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An introduction to functions (Algebra) for CAT exam

Tuesday, October 9th, 2018

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Quantitative Aptitude section of CAT has always been proven to be most difficult and challenging especially for all the mathematics haters. Yes, all those who have been running away from Math’s for all these years must now face their biggest fear and there’s no more hiding from it now. Quantitative Aptitude is all about high school elementary math’s, those silly geometry lessons which made no sense then or the Time, Speed and distance problems or Time and work (so confusing).  Also, who can forget the infamous Trigonometry which is still a mystery for most of us? Today in this blog, we will learn and work with one such topic which is Functions. Many of you who didn’t opt for math’s in your class 11^th might not be acquainted with this topic some of you might have. But to many, the basic concept of functions is still not clear. Students have this wrong picture in their mind where functions are just one theoretical chapter of class 12^th that have no significance. But unlike that, it’s really very important. So, let’s understand what they really are! and how to deal with them.

Before directly jumping on to the types and examples of Functions first understand what they truly are which will help us understand the interrelated concepts and topics and will make solving questions easy. So, let’s begin with the Formal definition of Functions.

Let A and B be two non-empty subsets. A relation f from A to B i.e. a sub set of A x B is called a function from A to B, If

1. For each, a ϵ A there exists b ϵ B such that (a, b) ϵ f.
2. (a, b) ϵ f and (a, c) ϵ f => b = c.

And that was the mathematical definition of Function now let’s learn what it means. In simple terms, if we have two set supposedly, for example

As given in above figure let there’s set A = {a, b, c, d} and set B = {1, 2, 3, 4} and let us call the set A be domain and B be codomain. Then, for each element present in A i.e. a, b, c, d there is a unique element corresponding to each of them in B. As you can see in the figure a goes to 4, b goes to 2, c goes to 1, and d goes to 3 i.e. for every element in domain there exist a unique element in codomain.

Now which of the above is a Function?

Fig 1 and Fig 3 are not functions (Why?). The answer lies in its definition. Therefore, Fig 1 is not a function reason being that in its Domain a has two images 2 and 4 which is not possible because as given in definition every element in its domain should have a unique image but here this is not the case. Similarly, Fig 2 is not a function with the same reason that here c in domain has got no image in codomain.

And, Fig 2 is a function as it satisfies the criteria of being a function that every element here in Fig 2 has got a unique image in its codomain.

Now that we have understood what conditions are necessary for functions. We can move to the basic properties and nature of functions.

Firstly, the basic notation of functions is Y = f(x)

Therefore, for any quadratic equation such as Y= x^2+x+5 can be re-written as f(x)= x^2+x+5 or g(x)= x^2+x+5.

Now, what is f (2) in the above function?

We will just replace x from above function by 2 and we will get the values. f(2) = 2^2+ 2+ 5 =4+ 2+ 5 =11

Domain and range of a function:

The set of all the inputs for a given function is called the domain of a function or if f: R→Z, here f is defined from real numbers to integer values. Then here the set of real numbers is a domain of a function. Here the set of integers Z is known as codomain.  Similarly, all the set of outputs (dependent variable) we get by plugging into the set of all inputs ( domain) in a given function is known as range of a function.

Before moving on to questions that determine the domain and range of a function. Look at this table given below.

Now we are in a position to solve a few questions on domain and range of function. Let’s start with a basic example:

Find range and domain of f(x)= x/ (x²+1)

Clearly, the domain of f(x)= R.

Let y= f(x). Then,

Y= f(x) => y= x/(x²+1)

x²y- x+ y= 0

x= (1±√1-4y²)/ 2y

Clearly, x will assume real values, if

1-4y²≥0 and y≠0 => 4y²-1≤0 and y≠0

⇒ y²-¼≤0 and y≠0

⇒ (y-½) (y+½) ≤0 and y≠0

⇒ -½≤y≤½and y≠0

⇒ y ϵ [-½, ½]- {0}

I hope now you have understood the basic way of finding out range and domain of functions. The above question was a type that comes in an exam but it was the basic and simple one, in an exam you will find a more complex question. Let’s do another such complex example.

Find the domain of f(x)= (√x-|x|)⁻¹

Since, |x|= {x, if x≥0

-x, if x<0}

⇒   x- |x|= {x-x= 0, if x≥0

x+ x= 2x, if x<0}

⇒ x- |x|≤0 for all x

⇒ 1/(√x-|x|) does not take any real values for any x ϵ R

⇒ f(x) is not defined for any x ϵ R

Hence, Domain (f)= φ.

Now that we have learned about domain and range of function we can move to another topic in function of importance which is Composition of Functions. But before directly hoping onto that we need to learn something about properties of functions given in following points.

1. Injective Functions: A function is called injective (or one-to-one) if for any a, b ϵ domain such that a ≠ b then, f(a) ≠ f(b).
2. Surjective Functions: A function is called surjective (or onto) if Range of the function = Codomain of the function.
3. Bijective Functions: A function is called bijective if it is both one-to-one and onto and is, therefore, invertible or the function has an inverse.

Composition of functions:

Suppose we have two functions f(x) and g(x), such that f: R→ R and G: R→ R the ꓱ a function fog(x) such that fog: R→ R or a function gof(x) such that gof: R→ R known as composition of function. In simple terms if output of one function becomes input of another it is known as composition.

Condition Necessary for composition of functions

1. fog(x) exists, if Range of f = Domain of g.
2. Similarly, gof(x) exists, if range of g= Domain of f.
3. If fog(x) = gof(x) = x then g is inverse of f and vice versa.

Let’s do one example now to understand the use of this concept.

If f(x)= 2x+5 and g(x)= x² then what is gof(x) and fog(x)?

Sol: gof(x)= g(f(x))= g(2x+5)= (2x+5)²= 4x²+20x+25.

Similarly, fog(x)= f(g(x))= f(x²)= 2x²+5.

You can solve any question on composition on similar lines as given above.

Odd and Even Functions:

1. Odd Function: A function is said to be odd if, f(-x)= -f(x) i.e. the graph of odd function is symmetric about origin. Examples of odd functions are x⁻¹, sin(x) etc.
2. Even Functions: A function is called even function if, f(-x) = f(x) i.e. the graph of even is symmetric about the y-axis. Examples of even functions are x², cos(x) etc.
3. Neither odd nor even Functions: There is a third category of functions called neither odd nor even that exists if the graph of function does not exhibit symmetry. Example of such type are f(x)= 2x.

Note: f(x)= 0 is both odd and even function.

Example:

If f(x) is a polynomial of degree 8 and f(x)f(1/x) = f(x) + f (1/x), then f(x) is odd, even or neither odd or even function?

Sol: Let f(x)= a₀x⁸ + a₁x⁷+….+a₇x + a₈

⇒ f(1/x)= a₀/x⁸ + a₁/x⁷+….+a₇/x + a₈

Now f(x)f(1/x) = f(x) + f(1/x)

⇒ (a₀x⁸ + a₁x⁷+….+a₇x + a₈) (a₀/x⁸ + a₁/x⁷+….+a₇/x + a₈)= (a₀x⁸ + a₁x⁷+….+a₇x + a₈) +( a₀/x⁸ + a₁/x⁷+….+a₇/x + a₈)

Comparing the coefficients of  like powers on both the sides, we get

a₀= ±1, a₈=1, a₁=a₂=…=a₇= 0

Therefore, f(x)= ±x⁸ + 1

⇒ f(-x) = ±x⁸ + 1

Hence, it is an even function.

I hope the above example would have helped you to understand the odd and even functions properly; there could thousands of different examples. And similarly, there could be variety of different questions can be formed on this concept using a mixture of any of these properties given above. But if we are thorough with them we can easily attempt any question. So, practice diverse questions of  functions of various forms to succeed in exam.

I believe the above blog would have benefit and guide you in your preparation. Do not dishearten if you are not able to solve questions in single attempt, give time and keep on trying because when there’s a will there’s a way.

You can also see How to Solve Questions on Seating and Circular Arrangement for CAT Exam

This article was contributed by Sejal Khurana. If you want to write for us, please email us on [email protected]

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