Class 12-science RD SHARMA Solutions Maths Chapter 2 - Functions
Functions Exercise MCQ
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,
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.
Solution 57
f: R → R is given by f(x) = 3x2 - 5
g: R → R is given by
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.
Solution 59
It is given that f: A → B and g: B → C are two bijective functions.
Consider,
Thus,
Solution 60
It is given that f: N → R as and g: Q → R as g(x) = x + 2.
Now,
Hence,
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
ANSWER PENDING
Solution 36
Solution 37
ANSWER PENDING
Solution 38
Solution 39
Solution 40
Solution 41
Solution 42
ANSWER PENDING
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
Solution 49
Solution 50
Solution 51
Solution 52
Solution 53
Solution 54
Functions Exercise Ex. 2.1
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 2
Solution 3
Solution 4
Solution 5(i)
Solution 5(ii)
Solution 5(iii)
Solution 5(iv)
Solution 5(v)
Solution 5(vi)
Solution 5(vii)
Solution 5(viii)
Solution 5(ix)
Solution 5(x)
Solution 5(xi)
Solution 5(xii)
Solution 5(xiii)
Solution 5(xiv)
Solution 5(xv)
Solution 5(xvi)
Solution 6
Solution 7
Solution 10
Solution 11
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 i.e, (0, ∞) then f becomes an onto function.
Solution 13
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.
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.
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 22
Solution 23
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.
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.
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.
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.
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.
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.
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.
Functions Exercise Ex. 2.2
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 14
Solution 13
Given:
Hence, fof(x) = x.
Functions Exercise Ex. 2VSAQ
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)}.
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,
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.
Solution 40
Given:
The function f is defined only when
Hence, the domain is [-5, 5].
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)}
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.
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)}.
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
Solution 45
Given: f(x) = 4 - (x - 7)3
Let y = 4 - (x - 7)3
Hence,
Solution 1
Solution 2
Solution 3
A = {1, 2, 3} B = {a, b}
The total number of functions is 8
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
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
Functions Exercise Ex. 2.3
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
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 1(vii)
Solution 1(viii)
Solution 1(ix)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11(i)
Solution 11(ii)
Solution 11(iii)
Solution 11(iv)
Solution 12
Functions Exercise Ex. 2.4
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
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)
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,
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,
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,
Solution 1
Thus, h is a bijection and is invertible.
Solution 2(i)
Solution 2(ii)
Solution 3
Solution 4
∴ f-1 = {(3, 1), (5, 2), (7, 3), (9, 4)}
f-1 0g-1 = {(7, 1), (23, 2), (47, 3), (79, 4)} ……. (A)
Now (A) & (B) we have (g0f)-1 = f-10g-1
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 20
Solution 21
Solution 22