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Correct Answer is (B) 2x

f(x) = 8x³

g(x) = x^1/3

\(\Rightarrow (gof)(x)=(f(x))^{1/3}\)\(=(8x^3)^{1/3}=2x\)

For f to be invertible, f should be bijective. 

One-One 

Let x₁, x₂ be any arbitrary value ∈ R, then 

f (x₁) = f (x₂) 

⇒ [4 – (x₁ – 7)³]^1/5 = [4 – (x₂ – 7)³]^1/5 

⇒ 4 – (x₁ – 7)³ = 4 – (x₂ – 7)³ (Putting both the sides to power 5) 

⇒ (x₁ – 7)³ = (x₂ – 7)³ 

⇒ x₁ – 7 = x₂ – 7 (Taking cube root of both the sides) 

⇒ x₁ = x₂ 

⇒ f is one-one.

Onto 

Let y = f (x) = [4 – (x – 7)³]^1/5 

⇒ y⁵ = 4 – (x – 7)³ ⇒ (x – 7)³ = 4 – y⁵

^{⇒ x - 7 = \(\sqrt[3]{4-y^5} \implies x = \sqrt[3]{4-y^5}\)} + 7 ... (i)

Thus for every y ∈ R, there exists an x = \(\big(\sqrt[3]{4-y^5} +7\big)\) ∈ R 

⇒ f is onto.

∴  From (i)  x = f ^-1(y) = \(\big(\sqrt[3]{4-y^5} +7\big)\) 

or, f ^-1(x) = \(\big(\sqrt[3]{4-y^5} +7\big)\) (substituting x for y).

\(f\big(\frac{3}{4}\big)\) = \(\big(\frac{3}{4}\big)^2\) = \(\frac{9}{16}\) as f (x) = x² when x > 0

\(f\big(-\frac{3}{4}\big)\) =  2 \(\times\) \(\big(\frac{-3}{4}\big)\) - 1

= \(-\frac{3}{2}\) -1 

= \(-\frac{5}{2}\) as f (x) = 2x – 1 when x ≤ 0

Given

 f(x) = \(\sqrt{x-2}\): x > and g(x) = \(\sqrt{x+2}\): x > 2

(i) To find: (f + g) (x) 

Domain(f) = (2, ∞) 

Range(f) = (0, ∞) 

Domain(g) = (2, ∞) 

Range(g) = (2, ∞) 

(f + g) (x) = f(x) + g(x)

= \(\sqrt{x-2}\) + \(\sqrt{x+2}\)

Therefore,

(f + g) (x) = \(\sqrt{x-2}\) + \(\sqrt{x+2}\)

(ii) To find:(f - g)(x)

Range(g) ⊆ Domain(f) 

Therefore, 

(f - g)(x) exists. 

(f - g)(x) = f(x) – g(x)

= \(\sqrt{x-2}\) + \(\sqrt{x+2}\)

Therefore,

(f - g) (x) = \(\sqrt{x-2}\) - \(\sqrt{x+2}\)

(iii) To find:(fg)(x) 

(fg)(x) = f(x).g(x)

= \(\sqrt{(x-2)}\) .\(\sqrt{(x+2)}\)

= \(\sqrt{(x-2)(x+2)}\)

= \(\sqrt{(x^2+2^2)}\) ( ∵  a² - b² = (a - b)(a + b))

= \(\sqrt{x^2-4}\)

Therefore,

(fg)(x) = \(\sqrt{x^2-4}\)

To find: g o f and f o g

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f(x) = 8x³

(ii) g(x) = x^1/3

We have,

g o f = g(f(x)) = g(8x³)

g o f = \((8x^3)^{1/3}\) = 2x

f o g = f(g(x)) = f(x^1/3)

f o g = \(8(x^{1/3})^3\) = 8x

g o f = 2x and f o g = 8x

For the function f to be invertible, f has to be one-one onto. 

One-One 

Let x₁, x₂ ∈ R⁺ such that f (x₁) = f (x₂) 

⇒ x₁² + 4 = x₂² + 4 

⇒ x₁ ² = x₂ ² 

⇒ (x₁ – x₂) (x₁ + x₂) = 0 

⇒ x₁ – x₂= 0 

⇒ x₁ = x₂ (∵  x₁ + x₂ ≠ 0 as x₁, x₂ ∈ R⁺) 

⇒ f is one-one

Onto 

Let y= f (x) = x² + 4 ⇒ x² = y – 4 

⇒ x = ±\(\sqrt{y-4}\) 

⇒ x = f ^–1(y) = \(\sqrt{y-4}\) 

⇒ f ^–1(x) = \(\sqrt{x-4}\) (∵ x ∈ R + so we ignore–ve value)

For every element y ∈ [4, ∞], there exists a pre image f ^–1(x) ∈ R ⁺. 

So f is onto. 

Hence f being one-one onto is invertible.

f(a) = \(log\big(\frac{1-a}{1+a}\big)\) , f(b) = \(log\big(\frac{1-b}{1+b}\big)\)

\(f\big(\frac{a+b}{1+ab}\big)\) = \(log\big(\frac{1- \frac{a+b}{1+ab}}{1+ \frac{a+b}{1+ab}}\big)\) 

⇒ \(f\big(\frac{a+b}{1+ab}\big)\) = \(log\big(\frac{\frac{1+ab-a-b}{1+ab}}{\frac{1+ab+a+b}{1+ab}}\big)\) 

= \(log\big(\frac{(1-a)+b(a-1)}{(1+a)+b(a+1)}\big)\) = \(log\big(\frac{(1-a)(1-b)}{(1+a)(1+b)}\big)\) 

= \(log\big(\frac{1-a}{1+a}\big)+log\big(\frac{1-b}{1+b}\big)\) (∵ log ab = log a + log b)

= f(a) + f(b).

We know that

f(x₁) = f(x₂)

It can be written as

2x₁ + 3 = 2x₂ + 3

On further calculation

2x₁ = 2x₂

So we get

x₁ = x₂

Hence, f is one-one.

Consider y = 2x + 3

It can be written as

y – 3 = 2x

So we get

x = (y – 3)/ 2

If y ∈ R, there exists x = (y – 3)/ 2 ∈ R

f (x) = f ([y-3]/2) = 2([y – 3]/ 2) +3 = y

f is onto

Here, f is one-one onto and invertible.

Take y = f(x)

It can be written as

y = 2x + 3

So we get

x = (y-3)/ 2

So f ^-1 (y) = (y – 3)/ 2

Hence, we define f ^-1: R → R: f ^-1(y) = (y – 3)/ 2 for all y ∈ R

Answer : (c) = \(\frac{6}{7}\)  

Given, f (x) = \(\frac{1}{\sqrt{x+2\sqrt{2x-4}}} + \frac{1}{\sqrt{x-2\sqrt{2x-4}}}\)

⇒ f(11) = \(\frac{1}{\sqrt{11+2\sqrt{22-4}}} + \frac{1}{\sqrt{11-2\sqrt{22-4}}}\) 

= \(\frac{1}{\sqrt{11+2\sqrt{18}}} + \frac{1}{\sqrt{11-2\sqrt{18}}}\) 

= \(\frac{1}{\sqrt{11+6\sqrt{2}}} + \frac{1}{\sqrt{11-6\sqrt{2}}}\)

= \(\frac{1}{\sqrt{(9+2\times 3\times \sqrt{2}+(\sqrt{2})^2)}} +\frac{1}{\sqrt{(9-2\times 3\times \sqrt{2}+(\sqrt{2})^2)}}\) 

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

= \(\frac{1}{{(3+\sqrt{2})}} +\frac{1}{{(3-\sqrt{2})}}\) 

= \(\frac{3-\sqrt{2} +3+ \sqrt{2}}{(3+\sqrt{2})(3-\sqrt{2})}\)  = \(\frac{6}{9-2}\) 

= \(\frac{6}{7}\)

To Show: that f is invertible

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A B is said to be a one-one function or injective mapping if different

elements of A have different images in B. Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ↔ x₁= x₂ or x₁ ≠ x₂ ↔ f(x₁) ≠  f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ Q and f(x) = \(\frac{(3x+1)}{2}\).So f(x₁) = f(x₂) → \(\frac{(3x_1+1)}{2}\)= \(\frac{(3x_2+1)}{2}\) → x₁=x₂

So f(x₁) = f(x₂) ↔ x₁= x₂, f(x) is one-one

Given co-domain of f(x) is R.

Let y = f(x) = \(\frac{(3x+1)}{2}\), So x = \(\frac{2y-1}{3}\)[Range of f(x) = Domain of y]

So Domain of y is R = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) = \(\frac{2y-1}{3}\)

\(f(-1) = \frac{-1-|-1|}{|-1|}\) 

= \(\frac{-1-1}{1}\) 

= - 2

We know that

f (2) = 7, f (3) = 9, f(4) = 11 and f (5) = 13

Here dom (f) = {2, 3, 4, 5} = A and range (f) = {7, 9, 11, 13} = B

So different element in A have different image.

f is one-one.

range (f) = B where f is onto

Hence, f is one-one onto and invertible.

We know that

f (2) = 7, f (3) = 9, f(4) = 11 and f (5) = 13

So we get

f ^-1 (7) = 2, f ^-1 (9) = 3, f ^-1 (11) = 4, f ^-1 (13) = 5

Therefore, f ^-1 {(7, 2), (9, 3), (11, 4), (13, 5)}.

Answer : (a) = 2x - 5 

Let f (x) = 2x – 5 f : R → R 

Then, for all x, y ∈ R, f (x) = f (y) 

⇒ 2x – 5 = 2y – 5 ⇒ x = y 

⇒ f is one-one 

Also, let z = f (x) = 2x – 5 

⇒ \(x= \frac{z+5}{2}\)

∴   For all z ∈ R, there exists an x ∈ R such that z = f (x) = 2x + 5 

⇒ f is onto. 

∴   f is one-one onto 

⇒ f is bijective. 

Now, check for other functions: Let g(x) = | x |, g : R → R

Then g(– 1) and g (1) = 1 

⇒ Both 1, and – 1 have the same image in R 

⇒ g is not one-one 

Similarly, for the functions x² + 1 and x² – x² + 1; 

– 1 and 1 have the same image in R, 

so both the functions are not one-one, 

hence bijective.

Answer: (b) = f (a)² 

 Given, 

f(x) = \(\frac{x}{x-1}\) 

⇒ \(\frac{f(a)}{f(a+1)} = \frac{\frac{a}{a-1}}{\frac{a+1}{a+1-1}}\) 

= \(\frac{a}{a-1} \times \frac{a}{a+1} = \frac{a^2}{a^2-1}\) 

= \(f(a^2)\) .

f (3) = \(\frac{3+1}{2} =2 \)  and  f(4) = \(\frac{4}{2}\) = 2

So, although 3 ≠ 4, f (3) = f (4) 

⇒ f is not one-one but many-one. 

For onto, consider any n ∈ N. 

When n is odd, then (2n – 1) is odd, so f (2n – 1) = \(\frac{2n-1+1}{2}\) = n

when n is even, then 2n is even, so f (2n) = \(\frac{2n}{2}\) = n

Thus for every n ∈ N, whether even or odd, there exists a pre-image n ∈ N, so the function f is onto. 

Hence the function f is many-one onto.

Answer: (b) one - one into 

 For all x₁, x₂ ∈ R, 

f (x₁) = f (x₂) 

⇒ e^x₁ = e^x₂ ⇒ x₁ = x₂ 

⇒ f is one-one. 

Let y = f (x) = e^x 

⇒ x = log_e y 

Now x is not defined for y < 0 

∴  y ∈ R does not have a pre-image in R when y is negative. 

∴  f is into.

To find: the function g : R → R : g o f = f o g = I_g

Formula used: (i) g o f = g(f(x))

(ii) f o g = f(g(x))

Given: f : R → R : f(x) = 10x + 7

We have,

f(x) = 10x + 7

Let f(x) = y

⇒ y = 10x + 7

⇒ y – 7 = 10x

\(\Rightarrow x= \frac{y-7}{10}\)

Let \(g(y)=\frac{y-7}{10}\) where g: R → R

g o f = g(f(x)) = g(10x + 7)\(=\frac{(10x+7)-7}{10}\)

= x

= I_^g

f o g = f(g(x)) =\(f(\frac{x-7}{10})\)

\(=10(\frac{x-7}{10})\)

\(=10(\frac{x-7}{10})+7\) 

= x – 7 + 7

= x

Clearly g o f = f o g = I_g

_{\(g(x)=\frac{x-7}{10}\)}

Many-one 

Let a₁, a₂ be any two arbitrary elements of R, 

then f (a₁) = f (a₂) ⇒ a₁² + a₁ = a₂² + a₂ 

⇒ a₁² – a₂² + a₁ + a₂ = 0 

⇒ (a₁ – a₂) (a₁ + a₂) + (a₁ – a₂) = 0 

⇒ (a₁ – a₂) (a₁ + a₂ + 1) = 0 

⇒ a₁ – a₂ = 0 or a₁ + a₂ + 1 = 0 

⇒ a₁ = a₂  or  a₁ + a₂ = – 1 ∈ R 

⇒ Both the inferences can be true. 

So, f (a₁) = f (a₂) does not necessarily imply a₁ = a₂ 

⇒ f is many-one. 

Into 

Let y = x² + x, then for all y ∈ R, there does not exist all x ∈ R, as for y = – 1, – 2, ..., etc. There is no pre-image in R. Hence f is an into function. 

⇒ f is many-one into function.

To state: Whether f is one-one

Given: f = {(1, 4), (2,5), (3, 6)}

Here the function is defined from A → B

For a function to be one-one if the images of distinct elements of A under f are distinct

i.e. 1,2 and 3 must have a distinct image.

From f = {(1, 4), (2, 5), (3, 6)} we can see that 1, 2 and 3 have distinct image.

Therefore f is one-one

f is one-one

One-One 

Let x, y be any two elements of A, then 

f (x) = f (y) ⇒ \(\frac{x-2}{x-3} = \frac{y-2}{y-3}\) 

⇒ (x – 2) (y – 3) = (y – 2) (x – 3) 

⇒ xy – 2y – 3x + 6 = xy – 2x – 3y + 6 

⇒ – x = – y 

⇒ x = y 

⇒ f is one-one.

Onto 

Let y be an element of B. Then 

f (x) = y 

⇒ y = \(\frac{(x-2)}{(x-3)}\) 

⇒ (x – 3)y = (x – 2) ⇒ xy – 3y = x – 2 

⇒ xy – x = 3y – 2 ⇒ x (y – 1) = 3y – 2 

⇒ x = \(\frac{3y-2}{y-1}\)  

For y ≠ 1, x = \(\frac{3y-2}{y-1}\) is a real number.

Also, A = R – {3} ⇒ \(\frac{3y-2}{y-1}\) ≠ 3,  because if we take \(\frac{3y-2}{y-1}\) = 3, then 3y – 2 = 3y – 3) 

⇒ 2 = 3 which is not true. 

Hence, f is onto. 

⇒ f is both one-one and onto.

f(x) = x! 

∴ Domain of f = set of whole numbers (W)

Given:

f (x) = \(\frac{5x+7}{x-3}\) , x ∈ R - {3}

Let a₁, a₂ be two arbitrary elements ∈R – {3} such that f (a₁) = f (a₂).

Then, f (a₁) = f(a₂) ⇒ \(\frac{5a_1+7}{a_1-3} = \frac{5a_2+7}{a_2-3}\)  

⇒ 5a₁a₂ + 7a₂ – 15a₁ – 21 = 5a₂a₁ – 21 + 7a₁ – 15a₂ 

⇒ – 22a₁ = – 22a₂ 

⇒ a₁ = a₂ 

⇒ f is one-one.

(ii) Given: f (x) = | x | 

⇒ f (1) = | 1 | = 1 and f (– 1) = |– 1| = 1 

Thus, f (1)= f (– 1) = 1 but 1 ≠ – 1 

∴  f is many-one. 

(iii) ∵ f (x) = [x] = n for all n ≤ x < n + 1, so, 1 ≤ x < 2 ⇒ f (x) = 1 

∴ The elements 1.1, 1.25, 1.4, 1.9 .... ∈ domain are mapped onto to the same value 1 in the range. 

Hence the function f is many-one.

Two functions f and g are said to be equal if and only if 

(i) domain of f = domain of g 

(ii) co-domain of f = co-domain of g 

(iii) f (x) = g(x) for every x belonging to their common domain. 

Ex. Let A = {1, 2} and B = {3, 6}. 

Then f : A → B given by f (x) = x² + 2 and g : A → B given by g(x) = 3x are equal as: 

• f (1) = 3 = g(1), f (2) = 6 = g(2) 

• Both f and g have the same domain and co-domain. 

Identity Function: 

The identity function maps each member of the domain onto itself, i.e., if I is the identity function I: R → R then for all x ∈ R, 

I(x) = x.

f(x) = ^5-xP_x-1

5 – x > 0, x – 1 ≥ 0, x – 1 ≤ 5 – x 

∴ x < 5, x ≥ 1 and 2x ≤ 6 

∴ x ≤ 3 

∴ Domain of f = {1, 2, 3}

Functions can be broadly categorized into two groups. 

(a) Algebraic Functions 

(b) Transcendental Functions 

 Algebraic Functions: 

A function that consists of a finite number of terms involving powers and roots of the independent variable x and the four fundamental operations of addition, subtraction, multiplication, and division is called an algebraic function.

e.g. 

5x³ – 7x² + 6, \(4\sqrt{7x-12}\) , \(2x^\frac{1}{2} \) +4x-7 , \(\frac{2x+1}{2x+3}\), etc.

The particular cases of algebraic functions are:

(i) Polynomial Functions: A function f (x) = a₀ + a₁x + a₂x² + ..... + a_nxⁿ, where n ∈ N and a₁, a₂, a₃, ..... , a_n ∈ R is called a polynomial function. 

Its 

Domain is the set of real numbers 

Range is the set of real numbers.

(ii) Rational Function:  A function f : A → R, f (x) = \(\frac{P(x)}{Q(x)}\) , where Q(x) ≠ 0 is called a rational function. Here A = {x : x ∈ R} such that Q(x) ≠ 0}, and P(x) and Q(x) are polynomial functions of x. 

Ex. 

(i) \(\frac{x^2+5x+6}{x^2-3x+2}\) is a polynomial function with domain = R – {1, 2}

(\(\because\) The roots of x2 – 3x + 2 are x = 1, x = 2)

(ii) f (x) = \(\frac{1}{x}\) : Domain = R – {0}, Range = R – {0}

(iii) f (x) = \(\frac{1}{x^2}\) : Domain = R – {0}, Range = Set of positive real numbers.

(iv) f (x) = \(\frac{1}{x^3}\) : Domain = R – {0}, Range = R – {0}.

(iii) Irrational Functions: 

Functions as \(\sqrt{3x^2-7x+4}\) , \(4x^2+\sqrt{3x}\) , \(\frac{1}{\sqrt[3]{5+3x}}\)

 i.e., involving radicals or non-integral powers of x are called irrational functions.

If f and g are real valued functions of x with domain set A and B respectively, then both f and g are defined in A ∩ B. Then,

(a) (f + g) x = f (x) + g(x) : Domain A ∩ B 

(b) (f – g) x = f (x) – g(x) : Domain A ∩ B 

(c) (fg) x = f (x) . g(x) : Domain A ∩ B

(d) \(\big[\frac{f}{g}\big]x = \frac{f(x)}{g(x)}\) : Domain A ∩ B 

(e) (f + k) x = f (x) + k : k is constant, so domain = A 

(f) (kf ) x = k f (x) 

(g) f ⁿ(x) = [f (x)]ⁿ

Transcendental Functions: A function that is not algebraic is called transcendental.

(i) Trigonometric functions as sin x, cos x, tan 3x, cosec 2x, etc. 

(ii) Inverse Trigonometric functions as sin^–1 x, tan^–1 2x, etc 

(iii) Exponential functions as y = a^x 

(iv) Logarithmic functions as the correspondence x → log x.

All variables in a program may not be accessible at all locations in that program. This depends on where you have declared a variable. The scope of a variable determines the portion of the program where you can access a particular identifier. There are two basic scopes of variables in Python :

(1) Global variables

 (2) Local variables

You can use the lambda keyword to create small anonymous functions. These functions are called anonymous because they are not declared in the standard manner by using the def keyword, lambda functions have their own local namespace and cannot access variables other than those in their parameter list and those in the global namespace.

Correct Answer is (B) one - one and into

f(x) = 2x

For One - One

f(x₁) = 2x₁

f(x₂) = 2x₂

put f(x₁) = f(x₂) we get

2x₁ = 2x₂

Hence, if f(x₁) = f(x₂), x₁ = x₂

Function f is one - one

For Onto

f(x) = 2x

let f(x) = y, such that y∈N

2x = y

\(\Rightarrow x=\frac{y}{2}\)

If y = 1

\(x=\frac{1}{2}=0.5\)

which is not possible as x∈N

Hence, f is not onto., f is into

Hence, option b is correct

\(f:A →A :f(x)=\frac{(x-2)}{(x-3)}\)

In this function

x = 3 and y = 1 are the asymptotes of this curve and these are not included in the functions of the domain and range respectively 

therefore the function f(x) is one one sice there are no different values of x which has same value of y .

and the function has no value at y = 1 here range = codomain

∴ f(x) is onto

Correct Answer is (D) one - one and onto

f(x) = e^x

Since the function f(x) is monotonically increasing from the domain R⁺ → R⁺

∴f(x) is one –one

Range of f(x) = (1,∞) = R⁺ (codomain)

∴f(x) is onto

∴ f : R⁺ → R⁺ : f(x) = e^x is one - one onto.

Correct Answer is (B)  one - one and onto

\(f:[\frac{-\pi}{2},\frac{\pi}{2}]\)→ [-1, 1] : f(x) = sin x

Here in this range, the function is NOT repeating its value,

Therefore it is one - one.

Range = Codomain

∴Function is onto

Hence, option B is the correct choice.

It is given that

f(x) = (2x + 1) and g(x) = (x² – 2)

(i) g o f = g o f (x) = g{f(x)} = g {2x + 1}

So we get

= (2x + 1)² – 2

It can be written as

= 4x² + 4x + 1 – 2

On further calculation

= 4x² + 4x – 1

(ii) f o g = f o g (x) = f {g(x)} = f{x² – 2}

So we get

= 2 {x² – 2} + 1

It can be written as

= 2x² – 4 + 1

On further calculation

= (2x² – 3)

(iii) f o f = f o f (x) = f{f(x)} = f(2x + 1)

So we get

= 2 (2x + 1) + 1

It can be written as

= 4x + 2 + 1

On further calculation

= 4x + 3

(iv) g o g = g o g (x) = g{g(x)} = g { x² – 2}

So we get

= (x² – 2)² – 2

On further calculation

= x⁴ – 4x² + 4 – 2

We get

= x⁴ – 4x² + 2

False

14 2/10 = 14.2

We know that for every negative and positive value of x we get the same value of f(x) which is not one-one that is many one.

x⁴ ≥ 0 for all values of f(x)

Thus, range [0, ∞) ⊂ R

Here, the function is into function.

Correct Answer is (C) (x² – 2)

\(f(x+\frac{1}{x})=x^2\frac{1}{x^2}\)\(=(x+\frac{1}{x})^2-2\)

\(\Rightarrow f(x)=x^2-2\)