fx xx +> 

then f is

(1) a function

(2) one one

(3) Onto

(4) One one onto

3. If f : R → R is defined by f(x) = x + 2 x ,then f is

(1) An injection

(2) Onto

(3) A bijection

(4) Only function

4. Set A has 3 elements. Set B has 4 elements. The number of surjections that can be defined from A to B is (1) 144 (2) 12 (3) 0 (4) 64

5. A = {x: –1 ≤ x ≤ 1}. f : A → A defined by f(x) = x|x|. Then f is (1) A bijection

(2) An injection but not surjection

(3) A surjection but not an injection

(4) neither an injection nor a surjection

6. If f(x) = |x – 1| + |x – 2| + |x – 3|, f : (2, 3) → R, then f is

(1) An onto function but not 1–1

(2) 1–1 function but not onto (3) bijection

(4) Neither 1–1 nor onto

7. If f : R → R defined by f(x) = x for x ≥ 2 and f(x) = 5x – 2 for x < 2, then f is

(1) One – one

(2) Onto only

(3) One – one onto

(4) Neither one – one nor onto

8. Strictly monotonic function is (1) onto definitely (2) bijection

(3) one – one definitely

(4) neither one – one nor onto

9. Let R be the set of real numbers. If f : R → R is a function defined by f(x) = x2, then f is

(1) injective but not surjective

(2) surjective but not injective

(3) bijective

(4) neither injective nor surjective

10. The number of surjections that can be defined from {1, 2, 3, 4, 5} onto {a,b} is

11. Let n(A) = 4 and n(B) = k. The number of all possible injections from A to B is 120, then k =____.

Answer Key

(1) 4 (2) 2 (3) 4 (4) 3 (5) 1 (6) 2 (7) 2 (8) 3 (9) 4 (10) 30 (11)5

20.7 COMPOSITION OF A FUNCTION

■ If : fAB → and : gBC → are two functions, the composite function of f and g denotes () gf  defined as () ()()()gfxgfx =  for all xA ∈

■ The function () ()gfx  is simply the g–image of f(x), where ()fx is the image of x in A.

■ The composite function exists only if the range of f is a subset of the domain of g.

■ If the range of f is not a subset of the domain of g, gf  is not defined.

x AfgBC f(x) g(f(x)) gof

Key Properties of Function Composition

■ fg  and gf  may not be always defined.

■ If ()fx and ()gx are onto functions, then gf  is onto, if exists.

■ If g is one-to-one, then gf  is one-to-one function.

■ If g is onto, then gf  is onto.

■ If : fAB → is any function, then AB fIIff == where IA and IB are identity functions on,A and B respectively.

■ If f and g are even functions, then gf  is even.

■ If f is even and g is an odd, then gf  is even

■ If f is odd and g is even, then gf  is even.

■ If f and g are odd functions, then gf  is odd.

■ Function composition is not commutative. i.e., f ° g ≠ g ° f

■ Function composition is associative i.e., ( f ° g ) ° h = f ° ( g ° h)

Solved example

39. If () 1 1 fxx x = + , then find ()ffx 

Sol: Given, () 1 1 fxx x = + ()()()ffxffx = 

(() ) 1 1 ffxfx x  = +  1 1 1 1 1 1 1 1 x fxx xx x  + = + + + () ffxx = 

Try yourself:

41. For () 23fxx=+ and () 2 1 gxx=−+ , then find () ()fgx 

Ans: 2 25 x −+

TEST YOURSELF

1. If () 2 1 fxx x = , then () ()fffx =  (1) 2 3 13 x x (2) 2 2 14 x x (3) 2 13 x x (4) 2 4 14 x x

2. The domain of () 3 2 5 log 5 1024 fxxx xx  =−+  −+  is

(1) (4, 5) (2) (6, ∞ ) (3) (4, 5) ∪ (6, ∞ ) (4) (4, 5] ∪ (6, ∞ )

3. The domain of 2 4 log 1 x x    +  is (1) (–2, 2) (2) (–1, ∞ ) (3) [–1, 2 ] (4) (–1, 2 )

4. If f(x) = logx and g(x) = x3 then f(g(a)) + f(g(b)) is

(1) f(g(a) + g(ab)) (2) f(g(ab)) (3) g(f(ab)) (4) g(f(a) + f(b))

CHAPTER 20: Functions

5. If f (x ) = | x – 2| and g (x ) = f (f (x )), then for x > 20, g(x) = (1) –x (2) x (3) x – 4 (4) 4 – x

6. The domain of function log10 log10 log10 log10 log10x is (1) (104, ∞ ) (2) (1010, ∞ ) (3) () 1010 10, ∞ (4) (10100, ∞ )

7. If f : [–6, 6] → R is defined by f(x) = x2 – 3, then ( f°f°f)(–1) + ( f°f°f)(0) + ( f°f°f)(1) = k, where k = ___.

8. If f(x) = sin2x + 2 sincoscos 33 xxx

and

5 1, 4 g  =  then () ()gfx =  ____.

9. For x ∈ R – {0, –1, –2}, let f 1( x ) = 1 x , f 2 ( x ) = 1 + x and 3 1 () 1 fx x = + be three functions. If a function j(x) satisfied j(x) = (f1 ° f2 ° f3)(x), then by j(1) =___.

Answer Key (1) 3 (2) 3 (3) 4 (4) 2 (5) 3 (6) 3 (7) 29 (8) 1 (9) 0.66

20.8 INVERSE OF A FUNCTION

■ A function : fAB → is invertible if it is both one-to-one (injective) and onto (surjective).

■ The inverse function : gBA → satisfies () () , A gfxI =  and () () B fgxI =  where IA and IB are identity functions

■ g(x) is inverse of ()fx and it is denoted by g(x) = f –1(x) then 1 gf = .

Conditions for Invertibility:

■ f(x) must be bijective.

■ If g = f–1, then

‰ g(b) = a, f(a) = b

‰ The domain of f becomes the range of g, and vice versa.

■ The graph of y = f ( x ) and y = f– 1( x ) are symmetric about y = x

■ If the graphs of y = f ( x ) and y = f– 1 ( x ) intersect, they do so along the line y = x, where f(x)= x

Finding the Inverse of a Function:

■ Verify whether f(x) is bijective.

■ If : fAB → is a bijection:

‰ Solve y = f(x) for x in terms of y

‰ Interchange x and y to obtain f–1(x)

Properties of Inverse Functions:

■ If f : A → B and g : B → C are two bijections, then (g ° f )–1 : f–1 ° g –1

■ If fggf =  then either 1 fg = or •f2=g2=I. In this case, •g = f–1

Solved example

40. If the function :[1,)[1,) f ∞→∞ is defined by ()() 1 2 fxxx = , then find () 1 fx

Sol: Given: ()() 1 2 fxxx = Suppose that () 1 2 xx y = Apply logarithm on both sides () 2 1log xxy −= It implies that 2 2 log0xxy−−= Therefore, 2 114log 2 y x ++ =

Try yourself:

42. Find the inverse of ()() 1 5 6 fxx =− Ans: 5 6 x

TEST YOURSELF

1. Domain of tan–13x is (1) (– ∞ , 0) (2) , 22 ππ

(3) (– ∞ , ∞ ) (4) R+

2. If f : [1, ∞) → [2, ∞) is given by f(x) = x + 1 x , then f–1(x) = (1) 2 4 2 xx+− (2) 2 1 x x + (3) 2 4 2 xx (4) 2 34 2 xx

3. If f : R → R is defined by f(x) = x – [ x], then the inverse function f–1(x) = (1) 1 xx  (2) [x] – x

(3) x + [x] (4) not defined

4. Let f(x) = sinx + cosx, g(x) = x2 – 1. Then g(f(x)) is invertible for x ∈

(1) ,0 2 π  

(3) , 44 ππ  

(2) , 2 π π

(4) 0, 2 π

5. Let f : [–1, ∞) → R be given by f(x) = (x + 1)2 – 1, x > –1, then f–1(x)

(1) 11 x −++ (2) 11 x −−+

(3) does not exists because f is not 1 – 1

(4) does not exists because f is not onto

Answer Key

(1) 3 (2) 1 (3) 4 (4) 3 (5) 4

# Exercises

JEE MAIN

Level – I

Real Valued Functions

Single Option Correct MCQs

1. Check whether the equation  y 2= 2 x+ 1 represents a function or not. (apply the vertical line test algebraically)

(1) It is a function (2) It is not a function (3) It can be both a function and not a function.

(4) Can’t determine

2. Given relation is function or not? Give reasons and also find the domain and range of the function. g = (2,1), (5,1), (8, ), (11,1)

(1) Yes, D = (2, 5, 8, 11), R=1

(2) Yes, D = (3, 2, 7, 12), R=1

(3) No

(4) None

3. Given relation is function or not? Give reasons and also find the domain and range of the function. f = (1,3), (1,5), (2,3), (2,5) (1) 3 and 5 (2) 2 and 7 (3) 3 and 8 (4) 4 and 9

4. If 23 1 (()) ,1,1 1 x fxfxx x  =≠− +  and f(x) ≠ 0, then {f(–2)} = ______. Here, {x} represents fractional part of x. (1) 2 3 (2) 1 3 (3) 1 2 (4) 0

5. For a real number x, [x] denotes the integral part of x. The value of 11112199 ..... 2210021002100

(1) 49 (2) 50 (3) 48 (4) 51

6. If f : R→R is given by () x x a fxxR aa =∀∈ + , then 121996 ... 199719971997 fff

(1) 998 (2) 997 (3) 1997 (4) 1998

7. A real valued function f ( x ) satisfies the functional equation f ( x – y ) = f ( x ) f ( y ) –f ( a – x ) f ( a + y ), where a is a given constant and f(0) = 1, f(2a–x) is equal to (1) f(–x) (2) f(a) +f(a–x)

(3) f(x) (4) – f(x)

8. If 11 ()33,()33 22 fxxxxx gx =+=− , then f(x) g(y)+f(y) g(x) =

(1) f(x + y) (2) g(x + y) (3) f(2x) (4) g(2x)

9. If f(x) = sin(log x), then ()2()cos(log) x ffxyfxy y  +−=

(1) 0 (2) sin(logx)

(3) cos(log x)

(4) cos (log y)

10. If f : R → R is defined by 1 ()[] 2 fxxx=−− for x∈ R where [x] is the greatest integer not exceeding x, then 1 :() 2 xRfx  ∈== 

(1) Z, the set of all integers. (2) N, the set of all natural numbers.

(3) f , the empty set.

(4) R

Numerical Value Questions

11. If f(x) and g(x) are periodic functions with periods 7 and 11, respectively, then the period of ()()()53 xxFxfxggxf =−

is ______.

12. The period of the function satisfying the relation f ( x ) + f ( x + 3) = 0, ∀ x ∈ R, is ______.

13. If f ( x ) and g ( x ) are any two real valued functions such that |f(x) + g(x)| ≥ |f(x)| + |g(x)| and g(x)≠ 0, f(x) g(x) ≤ 0, then the value of 100 1 () r fr = ∑= ______.

14. The number of positive integers x, for the which the function ()() 2 21 1011 x fx xx = is not defined, is _____.

15. If f ( x + y ) = f ( x ) f ( y ) and f (5) = 32, then f(7) = _____.

Methods of Finding Domain and Range

Single Option Correct MCQs

16. The domain of the function + 1010 1010 xx xx is

(1) R (2) R – {0}

(3) R – {1} (4) R+

17. The domain of () 0.3 log2 x fx x = is

(1) [1, 2] (2) [2, 3]

(3) [1, 2) ∪ (2, 3] (4) [0, 3]

18. The domain of the function

()()() 2 1 log44 x fxxx =++ is

(1) [–3, –1] ∪ [1, 2]

(2) (–2, –1] ∪ [2, ∞ )

(3) (– ∞ , –3] ∪ (–2, –1] ∪ (2, ∞ )

(4) [–3, –2] ∪ [0, 2]

19. If ()   =   1 log x x fx x , where [ ] is greatest integer function, then the domain and range are

(1) (2, ∞ ), (0, 1) (2) [3, ∞ ), {0}

(3) [3, ∞ ), {0, 1} (4) (– ∞ , ∞ ); {0}

20. The domain of (){} =−+ 2 2{}31fxxx , w here { . } denotes the fractional part in [–1, 1], is

(1) 1 1,1~,1 2 

(2) {} 11 1,0,1 22  −−∪∪  

(3) 1 1, 2   

(4) 1 ,1 2   

21. The range of the function ()= + xx xxfxee ee is

(1) (– ∞ , ∞ ) (2) [0, 1)

(3) (–1, 0] (4) (–1, 1)

22. A = {x/x ∈ R, x ≠ 0, –4 ≤ x ≤ 4} and f : A → R is defined by ()= x fx x for x ∈ A. Then, the range of f is

(1) {1, –1} (2) {x : 0 ≤ x ≤ 4}

(3) {1} (4) {x : –4 ≤ x ≤ 0}

23. The domain of ()() () () = 1 123 fx xxx is

(1) (– ∞ , 1) ∪ (3, ∞ )

(2) (1, 2) ∪ (3, ∞ )

(3) (– ∞ , 2)

(4) R

24. The domain of + ++32 21 23 x xxx is

(1) (– ∞ , –1)

(2) (0, ∞ )

(3) (– ∞ , –1) ∪ (0, ∞ )

(4) R

25. The domain of     1 log x x is

(1) (0, ∞ ) (2) (1, ∞ )

(3) (0, 1 ) ∪ (1, ∞ ) (4) [1, ∞ )

26. Domain of 1 log x is

(1) R – {0, 1, –1}

(2) R – {0}

(3) R – {–1, 1}

(4) R

27. The domain of ()=+− 2 2 1 4 9 fxx x is

(1) (–4, –2) ∪ (2, 4)

(2) (–3, –2] ∪ [2, 3)

(3) (– ∞ , –3) ∪ (2, ∞ )

(4) (– ∞ , ∞ )

28. If x ∈ R and = −+ 2 4224 x P xx , then P lies in the interval

(1)  

CHAPTER 20: Functions

(3)    1 0, 3 (4)    1 0, 4

29. Domain of   2 1 2 xx is (where [.] is the greatest integer less than or equal to x)

(1) R – [–1, 3) (2) (– ∞ , 3) ∩[3,∞) (3) [2, ∞ ) (4) (– ∞ , 3]

30. The domain of ()() =−+ 1 2 log4 fxx x is

(1) [2, ∞ ] (2) (– ∞ , 4) (3) [2, 3) ∪ (3, 4) (4) [3, ∞ )

31. The domain of the function ()() + =− 2 3 log1 x fxx is (1) (–3, –1) ∪ (1, ∞ )

(2) [–3, –1) ∪ [1, ∞ )

(3) (–3, –2) ∪ (–2, –1) ∪ (1, ∞ ) (4) [–3, 2) ∪ (–2, –1) ∪ [1, ∞ )

32. The domain of ()() + = ++ 2 2 log3 32 x fx xx is

(1) R – {–1, –2}

(2) (–2, ∞ )

(3) R – {–1, –2, –3} (4) (–3, ∞ ) – {–1, –2}

33. If the functions are defined as ()()and1, fxxgxx ==− then what is the common domain of the following functions f + g, f – g, f/g, g/f, g – f where

ffx x ggx ?

(1) 0 < x < 1 (2) 0 ≤ x ≤ 1

(3) 0 < x ≤ 1 (4) 0 ≤ x < 1

34 , 45

1 0, 2 (2) 

34. The functions f(x) = log(x – 1) – log(x – 2)

and () =  1 log 2 x gx x are identical on (1) [1, 2] (2) [2, ∞ ] (3) (2, ∞ ) (4) R

35. If A is the set ot real values of x such that < 1 1 1 ex , then A = (1) (– ∞ , 0) ∪(1,∞) (2) (– ∞ , 0) (3) ( 1,∞) (4) (0, 1 )

36. If ex + ef(x) = e, then the domain of f(x) is (1) (– ∞ , 0] (2) [0, 1] (3) (– ∞ , 1) (4) (1, ∞ )

37. If f : R → R is defined by

for x ∈ R, where [y] denotes the greatest integer not exceeding y, then {f(x) : | x | < 71} = (1) {–14, –13, ....., 0, ....., 13, 14}

(2) {–14, –13, ....., 0, ....., 13, 14, 15}

(3) {–15, –14, ....., 0, ....., 13, 14}

(4) {–15, –14, ....., 0, ....., 14, 15}

38. The range of [x] – x is A and x – [x] is B, then A ∩ B = (1) {0} (2) (–1, 1) (3) (0, 1) (4) f

Numerical Value Questions

39. If the domain of ()() =−− 3 2log1fxx is(k, 10], then k is _______.

40. If the domain of f ( x ) is [0, 5], then the number of integers in the dom ain of f(log2(x2 + 2x – 3)) is ______

Algebra of Functions

Single Option Correct MCQs

41. If f ( x ) = 4 x – x 2 , x ∈ R , then f ( a + 1) –f(a – 1) is equal to (1) 2(4 – a) (2) 4(2 – a) (3) 4(2 + a) (4) 2(4 + a)

42. If f(x) is a polynomial function such that f(x) 11 () ffxf xx

=+

and f(3) = –80, then 1 () fxf x  −= 

(1) 4 4 1 x x + (2) 4 4 1 x x (3) 4 4 1 x x (4) 4 4 1 x x

43. If f(x) is a function such that f(x+y) = f(x) + f (y) and f(1) = 7, then 1 () n r fr = ∑= (1) 7 2 n (2) ()71 2 n + (3) 7n(n + 1) (4) ()71 2 nn +

44. If f(x) is a function such that f(xy) = f(x)+ f(y) and f(2) = 1, then f(x) (1) x2 (2) 2x (3) log2x (4) logx 2

45. Let f(x)= ax2+bx + c be such that f(1) = 3, f(–2)= l and f(3) = 4. If f(0) +f(1) + f(–2) + f(3) = 14, then l is

(1) 2 (2) 4 (3) –4 (4) 23 2

46. Let f(1) = 1 and 1 11 ()2(), then () nm rn fnfrfn=∑∑= (1) 3m –1 (2) 3m (3) 3m–1 (4) 3m–2

47. If f : R →R  satisfies f(x + y) = f(x) + f(y)∀x, y ∈ R and f(1) = 6, then 10 1 () r fr = ∑=

(1) 331 (2) 330 (3) 334 (4) 333

48. The function y = f(x) satisfying the condition f(x + 1/x) = x3+1/ x3 is

(1) f(x) = x2 (2) f(x) = x2–2

(3) f(x) = x2+2 (4) f(x) = x3–3x

49 If 32 32 11 ()413 fxxx xx  =+−++   , then the value of (23) f += (1) 9 (2) 10 (3) 11 (4) 12

50. If f :R →R is defined by f(x–1) = x2 + 3x + 2, then f(x – 2) is (1) x2 + x (2) x2 – 3x + 2 (3) x2 + 2x (4) x2 – x

Numerical Value Questions

51. If f(x) is a polynomial function such that ()()11 fxffxf xx  =+   and f(3) = –80, then 11 () n n fxfx xx  −=− 

. Then, 4n + 5 is _____.

52. If f is defined on the natural numbers as f(1) = 1 and for n > 1, [f(n) = f [f(n – 1)]+f[n–f ( n –1)] then the value of 20 1 1 () 30 r fr = ∑ is l , then 7 λ is ______.

53. The natural number ‘ a ’ for which 22 2 1 2()31 fxfx x  +=−   , when the function ‘f’ satisfies f(x + y) = f(x) f(y) or all natural numbers x, y and further f(1) = 2, is ,___.

54. If 22 2 1 2()31 fxfx x  +=−   for x∈R–{0}, then 1 5 2 f    is equal to ______.

55. The function f satisfies the functional equation 59 3()2 1030 1 fxfxx x +  +=+   for all real x ≠ 1. Then, the value of f(7) is ______.

56. The function f : R → R satisfies the condition af(x – 1) + bf(–x) = 2|x| + 1. If f(–2) = 5 and f(1) = 1, then 8(b – a) = _______.

Introduction to Functions

Single Option Correct MCQs

57. The domain of 2 2 1 312xx xx +−− is (1) 1 ,1 2

(2) 1 ,1 2

58. The domain of () 2 43fxxx=−+ is

(1) (– ∞ , 3]

(2) [1, ∞ )

(3) (– ∞ , –3] ∪ [–1, ∞ )

(4) (– ∞ , –3] ∪ [–1, 1 ) ∪ [3, ∞ )

59. If f ( x ) + g ( x ) = e –x, where f ( x ) is an even function and g(x) is an odd function, then f(x) = (1) 2

60. If f(x) is an even polynomial function, then sin (f(x) – 3x) is (1) an even function (2) an odd function

(3) neither even nor odd (4) a periodic function

61. If f(x + y, x – y) = xy, then the arithmetic mean of f(x, y) and f(y, x) is

(1) x (2) y (3) 0 (4) xy

62. If f(x) = ax7 + bx3 + cx – 5, a, b, c are real constants, and f(–7) = 7, then the range of f(7) + 17cosx is (1) [–34, 0]

(2) [0, 34]

(3) [–34, 34]

(4) None of these

63. The range of x 2 + 4 y 2 + 9 z 2 –6 yz –3 xz – 2xy is

(1) f (2) R

(3) [0, ∞ ) (4) (– ∞ , 0)

64. The range of () 2 4 sin1 1 x fx x

= + is

(1) R (2) [–1, 1]

(3) {0, 1} (4) {0}

65. If f(x) = cos[π2]x + cos[–π2]x, where [x] is the step function, then

(1) f(0) = 1 (2) 2 4 f π 

(3) 1 2 f π 

Numerical Value Questions

(4) f(π) = 1

66. Let f(x) be a polynomial of degree 4 with leading coefficient 1 satisfying f(1) = 10, f(2) = 20, f(3) = 30, where f(12) + f(–8) = 3968λ. Then the value of λ is ___.

67. If f(3x +2 ) + f(3x + 29) = 0 ∀ x ∈ R, then the period of f(x) = ___.

Kinds of Functions

Single Option Correct MCQs

68. y = f(x) = 1 x x + , x ∈ R, y ∈ R is

(1) one – one

(2) onto but not one – one (3) one–one but not onto (4) neither one–one not onto

69. If f : R → (0, 1] defined by () 2 1 1 fx x = + , then f is

(1) Not one – one (2) Not onto (3) Not one – one but onto (4) One – one but not onto

70. If f : R → R defined by f ( x ) = x 2 – 2 x – 3, then f is

(1) A  function (2) One one (3) Onto (4) One one onto

71. If f : R → R is defined by () 2 2 4 1 fxx x = + , then f(x) is (1) one – one and not onto. (2) one – one and onto. (3) not one – one but onto. (4) neither one – one nor onto.

72. If A = { a 1, a 2, a 3,…, a 10} and B = { b 1, b 2, b3,…, b10}, then the number of bijections that can be defined from A to B is (1) 1010 (2) 10! (3) 103 (4) 109

73. The function f : R → R defined by f(x) = x – [x], x ∈ R is (1) one–one (2) onto (3) both one–one and onto (4) neither one–one nor onto

74. Let the function f : R → R be defined by f(x) = 2x + sinx for x ∈ R. Then, f is (1) one-to-one and onto (2) one-to-one but not onto (3) onto but not one-to-one (4) neither one-to-one nor onto

75. () 2 2 430 :, 818fRRfxxx xx ++ →= −+ is (1) one–one and onto (2) many–one and onto (3) one–one and into (4) many–one and into

Numerical Value Questions

76. If n ( A ) = 4, n ( B ) = 5, then number of functions from A to B such that the range contains exactly 3 elements is k, then the value of 60 k is __.

77. The number of constant functions that can be defined from {1, 2,…, 100} to {a,b,c,…z} is ____.

Composition of a Function Single Option Correct MCQs

78. If f : R → R and g : R → R are defined by f(x) = x – [x] and g(x) = [x] for all x ∈ R, then f(g(x)) is (1) x (2) 0 (3) f(x) (4) g(x)

79. If g(x) = 1 + x and f(g(x)) = 3 + 2 x + x, then f(x)= _____.

(1) 1 + 2x2 (2) 2 + x2

(3) 1 + x (4) 2 + x

80. If g(x) = x2 + x – 1 and (g ° f)(x) = 4x2 – 10x + 5, then 5 4 f

is equal to

(1) 1 2 (2) 1 2

(3) 3 2 (4) 3 2

81. If, for x ∈ [0, ∞), g[f(x)] = |sinx|, f[g(x)] = (sin x )2, then

(1) f(x) = sin2x, g(x) = x

(2) f(x) = sinx, g(x) = |x|

(3) f(x) = sin2x, g(x) = sin x

(4) f and g cannot be determined

82. The domain of the function () 2 111 fxx =−−− is

(1) {x| x < 1} (2) {x| x > –1}

(3) [0, 1] (4) [–1, 1]

Numerical Value Questions

83. Let A = {1, 2, 3, 4}. The number of functions

f : A → A, satisfying

f(f(i)) = i, ∀ i ∈ A, is ___.

84. If () 2010165 ,0, 1652010 fxxx x + => and 2010 165 x ≠ then the least value of

f(f(x)) + f(f( 4 x )) is______.

Inverse of a Function

Single Option Correct MCQs

85. The domain of () 1 3 sin 42sin fx x  =  +  is

(1) 5 , 63 ππ 

(3) , 6 π π

(2) 7 , 66 ππ 

(4) , 22 ππ  

86. The domain of () 1 31 sincossin 2 x x + is

(1) [–1, 1] (2) 1 ,0 3 

(3) (0, 1] (4) 1 ,1 3 

Numerical Value Questions

87. f : [0, ∞) → [4, ∞) is defined by f(x) = x2 + 4 then f –1(13) = ______.

88. If f : R → R, g : R → R are defined by f(x)= 5x – 2, g(x) = x2 + 3, then (g ο f –1)(3) = ______.

89. If a non-zero function f(x) is symmetrical about the line y = x, then the value of λ (constant) such that f 2( x ) = ( f –1( x )) 2 –λxf(x)f –1(x) + 3x2 f(x) ∀ x ∈ R+ , is ____.

Level – II

1. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).

Assertion (A) : If { x } and [ x ] represent fractional part and integral part of x, then {} 1000 1 1000 b xb xx = + +=

Reason (R) : {x} = x – [x] and [x + I] = [ x ] + I , where I is an integer

In the light of above statements, choose the correct answer from the options given below.

(1) Both (A) and (R) are true and (R) is the correct explanation of (A).

(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).

(3) (A) is true but (R) is false.

(4) Both (A) and (R) are false.

2. Let N be the set of all natural numbers and f: N → N be such that 1990 < f(1990) < 2100 and satisfies the equation

1990 1990

, where y denotes the greatest integer less than or equal to  y . Then the number of possible values of f (1990) is

(1) 1 (2) 2 (3) 3 (4) 4

3. Let [t] denote the greatest integer ≤ t. Then, the equation in x, [x]2 + 2[ x + 2] – 7 = 0 has (1) infinitely many solutions.

(2) exactly four integral solutions.

(3) exactly two solutions.

(4) no integral solution.

4 The domain of the function () 2 1 4109 fx xxx = −−+ is

(1) () 740,740 −+

(2) () 0,740 + (3) () 740,−∞

(4) ()() 740,33,740 −∪+

5. Domain of +  2 4 2 x x is

(1) (– ∞ , –2) ∪ (2, ∞ )

(2) R – {–2}

(3) [–1, 2)

(4) (– ∞ , –2) ∪ [1, 2 ]

6 If f : R → R is defined by f(x) = [2x] – 2[x] for x ∈ R, where [x] is the greatest integer not exceeding x, then the range of f is

(1) {x ∈ R : 0 ≤ x ≤ 1}

(2) {0, 1}

(3) {x ∈ R : x > 0}

(4) {x ∈ R : x < 0}

7. If [x] represents the greatest integer ≤ x and [a,b] is the set of all real values of x for which the real function ()++− = +  33 2 fxxx x is defined, then f 2( a + 1) + 5 f 2( b ) =

(1) 0 (2) 36/5

(3) 12 (4) 1

8. The domain of the function log10 log10 log10 log10 log10x is

(1) (104, ∞ ) (2) (1010, ∞ )

(3) () ∞ 1010 10, (4) (10100, ∞ )

9. Let f : (1, 3) → R be a function defined by () 2 1 xx fx x

 = + , where [x] denotes the greatest integer ≤ x. Then the range of f is ______.

(1) 2134 ,, 5255 

(2) 137 0,, 255

(3) 24 ,11, 55

(4) 124 0,, 355

Single Option Correct MCQs

10. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).

Assertion (A) : f(x)= log x3 and g(x) = 3log x are equal functions.

Reason (R) : Two functions f and g are

CHAPTER 20: Functions

said to be equal if their domains and co-domains are equal and f ( x ) = g(x) ∀ x

In light of the above statements, choose the correct answer from the options given below.

(1) Both (A) and (R) are true and (R) is the correct explanation of (A).

(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).

(3) (A) is true but (R) is false.

(4) Both (A) and (R) are false.

11. f and  h are from  A into  B, where A = {a, b , c , d }, B = { s , t , u }, defined as  f ( a ) = t , f(b) = s, f(c) = s, f(d) = u ; h(a) = s, h(b) = t, h(c) = s, h(a) = u, h(d) = u. Which one of the following statements is true?

(1) f and h are functions.

(2) f is a fucntion and h is not a function.

(3) f and h are not functions.

(4) f  is not a function but h is a function.

12. If f : N → Z is defined by () 23 10 31 032 ifnkkz fnnifnkkz ifnkkz =∈  =−=+∈  =+∈  , then

{n ∈ N, f(n)>2 } = (1) {3, 6, 4} (2) {1, 4, 7}

(3) {4, 7} (4) {7}

13. If f(x + y, x − y) = xy, then the arithmetic mean of f(x, y) and f(y, −x) is (1) 0 (2) x (3) y (4) 1

14. If a function F is such that  F (0) = 2, F (1) = 3, F ( x + 2) = 2 F ( x )− F ( x + 1) for  x ≥ 0, then F(5) is equal to (1) –7 (2) –3 (3) 17 (4) 13

15. If  f(x + 1) = x2 − 3x + 2, then  f(x) is equal to:

(1) x2 − 5x − 6

(2) x2 + 5x − 6

(3) x2 + 5x + 6

(4) x2 − 5x + 6

16. If f  and g are real valued functions defined by f(x) = 2x –1 and g(x) = x2, then (fg)x =

(1) 2x2 + 2x –1

(2) 2x3 – x2

(3) 2x – x2

(4) 2x3 – 4

Numerical Value Questions

17. If [α] denote the greatest integer ≤ α then [1][2][3][120] +++……+ is equal to ______.

18. If  f ( x ) is an odd periodic function with period 2, then f(4) = ______.

19. If the domain of the function

then k is ____.

20. Let f : A → B be a function, where n(A) = 3, n(B) = 2, then the total number of functions from set A to set B is _____.

21. If f : {1, 2, 3,......} → {0 ± 1, ± 2,.....} is defined by () ,ifiseven 2 1 ,ifisodd 2 n n

then f –1(–100) is ______.

Introduction to Functions

Single Option Correct MCQs

22. If f(x) and g(x) are two functions with all real numbers as their domains, then h(x) = [f(x) + f(–x)][g(x) – g(–x)] is

(1) always an odd function

(2) an odd function, where both f and g are odd

(3) an odd function, where f is even and g is odd

(4) always an even function

23. A function whose graph is symmetrical about the y-axis is given by

(1) ()() 2 coslog1fxxx

(2) ()

44 34 seccosec cot fxxx xxx + = +

(3) f(x + y) = f(x) + f(y) ∀ x,y ∈ R

(4) () 44 34 seccosec cot fxxx xxx =

24. If f is an even function defined on the interval [–5, 5], then the real values of x satisfying the equation () 1 2 fxfx x +  = +

are

(1) 1535 , 22 −±−±

(2) 3545 , 22 −±−±

(3) 2545 , 22 −±−±

(4) 4515 , 22 −±−±

25. The range of the function () 2 4 1 fxx x = + is

(1) 1 0, 2

(2) 1 0, 2 

(3) [0, ∞ ) (4) [0, 2]

26. If f(x) is a polynomial function such that

11fxffxf

and f(3) = –80, then ()

Numerical Value Questions

27. If f(x) = sgn(x2 – 2x + 3), then the value of f(x) is ____.

28. If f 3(x) – 3f 2(x) + 3f(x) –1 = x6, then the value of f(3) is ____.

Kinds of Functions

Single Option Correct MCQs

29. If f : R → R is defined by ()

=

, then f(x) is (1) one–one and not onto (2) one–one and onto (3) not one–one but onto (4) neither one–one nor onto

30. f : R + → R , defined by f ( x ) = log e x , x ∈ (0,1), f(x) = 2loge x, x ∈ [1, ∞) is (1) onto (2) one–one (3) not one–one (4) a bijection

31. If f : R → R is defined by ()

then f is

(1) one–one but not onto (2) not one–one but onto (3) one–one and onto (4) neither one–one nor onto

32. The function f:(–∞, –1) → (0, e5], defined by () 3 32xx fxe −+ = , is (1) many one and onto (2) many one and into (3) one–one and onto (4) one–one and into

33. If for real x, f(x) = x3 + 5x + 1, then (1) f is one–one but not onto on R (2) f is onto on R but not one–one (3) f is one–one and onto on R

(4) f is neither one–one nor onto on R

34. f : N → N, where f(x) = x – (–1)x, then f is (1) one-one and into (2) many-to-one and into (3) one-one and onto (4) many-to-one and onto

35. f : R+ → R defined by f(x) = 2x , x ∈ (0, 1), f(x) = 3x , x ∈ [1, ∞) is (1) onto (2) one–one

(3) neither one–one nor onto (4) one–one and onto

36. f : R → R is a function defined by

() . xx xxfxee ee = + Then f is (1) one – one and onto. (2) one – one but not onto. (3) onto but not one – one. (4) neither one – one nor onto.

37. Let N be the set of all natural numbers, Z be the set of all integers, and s : N → Z

be defined by () ,ifiseven 2 1 ,ifisodd 2 n n n n n

(1) s is one–one but not onto. (2) s is into but not one–one.

(3) s is one–one and onto.

(4) s is neither one–one nor onto.

Numerical Value Questions

38. The function f : [2, ∞) → Y defined by f(x) = x2 – 4x + 5 is both one-one and onto if Y = [a, ∞). Then the value of a is _____.

39. Let f : R → [2, ∞) be a function defined as f(x) = x2 – 12ax + 15 – 2a + 36a2. If f(x) is surjective on R , then the value of 2a is ______.

40. Let f be a one-one function with domain {x,y,z} and range {1, 2, 3}. It is given that exactly one of the following statements is true and the remaining two are false: f(x) = 1, f(y) ≠ 1, f(z) ≠ 2. Then, f(y) is

Composition of a Function

Single Option Correct MCQs

41. If () 1fxx x α = + , x ≠ –1, then the value of α, for which f(f(x)) = x, is

(1) 2 (2) 2

(3) 1 (4) –1

42. If f ( n ) = (–1) n–1( n – 1), G ( n ) = n – f ( n ) for every n ∈ N, then (GOG)(n) = ___.

(1) n (2) n – 1

(3) 1 (4) 2

44. Let ()()() 2 1 ,; 1 x fxffxfx x + ==  ()()() 3 , fffxfxffffx =   = f 4(x).....then f 2008(x) = ___.

(1) x (2) f(x) (3) 0 (4) 1 x

45. The range of f(x) = ()()() 1cos1cos1cos.... xxx −−−∞ is

(1) [0, 1] (2) 1 0, 2 

(3) [0, 2] (4) 1 ,1 2 

Numerical Value Questions

46. The functions f : R → R and g : R → R are defined as () 0whenisrational 1whenisirrational x fx x 

=

() 1whenisrational 0whenisirrational x gx x

 =   then

(f ° g)(π) + (g ° f)(e) + 10 = ___.

47. A continuous function f satisfies the property f ( f ( x )) = 1 – x . The value of 13 44ff +   is ___.

Inverse of a Function Single Option Correct MCQs

43. If f(x) is defined on [0, 1] as () ,if 1,if fxxxQxxQ ∈  =−∉ 

(f ° f)(x)= ___.

then

(1) 1 (2) x (3) 1 – x (4) 1 + x

48. If ()(){} 1 4 3 1 , sin36fxx=−− then f –1(x) = (1) 4 3 63sin x ++ (2) 4 3 63sin x +−

(3) 4 63sin x ++ (4) 4 63sin x +−

49. Let f be an injective map with domain { x,y,z } and range {1, 2, 4} such that exactly one of the following statements is correct and the remaining are false:

f(x) = 1, f(y) ≠ 1, f(z) ≠ 2. The value of f –1(1) is (1) y (2) x (3) z (4) 0

50. Let f : R → R be given by f(x) = (x + 1 )2 – 1, x ≥ –1. Then the set of values of x, for which f(x) = f –1(x), is given by (1) {0}

(2) {–1, 0}

(3) {–1}

(4) {0, 1}

51. If f : R → R and g : R → R are defined by f(x) = 3x – 4, g(x) = 2 + 3x and 2(g–1 ° f–1)(x) > (f–1 ο g–1)(x), then x ∈ (1) (1, ∞ ) (2) 1 , 2  ∞

(3) 1 , 2  −∞   (4) (14, ∞ )

Numerical Value Questions

52. If f : {1, 2, 3,…..} → {0, ±1, ±2,…..}is defined by f(n) = if is even 2 1 if is odd 2 n n n n    

then f –1(–100) is ______.

53. Let f be a real–valued invertible function such that 23 2 fx x    = 5x –2, x ≠ 2.

CHAPTER 20: Functions

Then the value of f –1(13) is ______.

Level-III

Single Option Correct MCQs

1 If Let f : R → R is defined by f(x) = 2x3 + 2x2 + 300x + 5sinx, then f is (1) one–one onto (2) one–one into (3) many one and onto (4) many one and into

2. Which of the following functions is not injective?

(1) f(x) = |x + 1|, x ∈ [–1, ∞ )

(2) ()() 1 ,0,gxxx x =+∈∞

(3) h(x) = x2 + 4x –5, x ∈ (0, ∞ )

(4) k(x) = e–x , x ∈ [0, ∞ )

3. The function y = f ( x ), satisfying the condition 3 3 11 , fxx xx  +=+  is

(1) f(x) = x2 (2) f(x) = x2 – 2

(3) f(x) = x2 + 2 (4) f(x) = x3 – 3x

4. If f(x) = cos(logx), then f(x2)f(y2) () 2 22 2 1 2 fxyfx y

(1) –2 (2) –1

(3) 1 2 (4) 0

5. If f(x) = sinx + cosx, g(x) = x2 – 1, then g(f(x)) is invertible in the domain

(1) 0, 2 π

(3) , 22 ππ

(2) , 44 ππ

(4) [0, π]

6. Let f : R → R be a function defined by f(x) = x3 + x – 1 and ‘g’ be the inverse of f. Then, the value of () () 9 9 g g ′ is (g’ is the dervative of g)

(1) 26 (2) 32 (3) 11 (4) 17

7. If the graph of f(x) = ||x – 2| – a|–3 has exactly 3 –intercepts on x-axis then the sum of all values of a ∈ [–10, 10], where and a is an integer is (1) 4 (2) 27 (3) 3 (4) 49

8. Let f : [–10, 10] → R , where () 2 sin fxxx a 

, be an odd function. Then the set of values of parameter a is (Where [.] represents GIF)

(1) (10, 10)– {0} (2) (0, 10) (3) [100, ∞ ) (4) (100, ∞ )

9. Let f(x) = |x – 2| and (() () ()) () () ntimes gxffffx =

If equation g ( x ) = k , k ∈ (0, 2) has 8 distinct solutions, then the value of n is equal to (1) 3 (2) 4 (3) 5 (4) 8

10. If (() )()(()) 1 7 7 ln11ln1 fxx +=−+ , then f(f(cosx)) is equal to (1) cos(ln(1 + |x|)) (2) cos7(ln(1 + |x|)) (3) cosx (4) cos7x

11. The number of integer values of ‘x’ for which (){} 2 3 log4log x xx   −+ is real is ___. ([ ] denotes GIF, { } denotes fraction part).

(1) 4 (2) 1 (3) 2 (4) 3

Numerical Value Questions

12. Let :, 36 fB ππ  →  be defined by () 2 2cos3sin21fxxx =++ , where B = [a,b] such that f –1(x) exists. Then a + b is _____.

13. A function is defined as srk(x –a) () 20242022 20242022 20242023 57 , 379 sgn 2020,, xx xa xx xxxa

where [.] is greatest integer function and sgn( x ) means signum function. If the number of real value(s) of ‘x’, where x ∈ [0, 3) satisfies the equation x(srk(x –0) – srk(x – 1)) – srk(x – 1) + 3srk(x – 2) = 2 394 2 xx−+ is ‘ λ’then the value of λ λ is _____.

14. If ()()32 47 35 3 a fxxaxx  =+−++   is one-to-one, where a ∈ [l, m], then the value of | l – m | is _____.

15. For x ∈ R, x ≠ 0, x ≠ 1, let f0(x) = 1 1 x and f n + 1 (x ) = f 0( f n( x )), n = 0, 1, 2,.... Then, the value of () 1001 2 23 3 32fff ++  i s equal to ______.

16. If () 1 7 47 1 4cos2cos2cos4 2 gxxxxx  =−−−   , then the value of g(g(100)) is _____.

19. If f is a function such that f(x) + 2f(1–x) = x2 + 1, then the value of f(3) is ______.

THEORY-BASED QUESTIONS

Statement Type Questions

1. S-I : Every relation is a function

S-II : Every function is a relation.

2. S-I : If any vertical line cuts the graph of a Relation at most one point, then that relation is called function.

S-II : In the function, no two ordered pairs have same second component.

3. S-I : Two functions f,g are said to be equal if and only if f ( x ) = g ( x ) for all x ∈ domain of f

S-II : Two functions f,g are said to be equal, if their domains are equal and their ranges also equal.

4. S-I : The domain of a constant function is set of all real numbers.

S-II : The range of a constant function is singleton set.

5. S-I : The range of modulus function is set of all non negative real numbers.

S-II : The range of greatest integer function is set of all integers.

6. S-I : The range of constant function is finite set

S-II : The range of signum function also finite set and it is {–1, 0, 1}

7. S-I : The function f(x) = loga x is negative, when 0 < x < 1, 0 < a < 1.

S-II : The function f(x) = loga x is positive, when x > 1, 0 < a < 1.

8. S-I : Suppose that the domain of the function f(x) is A, then the domain

of ()fx is {()} :,0xxRfx∈≥

S-II : The rule to find the domain of a square root function is the function under square root must be nonnegative.

9. S-I : The derivative of odd function is an even function.

S-II : The derivative of an even function is an odd function.

10. S-I : If n(A) < n(B), then the number of onto functions from the set A to the set B is zero.

S-II : If n(A) > n(B), then the number of one–one functions from the set A to the set B is zero.

11. S-I : The function f : R → R, defined as f(x) = x + 12, is a surjection.

S-II : The function f : R → R is defined as f(x) = x + 12 is a bijection.

12. S-I : If f : A → B and g : B → C are two bijections, then g ° f = A → C is bijection and (g ° f )–1 = f –1° g –1

S-II : If f ° g = g ° f, then either f –1 = g or g–1 = f; in this case, f ° g (x) = g ° f (x) = x.

Assertion and Reason Questions

In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as

(1) if both (A) and (R) are true and (R) is the correct explanation of (A),

(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),

(3) if (A) is true but (R) is false,

(4) if both (A) and (R) are false.

13. (A) : The function which associates each real number to it self, then that function is called identity function.

(R) : The graph of identity function is a straight line passing through the origin and making equal angles with axes.

14. (A) : The fractional part function {x} and x–[x] are two equal functions.

(R) : Two functions are said to be equal, if the images of two functions for every element of its domain must be equal and their domains are equal, ranges must be equal.

15. (A) : The functions f ( x ) = x 2 and ()= fxx are not equal functions.

(R) : The range of f(x) = x2 is same as that of ()= fxx

16. (A) : The function f ( x ) = | x | and () 2 = fxx are two equal functions.

(R) : The domain of f(x) = |x| is equal to domain of () 2 = fxx and the range of f(x) = |x| is equal to range of () 2 = fxx

17. (A) : log =⇔= y a xyxa

(R) : logarithmic function and exponential functions are inverse functions to each other.

18. (A) : If p is a period of a function f(x) then np also period of the function f(x)

(R) : A function f : A → R is said to be periodic function if there exists a positive real number p such that

f(x+p) = f(x) for all x ∈ A. The least positive real number p is called the fundamental period of f(x).

19. (A) : A sub set of the domain of the function ()() log gx fx is {x : f(x) ≥ 1, g(x) > 1}.

(R) : The domain of the function ()() log gx fx is intersection of two sets {x : f(x) ≥ 1, g(x) > 1} and {x : 0 < f(x) ≤ 1, 0 < g(x) < 1}.

20. (A) : The range of the function () 1 cossin = ++ fx axbxc is 2222 11 ,

(R) : Range of f(x) = a cos x + b sin x + c is 2222 ,  −+++  cabcab .

21. (A) : Let ()()11  += fxffxf xx and f(2) = 9 then f(x) = x3 + 1.

(R) : If ()()11  += fxffxf xx then f(x) = 1 + xn

22. (A) : If A= {a, b, c, d} and B= {2, 3} then the number of onto functions from the set A to the set B is 14.

(R) : The number of onto functions defined from the set A to the set B is () 22, nA where n(B) = 2.

23. (A) : The domain of the function () 2 9 fxx x = is (–3, 3).

20: Functions

(R) : Denominator value must not be equal to zero and the value under the root must be non-negative, so that 2 90 x −>

24. (A) : The range of the function f : R → R given by () 2 2 fxx x + = + is (–1, 1)

(R) : |x + 2| is equal to either x + 2 or x – 2 only.

25. (A) : Every function can be expressed as the sum of an even function and an odd function uniquely.

(R) : (()() ) 1 2 fxfx +− is an even function and (()() ) 1 2 fxfx an is odd function.

26. (A) : The function f : R → R, defined as () 2 , 1 fxx x = is not a one–one function.

(R) : () 4 4 17 f = and 14 417 f  = 

27. (A) : The function f : R → R, is defined as () 3 1, fxx=+ is a bijective function.

(R) : Every polynomial function defined on the set of real numbers is a bijective function.

28. (A) : If a function is a strictly increasing function, then that is one-to-one function.

(R) : Any line parallel to the x-axis does not intersect the continuous graph of increasing function at more than one point.

JEE ADVANCED LEVEL

Multiple Option Correct MCQs

1. The domain of ()=  1 15 fx x

(where [.] GIF) is

(1) [–7, 7] (2) (– ∞ , 7]

(3) (– ∞ , –7] (4) [7, ∞ )

2. If ex + ef(x) = e, then for f(x),

(1) domain = (– ∞ , 1)

(2) range = (– ∞ , 1)

(3) domain = (– ∞ , 0]

(4) range = (– ∞ , 1]

3. Let (){} =+−+

1/2 22 1 fxnnx , where [.] is the greatest integer function and n ∈ N . Then the value of x for which f(x) is defined is

(1) k , where k = 0, 1, 2, ..... n

(2) [–n2, 2n + 1]

(3) [0, 2n + 1]

(4) N

4. If f ( x ) = f (– x ) and and g ( x ) is an odd function and satisfies the relation ()()  −=   2 1 2 xfxfgx x , then

(1) f(x) = 0, ∀ x ∈ R

(2) f(2009) = 0

(3) f is a constant function

(4) g(x) = 0, ∀ x ∈ R

5. Let ()()()() +=−+−22 11[ fxfyfxyyxfx ()()()() +=−+−22 11[ fxfyfxyyxfx is not identically zero]. Then,

(1) f(4x3 - 3x) + 3f(x) = 0,

(2) f(4x3 - 3x) = 3f(x),

(3) ()()−+= 2 2120fxxfx , (4) ()() −= 2 212 fxxfx ,

6. Let f(x) be a polynomial with positive degree satisfying the relation f(x)g(y) = f(x) + f(y)

+ f(xy) – 2 and f(4) = 65. Then,

(1) f(2) = 9.

(2) ()=3327 f .

(3) f(5) = 126.

(4) the roots of the equation f(x) = 2x2 are all real.

7. Let f(x) is a polynomial of degree n such that ()()() 1 00,1, 21 fffnn n ==…= +

Then the value of f(n + 1) is

(1) 1 when n is odd (2) + 2 n n when n is even (3) + 1 n n when n is odd

(4) –1 when n is even

8. Let :, 22 fR ππ  −→

be given by f(x) = (log (sec x + tan x))3. Then,

(1) f(x) is an odd function

(2) f(x) is a one–one function

(3) f(x) is an onto function

(4) f(x) is an even function

9. Let f : (–1,1) → R be such that () 2 2 cos4 2sec f θ θ = for 0,, 442

. Then, the value of

is

(1) 3 1 2 (2) 3 1 2 + (3) 2 1 3 (4) 2 1 3 +

10. If f(x) = sin x + cos x, g(x) = x2 – 1, then g(f(x)) is invertible in the domain

(1) 0, 2

(2) , 44

(3) , 22

(4) [0, π ]

11. Let f be a function defined by () 5 , 3 fxx x = x ≠ 3, and 2 f k(x) denotes the composition of f with itself taken k times i.e., f3(x)=f(f(f(x))). Then,

(1) f2012(2009) = 2009

(2) () 2009 2005 2010 2007 f =

(3) () 2009 1002 2009 1003 f =

(4) f2012(2012) = 2012

12. Let f(x) = ln(2x – x2)+sin 2 xπ 

. Then, (1) graph of f is symmetrical about the line x = 1

(2) graph of f is symmetrical about the line x = 2

(3) maximum value of f is 1

(4) minimum value of f does not exist

13. Let f : R → R be a function defined by ()() () 5 1. 3 fxfxxR fx ∀ +=∈ Then, which of the following st atements is/are true?

(1) f(2008) = f(2004)

(2) f(2006) = f(2010)

(3) f(2006) = f(2002)

(4) f(2006) = f(2018)

14. f ( x ) = x 2 – 2 ax + a ( a +1), f : [ a , ∞ ) → [ a , ∞ ). If one of the solutions of the equation f(x) = f –1(x) is 5049, then the other may be

(1) 5051 (2) 5048

(3) 5052 (4) 5050

15. Consider the function y = f(x) satisfying the condition () 2 2 11 0. fxxx xx

+=+≠

Then the

(1) domain of f(x) is R (2) domain of f(x) is R-(–2,2) (3) range of f(x) is [–2, ∞ ) (4) range of f(x) is [2, ∞ )

16. If ()() , 0, and , 0, , fxxxQxQ gx xQxxQ

)() , 0, and , 0, , xxQxQ gx xQxxQ

then f – g is

(1) one–one and into

CHAPTER 20: Functions

(2) neither one–one nor onto (3) many–one

(4) one–one and onto

17. For real x, let f(x) = x3 + 5x + 1. Then,

(1) f is one–one but not onto R

(2) f is onto R but not one–one

(3) f is one–one and onto R

(4) f is neither one–one nor onto R

18. Let f be the greatest integer function and g be the modulous functions. Then,

(1) () 5 1 3 gffog

(2) (f+2g)(–1) = 1

(3) () 5 0 3 gffg

(4) (f + 2g)(1) = 1

19. If f : R + → R + is a polynomial function satisfying the functional equation f(f(x)) = 6x – f(x), then f(17) is equal to (1) 17 (2) –51

(3) 34 (4) –34

Numerical/Integer Value Questions

20. If [x] stands for greatest integer function,

21. Let

If the range of f(x) is [a, b), then [b] = ______, where [.] is GIF.

22. If f(x) is an odd function, f(1) = 3, and f(x + 2) = f(x) + f(2), then f(3) = _________.

23. If f : R → [0, ∞ ) is a function such that () (-1)(1) 31fxfxfx++=→ , then the period f(x) is ______.

24. If ()+ =−<<   1 log,11, 1 fxxx x and + −= ++  3 22 32 131 xxx ffM xxf(x), then the value of M is _____.

25. If f(x) is a polynomial such that f ( x ) f ( y ) = f ( xy ) – f ( x ) – f ( y ) ∀ x , y and f(2) = 7, then |f ( – 2) | = ________.

26. If the domain of f(x) = log[x]x2 is [k, ∞), then the value of k is _____ (where [ ] represents greatest integer function).

27. Minimum value of f(x) = |x – 1| + |x – 2| + |x – 3| is _______.

28. If f(x) = x2 and ()() () = 1.11 1.11 ffk , then [k] = ______. ([ ] is greatest integer function)

29. If ()() + =∈− 10 ,10,10 10 fxx ex x and f(x) = k 2 200 100 fx x 

, then 2k = ______.

30. f : {1,2,3,4} → {1,2,3,4} such that f ( i ) = even if i is even and f(i) ≤ f(j) ∀ i < J. If the number of such functions possible is n, then the sum of the digits in n ______.

31. If () 2010163 ,0 1652010 fxxx x + => and 2010 165 x ≠ , then the least value of (()

32. If () 1 log, 1 fxx x +  =  –1 < x < 1, and 3 22 32 131 xxxff xx

+

= Mf ( x ), then the value of M is _____.

33. If the range of the function ()() ()24 3 11xxx fx x +++ = when x > 0 is [k , ∞ ), then 3 k += ______.

34. Let g(x) = 1+x–[x], [x]is the greatest integer not greater than x. If () 1,0 0,0, 1,0 x fxx x  −<  ==

then for all x, f(g(x)) = ______.

35. Number of positive integers in the domain of the function () 2 loglog0.56 4 fxxx x  + =  +  is _____.

36. The function () 3 1 1 fxx x + = + can be expressed as the sum of even function g ( x ) and odd function h ( x ). The value of |g(0)| = ______

37. If f ( x ) is odd function, f (1) = 3 and f(x + 2) = f(x) + f(2)then f(3) = _____.

is ______ (where [.] denotes greatest integer function).

38. Let f:[–5,6] → R be a real valued function whose graph is as given blow. Then, the number of solutions of f ( f ( f ( x ))) = 3 is ____.

39. Number of solutions of the equation 2[ x ] = x + { x } is _____ (where [ ] is greatest integer function, { } is fractional point).

40. Let f ( x ) be a real valued function such that () 2024 2023 2023, 2024,' fxxxQ

If f(f(x)) = 2025 a + b2025x, then the sum of distinct real value(s) of b is _____.

Passage-based Questions

Q (41 - 42)

Let f : N → R be a function satisfying the following conditions: = 1 (1) 2 f and f(1) + 2. f(2) + 3. f(3) + ...... + n. f(n) = n(n + 1)f(n) for all n ∈ N, n ≥ 2.

41. The value of f(1003) is 1 k , where k equals (1) 1003 (2) 2003 (3) 2005 (4) 2006

42. The value of f(999) is 1 k , where k equals (1) 999 (2) 1000 (3) 1998 (4) 2000

Q (43 - 44)

Let f : R → R be a function satisfying f(2 – x) = f(2 + x) and f(20 – x) = f(x) ∀ x ∈ R. For this function f, answer the following.

43. If f (0 ) = 5, then the minimum possible number of values of x satisf ying f ( x ) = 5 for x ∈[0,170], is (1) 21 (2) 12 (3) 11 (4) 22

CHAPTER 20: Functions

44. The graph of y = f(x) is not symmetric about (1) x = 2 (2) x = 10 (3) x = 8 (4) none of these (Q.45–47)

Let f :R → R be a function satisfying f(2–x) = f(2+x) and f(20–x) = f(x) ∀ x ∈ R. For this function f, answer the following.

45. If f(0) = 5, then the minimum possible number of values of x satisfying f(x) = 5 for x ∈ [0, 170] is _______ (1) 21 (2) 12 (3) 11 (4) 22

46. The graph of y = f(x) is not symmetrial about (1) x = 2 (2) x = 10 (3) x = 8 (4) none of these.

47. If f(2) ≠ f(6), then the (1) fundamental period of f(x) is 1 (2) fundamental period of f(x) may be 1 (3) period of f(x) cannot be 1 (4) fundamental period of f(x) is 8

(Q.48–49)

Let ()() 2 log1 e fxxx=++ domain of f is where f(x) is defined for real values of x. If f is bijective, then f –1(x)exists

48. f –1(x) is defined on (1) (0, ∞ ) (2) (– ∞ , ∞ ) (3) [0, e] (4) (– ∞ , 0)

49. The inverse of f is positive on (1) (0, ∞ ) (2) (– ∞ , ∞ ) (3) [0, e] (4) (– ∞ , 0)

(Q.50–52)

A function f from a set X to Y is called onto if every y ∈ Y, ∃ x ∈ X, such that f(x) = y. Unless the contrary is specified, a real function is onto

if it takes all real values. Otherwise, it is called an into function. Thus, if X and Y are finite sets, then f cannot be onto, if Y contains more elements than x . Now answer the following questions.

50. The polynomial function

f(x) = a0xn + a1xn–1 + a2 xn–2 + ..... + a n, where a0 ≠0 is onto, for

(1) all positive integers n

(2) all even positive integers n

(3) all odd positive integers n

(4) no positive integer

51. The function () 2 2 2 43 fxxxc xxc ++ = ++ is onto, if

(1) 0 < c < 2 (2) 0 < c < 4

(3) 11 22 c << (4) 0 < c < 1

52. Which of the following is not true?

(1) A one–one function from the set {a, b, c} to { a , b , γ } is also onto.

(2) An onto function from an infinite set to a finite set cannot be one–one.

(3) An onto function is always invertible.

(4) The functions tanx and cotx are onto.

(Q.53–55)

Let () 2 2,1 31 xax fxbxx +≥−  =+<−  and () 4,04 3220 xx gx xx +≤≤

53. g(f(x)) is not defined if

(1) a ∈ (10, ∞ ), b ∈ (5, ∞ )

(2) a ∈ (4, 10), b ∈ (5, ∞ )

(3) a ∈ (10, ∞ ), b ∈ (0,1)

(4) a ∈ (4, 10), b ∈ (1,5)

54. If the domain of g(f(x)) is [–1,4] then (1) a = 1, b > 5 (2) a = 2, b > 7

(3) a = 2, b > 10 (4) a = 0, b ∈ R

55. If a = 2 and b = 3, then the range of g(f(x)) is

(1) (–2, ∞ ) (2) (0, ∞ ) (3) [4, ∞ ) (4) [–1, ∞ )

(Q.56–57)

For x ≠ 0,1, define f1(x) = x, f2(x) = 1 x ,

f3(x) = 1–x, 4 1 () 1 fx x = , f5(x) = 1 x x ,

f6(x) = 1 x x .

This family of functions is closed under composition that is, the composition of any two of these functions is again one of these.

56. Let G be a function such that GOf 3 = f 6 Then, G is equal to (1) f5 (2) f4 (3) f3 (4) f2

57. If H is a function such that f4OH = f5, then H is equal to (1) f6 (2) f4 (3) f5 (4) f3

(Q.58–60)

f : A → B is said to be injective if distinct elements in A have distinct images in B and surjective if f(A) = B. Now answer the following.

58. If f : A → B defined by () 1cos2 2 fxx + = is injective, then A can be (1) [0, π ] (2) [– π , π ] (3) ,0 2 π    (4) [– π , 0]

59. If the function f : R → B defined by () 2 fxx = is surjective, then B is (1) [– ∞ , 0) (2) [0, ∞ )

(3) (0, ∞ ) (4) 

60. If the function f : R → B, defined by f ( x ) = [ x ] + [– x ], (where [.] is GIF) is surjective, then B = (1) R (2) [0, 1]

(3) [–1, 0] (4) {–1, 0}

Matrix Matching Questions

61. Let f : R → R and g : R → R be functions such that f(g(x))is a one–one function.

List I List II

(A) g(x) I) must be one–one

(B) f(x) II) may not be oneone

(C) If g(x) is onto, then f(x) III) may be many–one

(D) If g(x) is into, then f(x) IV) must be many–one

(A) (B) (C) (D)

(1) I II,III I II, r

(2) I II III IV

(3) II III I IV

(4) II I III IV

62. List I gives functions and List II the nature of the functions. Choose the correct option.

List I List II

(A) f :[0, ∞)→ [0, ∞), () 1 fxx x = + I) one–one onto

(B) f: R–{0}→R, () 1 fxx x =− II) one–one but not onto

CHAPTER 20: Functions

(C) f: R–{0}→R, 1 () fxx x =+ III) onto but not one–one

(D) f : R→R, f(x)=2 x+sinx IV) neither one–one nor onto

(A) (B) (C) (D)

(1) I II III IV

(2) II III IV I

(3) III II I IV

(4) I IV III II

63. Match the following functions in List I with their inverse function in List II

List I List II

(A) () 2 xx xxfxee ee =+ + I) 2 1 (114log) 2 x ++

(B) f : (2,4) → (1,3) defined by () 2 fxxx =−  II) x+(–1)x–1

(C) f : [1,∞)→[1,∞) defined by f(x) = 2x(x–1) III) 1/2 1 log 3 e x x  

(D) f : N→N defined by f(x) = x+(–1)x–1 IV) x+1

(A) (B) (C) (D)

(1) IV III II I

(2) III IV II I

(3) II III I IV

(4) III IV I II

64. let R denote the set of all real numbers and R + denote the set of all positive real numbers. For the subsets A and B of R define f : A → B by f(x) = x2 for x∈A observe the two lists given below.

List I List II

(A) f is one–one and onto if I) A = R+ , B = R

(B) f is one–one but not onto if II) A = B = R

(C) f is onto but not one–one if III) A = R, B = R+  {0}

(D) f is neither one–one nor onto if IV) A = B = R+

(A) (B) (C) (D)

(1) IV I III II

(2) I II III IV

(3) III IV II I

(4) IV III II I

65. Let () 1 log 1 fxx x + = and () 3 2 3 13 xx gx x + = + , then match List I with List II.

List I List II

(A) 1 1 foge e   +  I) 3 (B) 1 1 gofe e   +  II) –3

(C) fog(0) III) 0 (D) 1 1 foge e   +  IV) 1

(A) (B) (C) (D)

(1) I II III IV

(2) II IV III I

(3) III II I IV (4) I III II IV

FLASHBACK (PREVIOUS JEE QUESTIONS)

JEE Main

1. Let f : R–{0, 1} → R be a function such that 1 ()1 1 fxfx x  +=+   . Then f(2) is (1st Feb 2023 Shift 2)

(1) 9 2 (2) 7 4 (3) 7 3 (4) 9 4

2. The number of integral solutions x of 2 7 2 7 log 0 23 x x x  +  

is (11 th Apr 2023 Shift 1)

(1) 7 (2) 5 (3) 6 (4) 8

3. The domain of the function () 2 1 310 fx xx =

is (where [ x ] denotes the great est integer less than or equal to x) (11th Apr 2023 Shift 2)

(1) (– ∞, –3] ∪ (5, ∞)

(2) (– ∞, –2] ∪ (5, ∞) (3) (– ∞, –2) ∪ [6, ∞) (4) (– ∞, –3] ∪ (6, ∞)

4. Let the sets A and B denote the domain and range, respectively of the function 1 () [] fx xx = , where [ x ] denotes the smallest integer greater than or equal to x Then among the statements.

(S1): A ∩ B = (1, ∞ )–N and (S2): A ∪ B = (1, ∞ ) (15th Apr 2023 Shift 1)

(1) only (S2) is true

(2) both (S1) and (S2) are true (3) neither (S1) nor (S2) is true (4) only (S1) is true

5. Let f(x) be a function such that f(x+y) = f(x) • f(y) for all x,y ∈ N. If f(1) = 3 and 1 ()3279, n kfk = ∑= then the value of n is (24th Jan 2023 Shift 2) (1) 6 (2) 8 (3) 7 (4) 9

6. If () 2 2 2 , 22 x fxx = + x ∈ R, then 12 2022 ..... 202320232023 fff

+++

is _____.

(24th Jan 2023 Shift 2)

(1) 2011 (2) 2010 (3) 1011 (4) 1010

7. The number of functions f: {1, 2, 3, 4}→ {a Z : |a| ≤ 8} satisfying 1 ()(1)1, fnfn n ++=

∀ n ∈ {1, 2, 3} is ___. (25th Jan 2023 Shift 2) (1) 1 (2) 3 (3) 2 (4) 4

8. Let f(x) = 2xn + λ, λ∈R, n ∈ N and f(4) = 133, f(5) = 255. Then the sum of all the positive integer divisors of (f(3) − f(2)) is (25 th Jan 2023 Shift 2) (1) 60 (2) 58 (3) 61 (4) 59

9. The domain of (1) 2log log(2) () (23) e x x x fx ex + = −+ is (29th Jan 2023 Shift 1)

(1) (–1, ∞ )–{3} (2) (2, ∞ )–{3} (3)  –{3} (4)  –{–1,3}

10. The range of the function ()32 fxxx =−++ (30 th Jan 2023 Shift 2)

(1) 2,7   (2) 5,10 

(3) 22,11 

(4) 5,13

11. Consider a function f:  → satisfying f(1) + 2f(2) + 3f(3)+....+ xf(x) = x(x + 1)f(x); x ≥ 2 with f(1) = 1. Then 11 (2022)(2028)ff + , is equal to (29th Jan 2023 Shift 2) (1) 8100 (2) 8000 (3) 8200 (4) 8400

12. Let R be a relation from the set {1, 2, 3,....60} to itself such that R = {(a,b): b = pq, where p,q ≥3 are prime numbers}. Then, the numbers of elements in R is (29th Jul 2022 Shift 1) (1) 600 (2) 660 (3) 540 (4) 720

13. If the domain of the function 2 [] (), 1 fxx x = + (where [x] is greatest integer) is [2, 6), then its range is (31th Jan 2023 Shift 1)

(1) 52 , 265

(2) 52 , 375

(3) 52927189 ,,,, 265291098953

(4) 52927189 ,,,, 375291098953

14. Let f(x) = ax2 + bx + c be such that f(1) = 3, f(–2) = λ, and f(3) = 4. If f(0) + f(1) + f(–2) + f(3) = 14, then λ is equal to ____. (28th Jul 2022 Shift 2) (1) –4 (2) 13/2 (3) 23/2 (4) 4

15. Let A = (1,2,3,4,5} and B = {1,2,3,4,5,6}. Then the number of functions f:A → B satisfying f(1) + f(2) = f(4) –1 is equal to ___. (11 th Apr 2023 Shift 2)

16. Suppose, f is a function satisfying f ( x+y ) = f(x)+f(y) for all x,y ∈  and 1 (1). 5 f = If 1 () 1 , (1)(2)12 m n fn nnn = = ∑++ then m is equal to ______. (29th Jan 2023 Shift 1)

17. Let f:R → R be a function defined 2 2 2 () x x fxe ee = + . Then, 123 99 100100100100 ffff

++++

is ____. (27 th Jun 2022 Shift 1)

18. Let c,k∈R. If f(x) = (c + 1)x2 + (1 – c2)x + 2k and f(x+y) = f(x) + f(y) – xy, for all x,y∈R, then the value of |2(f(1) + f(2) + f(3)+.....+ f(20)|is equal to _____.

(29th Jun 2022 Shift 1)

19. The number of functions f, from the set A = {x ∈ N : x2 – 10x + 9 ≤ 0} to the set B = {n2 : n ∈ N} such that f(x) ≤ (x – 3)2 + 1, for every x ∈ N, is ___.

(27th Jul 2022 Shift 2)

20. The function f : N – {1} → N ; defined by f(n)= the highest prime factor of n, is :

(27th Jan 2024 Shift 1) (1) both one–one and onto. (2) one–one only. (3) onto only. (4) neither one–one nor onto.

21. Let 1 : 2 fRR  −→ 

and 5 : 2 gRR

−→

be defined as () 23 21fxx x + = + and () 1 25 x gx x + = + Then the domain of the function f ο g is: (27th Jan 2024 Shift 2)

(1) 5 2 R

(2) R (3) 7 4 R

(4) 57 , 24 R

22. If () 22,10 ; 1,03 3 xx fxx x +−≤< 

() ,30 , ,01 xx gx xx

then range of (f ο g(x)) is (29th Jan 2024 Shift 1)

(1) (0,1] (2) [0,3) (3) [0,1] (4) [0,1)

23. If () 432 , 643 x fxx x + =≠ and ( f ο f ) ( x ) = g(x), where 22 : 33 g 

, then (g ο g ο g) (4) is equal to (31st Jan 2024 Shift 1)

(1) 19 20 (2) 19 20

(3) –4 (4) 4

24. Let f : R → R and g : R → R be defined as () log,0 ,0 e x xx fx ex > 

and () ,0 . ,0 x xx gx ex

Then, gοf : R → R is (1st Feb 2024 Shift 1)

(1) one–one but not onto (2) neither one–one nor onto (3) onto but not one–one (4) both one–one and onto

25. If the domain of the function ()() 2 2 25 4 fxx x = +log10(x2 +2x –15) is (–∞, α) ∪ [β,∞), then α2 + β3 is equal to: (1st Feb 2024 Shift 2) (1) 140 (2) 175 (3) 150 (4) 125

CHAPTER TEST – JEE MAIN

Section – A

1. Let Z be set of integers. If A={x ∈ Z: 2(x+2)(x2-5x+6)=1} and B = {x ∈ Z: –3 < 2x – 1 < 9}, then the number of subsets of the set A×B, is

(1) 215 (2) 218 (3) 212 (4) 210

2. Let ‘f ’ be a real valued function defined for all  x ∈ R such that for some fixed a > 0, ()()() 2 1 () 2 fxafxfx +=+− for all x, then the period of f(x) is (1) a/4 (2) a/3 (3) 2a (4) 3a

3 A function f well defined. x,y ∈R is such that f(1) = 2, f(2) = 8 and f(x+y) – kxy = f(x) + 2y2 where ‘k’ is some constant, then f(x) is (1) x2 (2) 3x2 (3) 2x2 (4) 4x2

4. The range of () 2 2 1 1 fxx x =+ + is

(1) [1, ∞ ] (2) [2, ∞) (3) 3 , 2 

(4) R

5. The range of f(x) = sin2x – 5sinx – 6 is (1) [–10,0] (2) [–1,1]

(3) [0, π ] (4) 49 ,0 4   

6. The range of () 1 xx fx xx   = −+   (where [.] is Greatest integer function) is

(1) 1 0, 2    (2) [0,1] (3) 1 0, 2    (4) 1 0, 2  

7. If () 2 2 log, 1 e fxxe x  + = +  then the range of f(x) is

(1) (0,1) (2) [0,1] (3) [0,1) (4) (0,1]

8. If f(x)is a polynomial satisfying f(x)f(1/x) = f ( x ) + f (1/ x ) and f (3) = 28, then f (4) is equal to

(1) 63 (2) 65

(3) 17 (4) 64

9. f:C → C is defined as () ,0fxaxbbd cxd + =≠ + , then f is a constant function when (1) a=c (2) b=d (3) ad=bc (4) ab=cd

10. The domain of ()sincos x

(1) 2,2, 2 nnnI π ππ  +∈

(2) 2,2, 2 nnnI π πππ

(3) 3 2,2, 2 nnnI π πππ

(4) 2,2, 22 nnnI ππ

11. Let f : R → R be defined as f (x) = 2x – 1 and g: R–{1} be defined as () 1 2 1 x gx x = , then the composite function f ( g ( x )) is defined as (1) both one - one and onto, (2) neither one one nor onto, (3) one one but not onto, (4) onto but not one one.

12. Let x denote the total number of one–one functions from a set A with 3 elements to a set B with 5 elements and y denotes the total number of one – one function from the set A to be set A × B, then (1) y = 91x (2) 2y = 273x (3) y = 273x (4) 2y = 91x

13. The domain of the function ()()() 1 2 4 2 21 3 9272193 x fxxx

(1) [–3, 3] (2) [3, ∞ )

(3) 5 , 2  ∞  (4) [0, 1]

14. The domain of the function

() 102142 fxxx =−− is

(1) [5, ∞ )

(2) 21,21  

(3) {}5,2121,50 −−∪∪ 

(4) (– ∞,-5 ]

15. Domain of the function

()() (() () ) 2 2423 loglogloglog423fxxx =+− is

(1) (–8, 4)

(2) ()() ,84, −∞−∪∞

(3) (–4, 8)

(4) ()() ,48, −∞−∪∞

16. The range of () 1 xx fx xx   = −+   , (where [ ] is GIF) is

(1) 1 0, 2    (2) [0, 1]

(3) 1 0, 2    (4) 1 0, 2

17. Let f :R→R defined by ()

fxee ee = + , then f(x) is (1) one–one but not onto (2) neither one–one nor onto (3) many-one but onto (4) one-one but not onto

18. If f ( x ) is an even function and satisfies the relation ()() 2 1 2 xfxfgx x  ⋅−⋅=

,

where g(x) is an odd function, then the value of f(5) is (1) 0 (2) 37 55 (3) 4 (4) 5

19. If )):1,2, f ∞→∞ is given by () 1 fxx x =+ then () 1 fx = (1) 2 3 4 xx±− (2) 2 4 2 xx±− (3) 2 4 2 xx+− (4) 2 4 2 xx

20. The domain of ()() {} 2 1010 log1log516fxxx =−−+ is (1) (2, 3) (2) (0, ∞ ) (3) [1, 3] (4) [2, 3]

Section – B

21. If () 22 4 1, fxxx=−+− then the maximum value of (f(x))2 is _____.

22. If a + α = 1, b + β = 2 and () 1 ,0afxfbxx xx β +α=+≠   , then the value of the expression () 1 1 fxf x x x  +  + is _____.

23. If f(x) = sgn(x2 – 2x + 3), then the value of f(x) is ______.

24. If f(3x + 2) + f(3x+29) = 0; xR∀∈ , then the period of f(x) is _____.

25. If f is a polynomial such that 1111 1111 xxxxffff xxxx −+−+  =+  +−+−  (where x ≠ 0,±1) and f(3)=28, then the value of (()) 10 1 1 1 605 n fn =    ∑ is ______.

CHAPTER TEST – JEE ADVANCED

2023 P1 Model

Section – A

[Multiple Option Correct MCQs]

1. The value of x in 2,2, ππ   for which the graph of the function 1sin sec 1sin x yx x + =− and 1sin sec, 1sin x yx x =−+ + coincide are:

(1) 33 2, ,2 22 ππ

(2) 33 ,, 2222 ππππ

(3) , 22 ππ

(4) 3 2,2, 22 ππ ππ

2. Consider the function () 2 1,fxxx =+− then which of the following is/are correct?

(1) range of f(x) is 1,2  

(2) f is neither even nor odd

(3) f is either even or odd.

(4) range of f(x) is identical to range of () 2cos 4 gxx

3. If f : R + → R + is a polynomial function satisfying the functional equation f(f(x)) = 6x-f(x), then f(17) is equal to (1) 17 (2) –51 (3) 34 (4) –34

CHAPTER 20: Functions

Section –B [Single Option Correct MCQs]

4. If {} 22 : ,:11 55 AxxByy π  =≤≤=−≤≤ 

and f(x) = cos(5x+2) then the mapping f:A → B is (1) one–one but not onto (2) onto but not one–one (3) both one–one and onto (4) neither one–one nor onto

5. Let g:R→R be given by g(x) = 3+4x. If gn(x) = gogo.....og(x)(n) times, and gn(x) = A+Bx, then A and B are (1) 2n+1–1, 2n+1 (2) 4n–1, 4n (3) 3n, 3n+1 (4) 5n–1, 5n

6. Let A = {1, 2, 3....10} and f : A → A be defined as () 1 if is odd if kis even kk fkk +  =   , then the number of possible functions g: A → A such that gof = f is (1) 10C5

(2) 55

(3) 105

(4) 5!

7. f : R → R is a function defined by () . xx xxfxee ee = + Then f is (1) one–one and onto (2) one–one but not onto (3) onto but not one–one (4) neither one–one nor onto

Section – C

[Integer Value Questions]

8. f ( x ) is a real valued function satisfying 2 f ( xy ) = ( f ( x )) y + f ( y )) x ; x,y ∈ R and () 1 1, 2 f = , then determine the value of () 1 i fi ∞ = ∑ ______.

9. For some a,b,c ∈ N, let f(x) = ax – 3 and g(x) = xb + c, x ∈ R. If ()() 1 3 1 7 2

then ( f ο g )( ac ) + ( g ο f )( b ) is equal to _____.

10. Let f(x) = 2x(2–x);0≤x≤2, then the number of solution of (() () ) 2 fffxx = is ___.

11. The number of integral solution(s) of the equation given by 2222 1cos1cos xxxx −+−=− is _____. (Given α is a positive solution of x2 = cos2x)

12. Let 13 :,,, 24 f

defined by () 42 2 1 , 1 fxxx xx ++ = −+ t hen the num ber of possible values of x sat isfying () 143 2 fxx−+− = is ______.

Section – D

[Matrix Matching Questions]

14. Match the following and choose the correct option.

List - I List - II

(A) The inverse function of sin(tan–1x) is (p) cot(cos–1x)

(B) The inverse function of f(x) = 1–2–x is (q) () 2 2 log log1 x x

(C) The inverse function of () 1 2 x fxx = (r) tan(sin–1x) (s) –log2(1–x) (t) 2 1 x x

(1) A → p,r; B → r; C → q

(2) A → r,s; B → s; C → t

(3) A → s,t; B → r; C → q

(4) A → p,r,t; B → s; C → q

15. If f: R → R is defined by () 4for4 32for44 4for4 xx fxxx xx +<−  =+−≤<  −≥  , then the correct matching of List I and List II is.

13. Let f be defined on the natural numbers as follow: f (1) = 1 and for n >1, f ( n ) = f [( n –1)]+ f [ f ( n –1)]+ f [ n – f ( n –1)], then the value of () 20 1 1 30 r fr = ∑ is l then 7 λ is ______.

List - I

List - II

(A) f(–5) + f(–4) (p) 14

(B) f(|f(–8)|) (q) 4

(C) f(f(–7) + f(3)) (r) –11

(D) f(f(f(f(0)))) + 1 (s) –4 (t) 1 (u) 0

(1) A-p, B-s, C-q, D-s

(2) A-r, B-s, C-q, D-t

(3) A-r, B-t, C-q, D-s

(4) A-r, B-t, C-p, D-s

16. Match the following and choose the correct option.

List - I List - II

(A) 2 2 1 x xx e ++ (p) Constant function

(B) () 2 log1 xx++ (q) Identity function

(C) For 3 < x < 5, the function |x – 2| + |x – 3| + |x – 5| (r) Even function

(D) f(x) = 2023 (s) Odd function

ANSWER KEY

(1) A-q, B-s, C-p, D-r

(2) A-r, B-p, C-s, D-q

(3) A-s, B-q, C-r, D-p

(4) A-r, B-s, C-q, D-p

17. Match the following and choose the correct option.

List - I List - II

(A) f(x) = e2x, g(x) = logx, then (fog)(x) = (p) (– ∞,1) (3,∞)

(B) f(x) = 2x–3, g(x) = x3+5, then (fog)–1 (x) (q) {0}

(C) Domain of log(x2 – 4x +3) (r) 1/3 7 2 x 

(D) Range of () 2 sin 1 x fx x

[.] represents G.I.F (s) x2

(1) A-s, B-r, C-q, D-p

(2) A-p, B-r, C-s, D-q

(3) A-s, B-r, C-p, D-q

(4) A-q, B-s, C-r, D-p

1155 (12)

Level – II

Level – III

Theory-based Questions

(21) 1 (22) 1 (23) 1 (24) 3

JEE Advanced Level

Flashback

Chapter Test-JEE Main

Chapter Test – JEE Advanced