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f: R → R

f(x) = x² + 2x – 1

Domain={… -2, -1. 0. 1, …}

Co-domain={… -1, -2, -1, 2,…}

Here, for two different elements -2 and 0 of domain, their Images are same In co-domain.

Therefore function f is many-one function.

f(x)= \(\sqrt{5600−4x}\)

Putting x = 1000,

f(4)= \(\sqrt{5600−4(1000)}\)

= \(\sqrt{5600−4000}\)

= \(\sqrt{1600}\) =40

f(x) = 20

∴ 20 = \(\sqrt{5600−4x}\)

∴ 400 = 5600 – 4x

4x = 5200

(20)² = 5600 – 4x

∴ 4x = 5600 – 400

∴ x = 1300

Hence, when x = 1300. then f(x) = 20.

f(x) = \(\frac{2x−4}{x+7}\)

Image = 0. Therefore, put f(x) = 0

0 = \(\frac{2x−4}{x+7}\)

∴ 0 = 2x – 4 ∴ 4 = 2x

Therefore, x = 2, the Image is zero.

f(x) = 6x³ – 5^x + 15

∴f(0) = 6 × 0 – 5⁰ + 15

= 0 – 1 + 15 = 14

∴f(0)=14

f(x) = 2x² + 1 / x

∴ f(3) + f( – 3) = 2 (3)² + 1 / 3 + 2 (- 3)2 + 1 / (−3)

= (2 × 9) + 1 / 3 + (2 × 9) – 1 / 3 = 18 + 18 = 36

Correct option is (b) Many-one function

f(x) = \(\frac{x^2−4}{x−2}=\frac{(x+2)(x−2)}{x−2}\) = x + 2

∴ f(0) + f(1) – f(-2)

= (0 + 2 + (1 + 2) – (-2 + 2)

= 2 + 3 – 0

= 5

Correct option is (b) g is not a function

f(x) = \(\sqrt{x^2−16}\); Domain = {4, 5}

Range R_f = {f(x) | x ∈ Domain}

= {f(4), f(5)}

Now, f(x) =\(\sqrt{x^2−16}\)

∴ f( 4) = \(\sqrt{(4)^2−16}\)

= \(\sqrt{16-16}\) = 0

f(5) = \(\sqrt{(5)^2−16}\)

= \(\sqrt{25-16}\) = \(\sqrt{9}\) = 3 (∵ x is positive.}

Hence, Range R_f = {0, 3}

g : N → N, i.e., N = {1, 2, 3, …};

N = {1, 2, 3,…}. Therefore, ‘subtract 2 from the elements of the domain’ this rule cannot be called a function.

Correct option is (a) f: {1, 2, 3, 4} → {3, 4, 5}, the rule ‘add 2 to the elements of domain’ is not a function.

k:X → Y

Domain X={t| t ∈ Z, -3 ≤ t ≤ 3)

= {-3, -2, -1, 0, 1, 2, 3}

k (t) = t² + 2

∴ k(- 3) = (-3)² + 2 = 9 + 2 = 11,

k (3) = (3)² + 2=11,

k (- 2) = (- 2)² + 2 = 4 + 2 = 6,

k (2) = (2)² + 2 = 6,

k(- 1) = (- 1)² + 2 = 1 + 2 = 3,k (1) = (1)² + 2 = 3,

k (0) = 0 + 2 = 2

Co-domain Y = {a|a ∈ N ⇒ 1, a ≤ 20}

= {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, …, 18, 19, 20}

Here, for two different elements (-3, 3) of domain, their images in co-domain are k (- 3) = k (3) = 11. Therefore, function k is many one function.

h : A → B

A = {1, 2, 3} and h (x) = x + 5

∴ h (1) = 1 + 5 = 6, h (2) = 2 + 5 = 7, h (3) = 3 + 5 = 8

B = {3, 4, 5, 6, 7, 8}

For the elements 1, 2, 3 of domain

h (1) ≠ h (2) ≠ h (3).

Therefore, function ft is one-one function.

(a) Pure function

If f: A → B. where A∈R then f is called a function of real variable.

If A and B are any two non-empty sets and each element of set A is related with one and only one element of set B by some rule, relation or correspondence, then it is called a function from set A to B and. is denoted by f: A → B.

f: N → N. ∴ N = {1, 2, 3, 4,…} and

N = {1, 2, 3, 4, …}

f(t) = t² + 1, t ∈ N

f(1)= 1 + 1 = 2,

f(2) = 2² + 1 = 5

Here, for two different elements of do-main

A, their images are different in co-domain. Therefore, function f is one-one function.

f: A → M

A = {x | x ∈ N, 1 ≤ x < 5}

= {1, 2, 3, 4}

M = {x | x G N, 1 ≤ x ≤ 20}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

f(x) = x² + 1

Range of f R_f= {f(x) | x ∈ A}

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

Now, f(x) = x² + 1

∴ f(1) = (1)² + 1 = 1 + 1 = 2

f(2) = (2)² + 1 = 4+ 1 = 5

f(3) = (3)² + 1 = 9 + 1 = 10

f(4) = (4)² + 1 = 16 + 1 = 17

Hence, Range of f R_f = {2, 5, 10, 17}

Domain of a function: Suppose, f: A → B, the set A is called domain of the function J. It is denoted by D_f.

Co-domain of a function : Suppose, f: A → B, the set B is called the co-domain of the function f.

f: Z → N.

∴ Z = {…-3,-2, -1, 0, 1, 2, 3,…} and N = {1, 2, 3, 4, …}

f(t) = t² + 1, t ∈ Z

∴ f(- 3) = 9 + 1 = 10, f(3) = 9 + 1 = 10, f(- 2) = 4 + 1 = 5 f(2) = 4 + 1 = 5

Here, for two different elements of domain A, their images are same in co-domain. Therefore, function is many-one function.

Suppose f: A → B. then a set of images or functional values of all the elements of set A is called range of function f It is denoted by f.

R_f = {f(x) | x ∈ A).

The range of a function may be co-domain itself or a subset of co-domain.

A = {x | x ∈ N, 1 < x ≤ 4}

A = {2, 3, 4} ⇒ Domain g [x] = x + 1

∴ g (2) = 2 + 1 = 3, g (3) = 3 + 1 = 4, g (4) = 4 + 1 = 5

Range of function R_f = {g (x) | x ∈ A}

= {g (2), g (3), g (4)} = {3, 4, 5}

Correct option is (a) f(A) = {f(x)|x ∈ A}

f: A → B, A = {10, 20, 30}

B = {18, 48. 98. 128. 148}

f(x) = 5x – 2

Domain: A = (10, 20, 30)

Co-domaln: B = {18, 48, 98, 128, 148}

Range: R_f ={f(x) | x ∈ A}

= {f(10), f(20), f(30)}

Now,f(10) = (5 × 10) – 2 = 50—2 = 48

f(20) = (5 × 20) – 2 = 100 – 2=98

f(30) = (5 × 30) – 2 = (150 – 2) = 148

∴ Range R_f = (48, 98, 148)

f: A → B, A = {- 3, -1, 1, 3};

B = {1, 0, 9}; f(x) = x². Therefore, f is a function.

Suppose, f: A → B. If for any two different elements x₁ and x₂. f(x₁) ≠ f(x₂), then function f is called one-one function.

Correct option is (c) Constant function

Correct option is (c) For any two different values of the domain their images are different.

k : R → R

∴ Domain = {…, -3, -2, -1, 0, 1, 2, …}

k (x) = x² + 3x – 12

∴ k (- 3) = (- 3)² + 3 (- 3) – 12 = – 12

k (- 2) = (- 2)² + 3 (- 2) – 12 = 4 – 6 – 12 = – 14

k (- 1) = (- 1 )² + 3 (- 1) – 12 = 1 – 3 – 12 = -14

k(0) = 0 + 0 – 12 = – 12

k(1) = 1 + 3 – 12 = -8

k (2) = 4 + 12 – 12 = 4

∴ Co-domain = {…, -12, -14, -14, -12, -8, 4, …}

Here, for the elements – 3 and 0, – 2 and – 1 of domain, their images are same. Therefore function k is many-one function.

Suppose, f: A → B. If for each element x₁, x₂, x₃,… of domain, the image is same in co-domain B, i.e.„ f(x₁) = f(x₂) = f(x₃) = … then, function f is called constant function.

Suppose, f: A → B. If for any two differents x₁, x₂ of domain A. f(x₁) = f(x₂), then function / is called many-one function.

Answer : (d) one-one onto function

\(\ f(x) =x|x| =\begin{cases}x^2, &\quad x \geq 0\\-x^2, &\quad x<0\end{cases}\)

Since – 1 ≤ x ≤ 1, therefore – 1 ≤ f (x) ≤ 1 

⇒ function is one-one onto.

Answer: (c) \(f(x) = x.\frac{a^x+1}{a^x -1}\)

(a) \(f(x) = \frac{a^x+1}{a^x -1}\) 

∴ f (– x) = \( \frac{a^{-x}+1}{a^{-x} -1}\) = \(\frac{\frac{1}{a^x}+1}{\frac{1}{a^x} -1}\) = \(\frac{1+a^x}{1-a^x}\) 

= \(- \frac{a^x+1}{a^x-1}\) = - f(x)

Hence f is odd.

(b) Similarly \(f(x) = \frac{a^x-a^{-x}}{a^x +a^{-x}}\) is an odd function.

(c) \(f(x) = x.\frac{a^x+1}{a^x -1}\) 

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

= (-x) \(\frac{1/a^x +1}{1/a^x -1}\) 

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

= x\(\big(\frac{a^x+1}{a^x-1}\big)\) = f(x) ⇒ f is even.

(d) f (x) = sin x f (– x) 

= sin (– x) 

= – sin x = – f (x) 

⇒ f is odd.

Answer : (d) 1

f : R → R given by f (x) = \(\sqrt{x}\)

∴ \(\frac{f(25)}{f(16)+f(1)} = \frac{5}{4+1} = \frac{5}{5}\) = 1.

Pure functions:

Pure functions are functions which will give exact result when the same arguments are passed. For example the mathematical function sin (0) always results 0. This means that every time you call the function with the same arguments, you will always get the same result. A function can be a pure function provided it should not have any external variable which will alter the behaviour of that variable.

Let us see an example

let square x

return: x * x

The above function square is a pure function because it will not give different results for same input. There are various theoretical advantages of having pure functions. One advantage is that if a function is pure, then if it is called several times with the same arguments, the compiler only needs to actually call the function once. Let’s see an example let i: = 0; if i < strlen (s) then – Do something which doesn’t affect s ++ i If it is compiled, strlen (s) is called each time and strlen needs to iterate over the whole of ‘s’.

If the compiler is smart enough to work out that strlen is a pure function and that ‘s’ is not updated in the loop, then it can remove the redundant extra calls to strlen and make the #loop to execute only one time. From these what we can understand, strlen is a pure function because the function takes one variable as a parameter, and accesses it to find its length.

This function reads external memory but does not change it, and the value returned derives from the external memory accessed. Impure functions: The variables used inside the function may cause side effects through the functions which are not passed with any arguments.

In such cases the function is called impure function. When a function depends on variables or functions outside of its definition block, you can never be sure that the function will behave the same every time it’s called. For example the mathematical function random Q will give different outputs for the same function call, let Random number let a : = random( ) if a > 10 then

return: a

else

return: 10 

Flere the function Random is impure as it is not sure what will be the result when we call the function.

(a) Definitions are expressions

(d) Definitions

(a) 1 – (iv) 2 – (iii) 3 – (ii) 4 – (i)

The Interface defines an object’s visibility to the outside world.

(c) Interface

(a) a : = (24) has an expression

Subroutines are the basic building blocks of a computer programs.

(a) Algorithm

A function is a unit of code that is often defined within a greater code structure. Specifically, a function contains a set of code that works on many kinds of inputs, like variants, expressions and produces a concrete output.

In programming languages, subroutines are called as Functions

The class template specifies the Interfaces to enable an object to be created and operated properly.