RD SHARMA Solutions for Class 12-science Maths Chapter 2 - Functions

Chapter 2 - Functions Exercise MCQ

Question 55

Let   be defined by   Then,

a.   

b.   

c. fof(x) = -x

d.   

Solution 55

Given:   be defined by

For inverse, we first need to check whether f is invertible.

Let   such that f(x) = f(y)

  

So, f is one-one.

Let y = f(x)

  

Therefore, for every   there exist y R such that f(x) = y.

Thus, f is bijective and so invertible.

Taking y = f(x), we get

  

Hence,

Question 56

Let f: R R be defined by   Then, f is

a. one-one

b. onto

c. bijective 

d. not defined

Solution 56

Given: f: R R be defined by

As x R, so it can take the value 0 as well.

At x = 0, f(x) is not defined.

Thus, f is not defined.

Question 57

Let f: R R be defined by f(x) = 3x2 - 5 and g: R R by   Then, (gof)(x) is

a.   

b.   

c.   

d. 

Solution 57

f: R R is given by f(x) = 3x2 - 5

g: R R is given by

  

Question 58

Which of the following functions from Z to Z are bijections?

a. f(x) = x2

b. f(x) = x + 2

c. f(x) = 2x + 1

d. f(x) = x2 + 1

Solution 58

a. If we take f: Z Z defined by f(x) = x2

Clearly, it is not one-one as considering f(x) = f(y) will give us

x = ±y

Therefore, it is not a bijection.

b. Consider f: Z Z defined by f(x) = x + 2

Let x, y Z such that f(x) = f(y)

x + 2 = y + 2

x = y

So, f is one-one.

For x in the domain Z, f(x) will also take integer values.

So, the range of f is Z.

Therefore, f is onto.

Thus, f is a bijection.

c. Consider f: Z Z defined by f(x) = x + 2

Let x, y Z such that f(x) = f(y)

x = y

So, f is one-one.

For x in the domain Z, f(x) = 2x + 1 will take odd integer values.

So, the range of f is not same as its co-domain.

Therefore, f is not onto.

Thus, f is not a bijection.

d. If we take f: Z Z defined by f(x) = x2 + 1

Clearly, it is not one-one as considering f(x) = f(y) will give us

x = ± y

Therefore, it is not a bijection.

Hence, option (b) is correct.

Question 59

Let f: A B and g: B C be the bijective functions. Then

a.   

b.   

c.   

d. 

Solution 59

It is given that f: A B and g: B C are two bijective functions.

Consider,

  

Thus,   

Question 60

Let f: N R be the function defined by   and g: Q R be another function defined by g(x) = x + 2. Then (g o f)(3/2) is

a. 1

b. 2

b.   

c. none of these

Solution 60

It is given that f: N R as   and g: Q R as g(x) = x + 2.

Now,

  

Hence,

Question 1

Let A = {x ε R : - 1 ≤ x ≤ 1} = B and C = {x ε R : x ≥ 0} and let S = {(x, y) ε A × B : x2 + y2 = 1} and s0 = {(x, y) ε A × C : x2 + y2 = 1}. Then


(a) S defines a function from A to B

(b) S0 defines a function from A To C

(c) S0 defines a function from A to B

(d) S defines a function from A to C

Solution 1

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
Given space that space space straight A space equals space left curly bracket straight x space straight epsilon space straight R space colon space minus space 1 space less or equal than space straight x space less or equal than space 1 right curly bracket space equals space straight B space and space straight C space equals space left curly bracket straight x space straight epsilon space straight R space colon space straight x space greater or equal than space 0 right curly bracket space and space straight S space equals space left curly bracket left parenthesis straight x comma space straight y right parenthesis space straight epsilon space straight A space cross times space straight B space colon space straight x to the power of 2 space end exponent plus space straight y to the power of 2 space end exponent equals space 1 right curly bracket space
and space straight s subscript 0 space equals space left curly bracket left parenthesis straight x comma space straight y right parenthesis space straight epsilon space straight A space cross times space straight C space colon space straight x squared space plus space straight y squared space equals space 1 right curly bracket
straight x squared space plus space straight y squared space equals space 1
rightwards double arrow space straight y squared space equals space 1 minus straight x squared
rightwards double arrow straight y equals square root of space 1 minus straight x squared end root
straight y element of straight B
Hence comma space straight S space defines space straight a space function space from space straight A space to space straight B. end style

Question 2

begin mathsize 12px style straight f space colon space straight R space rightwards arrow space straight R space given space by space fxx right parenthesis space equals space straight x space plus space square root of straight x squared end root space is
end style

 

(a)  injective

(b) surjective

(c) bijective

(d) none of these

Solution 2

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space function space is space straight f colon straight R rightwards arrow straight R space given space by space straight f left parenthesis straight x right parenthesis equals straight x plus square root of straight x squared end root
For space this space function space if space we space take space straight x equals 2 space and space straight x equals negative 2 space we space will space get space two space values space of space straight f left parenthesis straight x right parenthesis.
Hence comma space it space is space not space straight a space function. end style

Question 3

If f : A → B given by 3f(x) + 2-x = 4 is a bijection, then

 

(a) A = {x ε R : - 1 < x < ∞], B = {x ε R ; 2 < x < 4]

(b) A = {x ε R : - 3 < x < ∞}, B = {x ε R ; 0 < x < 4}

(c) A = {x ε R : - 2 < x ε R : 0 < x < 4}

(d) none of these


Solution 3

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space that space space straight f space colon space straight A space rightwards arrow space straight B space given space by space 3 to the power of straight f left parenthesis straight x right parenthesis end exponent space plus space 2 to the power of negative straight x end exponent space equals space 4 space is space straight a space bijection.
3 to the power of straight f left parenthesis straight x right parenthesis end exponent space plus space 2 to the power of negative straight x end exponent space equals space 4
3 to the power of straight f left parenthesis straight x right parenthesis end exponent space equals 4 minus 2 to the power of negative straight x end exponent
4 minus 2 to the power of negative straight x end exponent greater or equal than 0
4 greater or equal than 2 to the power of negative straight x end exponent
2 greater than negative straight x
straight x greater than negative 2
straight x element of open parentheses negative 2 comma space infinity close parentheses
But comma space for space straight x equals 0 space rightwards double arrow straight f left parenthesis straight x right parenthesis equals 1
Hence comma space the space correct space option space is space none space of space these.
end style

Question 4

 The function f : R → R defined by f(x) = 2x + 2|x| is


(a) one-one and onto

(b) many-one and onto

(c) one-one and into

(d) many-one and into

Solution 4

begin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
The space function space straight f space colon space straight R space rightwards arrow space straight R space defined space by space straight f left parenthesis straight x right parenthesis space equals space 2 to the power of straight x space plus space 2 to the power of vertical line straight x vertical line end exponent space
Here comma space for space each space value space of space straight x space we space will space get space different space values space of space straight f left parenthesis straight x right parenthesis.
Hence space it space is space one minus one space function.
Also comma space each space element space of space codomain space is space mapped space to space at space most space one space element space of space the space domain.
Function space is space one minus one space and space into.


end style

Question 5

begin mathsize 12px style let space the space function space straight f space semicolon space straight R space minus space left curly bracket negative space straight b right curly bracket space rightwards arrow space straight R space minus space open curly brackets 1 close curly brackets space be space defined space by space straight f open parentheses straight x close parentheses equals fraction numerator straight x plus straight a over denominator straight x plus straight b end fraction comma space straight a not equal to straight b comma space then end style

 

(a) f is one-one but not onto

(b) f is onto but not one-one

(c) f is both one-one and into

(d) none of these

Solution 5

begin mathsize 12px style Correct space option space colon space left parenthesis straight c right parenthesis
Given space function space is space straight f colon space straight R minus left curly bracket negative straight b right curly bracket space rightwards arrow straight R minus left curly bracket 1 right curly bracket space be space defined space by space straight f left parenthesis straight x right parenthesis equals fraction numerator straight x plus straight a over denominator straight x plus straight b end fraction comma space straight a space not equal to straight b.
Here comma space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight y right parenthesis rightwards double arrow straight x equals straight y space hence space it space is space one minus one space function.
straight f left parenthesis straight x right parenthesis equals straight y
rightwards double arrow fraction numerator straight x plus straight a over denominator straight x plus straight b end fraction equals straight y
rightwards double arrow straight x plus straight a equals yx plus by
rightwards double arrow straight x open parentheses 1 minus straight y close parentheses equals by minus straight a
rightwards double arrow straight x equals fraction numerator by minus straight a over denominator 1 minus straight y end fraction element of straight R minus open curly brackets negative straight b close curly brackets
Hence comma space straight f space is space onto.
end style

Question 6

The function f : A → B defined by f(x) = -x2 + 6x - 8 is a bijection, if

 

(a) A = ( -∞, 3] and B = ( -∞, 1]

(b) A = [-3, ∞) and B = (-∞, 1]

(c) A = ( -∞, 3] and B = [1, ∞)

(d) A = [3, ∞) and B = [1, ∞)

Solution 6

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
Given space function space is space space straight f space colon space straight A space rightwards arrow space straight B space defined space by space straight f left parenthesis straight x right parenthesis space equals space minus straight x to the power of 2 space end exponent plus space 6 straight x space minus space 8 space is space straight a space bijection.
straight f left parenthesis straight x right parenthesis space equals space minus straight x to the power of 2 space end exponent plus space 6 straight x space minus space 8
straight f left parenthesis straight x right parenthesis equals negative open parentheses straight x squared minus 6 straight x plus 8 close parentheses
straight f left parenthesis straight x right parenthesis equals negative open parentheses straight x squared minus 6 straight x plus 8 plus 1 minus 1 close parentheses
straight f left parenthesis straight x right parenthesis equals negative left parenthesis straight x squared minus 6 straight x plus 9 minus 1 right parenthesis
straight f left parenthesis straight x right parenthesis equals negative open square brackets open parentheses straight x minus 3 close parentheses squared minus 1 close square brackets space
Hence comma space straight x element of open parentheses negative infinity comma 3 close parentheses space and space straight y element of open parentheses negative infinity comma 1 close parentheses end style

Question 7

Let A = {x ε R : - 1 ≤ x ≤ 1} = B. then, the mapping f : A → B given by f(x) = x|x| is 


(a) injective but not surjective

(b) surjective but not injective

(c) bijective

(d) none of these

Solution 7

begin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
Given space function space is space straight A space equals space left curly bracket straight x space colon space minus 1 space less or equal than space straight x space less or equal than space 1 right curly bracket space and space straight f semicolon space straight A space rightwards arrow space straight A space such space that space straight f left parenthesis straight x right parenthesis space equals space straight x vertical line straight x vertical line
For space the space mod space function space we space have space to space check space three space cases space as space straight x less than 0 comma space straight x equals 0 comma space straight x greater than 0.
For space example comma
straight x less than 0
straight f left parenthesis straight x right parenthesis equals straight x open vertical bar straight x close vertical bar less than 0
open vertical bar straight x close vertical bar equals negative straight x
straight y equals negative straight x squared
straight x equals negative square root of negative straight y end root space which space is space not space posible space for space straight x greater than 0
Hence comma space straight f space is space onto.
rightwards double arrow straight f space is space bijection. end style

Question 8

Let f : R  → R be given by f(x) = [x]2 + [x + 1] - 3, where [x] denotes the greatest integer less than or equal to x. Then, f(x) is

 

(a) many-one and onto

(b) Many-one and into

(c) one-one and into

(d) one-one and onto

Solution 8

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Question 9

let m be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then the function f : M → R defined by f(A) = |A| for every A ε M, is

 

(a)  one-one and onto

(b) neither one-one nor onto

(c) one-one but-not onto

(d) onto but not one-one

Solution 9

begin mathsize 12px style Correct space option colon left parenthesis straight d right parenthesis
Given space that space straight m space be space the space set space of space all space 2 space cross times space 2 space matrices space with space entries space from space the space set space straight R space of space real space numbers. space
Then space the space function space straight f space colon space straight M space rightwards arrow space straight R space defined space by space straight f left parenthesis straight A right parenthesis space equals space vertical line straight A vertical line space for space every space straight A space straight epsilon space straight M.
If space straight f left parenthesis straight A right parenthesis equals straight f left parenthesis straight B right parenthesis
rightwards double arrow space space open vertical bar straight A close vertical bar equals open vertical bar straight B close vertical bar
but space we space can space not space say space that space straight A equals straight B.
Hence comma space it space is space not space one minus one.
As space straight A not equal to straight B space but space space open vertical bar straight A close vertical bar equals open vertical bar straight B close vertical bar
Function space is space onto. end style

Question 10

begin mathsize 12px style The space function space straight f space colon space left square bracket 0 comma space infinity right parenthesis space rightwards arrow space straight R space given space by space straight f left parenthesis straight x right parenthesis space equals space fraction numerator straight x over denominator straight x plus 1 end fraction is end style

 

(a) one-one and onto

(b) one-one but not onto

(c) onto but not one-one

(d) neither one-one nor onto

Solution 10

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space is space straight f left parenthesis straight x right parenthesis equals fraction numerator straight x over denominator straight x plus 1 end fraction space on space straight f colon left square bracket 0 comma infinity right parenthesis rightwards arrow straight R
If space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight y right parenthesis
rightwards double arrow fraction numerator straight x over denominator straight x plus 1 end fraction equals fraction numerator straight y over denominator straight y plus 1 end fraction
rightwards double arrow xy plus straight x equals xy plus straight y
rightwards double arrow straight x equals straight y
Hence space straight f space is space one minus one.
If space straight y equals straight f left parenthesis straight x right parenthesis
straight y equals fraction numerator straight x over denominator straight x plus 1 end fraction
xy plus straight y equals straight x
xy minus straight x equals negative straight y
straight x left parenthesis straight y minus 1 right parenthesis equals negative straight y
straight x equals fraction numerator negative straight y over denominator straight y minus 1 end fraction not equal to straight f left parenthesis straight x right parenthesis
It space is space not space onto.

end style

Question 11

The range of the function f (x) = 7-xPx-3 is

 

(a) {1, 2, 3, 4, 5}

(b) {1, 2, 3, 4, 5, 6}

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

(d) {1, 2, 3}

Solution 11

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space function space is space straight f space left parenthesis straight x right parenthesis space equals space large P presuperscript 7 minus straight x end presuperscript subscript straight x minus 3 end subscript
Here comma space 7 minus straight x greater or equal than straight x minus 3
rightwards double arrow 10 greater or equal than 2 straight x
rightwards double arrow straight x less or equal than 5
domain space equals space open curly brackets 3 comma 4 comma 5 close curly brackets
Range equals left curly bracket large P presuperscript 4 subscript 03 comma space large P presuperscript 3 subscript 1 comma space large P presuperscript 2 subscript 2 large }= open curly brackets 1 comma 3 comma 2 close curly brackets


end style

Question 12

A function f from the set of natural numbers to integers defined by

begin mathsize 12px style straight f open parentheses straight n close parentheses equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight n minus 1 over denominator 2 end fraction comma space when space straight n space is space odd end cell row cell negative straight n over 2 comma space when space straight n space is space even end cell end table close end style

is

 

(a) neither one-one nor onto

(b) one-one but not onto

(c) onto but not one-one

(d) one-one and onto both

Solution 12

begin mathsize 12px style Correct space option colon left parenthesis straight d right parenthesis
Given space function space is space
straight f left parenthesis straight n right parenthesis equals fraction numerator straight n minus 1 over denominator 2 end fraction space space space space space space space space space for space straight n space is space odd
space space space space space space space equals negative straight n over 2 space space space space space space space space space space space space for space straight n space is space even
For space straight n space is space odd comma
If space straight f left parenthesis straight n right parenthesis equals straight f left parenthesis straight m right parenthesis space then space
fraction numerator straight n minus 1 over denominator 2 end fraction space equals space fraction numerator straight m minus 1 over denominator 2 end fraction
rightwards double arrow straight n equals straight m
Also comma space for space straight n space is space even space if space straight f left parenthesis straight n right parenthesis equals straight f left parenthesis straight m right parenthesis space then space straight n equals straight m
Hence comma space straight f space is space one minus one.
Also comma space each space element space of space straight y space is space associated space with space at space least space one space element space of space straight x comma space
straight f space is space onto.
end style

Question 13

Let f be an injective map with domain {x, y, z} and range {1, 2, 3} 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

 

(a) x

(b) y

(c) z

(d) none of these

Solution 13

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space that space straight f space be space an space injective space map space with space domain space left curly bracket straight x comma space straight y comma space straight z right curly bracket space and space range space left curly bracket 1 comma space 2 comma space 3 right curly bracket space
straight f left parenthesis straight x right parenthesis space equals space 1 comma space straight f left parenthesis straight y right parenthesis space not equal to space 1 comma space straight f left parenthesis straight z right parenthesis space not equal to space 2.
As space straight f left parenthesis straight x right parenthesis equals 1 space rightwards double arrow straight f to the power of negative 1 end exponent left parenthesis 1 right parenthesis equals straight y end style

Question 14

Which of the following functions from Z to itself are bijections?

 

(a) f(x) = x3

(b)  f(x) = x + 2

(c) f(x) = 2x + 1

(d) f(x) = x2 + x

Solution 14

begin mathsize 12px style Correct space option colon left parenthesis straight b right parenthesis
Option space left parenthesis straight a right parenthesis colon space function space is space not space one minus one
Option space left parenthesis straight b right parenthesis colon space one minus one space and space onto space rightwards double arrow Bijective
Option space left parenthesis straight c right parenthesis colon space One minus one space but space not space onto
Option space left parenthesis straight d right parenthesis colon not space one minus one end style

Question 15

Which of the following from A = {x : -1 ≤ x ≤ 1} to itself are bijections?

 

begin mathsize 12px style open parentheses straight a close parentheses space straight f open parentheses straight x close parentheses equals straight x over 2
open parentheses straight b close parentheses space straight g open parentheses straight x close parentheses equals sin open parentheses nx over 2 close parentheses
open parentheses straight c close parentheses space straight h open parentheses straight x close parentheses space equals space open vertical bar straight x close vertical bar
open parentheses straight d close parentheses space straight k open parentheses straight x close parentheses equals straight x squared end style

Solution 15

begin mathsize 12px style
end stylebegin mathsize 12px style
end styleError converting from MathML to accessible text.

Question 16

Let A = {x : -1 ≤ x ≤ 1} and f; A → A such that f(x) = x|x|, then f is

 

(a) a bijection 

(b) injective but not surjective

(c) surjective but not injective

(d) neither injective nor surjective

Solution 16

begin mathsize 12px style Corect space option colon space left parenthesis straight a right parenthesis
Given space function space is space straight A space equals space left curly bracket straight x space colon space minus 1 space less or equal than space straight x space less or equal than space 1 right curly bracket space and space straight f semicolon space straight A space rightwards arrow space straight A space such space that space straight f left parenthesis straight x right parenthesis space equals space straight x vertical line straight x vertical line
For space the space mod space function space we space have space to space check space three space cases space as space straight x less than 0 comma space straight x equals 0 comma space straight x greater than 0.
For space example comma
straight x less than 0
straight f left parenthesis straight x right parenthesis equals straight x open vertical bar straight x close vertical bar less than 0
open vertical bar straight x close vertical bar equals negative straight x
straight y equals negative straight x squared
straight x equals negative square root of negative straight y end root space which space is space not space posible space for space straight x greater than 0
Hence comma space straight f space is space onto.
rightwards double arrow straight f space is space bijection.

end style

Question 17

begin mathsize 12px style if space the space function space straight f colon straight R space rightwards arrow straight A space given space by space straight f open parentheses straight x close parentheses equals fraction numerator straight x squared over denominator straight x squared space plus 1 end fraction is space straight a space surjection comma space then space straight A space equals end style

 

(a) R

(b) [0, 1]

(c) (0, 1]

(d) [0, 1)

Solution 17

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight f left parenthesis straight x right parenthesis equals fraction numerator straight x squared over denominator straight x squared plus 1 end fraction space is space straight a space surjection.
straight f left parenthesis straight x right parenthesis equals straight y
rightwards double arrow straight y equals fraction numerator straight x squared over denominator straight x squared plus 1 end fraction
rightwards double arrow straight y open parentheses straight x squared plus 1 close parentheses equals straight x squared
rightwards double arrow yx squared plus straight y equals straight x squared
rightwards double arrow straight y equals straight x squared minus yx squared
rightwards double arrow straight y equals straight x squared open parentheses 1 minus straight y close parentheses
rightwards double arrow straight x squared equals fraction numerator straight y over denominator 1 minus straight y end fraction
rightwards double arrow straight x equals square root of fraction numerator straight y over denominator 1 minus straight y end fraction end root
Here comma space fraction numerator straight y over denominator 1 minus straight y end fraction greater or equal than 0
straight y element of left square bracket 0 comma 1 right parenthesis end style

Question 18

if a function f : [2, ∞) → B defined by f(x) = x2 - 4x + 5 is a bijection, then B = 

 

(a) R

(b) [1, ∞)

(c) [4, ∞)

(d) [5, ∞)

Solution 18

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space straight f colon left square bracket 2 comma infinity right parenthesis rightwards arrow straight B space defined space by space straight f left parenthesis straight x right parenthesis equals straight x squared space minus space 4 straight x space plus space 5 space is space straight a space bijection.
From space above space we space will space put space straight x equals 2 space in space straight f left parenthesis straight x right parenthesis
space straight f left parenthesis 2 right parenthesis equals 2 squared space minus space 4 cross times 2 space plus space 5 equals 1
Hence comma space straight B equals left square bracket 1 comma infinity right parenthesis end style

Question 19

The function f : R → R defined by f(x) = (x - 1)(x - 2)(x - 3) is

 

(a) one-one but not onto

(b) onto but not one-one

(c) both one and onto

(d) neither one-one nor onto

Solution 19

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space is space straight f left parenthesis straight x right parenthesis equals left parenthesis straight x space minus space 1 right parenthesis left parenthesis straight x space minus space 2 right parenthesis left parenthesis straight x space minus space 3 right parenthesis
If space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight y right parenthesis space then space
left parenthesis straight x space minus space 1 right parenthesis left parenthesis straight x space minus space 2 right parenthesis left parenthesis straight x space minus space 3 right parenthesis equals left parenthesis straight y space minus space 1 right parenthesis left parenthesis straight y space minus space 2 right parenthesis left parenthesis straight y minus space 3 right parenthesis
rightwards double arrow straight f left parenthesis 1 right parenthesis equals straight f left parenthesis 2 right parenthesis equals straight f left parenthesis 3 right parenthesis equals 0
It space is space not space one minus one.
straight y equals straight f left parenthesis straight x right parenthesis
straight x element of straight R space also space straight y element of straight R space hence space straight f space is space onto. end style

Question 20

begin mathsize 12px style The space function space straight f space colon space open square brackets fraction numerator negative 1 over denominator 2 end fraction comma 1 half close square brackets rightwards arrow open square brackets fraction numerator negative straight pi over denominator 2 end fraction comma straight pi over 2 close square brackets space defined space by space straight f open parentheses straight x close parentheses equals space sin to the power of negative 1 end exponent open parentheses 3 straight x space minus space 4 straight x cubed close parentheses space is end style

 

(a) bijection

(b) injection but not a surjection

(c) surjection but not an injection

(d) neither an injection nor a surjection

Solution 20

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
Given space function space is space straight f left parenthesis straight x right parenthesis equals sin to the power of negative 1 end exponent left parenthesis 3 straight x minus 4 straight x cubed right parenthesis
straight x equals sin space straight theta
straight f left parenthesis straight x right parenthesis equals sin to the power of negative 1 end exponent left parenthesis 3 sinθ minus 4 sin cubed straight theta right parenthesis
straight f left parenthesis straight x right parenthesis equals sin to the power of negative 1 end exponent left parenthesis sin 3 straight theta right parenthesis
straight f left parenthesis straight x right parenthesis equals 3 straight theta
straight f left parenthesis straight x right parenthesis equals 3 sin to the power of negative 1 end exponent straight x
If space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight y right parenthesis space then
3 sin to the power of negative 1 end exponent straight x equals 3 sin to the power of negative 1 end exponent straight y
straight x equals straight y
straight f space is space one minus one.
straight y equals 3 sin to the power of negative 1 end exponent straight x
straight y over 3 equals sin to the power of negative 1 end exponent straight x
sin open parentheses straight y over 3 close parentheses equals straight x
straight F space is space onto space as space straight x element of straight R space and space straight y element of straight R.
straight f space is space bijection.
end style

Question 21

begin mathsize 12px style let space straight f space colon space straight R rightwards arrow straight R space be space straight a space function space defined space by space straight f left parenthesis straight x right parenthesis space equals fraction numerator straight e to the power of open vertical bar straight x close vertical bar end exponent minus straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x plus straight e to the power of negative straight x end exponent end fraction. Then comma end style

 

(a) f is a bijection

(b) f is an injection only

(c) f is surjection on only

(d) f is neither an injection nor a surjection

Solution 21

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space function space is space straight f left parenthesis straight x right parenthesis equals fraction numerator straight e to the power of open vertical bar straight x close vertical bar end exponent minus straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x plus straight e to the power of negative straight x end exponent end fraction
Here comma space straight e to the power of open vertical bar straight x close vertical bar end exponent space is space always space positive space whether space straight x space is space negative space or space positive.
So comma space we space will space get space same space values space of space straight f left parenthesis straight x right parenthesis space for space different space values space of space straight x.
Hence comma space it space is space not space one minus one.
Also comma space by space definition space it space is space not space surjective. end stylebegin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space function space is space straight f left parenthesis straight x right parenthesis equals fraction numerator straight e to the power of open vertical bar straight x close vertical bar end exponent minus straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x plus straight e to the power of negative straight x end exponent end fraction
Here comma space straight e to the power of open vertical bar straight x close vertical bar end exponent space is space always space positive space whether space straight x space is space negative space or space positive.
So comma space we space will space get space same space values space of space straight f left parenthesis straight x right parenthesis space for space different space values space of space straight x.
Hence comma space it space is space not space one minus one.
Also comma space by space definition space it space is space not space surjective. end style

Question 22

begin mathsize 12px style Let space straight f space colon space straight R space minus space open curly brackets straight n close curly brackets space rightwards arrow straight R space be space straight a space function space defined space by space straight f left parenthesis straight x right parenthesis equals fraction numerator straight x minus straight m over denominator straight x minus straight n end fraction comma where space straight m not equal to straight n. space Then comma end style

 

(a) f is one-one onto

(b) f is one-one into

(c) f is many one onto

(d) f is many one into

Solution 22

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space straight f colon straight R minus open curly brackets straight n close curly brackets rightwards arrow straight R space be space straight a space function space defined space by space straight f left parenthesis straight x right parenthesis equals fraction numerator straight x minus straight m over denominator straight x minus straight n end fraction comma space straight m not equal to straight n
If space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight y right parenthesis space then
fraction numerator straight x minus straight m over denominator straight x minus straight n end fraction equals fraction numerator straight y minus straight m over denominator straight y minus straight n end fraction
rightwards double arrow open parentheses straight x minus straight m close parentheses open parentheses straight y minus straight n close parentheses equals open parentheses straight y minus straight m close parentheses open parentheses straight x minus straight n close parentheses
After space solving space this space we space will space get space straight x equals straight y
Hence comma space it space is space one minus one.
straight f left parenthesis straight x right parenthesis equals fraction numerator straight x minus straight m over denominator straight x minus straight n end fraction comma space straight m not equal to straight n
straight y equals fraction numerator straight x minus straight m over denominator straight x minus straight n end fraction
rightwards double arrow straight y open parentheses straight x minus straight n close parentheses equals straight x minus straight m
rightwards double arrow yx minus ny equals straight x minus straight m
rightwards double arrow yx minus straight x equals ny minus straight m
rightwards double arrow straight x open parentheses straight y minus 1 close parentheses equals ny minus straight m
rightwards double arrow straight x equals fraction numerator ny minus straight m over denominator straight y minus 1 end fraction
Here comma space for space straight y equals 1 space we space can space not space define space straight x.
Hence comma space it space is space not space onto.
end style

Question 23

begin mathsize 12px style Let space straight f colon space straight R space rightwards arrow space straight R space be space straight a space function space defined space by space straight f left parenthesis straight x right parenthesis space equals space fraction numerator straight x squared minus 8 over denominator straight x squared plus 2 end fraction. space Then comma space straight f space is end style

 

(a) one-one but not onto 

(b) one-one and onto

(c) onto but not one-one

(d) neither one-one nor onto

Solution 23

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space function space is space straight f colon straight R rightwards arrow straight R space be space straight a space function space straight f left parenthesis straight x right parenthesis equals fraction numerator straight x squared minus 8 over denominator straight x squared plus 2 end fraction
Here comma space we space can space see space that space for space negative space and space positive space values space of space straight x space we space will space get space same space values space of space straight f left parenthesis straight x right parenthesis.
Hence comma space it space is space not space one minus one.
straight y equals straight f left parenthesis straight x right parenthesis
straight y equals fraction numerator straight x squared minus 8 over denominator straight x squared plus 2 end fraction
straight y open parentheses straight x squared plus 2 close parentheses equals straight x squared minus 8
After space solving space this space we space will space get space straight x equals square root of fraction numerator 2 straight y plus 1 over denominator 1 minus straight y end fraction end root
Here comma space for space straight y equals 1 space we space can space not space define space straight x space hence space it space is space not space onto. end style

Question 24

begin mathsize 12px style straight f colon space straight R rightwards arrow straight R space is space defined space by space straight f left parenthesis straight x right parenthesis equals fraction numerator straight e to the power of straight x squared end exponent minus straight e to the power of negative straight x squared end exponent over denominator straight e to the power of straight x squared end exponent plus straight e to the power of negative straight x squared end exponent end fraction space is end style

 

(a) one-one but not onto

(b) many-one but onto

(c) one-one and onto

(d) neither one-one nor onto

Solution 24

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
Given space function space is space straight f colon space straight R rightwards arrow straight R space is space defined space by space straight f left parenthesis straight x right parenthesis equals fraction numerator straight e to the power of straight x squared end exponent minus straight e to the power of negative straight x squared end exponent over denominator straight e to the power of straight x squared end exponent plus straight e to the power of negative straight x squared end exponent end fraction space
As space straight x squared space is space always space positive space we space will space get space same space values space of space straight f left parenthesis straight x right parenthesis space for space different space values space of space straight x.
hence comma space it space is space not space one minus one.
By space definition space of space onto space each space element space of space straight y space is space not space mapped space to space by space at space least space one space element space of space straight x.
Hence comma space it space is space not space onto.
end style

Question 25

The function f ; R → R, f(x) = x2 is

 

(a) injective but not surjective

(b) surjective but not injective

(c) injective as well as surjective

(d) neither injective nor surjective

Solution 25

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space function space is space straight f colon straight R rightwards arrow straight R comma space straight f left parenthesis straight x right parenthesis equals straight x squared
If space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight y right parenthesis space then
straight x squared equals straight y squared
rightwards double arrow straight x equals plus-or-minus straight y
Hence comma space it space is space not space one minus one space or space injective.
straight f left parenthesis straight x right parenthesis equals straight y
straight y equals straight x squared
straight x equals plus-or-minus square root of straight y
But space co minus domain space is space straight R.
hence comma space it space is space not space onto space or space surjective.

end style

Question 26

begin mathsize 12px style straight A space function space straight f space from space the space set space of space natural space numbers space to space the space set space of space integers space defined space by
straight f left parenthesis straight n right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight n minus 1 over denominator 2 end fraction comma space when space straight n space is space odd end cell row cell negative straight n over 2 comma space when space straight n space is space even end cell end table close end style

 

(a) neither one-one nor onto

(b) one-one but not onto

(c) onto but not one-one

(d) one-one and onto both

Solution 26

Error converting from MathML to accessible text.

Question 27

Which of the following functions from A = {x ε R : - 1 ≤ x  ≤ 1} to itself are bijections?

 

begin mathsize 12px style open parentheses straight a close parentheses space straight f open parentheses straight x close parentheses equals open vertical bar straight x close vertical bar
open parentheses straight b close parentheses straight f open parentheses straight x close parentheses equals sin πx over 2
open parentheses straight c close parentheses space straight f open parentheses straight x close parentheses equals space sin πx over 4
open parentheses straight d close parentheses space space none space of space these end style

Solution 27

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
space It space is space bijective space as space it space is space one minus one space and space onto. space Range space of space straight f left parenthesis straight x right parenthesis space element of open square brackets negative 1 comma 1 close square brackets end style

Question 28

begin mathsize 12px style Let space straight f space colon space straight Z space rightwards arrow space straight Z space be space given space by space straight f open parentheses straight x close parentheses space equals space open curly brackets table attributes columnalign left end attributes row cell straight x over 2 comma if space straight x space is space even end cell row cell 0 comma space if space straight x space is space odd end cell end table close. space
Then comma space straight f space is end style

 

(a) onto but not one-one

(b) one-one but not onto

(c) one-one and onto

(d) neither one-one nor onto

Solution 28

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
Given space function space is space space space
straight f left parenthesis straight x right parenthesis equals straight x over 2 space space space space space space if space straight x space is space even
space space space space space space space equals 0 space space space space space space space space if space straight x space is space odd
For space straight f left parenthesis 3 right parenthesis equals 0 space and space straight f left parenthesis 4 right parenthesis equals 0
rightwards double arrow straight f left parenthesis 3 right parenthesis equals straight f left parenthesis 4 right parenthesis
But space 3 not equal to 4
Hence comma space it space is space not space one minus one.
straight x element of straight R space rightwards double arrow straight y element of straight R
Here comma space Domain equals range space of space straight f
Hence comma space it space is space onto.
end style

Question 29

The function f ; R → R defined by f (x) = 6x + 6|x| is

 

(a) one-one and onto

(b) many one and onto

(c) one-one and into

(d) many one and into

Solution 29

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space function space is space straight f space colon space straight R space rightwards arrow space straight R space defined space by space straight f space left parenthesis straight x right parenthesis space equals space 6 straight x space plus space 6 vertical line straight x vertical line
If space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight y right parenthesis space then
6 straight x space plus space 6 vertical line straight x vertical line equals 6 straight y space plus space 6 vertical line straight y vertical line
straight x plus open vertical bar straight x close vertical bar equals straight y plus open vertical bar straight y close vertical bar
Here comma space open vertical bar straight x close vertical bar equals open vertical bar straight y close vertical bar
straight x equals plus-or-minus straight y
Hence comma space it space is space many space one.
Here comma space domain not equal to co minus domain
Hence comma space it space is space into.

end style

Question 30

Let f(x) = x2 and g(x) = 2x. Then the solution set of the equation fog (x) = gof (x) is 

 

(a) R

(b) {0}

(c) {0, 2}

(d) none of these

Solution 30

begin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
Given space fnctions space are space straight f left parenthesis straight x right parenthesis space equals space straight x squared space and space straight g left parenthesis straight x right parenthesis space equals space 2 to the power of straight x.
Also comma space straight f ring operator straight g left parenthesis straight x right parenthesis equals straight g ring operator straight f left parenthesis straight x right parenthesis
straight f open square brackets straight g open parentheses straight x close parentheses close square brackets equals straight g open square brackets straight f open parentheses straight x close parentheses close square brackets
rightwards double arrow straight f open parentheses 2 to the power of straight x close parentheses equals straight g open parentheses straight x squared close parentheses
rightwards double arrow open parentheses 2 to the power of straight x close parentheses squared equals 2 to the power of straight x squared end exponent
rightwards double arrow 2 to the power of 2 straight x end exponent equals 2 to the power of straight x squared end exponent
rightwards double arrow 2 straight x equals straight x squared
rightwards double arrow straight x squared minus 2 straight x equals 0
rightwards double arrow straight x open parentheses straight x minus 2 close parentheses equals 0
rightwards double arrow straight x equals 0 space or space straight x equals 2 end style

Question 31

if f : R → R is given by f(x) = 3x - 5, then f-1(x)

 

begin mathsize 12px style open parentheses straight a close parentheses space is space given space by space fraction numerator 1 over denominator 3 straight x minus 5 end fraction
open parentheses straight b close parentheses space is space given space by space fraction numerator straight x plus 5 over denominator 3 end fraction
open parentheses straight c close parentheses space does space not space exist space because space straight f space is space not space one minus one
open parentheses straight d close parentheses space does space not space exist space because space straight f space is space not space into end style

Solution 31

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space is space straight f space colon space straight R space rightwards arrow space straight R space is space given space by space straight f left parenthesis straight x right parenthesis space equals space 3 straight x space minus space 5
To space find space straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis comma
straight y equals straight f left parenthesis straight x right parenthesis
rightwards double arrow straight y equals 3 straight x minus 5
rightwards double arrow straight y plus 5 equals 3 straight x
rightwards double arrow straight y equals fraction numerator straight y plus 5 over denominator 3 end fraction
Hence comma space straight f to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator straight x plus 5 over denominator 3 end fraction
end style

Question 32

begin mathsize 12px style If space straight g open parentheses straight f open parentheses straight x close parentheses close parentheses equals open vertical bar sin space straight x close vertical bar space and space straight f open parentheses straight g open parentheses straight x close parentheses close parentheses equals space open parentheses sin square root of straight x close parentheses squared comma space then

open parentheses straight a close parentheses space straight f open parentheses straight x close parentheses space equals space sin 2 straight x comma space straight g open parentheses straight x close parentheses equals square root of straight x
open parentheses straight b close parentheses space straight f open parentheses straight x close parentheses space equals space sin space straight x comma space straight g open parentheses straight x close parentheses equals open vertical bar straight x close vertical bar
open parentheses straight c close parentheses space straight f open parentheses straight x close parentheses space equals space straight x squared comma space straight g open parentheses straight x close parentheses equals sin square root of straight x
open parentheses straight d close parentheses space straight f space and space straight g space cannot space be space determind. space end style

Solution 32

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
straight f left parenthesis straight x right parenthesis equals sin squared straight x space and space straight g left parenthesis straight x right parenthesis equals square root of straight x
straight f open square brackets straight g open parentheses straight x close parentheses close square brackets equals straight f open square brackets square root of straight x close square brackets
space space space space space space space space space space space space space equals space sin squared square root of straight x space space space
space space space space space space space space space space space space space equals open parentheses space sin square root of straight x space close parentheses squared
end style

Question 33

begin mathsize 12px style The space inverse space of space the space function space straight f colon space straight R rightwards arrow open curly brackets straight x element of straight R semicolon space straight x less than 1 close curly brackets space given space by space straight f open parentheses straight x close parentheses space equals space fraction numerator straight e to the power of straight x minus straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x minus straight e to the power of negative straight x end exponent end fraction comma space is
left parenthesis straight a right parenthesis space 1 half log fraction numerator 1 plus straight x over denominator 1 minus straight x end fraction
left parenthesis straight b right parenthesis space 1 half log fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction
left parenthesis straight c right parenthesis space 1 half log fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction
left parenthesis straight d right parenthesis space none space of space these end style

Solution 33

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
Given space function space is space straight f colon space straight R rightwards arrow open curly brackets straight x element of straight R colon straight x less than 1 close curly brackets space given space by space straight f left parenthesis straight x right parenthesis equals fraction numerator straight e to the power of straight x minus straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x plus straight e to the power of negative straight x end exponent end fraction
straight y equals straight f left parenthesis straight x right parenthesis
straight y equals fraction numerator straight e to the power of straight x minus straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x plus straight e to the power of negative straight x end exponent end fraction
rightwards double arrow straight y equals fraction numerator straight e to the power of 2 straight x end exponent minus 1 over denominator straight e to the power of 2 straight x end exponent plus 1 end fraction
rightwards double arrow straight y open parentheses straight e to the power of 2 straight x end exponent plus 1 close parentheses equals straight e to the power of 2 straight x end exponent minus 1
rightwards double arrow ye to the power of 2 straight x end exponent plus straight y equals straight e to the power of 2 straight x end exponent minus 1
rightwards double arrow straight y plus 1 equals straight e to the power of 2 straight x end exponent minus ye to the power of 2 straight x end exponent
rightwards double arrow straight y plus 1 equals straight e to the power of 2 straight x end exponent open parentheses 1 minus straight y close parentheses
rightwards double arrow fraction numerator straight y plus 1 over denominator 1 minus straight y end fraction equals straight e to the power of 2 straight x end exponent
rightwards double arrow log space open parentheses fraction numerator straight y plus 1 over denominator 1 minus straight y end fraction close parentheses equals 2 straight x
rightwards double arrow straight x equals 1 half log space open parentheses fraction numerator straight y plus 1 over denominator 1 minus straight y end fraction close parentheses
rightwards double arrow straight y equals 1 half log space open parentheses fraction numerator straight x plus 1 over denominator 1 minus straight x end fraction close parentheses


end style

Question 34

Let A = {x ε R : x ≥ 1}. The inverse of the function f ; A → A given by f(x) = 2x(x - 1), is 

 

begin mathsize 12px style open parentheses straight a close parentheses space open parentheses 1 half close parentheses to the power of straight x open parentheses straight x minus 1 close parentheses end exponent
open parentheses straight b close parentheses space 1 half open curly brackets 1 plus square root of 1 plus 4 log subscript 2 straight x end root close curly brackets
open parentheses straight c close parentheses space 1 half open curly brackets 1 minus square root of 1 plus 4 log subscript 2 straight x end root close curly brackets
open parentheses straight d close parentheses space not space defined space end style

Solution 34

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space is space straight A space equals space left curly bracket straight x space straight epsilon space straight R space colon space straight x space greater or equal than space 1 right curly bracket.
space The space inverse space of space the space function space straight f space colon space straight A space rightwards arrow space straight A space given space by space straight f left parenthesis straight x right parenthesis space equals space 2 to the power of straight x left parenthesis straight x space minus space 1 right parenthesis end exponent
straight f left parenthesis straight x right parenthesis equals straight y
space 2 to the power of straight x left parenthesis straight x space minus space 1 right parenthesis end exponent equals straight y
straight x left parenthesis straight x minus 1 right parenthesis equals log subscript 2 straight y
straight x squared plus straight x equals log subscript 2 straight y
straight x squared plus straight x plus 1 fourth equals log subscript 2 straight y plus 1 fourth
open parentheses straight x minus 1 half close parentheses squared equals fraction numerator 4 log subscript 2 straight y plus 1 over denominator 4 end fraction
straight x minus 1 half equals plus-or-minus square root of fraction numerator 4 log subscript 2 straight y plus 1 over denominator 4 end fraction end root
straight x equals 1 half plus-or-minus square root of fraction numerator 4 log subscript 2 straight y plus 1 over denominator 4 end fraction end root
straight x equals 1 half plus square root of fraction numerator 4 log subscript 2 straight y plus 1 over denominator 4 end fraction end root
straight f to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator 1 plus square root of 4 log subscript 2 straight y plus 1 end root over denominator 2 end fraction
end style

Question 35

Let A = {x ε R : x ≤ 1} and f : A → 1 A be defined as f (x) = x(2 - x). Then, f-1 (x) is

 

begin mathsize 12px style open parentheses straight a close parentheses space 1 space plus space square root of 1 space minus straight x end root
open parentheses straight b close parentheses space 1 space minus space square root of 1 space minus straight x end root
open parentheses straight c close parentheses space square root of 1 minus straight x end root
open parentheses straight d close parentheses space 1 plus-or-minus square root of 1 minus straight x end root end style

Solution 35

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space is space straight A space equals space left curly bracket straight x space straight epsilon space straight R space colon space straight x space less or equal than space 1 right curly bracket space and space straight f space colon space straight A space rightwards arrow space 1 space straight A space be space defined space as
space straight f space left parenthesis straight x right parenthesis space equals space straight x left parenthesis 2 space minus space straight x right parenthesis. space Then comma space straight f to the power of negative 1 end exponent space left parenthesis straight x right parenthesis
Let comma space straight f left parenthesis straight y right parenthesis equals straight x
straight y left parenthesis 2 minus straight y right parenthesis equals straight x
2 straight y minus straight y squared equals straight x
straight y squared minus 2 straight y equals negative straight x
straight y squared minus 2 straight y plus 1 equals 1 minus straight x
open parentheses straight y minus 1 close parentheses squared equals 1 minus straight x
straight y minus 1 equals plus-or-minus square root of 1 minus straight x end root
straight y equals 1 plus-or-minus square root of 1 minus straight x end root
Given space that space straight y less or equal than 1
rightwards double arrow straight y equals 1 minus square root of 1 minus straight x end root end styleANSWER PENDING

Question 36

begin mathsize 12px style Let space straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator 1 minus straight x end fraction. Then comma space open curly brackets straight f ring operator open parentheses straight f ring operator space straight f close parentheses close curly brackets open parentheses straight x close parentheses

open parentheses straight a close parentheses space straight x space for space all space straight x space element of space straight R
open parentheses straight b close parentheses space straight x space for space all space straight x space element of space straight R space minus space left curly bracket 1 right curly bracket
open parentheses straight c close parentheses space straight x space for space all space straight x space element of space straight R space minus space left curly bracket 0 comma space 1 right curly bracket
open parentheses straight d close parentheses space none space of space these end style

Solution 36

begin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
straight f ring operator straight f ring operator straight f left parenthesis straight x right parenthesis equals straight f ring operator straight f open parentheses fraction numerator 1 over denominator 1 minus straight x end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses straight x not equal to 1 close parentheses
space space space space space space space space space space space space space space space space space equals straight f open parentheses fraction numerator 1 over denominator 1 minus fraction numerator 1 over denominator 1 minus straight x end fraction end fraction close parentheses
space space space space space space space space space space space space space space space space space equals straight f open parentheses fraction numerator 1 minus straight x over denominator negative straight x end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open parentheses straight x not equal to 0 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
space space space space space space space space space space space space space space space space space equals straight f open parentheses fraction numerator straight x minus 1 over denominator straight x end fraction close parentheses
space space space space space space space space space space space space space space space space space equals fraction numerator 1 over denominator 1 minus fraction numerator straight x minus 1 over denominator straight x end fraction end fraction
space space space space space space space space space space space space space space space space space equals straight x
Hence comma space option space is space straight x space for space all space straight x element of straight R minus open curly brackets 0 comma 1 close curly brackets. end style

Question 37

If the function f: R→R be such that f(x) = x - [x], where [x] denotes the greatest integer less than or equal to x, then f-1(x) is 

 

begin mathsize 12px style open parentheses straight a close parentheses space fraction numerator 1 over denominator straight x minus left square bracket straight x right square bracket end fraction
open parentheses straight b close parentheses space left square bracket straight x right square bracket space minus space straight x
open parentheses straight c close parentheses space not space defined
open parentheses straight d close parentheses space none space of space these end style

Solution 37

Error converting from MathML to accessible text.ANSWER PENDING

Question 38

begin mathsize 12px style If space straight f colon space left square bracket 1 comma space infinity right parenthesis space is space given space by space straight f left parenthesis straight x right parenthesis space equals space straight x plus space 1 over straight x comma space then space straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals.

open parentheses straight a close parentheses space fraction numerator straight x plus square root of straight x squared space minus space 4 end root over denominator 2 end fraction
open parentheses straight b close parentheses space fraction numerator straight x over denominator 1 plus straight x squared end fraction
open parentheses straight c close parentheses space fraction numerator straight x minus square root of straight x squared minus 4 end root over denominator 2 end fraction
open parentheses straight d close parentheses space 1 plus space square root of straight x squared minus 4 end root end style

Solution 38

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
Given space that space straight f left parenthesis straight x right parenthesis equals straight x plus 1 over straight x
straight y equals straight x plus 1 over straight x
straight y equals fraction numerator straight x squared plus 1 over denominator straight x end fraction
xy equals straight x squared plus 1
straight x squared minus xy equals negative 1
straight x squared minus xy plus straight y squared over 4 equals straight y squared over 4 minus 1
open parentheses straight x minus straight y over 2 close parentheses squared equals fraction numerator straight y squared minus 4 over denominator 4 end fraction
straight x minus straight y over 2 equals square root of fraction numerator straight y squared minus 4 over denominator 4 end fraction end root
straight x minus straight y over 2 equals fraction numerator square root of straight y squared minus 4 end root over denominator 2 end fraction
straight x equals straight y over 2 plus fraction numerator square root of straight y squared minus 4 end root over denominator 2 end fraction
rightwards double arrow straight y equals straight x over 2 plus fraction numerator square root of straight x squared minus 4 end root over denominator 2 end fraction

end style

Question 39

Error converting from MathML to accessible text.

 

(a) x

(b) 1

(c) f(x)

(d) g(x)

Solution 39

Error converting from MathML to accessible text.

Question 40

begin mathsize 12px style Let space straight f left parenthesis straight x right parenthesis space equals space fraction numerator straight alpha space straight x over denominator straight x plus 1 end fraction comma space straight x not equal to negative 1. space then comma space for space what space value space of space straight alpha space is space straight f open parentheses straight f open parentheses straight x close parentheses close parentheses equals space straight x ?

open parentheses straight a close parentheses space square root of 2
open parentheses straight b close parentheses space minus square root of 2
open parentheses straight c close parentheses space 1
open parentheses straight d close parentheses space minus 1 end style

Solution 40

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis

Given space function space is space space straight f left parenthesis straight x right parenthesis space equals space fraction numerator straight alpha space straight x over denominator straight x plus 1 end fraction comma space straight x not equal to negative 1
Also space straight f left parenthesis straight f left parenthesis straight x right parenthesis right parenthesis equals straight x
straight f open parentheses fraction numerator straight alpha space straight x over denominator straight x plus 1 end fraction close parentheses equals straight x
fraction numerator straight alpha open parentheses fraction numerator straight alpha space straight x over denominator straight x plus 1 end fraction close parentheses over denominator fraction numerator straight alpha space straight x over denominator straight x plus 1 end fraction plus 1 end fraction equals straight x
fraction numerator straight alpha squared straight x over denominator αx plus straight x plus 1 end fraction equals straight x
straight alpha squared equals αx plus straight x plus 1
straight alpha squared equals open parentheses straight alpha plus 1 close parentheses straight x plus 1
Comparing space on space both space sides comma space
straight alpha plus 1 equals 0 rightwards double arrow straight alpha equals negative 1
end style

Question 41

The distinct linear functions which map [-1, 1] onto [0, 2] are

 

(a) f(x) = x + 1, g(x) = -x + 1

(b) f(x) = x - 1, g(x) = x + 1

(c) f(x) = - x - 1, g(x) = x - 1

(d) none of these

Solution 41

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
In space option space left parenthesis straight a right parenthesis
space straight f left parenthesis straight x right parenthesis space equals space straight x space plus space 1 comma space straight g left parenthesis straight x right parenthesis space equals space minus straight x space plus space 1
straight f left parenthesis negative 1 right parenthesis equals negative 1 plus 1 equals 0
straight g left parenthesis negative 1 right parenthesis equals negative open parentheses negative 1 close parentheses plus 1 equals 2
Also comma
straight f left parenthesis 1 right parenthesis equals 1 plus 1 equals 2
straight g left parenthesis 1 right parenthesis equals negative 1 plus 1 equals 0

end style

Question 42

Let f: [2, ∞) → X be defined by f(x) = 4x - x2. Then, f is invertible, if X=

 

(a) [2, ∞)

(b) ( -∞, 2]

(c) (-∞, 4]

(d) [4, ∞)

Solution 42

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
Given space function space is space straight f colon space left square bracket 2 comma space infinity right parenthesis space rightwards arrow space straight X space be space defined space by space straight f left parenthesis straight x right parenthesis space equals space 4 straight x space minus space straight x squared
straight y equals 4 straight x space minus space straight x squared
minus straight y equals straight x squared minus 4 straight x
4 minus straight y equals straight x squared minus 4 straight x plus 4
4 minus straight y equals open parentheses straight x minus 2 close parentheses squared
straight x minus 2 equals plus-or-minus square root of 4 minus straight y end root
straight x equals 2 plus-or-minus square root of 4 minus straight y end root
rightwards double arrow straight y equals 2 plus-or-minus square root of 4 minus straight x end root
Here comma space straight x space should space be space less than 4.
Hence comma space left parenthesis infinity comma 4 right square bracket. end styleANSWER PENDING

Question 43

begin mathsize 12px style If space straight f colon straight R space rightwards arrow open parentheses negative 1 comma space 1 close parentheses space is space defined space by space straight f open parentheses straight x close parentheses equals fraction numerator negative straight x open vertical bar straight x close vertical bar over denominator 1 plus straight x squared end fraction comma space then space straight f to the power of negative 1 end exponent open parentheses straight x close parentheses space equals end style

begin mathsize 12px style open parentheses straight a close parentheses space square root of fraction numerator open vertical bar straight x close vertical bar over denominator 1 minus open vertical bar straight x close vertical bar end fraction end root
open parentheses straight b close parentheses space minus Sgn open parentheses straight x close parentheses square root of fraction numerator open vertical bar straight x close vertical bar over denominator 1 minus open vertical bar straight x close vertical bar end fraction end root
open parentheses straight c close parentheses minus square root of fraction numerator straight x over denominator 1 minus straight x end fraction end root
open parentheses straight d close parentheses space none space of space these end style

Solution 43

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space is space straight f colon straight R space rightwards arrow open parentheses negative 1 comma space 1 close parentheses space is space defined space by space straight f open parentheses straight x close parentheses equals fraction numerator negative straight x open vertical bar straight x close vertical bar over denominator 1 plus straight x squared end fraction
Here comma space for space mod space function space we space will space have space to space consider space three space cases space as comma
straight x less than 0 comma space straight x equals 0 comma space straight x greater than 0
left parenthesis straight i right parenthesis space straight x less than 0 space rightwards double arrow open vertical bar straight x close vertical bar equals negative straight x
straight f left parenthesis open vertical bar blank close vertical bar straight x right parenthesis equals fraction numerator negative straight x open parentheses negative straight x close parentheses over denominator 1 plus straight x squared end fraction
straight y equals fraction numerator straight x squared over denominator 1 plus straight x squared end fraction
straight y open parentheses 1 plus straight x squared close parentheses equals straight x squared
straight y plus yx squared equals straight x squared
straight y equals straight x squared minus yx squared
straight y equals open parentheses 1 minus straight y close parentheses straight x squared
straight x squared equals fraction numerator straight y over denominator 1 minus straight y end fraction
straight x equals negative square root of fraction numerator straight y over denominator 1 minus straight y end fraction end root
rightwards double arrow straight x equals negative square root of fraction numerator open vertical bar straight y close vertical bar over denominator 1 minus open vertical bar straight y close vertical bar end fraction end root space space space space straight x less than 0
Also space you space can space check space for space the space cases space straight x equals 0 space and space straight x greater than 0 space that space straight x equals negative square root of fraction numerator open vertical bar straight y close vertical bar over denominator 1 minus open vertical bar straight y close vertical bar end fraction end root
minus sgn left parenthesis straight x right parenthesis square root of fraction numerator open vertical bar straight y close vertical bar over denominator 1 minus open vertical bar straight y close vertical bar end fraction end root end style

Question 44

Error converting from MathML to accessible text.

 

begin mathsize 12px style open parentheses straight a close parentheses space fogoh space open parentheses straight x close parentheses equals straight pi over 2
open parentheses straight b close parentheses space fogoh space open parentheses straight x close parentheses space equals space straight pi
open parentheses straight c close parentheses space hofog equals hogof
open parentheses straight d close parentheses space hofog not equal to hogof end style

Solution 44

begin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
straight h ring operator straight g ring operator straight f left parenthesis straight x right parenthesis
equals straight h left parenthesis straight f left parenthesis straight g left parenthesis straight x right parenthesis right parenthesis right parenthesis
equals straight h left parenthesis straight f left parenthesis left square bracket straight x right square bracket right parenthesis right parenthesis
equals straight h left parenthesis sin to the power of negative 1 end exponent left square bracket straight x right square bracket right parenthesis
equals 2 sin to the power of negative 1 end exponent left square bracket straight x right square bracket space
equals 2 cross times 0 equals 0
straight f left parenthesis straight x right parenthesis equals sin to the power of negative 1 end exponent straight x
straight h ring operator straight g ring operator straight f left parenthesis straight x right parenthesis equals straight h ring operator straight g ring operator left parenthesis straight x right parenthesis equals 0
end style

Question 45

begin mathsize 12px style If space straight g open parentheses straight x close parentheses space equals space straight x squared space plus space straight x space minus space 2 space and space 1 half g ring operator f open parentheses straight x close parentheses space equals space 2 straight x squared space minus space 5 straight x space plus space 2 comma space then space straight f open parentheses straight x close parentheses space is space equal space to space end style

(a) 2x - 3

(b) 2x + 3

(c) 2x2 + 3x + 1

(d) 2x2 - 3x - 1

Solution 45

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
Given space function space is space straight g open parentheses straight x close parentheses space equals space straight x squared space plus space straight x space minus space 2 space and space 1 half g ring operator f open parentheses straight x close parentheses space equals space 2 straight x squared space minus space 5 straight x space plus space 2
1 half g ring operator f open parentheses straight x close parentheses
equals 1 half g open parentheses 2 x minus 3 close parentheses
equals 1 half open square brackets open parentheses 2 x minus 3 close parentheses squared plus 2 x minus 3 minus 2 close square brackets
equals 1 half open square brackets 4 x squared minus 12 x plus 9 plus 2 x minus 5 close square brackets
equals 1 half open square brackets 4 x squared minus 10 x plus 4 close square brackets
equals 2 x squared minus 5 x plus 2
Hence comma space answer space is space straight g left parenthesis straight x right parenthesis equals 2 straight x minus 3 end style

Question 46

begin mathsize 12px style If space straight f open parentheses straight x close parentheses space equals space sin squared straight x space and space the space composite space function space straight g open parentheses straight f open parentheses straight x close parentheses close parentheses equals open vertical bar sin space straight x close vertical bar comma space then space straight g open parentheses straight x close parentheses space is space equal space to space

open parentheses straight a close parentheses space square root of straight x minus 1 end root
open parentheses straight b close parentheses space square root of straight x
open parentheses straight c close parentheses space square root of straight x plus 1 end root
open parentheses straight d close parentheses space minus square root of straight x end style

Solution 46

begin mathsize 12px style Correct space option colon left parenthesis straight b right parenthesis
Given space that space straight f open parentheses straight x close parentheses space equals space sin squared straight x space and space the space composite space function space straight g open parentheses straight f open parentheses straight x close parentheses close parentheses equals open vertical bar sin space straight x close vertical bar
We space will space do space it space using space trial space and space error space method.
If space we space take space straight g left parenthesis straight x right parenthesis equals negative square root of straight x space and space straight f left parenthesis straight x right parenthesis equals space sin squared straight x
straight g left parenthesis straight f left parenthesis straight x right parenthesis right parenthesis equals straight g left parenthesis sin squared straight x right parenthesis
space space space space space space space space space space space equals negative sinx space
which space contradicts space to space the space straight g open parentheses straight f open parentheses straight x close parentheses close parentheses equals open vertical bar sin space straight x close vertical bar
Hence comma space we space take space straight g left parenthesis straight x right parenthesis equals square root of straight x
straight g left parenthesis straight f left parenthesis straight x right parenthesis right parenthesis equals straight g left parenthesis sin squared straight x right parenthesis
space space space space space space space space space space space equals square root of sin squared straight x end root equals open vertical bar sinx close vertical bar
end style

Question 47

begin mathsize 12px style If space straight f colon space straight R space rightwards arrow space straight R space is space given space by space straight f left parenthesis straight x right parenthesis space equals space straight x cubed space plus space 3 comma space then space straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space is space equal space to

open parentheses straight a close parentheses space straight x to the power of 1 divided by 3 end exponent minus 3
open parentheses straight b close parentheses space straight x to the power of bevelled 1 third end exponent plus 3
left parenthesis straight c right parenthesis space open parentheses straight x space minus space 3 close parentheses to the power of bevelled 1 third end exponent
left parenthesis straight d right parenthesis space straight x plus 3 to the power of bevelled 1 third end exponent end style

Solution 47

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
Given space function space is space straight f colon space straight R space rightwards arrow space straight R space is space given space by space straight f left parenthesis straight x right parenthesis space equals space straight x cubed space plus space 3
straight y equals space straight x cubed space plus space 3
straight y minus 3 equals straight x cubed
cube root of straight y minus 3 end root equals straight x
rightwards double arrow straight y equals cube root of straight x minus 3 end root
rightwards double arrow straight y equals open parentheses straight x minus 3 close parentheses to the power of bevelled 1 third end exponent end style

Question 48

Let f(x) = x3 be a function with domain {0, 1, 2, 3}. Then domain of f-1 is

 

(a) {3, 2, 1, 0}

(b) {0, -1, -2, -3}

(c) {0, 1, 8, 27}

(d) {0, -1, -8, -27}

Solution 48

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
Given space function space is space straight f left parenthesis straight x right parenthesis space equals space straight x cubed space be space straight a space function space with space domain space left curly bracket 0 comma space 1 comma space 2 comma space 3 right curly bracket.
Range space equals left curly bracket 0 comma 1 cubed comma 2 cubed comma 3 cubed right curly bracket equals left curly bracket 0 comma 1 comma 8 comma 27 right curly bracket
straight f space can space be space written space as
left curly bracket left parenthesis 0 comma 0 right parenthesis comma space left parenthesis 1 comma 1 right parenthesis comma space left parenthesis 2 comma 8 right parenthesis comma space left parenthesis 3 comma 27 right parenthesis right curly bracket
Hence comma space straight f to the power of negative 1 end exponent space can space be space written space as
left curly bracket left parenthesis 0 comma 0 right parenthesis comma space left parenthesis 1 comma 1 right parenthesis comma space left parenthesis 8 comma 2 right parenthesis comma space left parenthesis 27 comma 3 right parenthesis right curly bracket
Domain space of space space straight f to the power of negative 1 end exponent space is space left curly bracket 0 comma 1 comma 8 comma 27 right curly bracket end style

Question 49

Let f : R → R be given by f(x) = x2 - 3. Then domain of f-1 is given by:

 

begin mathsize 12px style open parentheses straight a close parentheses space square root of straight x plus 3 end root
open parentheses straight b close parentheses space square root of straight x space plus space 3
open parentheses straight c close parentheses space straight x plus square root of 3
open parentheses straight d close parentheses space none space of space these end style

Solution 49

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
Given space function space is space space straight f space colon space straight R space rightwards arrow space straight R space be space given space by space straight f left parenthesis straight x right parenthesis space equals space straight x squared space minus space 3.
straight y equals straight x squared minus 3
straight y plus 3 equals straight x squared
straight x equals plus-or-minus square root of straight y plus 3 end root
rightwards double arrow straight y equals plus-or-minus square root of straight x plus 3 end root
end style

Question 50

begin mathsize 12px style Let space straight f colon straight R space rightwards arrow straight R space be space given space by space straight f open parentheses straight x close parentheses equals tanx. space Then comma space straight f to the power of negative 1 end exponent open parentheses 1 close parentheses space is
open parentheses straight a close parentheses space straight pi over 4
open parentheses straight b close parentheses space open curly brackets nπ plus straight pi over 4 colon straight n element of straight Z close curly brackets
left parenthesis straight c right parenthesis space does space not space exist
open parentheses straight d close parentheses space none space of space these end style

Solution 50

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
Given space function space is space straight f colon straight R space rightwards arrow straight R space be space given space by space straight f open parentheses straight x close parentheses equals tanx
straight y equals tanx
tan to the power of negative 1 end exponent straight x equals straight y
Hence comma space tan to the power of negative 1 end exponent open parentheses 1 close parentheses equals straight y
nπ plus straight pi over 4 semicolon space space straight n element of straight Z end style

Question 51

begin mathsize 12px style Let space straight f colon straight R space rightwards arrow space straight R space be space defined space as space straight f open parentheses straight x close parentheses equals open curly brackets table row cell 2 straight x comma space if space straight x space greater than space 3 end cell row cell straight x squared comma space if space 1 space less than space straight x space less or equal than space 3 end cell row cell 3 straight x comma space if space straight x space less or equal than space 1 end cell end table close
Then comma space find space straight f open parentheses negative 1 close parentheses plus straight f open parentheses 2 close parentheses plus straight f open parentheses 4 close parentheses

open parentheses straight a close parentheses space 9
open parentheses straight b close parentheses space 14
open parentheses straight c close parentheses space 5
open parentheses straight d close parentheses space none space of space these end style

Solution 51

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
Given space function space is space straight f colon straight R space rightwards arrow space straight R space be space defined space as space straight f open parentheses straight x close parentheses equals open curly brackets table row cell 2 straight x comma space if space straight x space greater than space 3 end cell row cell straight x squared comma space if space 1 space less than space straight x space less or equal than space 3 end cell row cell 3 straight x comma space if space straight x space less or equal than space 1 end cell end table close
To space find space straight f left parenthesis negative 1 right parenthesis plus straight f left parenthesis 2 right parenthesis plus straight f left parenthesis 4 right parenthesis comma
for space straight f left parenthesis negative 1 right parenthesis rightwards double arrow negative 1 less than 1 space
Hence comma space straight f left parenthesis straight x right parenthesis equals 3 straight x
rightwards double arrow straight f left parenthesis negative 1 right parenthesis equals negative 3
for space straight f left parenthesis 2 right parenthesis rightwards double arrow 2 less than 3
Hence comma space straight f left parenthesis straight x right parenthesis equals straight x squared
rightwards double arrow straight f left parenthesis 2 right parenthesis equals 4
for space straight f left parenthesis 4 right parenthesis rightwards double arrow 4 greater than 3
Hence comma space straight f left parenthesis straight x right parenthesis equals 2 straight x
rightwards double arrow straight f left parenthesis 4 right parenthesis equals 8
straight f left parenthesis negative 1 right parenthesis plus straight f left parenthesis 2 right parenthesis plus straight f left parenthesis 4 right parenthesis equals negative 3 plus 4 plus 8 equals 9 end style

Question 52

Let A = {1, 2, ..., n} and B = {a, b}. Then the number of subjections  from A into B is 

 

begin mathsize 12px style open parentheses straight a close parentheses space to the power of straight n straight P subscript 2
open parentheses straight b close parentheses space 2 to the power of n minus 2
open parentheses straight c close parentheses space 2 to the power of n minus 1
open parentheses straight d close parentheses space to the power of straight n straight C subscript 2
space end style

Solution 52

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
The space number space of space functions space from space straight a space set space with space straight n space number space of space elements space into space straight a space set space with space 2 space number space of space elements equals 2 to the power of straight n
But space two space functions space can space be space many minus one space into space functions.
hence comma space answer space is space 2 to the power of straight n minus 2. end style

Question 53

If the set A contains 5 elements and the set B 6 elements, then the number of one-one and onto mappings from A to B is 

(a) 720

(b) 120

(c) 0

(d) none of these

Solution 53

begin mathsize 12px style blank end stylebegin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
The space set space straight A space contains space 5 space elements space and space the space set space straight B space 6 space elements.
Number space of space one minus to minus one space and space onto space mappings space from space straight A space to space straight B space means
the space number space of space bijections space from space straight A space to space straight B.
Number space of space bijections space possible space only space when space straight n left parenthesis straight B right parenthesis less or equal than straight n left parenthesis straight A right parenthesis.
Here comma space straight n left parenthesis straight A right parenthesis less than straight n left parenthesis straight B right parenthesis.
Hence comma space the space number space of space one minus one space and space onto space mappings space from space straight A space to space straight B space is space 0. end style

Question 54

If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is 

 

begin mathsize 12px style open parentheses straight a close parentheses space to the power of 10 straight C subscript 7
left parenthesis straight b right parenthesis space to the power of 10 straight C subscript 7 space cross times space 7 factorial
left parenthesis straight C right parenthesis space 7 to the power of 10
left parenthesis straight d right parenthesis space 10 to the power of 7 end style

Solution 54

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
The space number space of space one minus one space functions space from space straight A space to space straight B space with space 10 space and space 7 space elements equals to the power of 10 straight C subscript 7 cross times 7 factorial end style

Chapter 2 - Functions Exercise Ex. 2.1

Question 1(i)

Give an example of function which is one-one but not onto.

Solution 1(i)

Question 1(ii)

Give an example of a function which is not one-one but onto.

Solution 1(ii)

Question 1(iii)

Given an example of a function which is neither one-one nor onto.

Solution 1(iii)

Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5(i)
Solution 5(i)
Question 5(ii)
Solution 5(ii)
Question 5(iii)
Solution 5(iii)
Question 5(iv)
Solution 5(iv)
Question 5(v)
Solution 5(v)
Question 5(vi)
Solution 5(vi)
Question 5(vii)
Solution 5(vii)
Question 5(viii)

Solution 5(viii)

Question 5(ix)
Solution 5(ix)
Question 5(x)

Solution 5(x)

Question 5(xi)

Classify the following functions as injection, surjection or bijection:

f : R → R, defined by f(x) = sin2x + cos2x

Solution 5(xi)

Question 5(xii)
Solution 5(xii)
Question 5(xiii)

Solution 5(xiii)

Question 5(xiv)

Solution 5(xiv)

Question 5(xv)

Solution 5(xv)
Question 5(xvi)
Solution 5(xvi)
Question 6
Solution 6
Question 7
Solution 7

Question 10
Solution 10
Question 11
Solution 11
Question 12

Solution 12

We have f : R → R given by f(x) = ex

let x, y  R, such that

f(x) = f(y)

⇒ ex = ey

⇒ ex-y = 1 = e°

⇒ x - y = 0

⇒ x = y

∴ f is one-one

clearly range of f = (0, ∞) ≠ R
∴ f is not onto

when co-domain is replaced by begin mathsize 12px style straight R subscript 0 superscript plus end style i.e, (0, ∞) then f becomes an onto function.

Question 13
Solution 13
Question 14

If A = { 1, 2, 3}, show that a ono-one function f : A → A must be onto.

Solution 14

Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f.

Hence, f has to be onto.

Question 15

If A = {1, 2, 3}, show that a onto function f : A → A must be one-one.

Solution 15

Suppose f is not one-one.

Then, there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same.

Also, the image of 3 under f can be only one element.

Therefore, the range set can have at most two elements of the co-domain {1, 2, 3}

i.e f is not an onto function,  a contradiction.

Hence, f must be one-one.

Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 22

Solution 22

Question 23

Solution 23

Question 5(xvii)

Classify the following function as injection, surjection or bijection:

f: R R, defined by

Solution 5(xvii)

Given: f: R R, defined by

Injective: Let x, y R such that f(x) = f(y)

  

This does not imply x = y

Therefore, f is not one-one.

 

Surjective: Let y R be arbitrary, then

f(x) = y

  

This does not give any value for x.

Therefore, f is not onto.

Thus, it is not bijective.

Question 8(i)

Let A = [-1, 1]. Then, discuss whether the following function from A to itself are one-one, onto or bijective.

  

Solution 8(i)

Given: f: A A, defined by

Injective: Let x, y A such that f(x) = f(y)

  

Therefore, f is one-one.

 

Surjective: For f(x), x [-1, 1]

  

Therefore, range is the subset of codomain.

Thus, f is not onto.

Hence, f is one-one but not onto.

Question 8(ii)

Let A = [-1, 1]. Then, discuss whether the following function from A to itself are one-one, onto or bijective.

  

Solution 8(ii)

Given: g: A A, defined by

Injective: Let x, y A such that g(x) = g(y)

  

Therefore, f is not one-one.

 

Surjective: For f(x), x [-1, 1]

  

Therefore, range is the subset of codomain.

Thus, f is not onto.

Hence, f is neither one-one nor onto.

Question 8(iii)

Let A = [-1, 1]. Then, discuss whether the following function from A to itself are one-one, onto or bijective.

  

Solution 8(iii)

Given: h: A A, defined by

Injective: Let x, y A such that h(x) = h(y)

  

Therefore, f is not one-one.

 

Surjective: For f(x) = x2, x [-1, 1]

x2 [0, 1]

So, the range is subset of co-domain.

Thus, f is not onto.

Hence, f is neither one-one nor onto.

Question 9(i)

Is the following set of ordered pairs function? If so, examine whether the mapping is injective or surjective.

{(x, y): x is a person, y is the mother of x}

Solution 9(i)

Given set of ordered pair is {(x, y): x is a person, y is the mother of x}

Now biologically, each person 'x' has only one mother.

So, the above set of ordered pairs is a function.

Also, more than one person may have same mother.

So the function is many-one and surjective.

Question 9(ii)

Is the following set of ordered pairs function? If so, examine whether the mapping is injective or surjective.

{(a, b): 'a' is a person, 'b' is an ancestor of 'a'}

Solution 9(ii)

Given set of ordered pair is {(a, b): 'a' is a person, 'b' is an ancestor of 'a'}

Clearly, a person can have more than one ancestors.

So, the above set of ordered pairs is not a function.

Question 21

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

a. an injective map from A to B

b. a mapping from A to B which is not injective

c. a mapping from A to B

Solution 21

Given: A = {2, 3, 4}, B = {2, 5, 6, 7}

a. Consider f: A B defined by f(x) = x + 3

Clearly, f is injective because each element of A has distinct output in B.

b. Consider g: A B defined by g(x) = 3

Here, g is a constant function.

Therefore, each element in A has the same output in B.

Thus, g is not injective.

c. Consider h: A B such that h = {(2, 2), (3, 5), (4, 7)}

Clearly, h is a mapping from A to B.

Chapter 2 - Functions Exercise Ex. 2.2

Question 1(i)
Solution 1(i)
Question 1(ii)

Solution 1(ii)


Question 1(iii)

Solution 1(iii)

f open parentheses x close parentheses equals x squared plus 8 space a n d space g open parentheses x close parentheses equals 3 x cubed plus 1
T h u s comma space g ring operator f open parentheses x close parentheses equals g open square brackets f open parentheses x close parentheses close square brackets
rightwards double arrow space g ring operator f open parentheses x close parentheses equals g open square brackets x squared plus 8 close square brackets
rightwards double arrow space g ring operator f open parentheses x close parentheses equals 3 open square brackets x squared plus 8 close square brackets cubed plus 1
S i m i l a r l y comma space f ring operator g open parentheses x close parentheses equals f open square brackets g open parentheses x close parentheses close square brackets
rightwards double arrow space space f ring operator g open parentheses x close parentheses equals f open square brackets 3 x cubed plus 1 close square brackets
rightwards double arrow space space f ring operator g open parentheses x close parentheses equals open square brackets 3 x cubed plus 1 close square brackets squared plus 8
rightwards double arrow space space f ring operator g open parentheses x close parentheses equals open square brackets 9 x to the power of 6 plus 1 plus 6 x cubed close square brackets plus 8
rightwards double arrow space space f ring operator g open parentheses x close parentheses equals 9 x to the power of 6 plus 6 x cubed plus 9

Question 1(iv)
Solution 1(iv)
Question 1(v)
Solution 1(v)
Question 1(vi)
Solution 1(vi)
Question 2
Solution 2
Question 3
Solution 3

Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10

Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and

h(z)  = sin z for all x, y, z ϵ N.

show tht ho (gof) = (hog) of.

Solution 10

Question 11

Given Examples of two functions f : N → N and g ; N → N such that gof is onto but f is not onto.

Solution 11


Question 12
Solution 12
Question 14
Solution 14
Question 13

If   then find fof(x).

Solution 13

Given:

  

Hence, fof(x) = x.

Chapter 2 - Functions Exercise Ex. 2VSAQ

Question 37

If f: [5, 6] [2, 3] and g: [2, 3] [5, 6] are given by f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}, find f o g.

Solution 37

We have f: [5, 6] [2, 3] and g: [2, 3] [5, 6] are given by:

f = {(5, 2), (6, 3)} and g = {(2, 5), (3, 6)}

Now, (f o g)(2) = f(g(2)) = f(5) = 2

And, (f o g)(3) = f(g(3)) = f(6) = 3

Therefore, fog = {(2, 2), (3, 3)}.

Question 38

Let f: R R be the function defined by f(x) = 4x - 3 for all x R. Then write f-1.

Solution 38

Let x, y R such that f(x) = f(y)

4x - 3 = 4y - 3

x = y

So, f is one-one.

For x R, 4x - 3 R

Therefore, range and co-domain are same.

So, f is onto.

Thus, f is bijective and so it is invertible.

Let y = f(x)

y = 4x - 3

  

Hence,   

Question 39

Which one of the following relations on A = {1, 2, 3} is a function?

f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}

Solution 39

Consider, f = {(1, 3), (2, 3), (3, 2)}

In f, every element in the domain A has unique image.

So, f is a function.

It is given that g = {(1, 2), (1, 3), (3, 1)}

For g, the element 1 in the domain A has two different images.

So, g is not a function.

Question 40

Write the domain of the real function f defined by   

Solution 40

Given:   

The function f is defined only when

  

Hence, the domain is [-5, 5].

Question 41

Let A = {a, b, c, d} and f: A A be given by f = {(a, b), (b, d), (c, a), (d, c)}, write f-1.

Solution 41

Given: A = {a, b, c, d} and f: A A is given by f = {(a, b), (b, d), (c, a), (d, c)}

Here, f-1 will be a function from A to A and given by

f-1 = {(b, a), (d, b), (a, c), (c, d)}

Question 42

Let f, g: R R be given by f(x) = 2x + 1 and g(x) = x2 - 2 for all x R, respectively. Then, find gof.

Solution 42

Given: f, g: R R

Therefore, gof: R R will be

gof(x) = g(f(x))

= g(2x + 1)

= (2x + 1)2 - 2

 = 4x2 + 4x - 1

Hence, gof(x) = 4x2 + 4x - 1.

Question 43

If the mapping f: {1, 3, 4} {1, 2, 5} and g: {1, 2, 5} {1, 3}, given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write fog.

Solution 43

Given: f: {1, 3, 4} {1, 2, 5} and g: {1, 2, 5} {1, 3}

fog(1) = f(g(1)) = f(3) = 5

fog(2) = f(g(2)) = f(3) = 5

fog(5) = f(g(5)) = f(1) = 2

Therefore, fog: {1, 2, 5} {5, 2}

Hence, fog = {(1, 5), (2, 5), (5, 2)}.

Question 44

If a function g = {(1, 1), (2, 3), (3, 5), (4, 7)} is described by g(x) = αx + β, find the values of α and β.

Solution 44

Given: g(x) = αx + β and g = {(1, 1), (2, 3), (3, 5), (4, 7)}

Now, g(1) = 1

∴ α + β = 1 … (i)

And, g(2) = 3

2α + β = 3 … (ii)

Solving (i) and (ii), we get

α = 2 and β = -1 

Question 45

If f(x) = 4 - (x - 7)3, write

Solution 45

Given: f(x) = 4 - (x - 7)3

Let y = 4 - (x - 7)3

  

Hence,   

Question 1

Figure (a)

 

Figure (b) 

 

Solution 1

Question 2

Figure (a)  

 

Figure (b) 

Solution 2

Question 3

If A = {1, 2, 3} B = {a, b}, write total number of functions from A to B.

Solution 3

A = {1, 2, 3} B = {a, b}

The total number of functions is 8

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Let space straight A space equals space left curly bracket straight x element of straight R space colon space minus 4 space less or equal than space straight x less or equal than 4 space and space straight x space not equal to 0 right curly bracket space and space straight f space colon space straight A space rightwards arrow space straight R space be space defined
by space straight f left parenthesis straight x right parenthesis space equals space fraction numerator open vertical bar straight x close vertical bar over denominator straight x end fraction.
Write space the space range space of space straight f.

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

rightwards double arrow straight f to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator straight x plus 7 over denominator 10 end fraction

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1,4), (2,5), (3,6)} be a function from A to B.

State the whether f is one - one or onto

Solution 36

It is given that

A = {1, 2, 3}, B = {4, 5, 6, 7} and f = {(1,4), (2,5), (3,6)}

The function f is one-one from A to B

Chapter 2 - Functions Exercise Ex. 2.3

Question 13

If f, g: R R be two functions defined as f(x) = |x| + x and g(x) = |x| - x for all x R. Then, find fog and gof. Hence, find fog(-3), fog(5) and gof(-2).

Solution 13

Given: f(x) = |x| + x and g(x) = |x| - x for all x R

  

  

fog(-3) = -4(-3) = 12, fog(5) = 0 and

gof(-2) = 0

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 2
Solution 2

Question 3
Solution 3
Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11(i)

Let f be a real function given by

Find each of the following

 

fof

Solution 11(i)

Question 11(ii)

Let f be a real function given by

Find each of the following

 

fofof

Solution 11(ii)

Question 11(iii)

Let f be a real function given by

Find each of the following

 

(f0f0f) (38)

Solution 11(iii)

Question 11(iv)

Let f be a real function given by

Find each of the following

 

f2Also, show that fof ≠ f2 .

Solution 11(iv)

Question 12

Solution 12

Chapter 2 - Functions Exercise Ex. 2.4

Question 15

Let f: N N be a function defined as f(x) = 9x2 + 6x - 5. Show that f: N S, where S is the range of f, is invertible. Find the inverse of f and hence find f-1(43) and f-1(163).

Solution 15

Given: f(x) = 9x2 + 6x - 5

Let x, y N such that f(x) = f(y)

9x2 + 6x - 5 = 9y2 + 6y - 5

9(x2 - y2) + 6(x - y) = 0

(x - y){9(x + y) + 6} = 0

As x, y N, 9(x + y) + 6 can't be zero.

x - y = 0

x = y

So, f is one-one.

Clearly, f is onto as range and co-domain are same.

Therefore, f is bijective.

Thus, f is invertible.

Let y = f(x) = 9x2 + 6x - 5

  

  

Question 16

Let   be a function defined as   Show that   is one-one and onto. Hence, find f-1.

Solution 16

Given:

Let x, y   such that f(x) = f(y)

  

So, f is one-one.

Clearly, f is onto as the co-domain of   is same as its range.

Therefore, f is bijective.

Thus, f is invertible.

Let y = f(x)

  

Question 17

Let A = R - {2} and B = R - {1}. If f: A B is a function defined by   show that f is one-one and onto. Find f-1.

Solution 17

Given:

Let x, y A such that f(x) = f(y)

  

So, f is one-one.

Let y B

  

So, for every x A, there is a y B such that f(x) = y.

Therefore, f is onto.

Thus, f is bijective and so it is invertible.

Hence, 

Question 18

Show that the function f: N N defined by f(x) = x2 + x + 1 is one-one but onto. Find the inverse of f: N S, where S is range of f.

Solution 18

Given: f: N N defined by f(x) = x2 + x + 1

Let x, y N such that f(x) = f(y)

x2 + x + 1 = y2 + y + 1

x2 - y2 + x - y = 0

(x - y)(x + y + 1) = 0

As x, y N, so (x2 + x + 1) can't be zero.

x - y = 0

x = y

So, f is one-one.

Since x N, x2 + x + 1 > 3 as the minimum value x can take is 1.

Therefore, range of f = (3, )

As the co-domain and range does not match, f is not onto.

But, f: N (3, ) is onto as the range matches with co-domain.

Thus, f: N S is a bijective function and so it is invertible.

Let y = f(x)

y = x2 + x + 1

x2 + x + 1 - y = 0

  

Hence,   

Question 19

Consider the bijective function f: R+ (7, ) given by f(x) = 16x2 + 24x + 7, where R+ is the set of positive real numbers. Find the inverse function of f.

Solution 19

Given: f: R+ (7, ) given by f(x) = 16x2 + 24x + 7

As f is a bijective function, so it is invertible.

Let y = f(x)

y = 16x2 + 24x + 7

16x2 + 24x + 7 - y = 0

  

Hence,   

Question 1

Solution 1

Thus, h is a bijection and is invertible.
 

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3

Consider f : left curly bracket 1 comma 2 comma 3 right curly bracket rightwards arrow left curly bracket a comma b comma c right curly bracket and g : open curly brackets a comma b comma c close curly brackets rightwards arrow open curly brackets a p p l e comma space b a l l comma space c a t close curly brackets defined as f open parentheses 1 close parentheses equals a comma space f open parentheses 2 close parentheses equals b comma space f open parentheses 3 close parentheses equals c comma space g open parentheses a close parentheses equals a p p l e comma space g open parentheses b close parentheses equals b a l l space a n d space g open parentheses c close parentheses equals c a t.

Show that f comma space g space a n d space g ring operator f are invertible. Find f to the power of minus 1 end exponent comma space g to the power of minus 1 end exponent space a n d space open parentheses g ring operator f close parentheses to the power of minus 1 end exponent and show that open parentheses g ring operator f close parentheses to the power of minus 1 end exponent equals f to the power of minus 1 end exponent ring operator g to the power of minus 1 end exponent.

Solution 3

G i v e n space t h a t space f : open curly brackets 1 comma 2 comma 3 close curly brackets rightwards arrow open curly brackets a comma b comma c close curly brackets space a n d space g : open curly brackets a comma b comma c close curly brackets rightwards arrow open curly brackets a p p l e comma space b a l l comma space c a t close curly brackets space s u c h space t h a t
f open parentheses 1 close parentheses equals a comma space f open parentheses 2 close parentheses equals b comma space f open parentheses 3 close parentheses equals c comma space g open parentheses a close parentheses equals a p p l e comma space g open parentheses b close parentheses equals b a l l space a n d space g open parentheses c close parentheses equals c a t
W e space n e e d space t o space p r o v e space t h a t space f comma space g space a n d space g ring operator f space a r e space i n v e r t i b l e.
I n space o r d e r space t o space p r o v e space t h a t space f space i s space i n v e r t i b l e m space i s space i s space s u f f i c i e n t space t o space s h o w space t h a t
f : open curly brackets 1 comma 2 comma 3 close curly brackets rightwards arrow open curly brackets a comma b comma c close curly brackets space i s space a space b i j e c t i o n.
f space i s space o n e minus o n e :
E a c h space a n d space e v e r y space e l e m e n t space o f space t h e space s e t space open curly brackets 1 comma 2 comma 3 close curly brackets space i s space h a v i n g space a n space i m a g e space i n space t h e space s e t space open curly brackets a comma b comma c close curly brackets
T h u s comma space f space i s space o n e minus o n e.
O b v i o u s l y comma space t h e space n u m b e r space o f space e l e m e n t space o f space t h e space s e t s space open curly brackets 1 comma 2 comma 3 close curly brackets space a n d space open curly brackets a comma b comma c close curly brackets space a r e space e q u a l space a n d space h e n c e
f space i s space o n t o.
T h u s comma space t h e space f u n c t i o n space f space i s space i n v e r t i b l e.
S i m i l a r l y comma space l e t space u s space o b s e r v e space f o r space t h e space f u n c t i o n space g :
g space i s space o n e minus o n e :
E a c h space a n d space e v e r y space e l e m e n t space o f space t h e space s e t space space open curly brackets a comma b comma c close curly brackets space i s space h a v i n g space a n space i m a g e space i n space t h e space s e t space open curly brackets a p p l e comma space b a l l comma space c a t close curly brackets
T h u s comma space g space i s space o n e minus o n e.
O b v i o u s l y comma space t h e space n u m b e r space o f space e l e m e n t space o f space t h e space s e t s space open curly brackets a comma b comma c close curly brackets space a n d space open curly brackets a p p l e comma space b a l l comma space c a t close curly brackets space space a r e space e q u a l space a n d space h e n c e
g space i s space o n t o.
T h u s comma space t h e space f u n c t i o n space g space i s space i n v e r t i b l e.