Explore topic-wise InterviewSolutions in Current Affairs.

Following are the objectives of ‘Project Tiger’:

• To protect the tigers from hunting and poaching practices.
• To ensure the survival of tiger population.
• To maintain the tiger population in country so that they do not get extinct.

Yes, ELISA is a highly sensitive and precise procedure and can detect antigens even in the range of a nanogram. So, it can be used to detect HIV in blood.

There are 50 tiger reserves in India which are governed under Project Tiger.

The family was shocked to see the grandfather alive as all of them wanted to share his belongings which they would not be able to do then.

(a) Mrs. Slater.

(b) As they are stealing grandfather’s bureau before the others come.

(c) The bureau of the grandfather.

(d) Amelia’s husband.

(i) f (x) = | x – 3 | is defined for all x ∈ R, so domain of f (x) = R. 

 Now | x – 3 | ≥ 0 for all x ∈ R ⇒ 0 ≤ | x – 3 | < ∞ for all x ∈ R 

⇒ f (x) ∈ [0, ∞) for all x ∈ R ⇒ Range of f (x) = | x – 3 | is [0, ∞)

(ii) Let y = f (x) = \(\sqrt{x-5}\)

f (x) is defined for x – 5 ≥ 0 ⇒ x ≥ 5 ⇒ x ∈ [5, ∞)

Now y = \(\sqrt{x-5}\)  

⇒ x – 5 = y² ⇒ x = y² + 5 

⇒ x > 0 for all value of y 

For x to be real and x ∈ [5, ∞), y ∈ [0, ∞). 

∴ Range of f is [0, ∞ ) 

(iii) Let y = f (x) = \(\sqrt{3x^2-4x+5}\) 

f (x) is defined if 3x² – 4x + 5 ≥ 0 

⇒ 3\(\big(x^2-\frac{4}{3}x+\frac{5}{3}\big)\)≥ 0

⇒ \(3\big(x^2-\frac{4}{3}x+\frac{4}{9}-\frac{4}{9}+\frac{5}{3}\big) \geq 0\)

⇒ 3 \(\big(\big(x-\frac{2}{3}\big)^2 +\frac{11}{9}\big) \geq 0\) , which is true for all real numbers, 

i.e., Domain of f is (– ∞, ∞)

y = \(\sqrt{3x^2-4x+5}\)  

⇒ 3x² - 4x +5 =y² 

⇒ 3x² - 4x +(5-y²) = 0

⇒ x = \(\frac{4 \,\pm \sqrt{(-4)^2 -4 \times 3 \times (5-y^2)}}{2 \times 3}\) 

For x to be real, (– 4)² – 4 × 3 × (5 – y²) ≥ 0 

⇒ 16 – 60 + 12y² ≥ 0 

⇒ – 44 + 12y² ≥ 0 

⇒ 12y² ≥ 44 

⇒ \(y^2 \geq \frac{11}{3} \implies y \geq \sqrt{\frac{11}{3}}\) 

∴ Range of y = [ \(\sqrt{\frac{11}{3}}\) ,  ∞ )

(iv) Let y = f(x) = \(\frac{x}{1+x^2}\) 

The given function f(x) = \(\frac{x}{1+x^2}\) is defined for all real numbers, so domain of f is R.

y = \(\frac{x}{1+x^2}\) 

⇒ x²y -x+y = 0  ⇒ x = \(\frac{-(-1)\, \pm \sqrt{1-4y^2}}{2y}\)

Now, \(\frac{1\, \pm \sqrt{1-4y^2}}{2y}\) will be a real number if and only if 

1 – 4y² ≥ 0 and y ≠ 0 

⇒ 4y² – 1 ≤ 0 and y ≠ 0 

⇒ \(\big( y^2 -\frac{1}{4}\big) \leq 0\) and y ≠ 0

⇒ \(\big(y-\frac{1}{2}\big)\big(y+\frac{1}{2}\big) \leq 0\) and y ≠ 0

⇒ \(-\frac{1}{2} \) \(\leq\) \(\frac{1}{2}\)  and  y ≠ 0

⇒ y ∈ [ \(-\frac{1}{2} \), 0 ) \(\cup\) ( 0, \(\frac{1}{2}\) ]

For x = 0, y = 0 

∴ range is [ \(-\frac{1}{2} \), \(\frac{1}{2}\) ]

Let y = \(\frac{x^2-2}{x^2-3}\)  

y = \(\frac{x^2-2}{x^2-3}\)  is not defined for x² – 3 = 0, i.e., x = ± \(\sqrt{3}\) 

∴ Domain of y = R - { \(-\sqrt{3}\) , +\(\sqrt{3}\) }

Now  y = \(\frac{x^2-2}{x^2-3}\)  ⇒ yx² – 3y = x² – 2 

⇒ x² (y – 1) 

= 3y – 2 

⇒ \(x^2\) = \(\frac{3y-2}{y-1}\) 

∵ LHS is a perfect square ⇒ \(\frac{3y-2}{y-1}\) ≥ 0

⇒ (3y – 2) ≥ 0 and (y – 1) ≥ 0 

or (3y – 2) ≤ 0, (y – 1) ≤ 0 and y ≠ 1 (Note) 

⇒ y \(\geq \frac{2}{3}\)  and y > 1 or y ≤ \(\frac{2}{3}\) , y < 1 

⇒ y ∈ (1, ∞)  or y ∈ (- ∞ , \(\frac{2}{3}\)]

⇒ y ∈   (- ∞ , \(\frac{2}{3}\)]  \(\cup\)  (1, ∞) 

For x = ± \(\sqrt{3}\),  y does not exist.

Let f (x) = y = x² – 6x + 7 

⇒ x² – 6x + 7 – y = 0 

⇒ x = \(\frac{6\,\pm \sqrt{36-4(7-y)}}{2} = 3\,\pm \sqrt{8+4y}\)  

For real values of x, 8 + 4y ≥ 0 

⇒ y ≥ – 2 

∴  For f (x) = x² – 6x + 7, the range is [– 2, ∞)

. Let y = \(\frac{x^2-4}{x-2}\) 

⇒ y = x + 2 when x ≠ 2

Now \(\frac{x^2-4}{x-2}\) is not defined only when x = 2, therefore

Domain of y = \(\frac{x^2-4}{x-2}\) is R - {2}. 

But y = x + 2 will take up all real values, but y = 4, when x = 2, (not possible) 

∴  Range of function is R – {4}.

Answer : (a) = x ∈ (– ∞, ∞)

∵  | x | ≥ 0 ∀ x ∈R 

∴  \(\sqrt{|x|}\) is defined for all real values of x. 

Hence the domain of \(\sqrt{|x|}\)  is R or (– ∞, ∞)

Answer : (c) =  R – {0}

Since log_ba is defined only when a > 0, b > 0 and b ≠ 1, 

log x² is defined only when x² > 0, which is true for all real values of x except x = 0. 

∴ Domain of log x² is R – {0}.

Answer : (c) =  0 ≤ x < ∞ 

Since \(\sqrt{x}\) is defined for only non-negative real numbers, 

therefore domain of f (x) = (\(\sqrt{x}\))²  is [0, ∞).

Answer : (c) [1, ∞) 

f(x) = \(\frac{1}{1-x^2}\) is not defined when 1 – x² = 0, i.e. x = ± 1

∴  Domain of f (x) = R – {– 1, 1}

Let y = \(\frac{1}{1-x^2}\) ⇒ (1-x²) = \(\frac{1}{y}\) 

⇒ x2 = 1 -  \(\frac{1}{y}\)  ⇒ \(x\,\pm\sqrt{1-\frac{1}{y}}\)

This shows that x will not be defined when  1 -  \(\frac{1}{y}\) < 0, i.e., y < 1 

∴  Range of f : y ≥ 1, i.e. y ∈ [1, ∞)

Let f (x) = \(\sqrt{x}\) and g(x) = \(\sqrt{3x-2}\) 

Let the domain of f (x) be A and that of g(x) be B 

Then, domain of f (x) = A = [0, ∞] (∵ \(\sqrt{x}\) is defined when x > 0 )

domain of g(x) = B = (\(\frac{2}{3}, \) ∞ ) (∵ \(\sqrt{3x-2}\) is defined when 3x-2 ≥ 0 ⇒ x ≥ \(\frac{2}{3}\) )

∴  Domain of \(\sqrt{x}\) + \(\sqrt{3x-2}\)  = A \(\cap\) B = [0, ∞ ) \(\cap\) [\(\frac{2}{3} \) , ∞ ) 

= [ \(\frac{2}{3} \), ∞)

(i) Since f is a real function, 

9 – x² ≥ 0 

⇒ (3 + x) (3 – x) ≥ 0 

⇒ – a ≤ x ≤ a 

∴  Domain of f (x) = \(\sqrt{9-x^2}\) is {x| – 3 ≤ x ≤ 3}. 

(ii) Let f (x) = 10^x 

a^x is defined for all real values of x when a > 0 Here a = 10, 

∴  Domain of f (x) = 10^x is R.

(iii) f (x) = \(\frac{2}{4x+7}\) 

Since f is real, x can take all real values except the value for which 4x + 7 = 0, 

i.e., x = \(-\frac{7}{4}\) 

⇒x ≠ \(-\frac{7}{4}\)

∴ Domain of f (x) = \(\frac{2}{4x+7}\)  = R - { \(-\frac{7}{4}\) }. 

(iv) f (x) = log (2 – 3x) 

For f (x) to be defined (2 – 3x) > 0 

⇒ 2 > 3x 

⇒ x < \(\frac{2}{3}\) 

∴  Domain of f (x) = log (2 – 3x)  =  −∞< x < \(\frac{2}{3}\) 

or x \(\in\) (−∞ \(\frac{2}{3}\)  )

Answer : (b) [ \(\frac{1}{3}\),1] 

f(x) = \(\frac{1}{2-cos\,3x}\) 

– 1 ≤ cos 3x ≤ 1 

⇒ – 1 ≤ – cos 3x ≤ 1 (Multiplying all terms by – 1) 

⇒ 2 – 1 ≤ 2 – cos 3x ≤ 3 (Adding 2 to each term) 

⇒ 1 ≤ 2 – cos 3x ≤ 3 

⇒ \(1\geq \frac{1}{2-3\,cos\,3x}\geq \frac{1}{3}\)

Range of f is  [ \(\frac{1}{3}\),1] 

Answer: (a) [1, ∞)

Given 

f(x) = x² + \(\frac{1}{x^2+1}\)  = \(\frac{x^4+x^2+1}{x^2+1}\)  

 = \(\frac{x^4+2x^2+1-x^2}{x^2+1} = \frac{(x^2+1)^2-x^2}{x^2+1}\)  

= \((x^2+1) - \frac{x^2}{x^2+1}\)  

= 1+ x² \(\big(1-\frac{1}{x^2+1}\big)\) \(\geq\) 1 \(\forall\) x \(\in\) R

∴  Range of f (x) = [1, ∞ ).

Answer : (a) [0, 1]

Let y = f(x) = \(\sqrt{(x-1)(3-x)}\) 

⇒ y² = (x – 1) (3 – x) = – x² + 4x – 3 

⇒ x² – 4x + (3 + y²) = 0 

This is a quadratic in x, so

\(x = \frac{+4\,\pm\sqrt{16-4(3+y^2)}}{2}\)  = \(\frac{4\,\pm\,2\sqrt{1-y^2}}{2}\) 

Since x is real, 1 – y² ≥ 0 

⇒ y² – 1 ≤ 0 

⇒ – 1 ≤ y ≤ 1 

But f (x) attains only non-negative values, 

so range of f = [0, 1]

Answer : (b) [0, 1) 

. Let y = \(\frac{x^2}{1+x^2}\)  ≥ 0  for all x ∈ R 

⇒ y (1 + x) ² ≥ x² 

⇒ y + yx² – x² ≥ 0 

⇒ (y – 1)x² + y ≥ 0 

Let y = \(\frac{x^2}{1+x^2}\)  and y ≥ 0 for all x ∈ R. 

⇒ \(\frac{x^2}{1+x^2}\) ≥ 0 

Also x² < x² + 1 

⇒ \(\frac{x^2}{1+x^2}\) < 1  

∴  Required range = [0, 1).

To find the range of a function y = f (x):

Step 1. Find the domain of the function y = f (x) 

Step 2. Solve the equation y = f (x) to find x in terms of y. 

Step 3. Find the real values of y for which x is real. The set of values of y so obtained makes up the range of f

Once the domain of a function f (x) is known, the set of all the values that f (x) can take is called the range of the function.

To find the domain of a function, following points should be kept in mind:

(i) For algebraic functions

(a) denominator should be non-zero 

(b) expression under an even root should be non-negative

(ii) For trigonometric functions

(a) sin x and cos x are defined for all values of x 

(b) tan x and sec x are defined for all real values of x except x =\((2n+1)\frac{\pi}{2}\) where n∈I. 

(c) cot x and cosec x are defined for all real values of x except x = n π, where n ∈ I.

(iii) Logarithmic functions:

log_b a is defined when a > 0, b > 1 and b ≠ 1.

(iv) Exponential functions: 

a^x is defined for all real values of x, where a > 0

The set of numbers x for which the function f (x) is defined is called the domain of the function.

Ex. 

(i) f (x) = \(\frac{1}{x}\) ; Domain = R – {0} as the function is defined for all real numbers except 0. 

(ii) f (x) = \(\sqrt{x}\) ; Domain = All non-negative real numbers as \(\sqrt{x}\) is not defined when x is a negative real numbers.

Structure assignment is possible only if both structure variables/objects are same type.

(a) Adidas # 9.99; Adidas # 1.11

Objects declared along with structure definition are called global objects.

A structure can be passed to a function through its object. Therefore, passing a structure to a function or passing a structure object to a function is the same because structure object represents the structure. Like a normal variable, structure variable(structure object) can be passed by value or by references / addresses. Similar to built-in data types, structures also can be returned from a function.

Structures are usually passed by reference method because it saves the memory space and executes faster.

(c) global objects

Output:

Details of Book No. 1

Book name : Programming

Book author : Dromy

Details of Book No. 2

Book Name : C++ Programming

Book Author : Bjame Stroustrup

1 Iprogrammingl Dromy

2| C++ Programming | Bjame Stroustmp

Errors:

1. ‘i’ is missing.

2. Spaces are missing at two places

3. Structure name given wrongly.

Corrected structure:

struct employee {int eno; char ename [20];

char dept;}

employee e, e2;

int main()

{

cout<<“Enter 1st distance:”<<end1;

cout<<“Enter feet:”; cin>>d1.feet;

cout <<“Enter inch:”; cin >> d1.inch;

cout << “\n information for 2nd distance:” << end1;

cout<<“Enter feet:”; cin >> d2.feet;

cout <<“Enter inch:”; cin >> d2.inch;

sum.feet = d1 . feet + d2.feet;

sum.inch = d1.inch + d2.inch;

if (sum.inch > 12)

{

++ sum.feet;

sum.inch = 12;

}

cout << end1; “Sum of distance = ” <<

sum.feet << “feet” << sum.inch <<“inches”;

return 0;

}

Output:

Enter 1 st distance

Enter feet: 6

Enter inch: 3.4

Enter 2nd distance

Enter feet: 5

Enter inch: 10.2

Sum of distances = 12 feet 1.6 inches

When a structure is passed as argument to a function using call by value method, any change made to the contents of the structure variable inside the function to which it is passed do not affect the structure variable used as an argument.

#include

using namespace std;

struct Employee

{

char name[50];

int age;

float salary;

};

void printData(Employee); // Function declaration

int main()

{

Employee p;

cout << "Enter Full name:";cin>>p.name;

cout << "Enter age :";cin>>p.age;

cout << "Enter salary:";cin>>p.salary;

// Function call with structure variable as an argument

printData(p);

return 0;

}

void printData(Employee q)

{

cout << “\nDisplaying Information.” << endl; 

cout << “Name:” << q.name << endl;

cout <<”Age:” << q.age << endl;

cout << “Salary:” << q.salary;

}

Output:

Enter Full name: Kumar

Enter age : 55

Enter salary : 34233.4

Displaying Information.

Name : Kumar

Age : 55

Salary : 34233.4

In the above example, a structure named Employee is declared and used. The values that are entered into the structure are name, age and salary of a Employee are displayed using a function named print Data(). The argument for the above function is the structure Employee. The input can be received through a function named readData().

In this method of passing the structures to functions, the address of a structure variable /object is passed to the function using address of(&) operator. So any change made to the contents of structure variable inside the function are reflected back to the calling function.

The declaration of a 2 – D array is data – type array_name[row – size] [col – size]; In the above declaration, data-type refers to any valid C++ data – type, array _ name refers to the name of the 2 – D array, row – size refers to the number of rows and col-size refers to the number of columns in the 2 – D array.

Number of bytes allocated for type of array x Number of elements.

(d) t. seconds

(d) struct sum {int num;};

Structure definition is terminated by ;

(d) 123 Balu

An array of strings is a two – dimensional character array. The size of the first Index (rows) denotes the number of strings and the size of the second index (columns) denotes the maximum length of each string. Usually, array of strings are declared in such a way to accommodate the null character at the end of each string. For example, the 2 – D array has the declaration: char name [7][10];

In the above declaration,

No. of rows = 7;

No. of columns =10;

We can store 7 strings each of maximum length 10 characters.

(a) cout <<e[0].empno << e[0] → ename;

The data elements in the structure are also known as objects

“An array is a collection of variables of the same type that are referenced by a common name”. An array is also a derived datatype in C++. There are different types of arrays used in C++.

They are:

1. One – dimensional arrays

2. Two – dimensional arrays

3. Multi – dimensional arrays

#include

using namespace std;

int main()

{

charname[5];

cout<<“Enter your name:”; cin >>name;

cout<< “My name is”<< name;

}

Output:

Enter your name: PRIYA

My name is PRIYA

Array is good to use.

Reasons:

1. All 100 numbers are integer type.

2. Array index helps to access the numbers quickly.

(b) it will allocate the memory

(b) Function

Structure is a user – defined which has the combination of data items with different data types. This allows to group of variables of mixed data types together into a single unit. The structure provides a facility to store different data types as a part of the same logical element in one memory chunk adjacent to each other.

A structure without a name/tag is called anonymous structure.

Example:

struct

{

long rollno; 

int age;

float weight;

};

The student can be referred as reference name to the above structure and the elements can be accessed like student.rollno, student.age and student.weight.