Class 12-commerce RD SHARMA Solutions Maths Chapter 2: Functions

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,

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.

f: R → R is given by f(x) = 3x² - 5

g: R → R is given by

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.

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

Consider,

Thus,  

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

Now,

Hence,

ANSWER PENDING

ANSWER PENDING

ANSWER PENDING

We have f : R → R given by f(x) = e^x

let x, y ∊ R, such that

f(x) = f(y)

⇒ e^x = e^y

⇒ e^x-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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

Given:

Hence, fof(x) = x.

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)}.

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,  

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.

Given:  

The function f is defined only when

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

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)}

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.

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)}.

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 

Given: f(x) = 4 - (x - 7)³

Let y = 4 - (x - 7)³

Hence,  

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

The total number of functions is 8

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

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

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

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)

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, 

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,  

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,  

Thus, h is a bijection and is invertible.
 

∴ 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