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Injectivity:

Consider x, y ∈ R where f(x) = f(y)

We get

x⁵ = y⁵

It can be written as

x⁵ – y⁵ = 0

By multiplying and dividing by 2 on both sides

(x^5/2)² – (y^5/2)² = 0

We know that

(x^5/2 + y^5/2) (x^5/2 – y^5/2) = 0

So we get

x^5/2 – y^5/2 = 0

where x = y

Hence, f(x) = f(y) is x = y for all x, y ∈ R

f is injective.

Surjectivity:

Consider y as an arbitrary element of R

We know that

f(x) = y

It can be written as

x⁵ = y

So we get

x⁵ – y = 0

Odd degree equation has one real root

Thus, for every real value of y

x⁵ – y = 0 has real root α where

α ⁵ – y = 0

So α ⁵ = y

Thus, f(α) = y

For every y ∈ R there exists α ∈ R where f(α) = y

f is surjective

Thus, f: R → R is bijective.

(i) We know that domain (g o f) = domain (f) = {3, 9, 12}

By substituting the values

(g o f) (3) = g{f(3)} = g{1} = 3

(g o f) (9) = g{f(9)} = g{3} = 3

(g o f) (12) = g{f(12)} = g{4} = 9

Hence, g o f = {(3, 3), (9, 3), (12, 9)}

(ii) We know that domain (f o g) = domain (g) = {1, 3, 4, 5}

By substituting the values

(f o g) (1) = f{g(1)} = f{3} = 1

(f o g) (3) = f{g(3)} = f{3} = 1

(f o g) (4) = f{g(4)} = f{9} = 3

(f o g) (5) = f{g(5)} = f{9} = 3

Hence, f o g = {(1, 1), (3, 1), (4, 3), (5, 3)}

Correct Answer is (C) (2x² + 1) 

f(x) = (x² – 1)

g(x) = (2x + 3)

∴(g o f) (x) = g(f(x))

\(\Rightarrow g(f(x))=2f(x)+3\)

\(\Rightarrow g(f(x))\)\(=2((x^2-1))+3\)\(=2x^2-2+3\)\(=2x^2+1\)

For any two distinct elements x₁ and x₂ in [0, π/2]

We know that

sin x₁ ≠ sin x₂ and cos x₁ ≠ cos x₂

So we get

f(x₁) ≠ f(x₂) and g(x₁) ≠ g(x₂)

Here, both f and g are one-one

We know that

(f + g) (x) = f( x) + g(x) = sin x + cos x

By substituting the values

(f + g) (0) = sin 0 + cos 0 = 1

(f + g) (π/2) = sin π/2 + cos π/2 = 1

We know that 0 ≠ π/2

Here, (f + g) (0) = (f + g) (π/2)

So, f + g is not one-one.

Correct Answer is (A) \(\frac{4y}{(4-3y)}\)

f(x) =\(\frac{4x}{(3x+4)}\)

⇒y =\(\frac{4x}{(3x+4)}\)

x ⟺ y

⇒x =\(\frac{4y}{(3y+4)}\)

⇒3yx + 4x = 4y

⇒y(3x - 4) = - 4x

⇒y = \(\frac{4x}{(4-3x)}\)

x ⟺ y

⇒x = \(\frac{4y}{(4-3y)}\)

We know that

range f = {1, 2, 3, 4} and domain (g) = {1, 2, 3, 4}

So range (f) ⊆ domain (g)

(i) We know that domain (g o f) = domain (f) = {1, 2, 3, 4}

By substituting the values

(g o f) (1) = g{f(1)} = g(4) = 4

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

(g o f) (3) = g{f(3)} = g(3) = 2

(g o f) (4) = g{f(4)} = g(2) = 1

Therefore, g o f = {(1, 4), (2, 3), (3, 2), (4, 1)}

(ii) We know that domain (f o g) = domain (g) = {1, 2, 3, 4}

By substituting the values

(f o g) (1) = f{g(1)} = f(3) = 3

(f o g) (2) = f{g(2)} = f(1) = 4

(f o g) (3) = f{g(3)} = f(3) = 2

(f o g) (4) = f{g(4)} = f(4) = 2

Therefore, f o g = {(1, 3), (2, 4), (3, 2), (4, 2)}

(iii) We know that domain (f o f) = domain (f) = {1, 2, 3, 4}

By substituting the values

(f o f) (1) = f{f(1)} = f(4) = 2

(f o f) (2) = f{f(2)} = f(1) = 4

(f o f) (3) = f{f(3)} = f(3) = 4

(f o f) (4) = f{f(4)} = f(2) = 1

Therefore, f o f = {(1, 2), (2, 4), (3, 3), (4, 1)}

Correct Answer is (A) x

f(x) = \(\frac{4x-6}{3x+4}\)

\(\Rightarrow f(f(x))=\frac{4f(x)+3}{6f(x)-4}=(fof)(x)\)

\(\Rightarrow f(f(x))=\frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4}\)

\(\Rightarrow f(f(x))=\frac{16x+12+18x-12}{24x+18-24x+16}\)\(=\frac{34x}{34}\)= x

Consider N as the set of all natural numbers

It is given that f: N → N: f(x) = x² Ɐ x ∈ N

We know that

f(x₁) = f(x₂)

It can be written as

x₁² = x₂²

So we get

x₁² – x₂² = 0

We get

(x₁ – x₂) (x₁ + x₂) = 0

Here, (x₁ – x₂) = 0 where x₁ = x₂

Hence, f is one-one.

Consider A = {1, 2, 3, 4} and B = {1, 4, 9, 16, 25}

Here, f: A → B: f(x) = x²

By substituting the values

f(1) = 1² = 1

f(2) = 2² = 4

f(3) = 3² = 9

f(4) = 4² = 16

Hence, every moment in A has unique image in B. Here, ∃ 25 ∈ B which has no pre-image in A f is into.

It is given that

f(x) = 1 + x²

By substituting values we get

f(-1) = 1 + (-1)² = 2

f(1) = 1 + 1² = 2

Hence, f is many one.

Consider y = 1 + x²

It can be written as

x = √(y – 1)

If y <1 then we know that √(y – 1) is imaginary

O ∈ R has no pre image in R

Hence, f is into.

Consider x₁, x₂ ∈ Z

We know that

f(x₁) = f(x₂)

So we get

x₁³ = x₂³

where x₁ = x₂

Hence, f is one-one

Consider 2 ∈ Z, there exists no x ∈ Z where x³ = 2

Here, 2 ∈ Z has no pre-image in Z

Hence, f is into.

We know that

f(x₁) = f(x₂)

It can be written as

1/x₁ = 1/x₂

So we get

x₁ = x₂

Hence, f is one-one.

Take y = 1/x

It can be written as

x = 1/y

Each y in co domain R_o there exists 1/y in domain R_o where f(1/y) = 1/(1/y) = y

Hence, f is onto.

To prove: function is one-one and into

Given: f : Z → Z : f (x) = x³

Solution: We have,

f(x) = x³

For, f(x₁) = f(x₂)

⇒ x₁³ = x₂³

⇒ x₁ = x₂

When, f(x₁) = f(x₂) then x₁ = x₂

∴ f(x) is one-one

f(x) = x³

Let f(x) = y such that \(y\in Z\)

⇒ y = x³

^{\(\Rightarrow x=\sqrt[3]{y}\)}

If y = 2, as \(y\in Z\)

Then we will get an irrational value of x, but \(x\in Z\)

Hence f(x) is into

Hence Proved

Correct Answer is (B) \(\frac{1}{2}(\sqrt{y-4}+3)\)

f(x) = 4x² + 12x + 15

⇒y = 4x² + 12x + 15

⇒y = (2x + 3)² + 6

⇒√(y - 6) = 2x + 3

⇒ \(\frac{1}{2}(\sqrt{y-4}+3)\) = x

f^-1(y) =\(\frac{1}{2}(\sqrt{y-4}+3)\)

To prove: function is one-one and onto

Given: \(f:R_0→R_0:f(x)=\frac{1}{x}\)

We have,

\(f(x)=\frac{1}{x}\)

For, f(x₁) = f(x₂)

\(\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}\)

⇒ x₁ = x₂

When, f(x₁) = f(x₂) then x₁ = x₂

∴ f(x) is one-one

\(f(x)=\frac{1}{x}\)

Let f(x) = y such that \(y\in R_0\)

\(\Rightarrow y=\frac{1}{x}\)

\(\Rightarrow x=\frac{1}{y}\)

Since \(y\in R_0\)

\(\Rightarrow \frac{1}{y}\in R_0\)

⇒ x will also \(\in R_0\), which means that every value of y is associated with some x

∴ f(x) is onto

Hence Proved

To prove: function is many-one into

Given: f : R → R : f(x) = 1 + x²

We have,

f(x) = 1 + x²

For, f(x₁) = f(x₂)

⇒ 1 + x₁² = 1 + x₂²

⇒ x₁² = x₂²

⇒ x₁² - x₂² = 0

⇒ (x₁ – x₂) (x₁ + x₂) = 0

⇒ x₁ = x₂ or, x₁ = –x₂

Clearly x₁ has more than one image

∴ f(x) is many-one

f(x) = 1 + x²

Let f(x) = y such that \(y\in R\)

⇒ y = 1 + x²

⇒ x² = y – 1

\(\Rightarrow x=\sqrt{y-1}\)

If y = 3, as \(y \in R\)

Then x will be undefined as we can’t place the negative value under the square root

Hence f(x) is into

Hence Proved

To find: f^-1

Given: \(f:R→R:f(x)=\frac{2x-7}{4}\)

We have,

\(f(x)=\frac{2x-7}{4}\)

Let f(x) = y such that \(y\in R\)

\(\Rightarrow y=\frac{2x-7}{4}\)

⇒ 4y = 2x – 7

⇒ 4y + 7 = 2x

\(\Rightarrow x=\frac{4y+7}{2}\)

\(\Rightarrow f^{-1}=\frac{4y+7}{2}\)

\(f^{-1}(y)=\frac{4y+7}{2}\) for all \(y\in R\) 

f(x) = 9x³ + 8

Let f(x₁ ) = f(x₂)

\(\therefore9x^3_1+8=9x^3_2+8\) 

\(\therefore\) x₁ = x₂

∴ f is a one-one function.

∴ f(x) = 9x³ + 8 = y, (say)

\(\therefore\) x = \(\sqrt[3]{\frac{y-8}9}\) 

∴ For every y we can get x.

∴ f is an onto function.

\(\therefore\) x = \(\sqrt[3]{\frac{y-8}9}\) = f^-1(y)

Replacing y by x, we get

f^-1(x) = \(\sqrt[3]{\frac{y-8}9}\)

f(x) = 8 = y (say)

For every value of x, the value of the function f is the same.

∴ f is not one-one i.e. (many-one) function.

∴ f does not have the inverse.

(i) Given function, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

Also given f(x) = 2x + 3 and g(x) = x² + 5

Now, (gof)(x) = g(f(x))

= g(2x + 3)

= (2x + 3)² + 5

= 4x² + 9 + 12x + 5

=4x² + 12x + 14

Then, (fog)(x) = f(g(x))

= f(x² + 5)

= 2(x² + 5) + 3

= 2x²+ 10 + 3

= 2x² + 13

(ii) Given function, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

f(x) = 2x + x² and g(x) = x³

(gof)(x)= g(f(x))

= g(2x + x²)

= (2x + x²)³

Now, (fog)(x) = f(g(x))

= f(x³)

= 2(x³) + (x³)²

= 2x³ + x⁶

(iii) Given function, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

f(x) = x² + 8  and g(x) = 3x³ + 1

(gof)(x) = g(f(x))

= g(x² + 8)

= 3(x² + 8)³ + 1

Then, (fog)(x) = f(g(x))

= f(3x³ + 1)

= (3x³ + 1)² + 8

= 9x⁶ + 6x³ + 1 + 8

= 9x⁶ + 6x³ + 9

The term string is defined as a character array, terminated with a null character [‘\0’]

Eval(s) evaluates an expression contained in a string s and returns the resulting object.

The fabs() mathematical function is used to return the absolute value of the given argument.

A list containing the program name and all the command-line arguments to a program.

No, This input will not work

Returns the zero.