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The output is “Computer Applicatios”. That is the second string is concatenated to the first string.

abs(): This function returns the absolute value.

There is an error. The error is , Y is a constant T must be initialised and cannot be changed during execution.

Functions in C++

Some functions that are already available in C++ are called pre defined or built in functions.

In C++, we can create our own functions for a specific job or task, such functions are called user defined functions.

A C++ program must contain a main( ) function. A C++ program may contain many lines of statements(including so many functions) but the execution of the program starts and ends with main( ) function.

Predefined functions 

To invoke a function that requires some data for performing the task, such data is called parameter or argument. Some functions return some value back to the called function.

String functions 

To manipulate string in C++ a header le called string.h must be included.

(a) strlen( )- to find the number of characters in a string(i.e. string length).

Syntax: strlen(string);

Eg- cout<<strlen(“Computer”);It prints 8.

(b) strcpy( )- It is used to copy second string into first string.

Syntax: strcpy(string₁ ,string₂);

Eg. strcpy(str,”BVM HSS”); cout<<str; It prints BVM HSS.

(c) strcat( )- It is used to concatenate second string into first one.

Syntax: strcat(string₁ ,string₂)

Eg. strcpy(str₁ ,’’Hello”); strcpy(str₂ ,” World”); strcat(str₁ ,str₂);

cout<<str₁ ; It displays the concatenated string “Hello World”

(d) strcmp( )- it is used to compare two strings and returns an integer.

Syntax: strcmp (string₁,string₂)

if it is 0 both strings are equal.

if it is greater than 0(i.e. +ve) string₁ is greater than string₂

if it is less than 0(i.e. -ve) string₂ is greater than string₁

Eg.

include

#include using namespace std; int main()

{

char str1 [10],str2[10];

strcpy(str₁ ,”Kiran”);

strcpy(str₂,”Jobi”);

cout<<strcmp(str₁ ,str₂);

}

It returns a +ve integer.

(e) strcmpi( )- It is same as strcmp( ) but it is not case sensitive. That means uppercase and lowercase are treated as same.

Eg. “ANDREA” and “Andrea” and “andrea” are the same.

# include

# include

using namespace std;

int main( )

{

char str1 [10],str2[10];

strcpy(str₁ ,”Kiran”);

strcpy(str₂ ,”KIRAN”);

cout<<strcmpi(str₁,str₂);

}

It returns 0. That is both are same.

#include

using namespace std;

int sum(int x=100,int y=50,int z=10)

{

retum(x+y+z);

}

int main( )

{

cout<<sum()<<endl; cout<<sum(1)<<endl;

cout<<sum(1,2)<<endl;

cout<<sum(1,2)<<endl;

}

# include

using namespace std;

float avg(float n₁ , float n₂ , float n₃)

{

return ((n₁ + n₂ + n₃)/3);

}

int main( )

{

float x, y, z;

cout<<"Enter 3 nos”; cin>> x>>y>>z;

cout<<“the average is”<<avg(x,y,z);

}

# include<iostream> 

# include<cstdio> 

# include<cstring> 

using namespace std; 

int main()

{

int m;

char str1[100],str2[100];

cout<<"Enter first string";

gets(str1);

cout<<"Enter second string";

gets(str2);

m=strcmp(str1,str2);

if(m<0)

cout<<str1<<"is less than"<<str2;

else if(m>0)

cout<<str1<<"is greater than"<<str2;

else

cout<<str1<<"is equal to"<<str2;

}

# include<iostream>

# include<cstdio>

using namespace std;

int main()

{

char str[40];

int space=0,i;

cout<<” Enter a string:”; 

gets(str);

for(i=0;str[i]!='\0'/;i++)

if(str[i]==32)

space++;

cout<<"The number of space is"<<space;

}

# include<iostream>

# include<cstdio>

# include<cctype>

using namespace std; 

int main()

{

char line(80);

int i,vowel=0;

puts(Enter a string");

gets(line);

for(i=0;line[i]!='\0';i++)

switch(to lower(line[i]))

{

case 'a':

case 'e':

case 'i':

case 'o':

case 'u':

vowel++;

}

cout<<"The number of vowels is"<<vowel;

}

gets() function can be used to accept a lengthy text.

(d) getchar()

# include<iostream>

# include<cstdio>

using namespace std;

int main()

{

int i,words=1;

char str[80];

cout<<” Enter a string\n”; 

gets(str);

for(i=0;str[i]!='\0'/;i++)

if(str[i]==32)

words++;

cout<<"The number of words is"<<words;

}

Put the functions in a file with name helloworld. 

py

(b) isalnum( );

python version 2.x print type(l/2) 

<type’int’>

print x//y = 2.0

On one line the text: 

helloworld

Statement is true.

 snippet print = 5.0

An exception is thrown.

(a) int len= strlen(str)

(c)  getchar()

getchar is a character function but gets is a string function. The header file cstdio must be included. It reads a character from the keyboard. 

Eg. 

char ch; 

ch=getchar();

cout<<ch;

gets is used to read a string from the key board. It reads the characters upto enter key. 

The header file 

cstdio must be included. 

char str[80];

cout<<” Enter a string”; 

gets(str);

gets( ) function is used to get a string from the key board including spaces.

puts( ) function is used to print a string on the screen. To use gets( ) and puts ( ) function the header file cstdio must be included.

(b) getchar()

To use getchar(), putchar(), gets() and puts(), cstdio header file is used

(c)  cstring

Given: f(x) = \(\frac{1}{(2x+1)}\), where x ≠ \(\frac{-1}2\)

Need to prove: f{f(x)} = \(\frac{2x +1}{2x+3}\) When x ≠ \(\frac{-3}2\)

Now placing f(x) in place of x

⇒ f{f(x)} = \(\frac{1}{2f(x) + 1}\)

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

⇒ f{f(x)} = \(\frac{1}{\frac{2+2x+1}{2x+1 }}\) = \(\frac{2x +1}{2x+3}\), Where x ≠ \(\frac{-3}2\)

[proved]

f(x) = x² + 3, x ≤ 2

= 5x + 7, x > 2

(i) f(3) = 5(3) + 7

= 15 + 7

= 22

(ii) f(2) = 2² + 3

= 4 + 3

 = 7

(iii) f(0) = 0² + 3 = 3

To prove: function is one-one and into

Given: f : N → N : f (x) = x²

Solution: We have,

f(x) = x²

For, f(x₁) = f(x₂)

⇒ x₁² = x₂²

⇒ x₁ = x₂

Here we can’t consider x₁ = -x₂ as \(x\in N\), we can’t have negative values

∴ f(x) is one-one

f(x) = x²

Let f(x) = y such that \(y\in N\)

⇒ y = x²

^{\(\Rightarrow x=\sqrt{y}\)}

If y = 2, as \(y\in N\)

Then we will get the irrational value of x, but \(x\in N\)

Hence f(x) is not into

Hence Proved

To prove: function is neither one-one nor onto

Given: f : R → R : f (x) = x²

Solution: We have,

f(x) = x²

For, f(x₁) = f(x₂)

⇒ x₁² = x₂²

⇒ x₁ = x₂ or, x₁ = -x₂

Since x₁ doesn’t has unique image

∴ f(x) is not one-one

f(x) = x²

Let f(x) = y such that \(y\in R\)

⇒ y = x²

⇒ \(x=\sqrt{y}\)

If y = -1, as \(y\in R\)

Then x will be undefined as we cannot place the negative value under the square root

Hence f(x) is not onto

Hence Proved

Surjective Function

A function f: A → B is said to be onto if every element in B has at least one pre-image in A. Thus, if f is onto then for each y ∈ B ∃ at least one element x ∈ A such that y = f(x)

Also, f is onto range f = B

Example:

Let N be the set of all natural numbers and let E be the set of all even natural numbers

Let f: N → E: f(x) = 2x Ɐ x ∈ N

Then, y = 2x => x = ½ y

Thus, for each y ∈ E there exists ½ y ∈ N such that

f(1/2 y) = (2 × ½ y) = y

Hence, f is onto.

Into Function

A function f: A → B is said to be into if there exists even a single element in B having no pre-image in A.

So, f is into range (f) ⊂ B.

Example:

Let A = {2, 3, 5, 7} and B = {0, 1, 3, 5, 7}

Let f: A → B: f(x) = (x – 2) then

f(2) = (2 – 2) = 0

f(3) = (3 – 2) = 1

f(5) = (5 – 2) = 3

f(7) = (7 – 2) = 5

Thus, every element in A has a unique image in B

Now ∃ 7 ∈ B having no pre-image in A

Hence, f is into.

Bijective Function

A one-one onto function is said to be bijective or a one-to-one correspondence.

Example:

If f(x) = x² from set of positive real numbers to positive real numbers which is both surjective and injective.

Hence, f is bijective function.

Injective function

A function f: A → B is said to be one-one if distinct elements in A have distinct images in B.

Example:

Let N be the set of all natural numbers.

Let f: N → N: f(x) = 2x Ɐ x ∈ N

Then, f(x₁) = f(x₂)

2x₁ = 2x₂

x₁ = x₂

Hence, f is one-one.

Definition: A relation R from a set A to a set B is called a function if each element of A has a unique image in B.

It is denoted by the symbol f:A→B which reads ‘f’ is a function from A to B ‘f’ maps A to B

Let f:A→B,then the set A is known as the domain of f & the set B is known as co - domain of f .The set of images of all the elements of A is known as the range of f.

Thus, Domain of f = {a | a ∈ A,(a ,f(a)) ∈ f )

Range of f = {f(a) | a∈ A ,f(a) ∈ B }

Example: The domain of y = sin x is all values of x i.e. R , since there are no restrictions on the values for x. The range of y is between −1 and 1. We could write this as −1 ≤ y ≤ 1.

Given as

f(x) = log_e (1 – x) and g(x) = [x]

As we know, f(x) takes real values only when 1 – x > 0

1 > x

x < 1, ∴ x ∈ (–∞, 1)

The domain of f = (–∞, 1)

Similarly, g(x) is defined for all real numbers x.

The domain of g = [x], x ∈ R

= R

(i) f + g

As we know, (f + g) (x) = f(x) + g(x)

(f + g) (x) = log_e (1 – x) + [x]

The domain of f + g = Domain of f ∩ Domain of g

The domain of f + g = (–∞, 1) ∩ R

= (–∞, 1)

∴ f + g: (–∞, 1) → R is given by (f + g) (x) = log_e (1 – x) + [x]

(ii) fg

As we know, (fg) (x) = f(x) g(x)

(fg) (x) = log_e (1 – x) × [x]

= [x] log_e (1 – x)

The domain of fg = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

∴ fg: (–∞, 1) → R is given by (fg) (x) = [x] log_e (1 – x)

(iii) f/g

As we know, (f/g) (x) = f(x)/g(x) 

(f/g) (x) = log_e (1 – x)/[x]
The domain of f/g = Domain of f ∩ Domain of g

= (–∞, 1) ∩ R

= (–∞, 1)

However, (f/g) (x) is defined for all real values of x ∈ (–∞, 1), except for the case when [x] = 0.

Now, we have, [x] = 0 when 0 ≤ x < 1 or x ∈ [0, 1)

If  0 ≤ x < 1, (f/g) (x) will be undefined as the division result will be indeterminate.

The domain of f/g = (–∞, 1) – [0, 1)

= (–∞, 0)

∴ f/g: (–∞, 0) → R is given by (f/g) (x) = log_e (1 – x)/[x] 

(iv) g/f

As we know, (g/f) (x) = g(x)/f(x) 

(g/f) (x) = [x]/log_e (1 – x)

However, (g/f) (x) is defined for all real values of x ∈ (–∞, 1), except for the case when log_e (1 – x) = 0.

log_e (1 – x) = 0 ⇒ 1 – x = 1 or x = 0

If x = 0, (g/f) (x) will be undefined as the division result will be indeterminate.

The domain of g/f = (–∞, 1) – {0}

= (–∞, 0) ∪ (0, 1)

∴ g/f: (–∞, 0) ∪ (0, 1) → R is given by (g/f) (x) = [x]/log_e (1 – x) 

(a) Here, we need to find (f + g) (–1).

We have, (f + g) (x) = log_e (1 – x) + [x], x ∈ (–∞, 1)

By substituting x = –1 in the above equation, we get

(f + g)(–1) = log_e (1 – (–1)) + [–1]

= log_e (1 + 1) + (–1)

= log_e2 – 1

∴ (f + g) (–1) = log_e2 – 1

(b) Here, we need to find (fg) (0).

We have, (fg) (x) = [x] log_e (1 – x), x ∈ (–∞, 1)

By substituting x = 0 in the above equation, we get

(fg) (0) = [0] log_e (1 – 0)

= 0 × log_e1

∴ (fg) (0) = 0

(c) Here, we need to find (f/g) (1/2) 

We have, (f/g) (x) = log_e (1 – x)/[x], x ∈ (–∞, 0)

However, 1/2 is not in the domain of f/g.

∴ (f/g) (1/2) does not exist.

(d) Here, we need to find (g/f) (1/2) 

We have, (g/f) (x) = [x]/log_e (1 – x), x ∈ (–∞, 0) ∪ (0, ∞)

By substituting x = 1/2 in the above equation, we get

(g/f) (1/2) = [x]/log_e (1 – x)

= (1/2)/log_e (1 – 1/2)

= 0.5/log_e (1/2)

= 0/log_e (1/2)

= 0

Thus, (g/f) (1/2) = 0

Given, f : R⁺ → R : f(x) = log_e x 

f(x) = -2 

log_e x = -2 

Taking antilog on both sides 

x = e^-2

Hence, the value of x for which f(x) = -2 is e^-2

x = 3^3log₃2

 = \(3^{log3(2^3)}\)

 = 2³....[a^log_ab = b]

 = 8

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

Now, f(x) = 2x + 3 and g(x) = x² + 5

gof(x) = g(2x + 3) = (2x + 3)² + 5

⇒ gof(x) = 4x² + 12x + 9 + 5 = 4x² + 12x + 14

fog (x) = f(g(x)) = f (x² + 5) = 2 (x² + 5) + 3

⇒ fog(x)= 2x² + 10 + 3 = 2x² + 13

Hence, gof(x) = 4x² + 12x + 14 and fog (x) = 2x² + 13

Since, f:R → R and g:R → R

fog:R → R and gof:R → R

f(x) = 2x + x² and g(x)=x³

Now, gof(x) = g(f (x)) =g(2x + x²)

gof (x)=(2x + x²)³ = x⁶ + 8x³ + 6x⁵ + 12x⁴

and fog(x)=f(g(x))= f(x³)

⇒ fog(x) = 2x³ + x⁶

So, gof(x) = x⁶ + 6x⁵ + 12x⁴ + 8x³ and fog(x) = 2x³ + x⁶

Correct Answer is (B) {(1, 3), (3, 1), (4, 3)}

g = {(2, 3), (5, 1), (1, 3)}

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

⇒(g o f) = {(1, 3), (3, 1), (4, 3)}

Correct Answer is (C) x

\(f(x)=\sqrt[3]{3-x^3}\)

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

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

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

Answer : (b) one-one but not onto

For all (x1, x2) ∈ [0, ∞) 

f (x1) = f (x2) 

⇒\(\frac{2x_1}{1+2x_1}\) = \(\frac{2x_2}{1+2x_2}\) 

⇒ 2x₁ + 4x₁x₂ = 2x₂ + 4x₁x₂

⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂ 

⇒ f is one-one.

Let  y = f(x) = \(\frac{2x}{1+2x}\) 

⇒ y + 2xy = 2x 

⇒ y = 2x – 2xy = 2x (1 – y) 

⇒ x = \(\frac{y}{2(1-y)}\) 

x is not defined when (1 – y) = 0, i.e., y = 1 ∈ [0, ∞) 

⇒ f is not onto 

 ∴  f is one-one, not onto.

Correct Answer is (D) None of these

f(x) = x² – 3x + 2

⇒f(x) = x² - 2x - x + 2 = x(x - 2) - 1(x - 2)

⇒f(x) = (x - 2)(x - 1)

⇒f(x) = (x - 2)(x - 1)

⇒f(f(x)) = ( f(x) - 2)( f(x) - 1)

⇒f(f(x)) = ((x - 2)(x - 1) - 2) ((x - 2)(x - 1) - 1)

⇒f(f(x)) = (x² – 3x + 2 - 2) (x² – 3x + 2 - 1)

⇒f(f(x)) = (x² – 3x) (x² – 3x + 1)

⇒f(f(x)) = x⁴ - 3x³ + x² - 3x³ + 9x² - 3x

⇒f(f(x)) = x⁴ - 6x³ + 10x² - 3x

Correct Answer is (A) 0

f(x) = x², g(x) = tan x and h(x) = log x

\(\Rightarrow g(f(x))=tan(f(x))=tan(x^2)\)

\(\Rightarrow h(g(f(x)))=log(g(f(x)))=log(tan(x^2))\)

\(\Rightarrow h(g(f(\sqrt{\frac{\pi}{4}})))\)\(=log(tan(\sqrt{\frac{\pi}{4}}))\)\(=log(tan(\frac{\pi}{4}))\)\(=log(1)=0\)

b² = ac

Taking log on both sides, we get

log b² = log ac

∴ 2 log b = log a + log c

∴ log a + log c = 2 log b