FUNCTIONS - Algebra - How to Prepare for Quantitative Aptitude for CAT

How to Prepare for Quantitative Aptitude for CAT (2014)

Block V: Algebra

...BACK TO SCHOOL

This block of chapters (consisting of Functions, Inequalities, Quadratic and other Equations and Logarithms) has always been a source of discomfort for students who have had negative experiences with Mathematics. At the same time students who have had a positive experience with mathematics throughout their school, junior college and college years find these chapters extremely easy.

Throughout my experience in training students preparing for the CAT exam and other aptitude exams like CMAT, SNAP, IIFT, MAT, XAT, etc. I have always critically tried to understand the key mental representations that differentiate the two categories of students I have mentioned above—how is it that two students who are basically similarly talented with similar IQs be so different when it comes to Higher Maths?

The reason I am talking about this issue is that Block V consists of some of the most critical areas of the entire QA Section for the CAT exam. Besides the XLRI entrance test, XAT also has a very high focus on this block of chapters. Hence, this is one area where you simply cannot allow yourselves to remain in the ‘I cannot solve Block V questions’ category if you want to seriously take a crack at the CAT Exam. However to enable students who are weak at Maths to move from the ‘I cannot’ to the ‘I can’ category in this block of chapters you will first have to believe that this can happen.

And the good news is that in my experience of having trained over 50000 students for aptitude examinations, I have seen a lot of mediocre students in Maths transit from being initially poor at the chapters of this block to becoming strong in the same.

If you belong to such a category of students what you first need to understand is that the reason for your inability in Higher Maths lies in the way you have approached it up to now. There is simply no reason that a better approach will not help you.

WHAT APPROACH CHANGE DO YOU NEED?

To put it simply—rather than trying to study these Higher Maths chapters in so called Mathematical language, you need to study the same in plain logical language (with as little Maths as possible). The following write up is an attempt at the same. My attempt in the following part of this section is to give you in as simple language as possible some of the missing technology that makes you consider yourself weak at Maths (esp. this block of chapters).

The following is a list of startup issues you need to resolve in your mind in order to make yourself comfortable with this block of chapters.

ISSUE 1: The X – Y Plot:

ISSUE 2: Locus of points

ISSUE 3: What is an equation?

ISSUE 4: Equations and Functions:

ISSUE 5:How inequalities relate to Graphs of functions?

Let us now start to look at these issues one by one

ISSUE 1: The X – Y Plot: The need for an X – Y plot essentially arises out of the need to measure anything which cannot be measured in a single dimension.

For instance, any object that can be represented on a straight line (like a rope, a tape etc.) could be easily represented on the number line itself. However, when you are required to represent any two dimensional figures (like table top, a book etc.) simply unidimensional representations (which can be done on the number line only) are not sufficient.

You are required to be able to represent things in two dimensions.

For this purpose, a vertical line is drawn from point ‘0’ of the number line. This line is perpendicular to the number line and is called the Y-axis (which in plain language terms can be understood as the axis which provides the vital second dimension of measurement which helps us describe plain figures better.)

In this situation, the horizontal line (which we know as the number line) is called the X-axis and the diagram which emerges represents what is known as the X – Y plane and is also called as the X – Y plot.

The number line (The X-axis) has a positive measurement direction on the right side of zero (and keeps going to a mathematically defined, in reality non-existent point called infinity.) On the left side of zero the X axis shows negative measurements and keeps going to negative infinity.

In a similar manner, the Y-axis starts from 0 and moves upwards to positive infinity and moves downwards to negative infinity.

The following figures illustrate this point.

Fig. 1 The number line (Also the X axis)

As can be seen the X – Y plot divides the plane into 4 parts (Quadrants). The origin is the point where both X and Y are zero.

Fig. 2 The X – Y plot

Plotting a Point on the X – Y Plot

Suppose we have to plot the point (2, 3) on the X – Y plot. In order to do so, we first need to understand what x = 2 means on the X – Y chart. On the X – Y plot if you start from the x = 2 point on the X-axis and draw a line parallel to the Y-axis you will get a set of points which represent a constant value of x (=2). Similarly for y = 3, draw a line parallel to the X-axis from the y = 3 point on the Y-axis.

In this situation, the point of intersection between these two lines will be the point (2, 3).

I would encourage the reader to try to draw the figure as described above.

ISSUE 2: Locus of Points In the above situation we have come across a set of points which represents x = 2. In mathematics lingo, a set of points is also referred to as a locus of points.

ISSUE 3: What is an equation? Consider the following:

(a)2x + 3 = 0

(b)x² – 5x + 6 = 0

In each of the above cases, we have a situation where there is an equality between what is written on the left side of the equality sign and the right side of the equality sign. i.e. LHS = RHS (Left Hand Side = Right Hand Side)

This is the basic definition of an equation.

All Equations have solutions

Consider the first equation above:

Here you have: 2x + 3 = 0 Æ 2x = –3 Æ x = –3/2

i.e. x = –1.5

This means that at the value x = –1.5, the condition of the equation (i.e. LHS = RHS) is satisfied. This value of x is called the ‘solution’ or the ‘root’ of the equation.

Note: The equation we have just seen had a highest power of ‘x’ as 1. Such equations are called as linear equations. [We shall come to this issue of why they are called as linear equations shortly]. What you need to remember at this stage is that linear equations have only one solution. (or Root as mathematicians prefer to call it.)

ISSUE 4: Equations and Functions: Before we move ahead into the second and third equations given above, let us first try to connect equations and functions.

If you remember we studied in school a chapter called as the equation of a straight line. For instance, parallel to the equation 2x + 3 = 0 we have y = 2x + 3 (which was defined as the equation of a straight line – hence 2x + 3 = 0 is called a linear equation).

In y = 2x + 3 we have a consistent relationship between the values of x and the corresponding values of y that we could get.

Thus for y = 2x + 3, we can create the following table giving us the specific points which satisfy the equation as below:

x	–3	–2	–1	0	1	2
y	–3	–1	+1	3	5	7

If you plot these points on the X – Y plane, you will get the following figure.

All the points in the table above will lie on this line. The locus of points which are commonly represented by y = 2x + 3 is represented by the line above.

In higher mathematics instead of calling expressions like y = 2x + 3 as linear equations, they are called as linear functions. (Perhaps the reason for this is that in higher maths, we are interested in identifying relationships and behaviors of interdependent variables especially how a change in one variable changes the other variables. Hence, while equations which are of the form f(x) = 0 deal with only one variable x, functions represent relationships between two or more variables. They are denoted in Maths as y = f(x) [Reads as: y is a function of x] or y = f(a, b) [Read as: y is function of a and b].

In general whenever we have any expression in x (denoted by f(x)) we can use it either as

y = f(x) [which is what a function is !!]

f(x) = 0 [An equation]

Other Principles Related to Functions and Equations

(a)Every function of the form y = f(x) can be plotted on the X – Y plane.

For instance, consider what we did with the function y = 2x + 3. [This form of writing a function is called as its analytical representation, while the graph plotted is called as the graphical representation of the function.]

We first created a table where we started putting values of x and getting the respective value for y. Thus, we got a locus of points (as shown in the table above) which represented the function y = 2x + 3. This representation of the function is called as the analytical representation of the function.

Further, the line representing y = 2x + 3 was plotted on the X – Y plane. This is called as the graphical representation of the function.

(b)The point/points where the graph of the function y = f(x) cuts the x-axis, represents the solution of the equation f(x) = 0.

Thus, if you noticed y = 2x + 3 cuts the x axis at x = –1.5. At the same time the root or solution of the equation 2x + 3 = 0 is x = – 1.5.

This is not accidental but will happen in each and every case. Thus, in general you can state that the point where the graph of a function y = f(x) cuts the x-axis is the solution of the equation f(x) = 0. [Note: This will be true for every expression in x-irrespective of the highest power of ‘x’ in the expression.]

It is a consequence of this logic, that linear functions will cut the x-axis only once (since linear equations have only one solution.)

(c)Quadratic Equations and Quadratic Functions: Let us now pick up the equation: x² – 5x + 6 = 0.

If you solve this equation you will get the solution of this equation at two values of x. viz. at x = 2 as well as at x = 3.

[Remember, the root/solution of an equation is that value of the variable which makes the LHS = RHS in the equation. You can easily see that at both x = 2 and x = 3 the expression x² – 5x + 6 yields a value of 0.

At x = 2, x² – 5x + 6 Æ 2² – 5 × 2 + 6 = 0

At x = 3, x² – 5x + 6 Æ 3² – 5 × 3 + 6 = 0

At this point you can predict that the graph of the expression y = x² – 5x + 6 will cut the x axis at both these points viz. x = 2 and x = 3.

If you now try to plot y = x² – 5x + 6, your tabular representation will be:

	x	–5	–4	–3	–2	–1	0	1	2	3	4	5
	y	56	42	30	20	12	6	2	0	0	2	6

This locus of points will look like the figure below:

Naturally, the graph cuts the x axis at x = 2 and x = 3.

Special Note on Quadratic Equations

Since all quadratic equations have two solutions, graph of the function for the same expression should cut the x-axis at two points.

However, this might not always be true.

For instance, look at the following graphs.

Case 1:

In this case the roots are said to be real and equal.

Case 2:

In this case the roots of the quadratic equation are imaginary with respect to the x-axis.

Note: You also need to know that quadratic functions can also look inverted as shown below. This occurs when the coefficient of x² is negative.

In general, quadratic equations then can take any of six shapes (with respect to the x-axis):

(d)How cubic functions look: Cubic functions have x³ as their highest power of x. Cubic equations have three solutions and hence cubic functions will cut the x-axis thrice.

They can look like:

CASE 1: When the Coefficient of x³ is Positive

(A)Cubic functions with 3 distinct real roots (e.g.: y = x³ – 6x² + 11x – 6)

(B)Cubic function with three real and equal roots. e.g.: y = (x – 1)³

(C)Cubic function with one real and two imaginary roots

CASE 2: When the Coefficient of x³ is Negative

(A)Cubic function with 3 distinct real roots.

(B)Cubic function with three real and equal roots.

(C)Cubic function with one real and two imaginary roots.

You can similarly visualize functions with highest power 4, 5 and so on to have shapes such that they can cut the x – axis 4 times, 5 times and so on. The following are standard figures for expressions with

(a)Degree 4 (i.e. x⁴) [4 roots]

(b)Degree 5 [5 roots]

A Key Note

Graphical representations and ability to understand how an expression in x will look on the X – Y plot is a very key factor in your moving towards proficiency and self dependence in this crucial block of chapters. Hence, I would encourage you to try to plot the graphs of as many y = f(x) functions as you can imagine. To give you a target, I would recommend a minimum of 100 graphs to be plotted by you. Only then would you be able to move to a level of comfort in visualizing graphs.

ISSUE 5: How Inequalities Relate to Graphs of Functions?

We have just seen the relationship between a function and an equation. While a function is written in the y = f(x) form, an equation is written in the f(x) = 0 form.

An inequality (or an inequation) can be written in the form f(x) > 0 or f(x) ≥ 0.

Thus, if we say x² – 5x + 6 > 0, we are looking for those values of x where the above expression yields values above 0.

Going back to the graph for x² – 5x + 6, it is obvious that the function is greater than 0 when x > 3 or x < 2.

In such a case, the solution of the inequality x² – 5x + 6 > 0 becomes x > 3 or x < 2.

Understand here that when we are solving an inequality all we are bothered about is to see at what values of x does the inequality hold true.

Pre-assessment Test

Directions for Questions 1 to 3: Answer the questions on the basis of the tables given below.

Two binary operations (+) and * are defined over the set {i, j, k, l, m} as per the following tables:

(+)	i	j	k	l	m
i	i	j	k	l	m
j	j	k	l	m	i
k	k	l	m	i	J
l	l	m	i	j	k
m	m	i	j	k	l

*	i	j	k	l	m
i	i	i	i	i	I
j	i	j	k	l	m
k	i	k	m	j	L
l	i	l	j	l	k
m	i	m	l	k	j

Thus according to the first table k (+) l = i while according to the second table l* m = k, and so on. Also, let k² = k * k, l³ = l * l * l, and so on.

1.	What is the smallest positive integer n such that kⁿ = l?
	(a) 4	(b) 5
	(c) 2	(d) 3
2.	Upon simplification m (+) [k * {k (+) (k * k)}] equals
	(a) j	(b) k
	(c) l	(d) m
3.	Upon simplification, {i¹⁰ * (k¹⁰ (+) l⁹)} (+) j⁸ equals
	(a) j	(b) k
	(c) l	(d) m
4.	What is the sum of n terms in the series: log m + log (m²/n) + log (m⁴/n³) +……..
	(a) log [n^(n–1) /m⁽ⁿ⁺¹⁾]ⁿ^/2
	(b) log [m^m/nⁿ]ⁿ^/2
	(c) log [m^(1–n)/n^(1–m)]ⁿ^/2
	(d) log [m⁽ⁿ⁺¹⁾/n^(n–1)]ⁿ^/2
5.	If n is such that 36 £ n £ 72, then x = (n² + 2 + 16)/(n + 4) satisfies:
	(a) 20 < x < 54	(b) 23 < x < 58
	(c) 25 < x < 64	(d) 28 < x < 60
6.	A real number x is such that it satisfies the condition: 2–1/m < x £ 4 + 1/m,for every positive integer m. Which of the following best describes the range of values that x can take:
	(a) 1 < x < 5	(b) 1 < x £ 4
	(c) 1 < x £ 5	(d) 1 £ x £ 4
7.	If 1/3 log ₃M + 3 log ₃N = 1 + log _0.0085, then:
	(a) M⁹ = 9/N	(b) N⁹ = 9/M
	(c) M³ = 3/N	(d) N⁹= 3/M
8.	If f (x) = log {(1 + x)/ (1 – x)}, then f(x) + f(y) is:
	(a) f (x + y)
	(b) f{(x + y)/ (1 + xy)}
	(c) (x + y) f {1/ (1 + xy)}
	(d) f(x) + f(y)/ (1 + xy)
9.	The nth element of a series is represented as:
	X_n= (–2)ⁿ X_n_–1
	If X₀= x and x > 0 then the following is always true.
	(a) X_n is positive if n is even
	(b) X_n is positive if n is odd
	(c) X_n is negative if n is even
	(d) None of these
10.	The number of roots common between the two equations x³ + 3x² + 4x + 7 = 0 and x³ + 2x² + 7x + 5 = 0 is:
	(a) 0	(b) 1
	(c) 2	(d) 3
11.	If \|y\| ≥ 1 and x = –\|a\| y, then which one of the following is necessarily true?
	(a) a – xy < 0	(b) a – xy ≥ 0
	(c) a – xy > 0	(d) a – xy £ 0
12.	Consider the sets T_n = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3, ….., 96. How many of these sets do not contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ….)?
	(a)16	(b) 15
	(c)14	(d) 13
13.	If 13m + 1 < 2n, and n + 3 = 5y², then:
	(a) m is necessarily less than n
	(b) m is necessarily greater than n
	(c) m is necessarily equal to n
	(d) None of the above is necessarily true
14.	If both a and b belong to the set {1, 2, 3, 4}, then the number of equations of the form
	ax² + bx + 1 = 0 having real roots is:
	(a)10	(b) 7
	(c) 6	(d) 12
15.	If three positive real numbers x, y and z satisfy y – x = z – y and xyz = 4, then what is the minimum possible value of y?
	(a) 2^1/3	(b) 2^2/3
	(c) 2^1/4	(d) 2^3/4
16.	There are 12 towns grouped into four zones with three towns per zone. It is intended to connect the towns with telephone lines such that every two towns are connect with three direct lines if they belong to the same zone, and with only one direct line otherwise. How many direct telephone lines are required?
	(a) 72	(b) 90
	(c) 96	(d) 144
17.	If log₁₀ x + log ₁₀x = 2 log_x 10, then a possible value of x is given by:
	(a) 10	(b) 1/10
	(c) 1/100	(d) None of these
18.	Find the area enclosed by the region in sq. units described by 0 = x = 5 and 0 £ \|y\| £ 8.
	(a) 40	(b) 80
	(c) 160	(d) 120
19.	The value of the expression [(1/2) + (1/100)] + [(1/2) + (2/100)] + [(1/2) + (3/100)] + ….[(1/2) + (103/100)] is equal to
	(a) 53	(b) 54
	(c) 52	(d) None of these
20.	The number of non-negative real roots of 2^x – x – 1 = equals:
	(a) 0	(b) 1
	(c) 2	(d) 3
21.	Let a, b, c and d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Given m, which one of the following is necessarily true?
	(a)The minimum possible value of a² + b² + c² + d² is 4m² – 2m + 1
	(b)The minimum possible value of a² + b² + c² + d² is 4m² + 2m + 1
	(c)The maximum possible value of a² + b² + c² + d² is 4m² – 2m + 1
	(d)The maximum possible value of a² + b² + c² + d² is 4m² + 2m + 1
22.	The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f …… is
	(a) u	(b) v
	(c) w	(d) x
23.	Find the area enclosed by the curve \|x\| + \|y\| = 4.
	(a) 24	(b) 28
	(c) 32	(d) 36
24.	Let y = min {(x + 8), (6 – x)}. If x e R, what is the maximum value of y ?
	(a) 8	(b) 7
	(c) 6	(d) 10
25.	Let S be the set of integers {3, 11, 19, 27, …… 451, 459, 467} and M be a subset of S such that no two elements of M add up to 470. The maximum possible number of elements in M is:
	(a) 32	(b) 28
	(c) 29	(d) 30

ANSWER KEY

1. (a)	2. (a)	3. (d)	4. (d)
5. (d)	6. (c)	7. (b)	8. (b)
9. (d)	10. (a)	11. (b)	12. (a)
13. (d)	14. (b)	15. (b)	16. (b)
17. (b)	18. (b)	19. (b)	20. (c)
21. (b)	22. (d)	23. (c)	24. (b)
25. (d)

Chapter 13. FUNCTIONS

Quantities of various characters such as length, area, mass, temperature and volume either have constant values or they vary based on the values of other quantities. Such quantities are called constant and variable respectively.

Function is a concept of mathematics that studies the dependence between variable quantities in the process of their change. For instance, with a change in the side of a square, the area of the square also varies. The question of how the change in the side of the square affects the area is answered by a mathematical relationship between the area of the square and the side of the square.

Let the variable x take on numerical values from the set D.

A function is a rule that attributes to every number x from D one definite number y where y belongs to the set of real numbers.

Here, x is called the independent variable and y is called the dependent variable.

The set D is referred to as the domain of definition of the function and the set of all values attained by the variable y is called the range of the function.

In other words, a variable y is said to be the value of function of a variable x in the domain of definition D if to each value of x belonging to this domain there corresponds a definite value of the variable y.

This is symbolised as y = f(x) where f denotes the rule by which y varies with x.

BASIC METHODS OF REPRESENTING FUNCTIONS

Analytical Representation

This is essentially representation through a formula.

This representation could be a uniform formula in the entire domain, for example, y = 3x²

by several formulae which are different for different parts of the domain.

Example: y = 3x² if x < 0

and y = x² if x > 0

In analytical representations, the domain of the function is generally understood as the set of values for which the equation makes sense.

For instance, if y = x² represents the area of a square then we get that the domain of the function is x > 0.

Problems based on the analytical representation of functions have been a favourite for the XLRI exam and have also become very common in the CAT over the past few years. Other exams are also moving towards asking questions based on this representation of functions.

Tabular Representation of Functions

For representing functions through a table, we simply write down a sequence of values of the independent variable x and then write down the corresponding values of the dependent variable y. Thus, we have tables of logarithms, trigonometric values and so forth, which are essentially tabular representations of functions.

The types of problems that appear based on tabular representation have been restricted to questions that give a table and then ask the student to trace the appropriate analytical representation or graph of the function based on the table.

Graphical Representation of Functions

This is a very important way to represent functions. The process is: on the coordinate xy plane for every value of x from the domain D of the function, a point P(x, y) is constructed whose abscissa is x and whose ordinate y is got by putting the particular value of x in the formula representing the function.

For example, for plotting the function y = x², we first decide on the values of x for which we need to plot the graph.

Thus we can take x = 0 and get y = 0 (means the point (0, 0) is on the graph).

Then for x = 1, y = 1 ; for x = 2, y = 4; for x = 3, y = 9 and for x = –1 y = 1; for x = –2 , y = 4, and so on.

EVEN AND ODD FUNCTIONS

Even Functions

Let a function y = f(x) be given in a certain interval. The function is said to be even if for any value of x

Æ f(x) = f(–x)

Properties of even functions:

(a)The sum, difference, product and quotient of an even function is also an even function.

(b)The graph of an even function is symmetrical about the y-axis.

However, when y is the independent variable, it is symmetrical about the x-axis. In other words, if x = f(y) is an even function, then the graph of this function will be symmetrical about the x-axis. Example: x = y².

Examples of even functions: y = x², y = x⁴, y = –3x⁸, y = x² + 3 , y = x⁴/5, y = | x | are all even functions.

The symmetry about the y-axis of an even function is illustrated below.

Odd Functions

Let a function y = f(x) be given in a certain interval. The function is said to be odd if for any value of x

Æ f(x) = – f(–x)

Properties of odd functions.

(a)The sum and difference of an odd function is an odd function.

(b)The product and quotient of an odd function is an even function.

(c)The graph of an odd function is symmetrical about the origin.

The symmetry about the origin of an odd function is illustrated below.

Examples of odd functions y = x³, y = x⁵, y = x³ + x, y= x/(x² + 1).

Not all functions need be even or odd. However, every function can be represented as the sum of an even function and an odd function.

Inverse of a Function

Let there be a function y = f(x), which is defined for the domain D and has a range R.

Then, by definition, for every value of the independent variable x in the domain D, there exists a certain value of the dependent variable y. In certain cases the same value of the dependent variable y can be got for different values of x. For example, if y = x², then for x = 2 and x = –2 give the value of y as 4.

In such a case, the inverse function of the function y = f(x) does not exist.

However, if a function y = f(x) is such that for every value of y (from the range of the function R) there corresponds one and only one value of x from the domain D, then the inverse function of y = f(x) exists and is given by x = g(y). Here it can be noticed that x becomes the dependent variable and y becomes the independent variable. Hence, this function has a domain R and a range D.

Under the above situation, the graph of y = f(x) and x = g(y) are one and the same.

However, when denoting the inverse of the function, we normally denote the independent variable by y and, hence, the inverse function of y = f(x) is denoted by y = g(x) and not by x = g(y).

The graphs of two inverse functions when this change is used are symmetrical about the line y = x (which is the bisector of the first and the third quadrants).

Graphs of Some Simple Functions The student is advised to familiarise himself/herself with the following figures.

Graphs of y = b, y = kx , y = kx + b , y = kx – b.

Note the shifting of the line when a positive number b is added and subtracted to the function’s equation.

SHIFTING OF GRAPHS

The ability to visualize how graphs shift when the basic analytical expression is changed is a very important skill. For instance if you knew how to visualize the graph of (x + 2)² – 5, it will definitely add a lot of value to your ability to solve questions of functions and all related chapters of block 5 graphically.

In order to be able to do so, you first need to understand the following points clearly:

(1)The relationship between the graph of y = f(x) and y = f(x) + c (where c represents a positive constant.): The shape of the graph of y = f(x) + c will be the same as that of the y = f(x) graph. The only difference would be in terms of the fact that f(x) + c is shifted c units up on the x – y plot.

The following figure will make it clear for you:

Example: Relationship between y = x² and y = x² + 2.

(2)The relationship between y = f(x) and y = f(x) – c: In this case while the shape remains the same, the position of the graph gets shifted c units down.

(3)The relationship between y = f(x) and y = f(x + c): In this case the graph will get shifted c units to the left. (Remember, c was a positive constant)

Example:

(4)The relationship between y = f(x) and y = f(x – c): In this case the graph will get shifted c units to the right on the x – y plane.

COMBINING MOVEMENTS

It is best understood through an example:

Visualizing a graph for a function like x² – 4x + 7.

First convert x² – 4x + 7 into (x – 2)² + 3

[Note: In order to do this conversion, the key point of your thinking should be on the –4x. Your first focus has to be to put down a bracket (x – a)² which on expansion gives – 4x as the middle term. When you think this way, you will get (x – 2)². On expansion (x – 2)² = x² – 4x + 4. But you wanted x² – 4x + 7. Hence add +3 to (x – 2)². Hence the expression x² – 4x + 7 is equivalent to (x – 2)² + 3.]

To visualize (x – 2)² + 3 shift the x² graph two units right [to account for (x – 2)²] and 3 units up [to account for the +3] on the x – y plot. This will give you the required plot.

Task for the student: I would now like to challenge and encourage you to think of how to add and multiply functions graphically.

INEQUALITIES

The Logical Graphical Process for Solving Inequalities

Your knowledge of the standard graphs of functions and how these shift can help you immensely while solving inequalities.

Thus, for instance if you are given an inequality question based on a quadratic function like ax² + bx + c < 0 (and a is positive) you should realize that the curve will be U shaped. And the inequality would be satisfied between the roots of the quadratic equation ax² + bx + c = 0. [Remember, we have already seen and understood that the solution of an equation f(x) = 0 is seen at the points where the graph of y = f(x) cuts the x axis.]

Similarly, for a cubic curve like the one shown below, you should realize that it is greater than 0 to the left of the point 1 shown in the figure. This is also true between points 2 and 3. At the same time the function is less than zero between points 1 and 2 and to the right of point 3. (on the x-axis.)

Another important point to note is that in the case of strict inequalities (i.e. inequalities with the ‘<’ or ‘>’ sign) the answer will also consist of strict inequalities only. On the other hand in the case of slack inequalities (inequalities having £ or ≥ sign) the solution of the inequality will also have a slack inequality sign.

LOGARITHMS

Graphical View of the Logarithmic Function

The typical logarithmic function is shown in the graph below:

Note the following points about the logarithmic function y = log x.

(1)It is only defined for positive values of x.

(2)For values of x below 1, the logarithmic function is negative. At the same time for x = 1, the logarithmic function has a value of 0. (Irrespective of the base)

(3)The value of log x becomes 2, when the value of x becomes equal to the square of the base.

(4)As we go further right on the x axis, the graph keeps increasing. However, this increase becomes more and more gradual and hence the shape of the graph becomes increasingly flatter as we move further on the x axis.

WORKED-OUT PROBLEMS

Problem 13.1 Find the domain of the definition of the function y = 1 /(x² – 2x)^1/2

	(a) (–•, –2)	(b) (–•, + •) except [0, 2]
	(c) (2, + •)	(d) (–•, 0)

Solution For the function to be defined, the expression under the square root should be non-negative and the denominator should not be equal to zero.

So, x² – 2x > 0 and (x² – 2x) π 0

or, x (x – 2) > 0 or (x² – 2x) π 0

So, x won’t lie in between 0 and 2 and x π 0, x π 2.

So, x will be x (–• , + •) excluding the range 0 £ x is £ 2.

In exam situations, to solve the above problem, you should check the options as below.

In fact, for solving all questions on functions, the student should explore the option-based approach.

Often you will find that going through the option-based approach will help you save a significant amount of time. The student should try to improve his/her selection of values through practice so that he/she is able to eliminate the maximum number of options on the basis of every check. The student should develop a knack for disproving three options so that the fourth automatically becomes the answer. It should also be noted that if an option cannot be disproved, it means that it is the correct option.

What I am trying to say will be clear from the following solution process.

For this question, if we check at x = 3, the function is defined. However, x = 3 is outside the ambit of option a and d. Hence, a and d are rejected on the basis of just one value check, and b or c has to be the answer.

Alternately, you can try to disprove each and every option one by one.

Problem 13.2 Which of the following is an even function?

	(a) \| x² \| – 5x	(b) x⁴ + x⁵
	(c) e^2x + e^–2x	(d) \| x \|²/x

Solution Use options for solving.

If a function is even it should satisfy the equation f(x) = f(–x).

We now check the four options to see which of them represents an even function.

Checking option (a) f(x) = | x² | – 5x

Putting –x in the place of x.

f (x) = | (–x)² | – 5(–x)

= | x² | + 5(x) π f(x)

Checking option (b) f(x) = x⁴ + x⁵.

Putting (–x) at the place of x,

f(–x) = (–x)⁴ + (–x)⁵ = x⁴ – x⁵ π f(x)

Checking option (c), f (x) = e^2x + e^–2x

Putting (–x) at the place of x.

f (–x) = e^–^2x + e^–^(-2x) = e^–^2x+ e^2x = f(x)

So (c) is the answer.

You do not need to go further to check for d. However, if you had checked, you would have been able to disprove it as follows:

Checking option (d), f(x) = |x|²/x

Putting f(x) at the place of x,

f(–x) = | –x |²/–x = | x² | /–x–π f (–x)

Directions for Questions 13.3–13.6:

Mark (a) if f (–x) = f (x)

Mark (b) if f (–x) = –f (x)

Mark (c) if neither (a) nor (b)

Problem 13.3

Solution The graph is symmetrical about the y-axis. This is the definition of an even function. So (a).

Problem 13.4

Solution The graph is symmetrical about the y-axis. This is the definition of an even function. So (a).

Problem 13.5

Solution The graph is symmetrical about origin. This is the definition of an odd function. So (b).

Problem 13.6

Solution The graph is neither symmetrical about the y-axis nor about origin. So (c).

Problem 13.7 Which of the following two functions are identical?

	(a) f(x) = x²/x	(b) g(x) =
	(c) h(x) = x
	(i) (a) and (b)	(ii) (b) and (c)
	(iii) (a) and (c)	(iv) None of these

Solution For two functions to be identical, their domains should be equal.

Checking the domains of f(x), g(x) and h(x),

f(x) = x²/x, x should not be equal to zero.

So, domain will be all real numbers except at x = 0.

g(x) = , x should be non-negative.

So, domain will be all positive real numbers.

h(x) = x, x is defined every where.

So, we can see that none of them have the same domain.

Hence, (d) is the correct option.

Problem 13.8 If f(x) = 1/x, g(x) = 1/(1 – x) and h(x) = x², then find fogoh (2).

	(a) –1	(b) 1
	(c) 1/2	(d) None of these

Solution fogoh (2) is the same as f(g(h(2)))

To solve this, open the innermost bracket first. This means that we first resolve the function h(2). Since h(2) = 4 we will get

f(g(h(2))) = f(g(4)) = f(–1/3) = –3. Hence, the option (d) is the correct answer.

Read the instructions below and solve Problems 11.9 and 11.10.

A * B = A³ – B³

A + B = A – B

A – B = A/B

Problem 13.9 Find the value of (3 * 4) – (8 + 12).

	(a) 9	(b) 9.25
	(c) –9.25	(d) None of these

Solution Such problems should be solved by the BODMAS rule for sequencing of operations.

Solving, thus, we get: (3 * 4) – (8 + 12)

= –37 – (–4). [Note here that the ‘–’ sign between –37 and –4 is the operation defined above.]

= 37/4 = 9.25

Problem 13.10 Which of the following operation will give the sum of the reciprocals of x and y and unity?

(a)(x + y) * (x – y)

(b)[(x * y) – x] – y

(c)[{(x * y) – (x + y)} – x] – y

(d)(x + y) – (x – y)

For solving questions containing a function in the question as well as a function in the options (where values are absent), the safest process for students weak at math is to assume certain convenient values of the variables in the expression and checking for the correct option that gives us equality with the expression in the question. The advantages of this process of solution is that there is very little scope for making mistakes. Besides, if the expression is not simple and directly visible, this process takes far less time as compared to simplifying the expression from one form to another.

This process will be clear after perusing the following solution to the above problem.

Solution The problem statement above defines the expression: (1/x) + (1/y) + 1 and asks us to find out which of the four options is equal to this expression. If we try to simplify, we can start from the problem expression and rewrite it to get the correct option. However, in the above case this will become extremely complicated since the symbols are indirect. Hence, if we have to solve through simplifying, we should start from the options one by one and try to get the problem expression. However, this is easier said than done and for this particular problem, going through this approach will take you at least two minutes plus.

Hence, consider the following approach:

Take the values of x and y as 1 each. Then,

(1/x) + (1/y) + 1 = 3

Put the value of x and y as 1 each in each of the four options that we have to consider.

Option (a) will give a value of –1 π 3. Hence, option (a) is incorrect.

Opton (b) will give a value of 0. 0 π 3. Hence, option (b) is incorrect.

Option (c) gives an answer of 3. Hence, option (c) could be the answer.

Option (d) gives an answer of 0. 0 π 3. Hence, option (d) is incorrect.

Now since options (a), (b) and (d) are incorrect and option (c) is the only possibility left, it has to be the answer.

[Note that in case there is a fourth option of ‘none of these’ then we have to be careful before marking the answer.]

LEVEL OF DIFFICULTY (I)

1.	Find the domain of the definition of the function y = \|x\|.
	(a) 0 £ x	(b) –• < x < + •
	(c) x < + •	(d) 0 £ x < + •
2.	Find the domain of the definition of the function y = .
	(a) –• < x < + •	(b) x £ 0
	(c) x > 0	(d) x ≥ 0
3.	Find the domain of the definition of the function y = .
	(a) x ≥ 0	(b) –• < x < + •
	(c) x > 0	(d) x < + •
4.	Find the domain of the definition of the function y = (x – 2)^1/2 + (8 – x)^1/2.
	(a) All the real values except 2 £ x £ 8
	(b) 2 £ x
	(c) 2 £ x £ 8
	(d) x £ 8
5.	Find the domain of the definition of the function y = (9 – x²)^1/2.
	(a) –3 £ x £ 3	(b) (–•, –3] » [3, •)
	(c) –3 £ x	(d) x £ 3
6.	Find the domain of the definition of the function y = 1/(x² – 4x + 3).
	(a) 1 £ x £ 3
	(b) (–•, –3) » (3, •)
	(c) x = (1, 3)
	(d) –• < x < •, excluding 1, 3
7.	The values of x for which the functions f(x) = x and g(x) = are identical is
	(a) – • < x < + •	(b) x ≥ 0
	(c) x > 0	(d) x £ 0
8.	The values of x for which the functions f(x) = x and g(x) = x²/x are identical is
	(a) Set of real numbers excluding 0
	(b) Set of real numbers
	(c) x ≥ 0
	(d) x > 0
9.	If f(x) = , then f(3x) will be equal to
	(a)	(b)
	(c)	(d) 3
10.	If f(x) = e^x, then the value of 7 f(x) will be equal to
	(a) e^7x	(b) 7e^x
	(c) 7 e^7x	(d) e^x
11.	If f(x) = log x² and g(x) = 2 log x, then f(x) and g(x) are identical for
	(a) –• < x < + •	(b) 0 £ x < •
	(c) –• x £ 0	(d) 0 < x < •
12.	If f(x) is an even function, then the graph y = f(x) will be symmetrical about
	(a) x-axis	(b) y-axis
	(c) Both the axes	(d) None of these
13.	If f(x) is an odd function, then the graph y = f(x) will be symmetrical about
	(a) x-axis	(b) y-axis
	(c) Both the axes	(d) origin
14.	Which of the following is an even function?
	(a) x^–⁸	(b) x³
	(c) x^–³³	(d) x⁷³
15.	Which of the following is not an odd function?
	(a) (x + 1)³	(b) x²³
	(c) x⁵³	(d) x⁷⁷
16.	For what value of x, x² + 10x + 11 will give the minimum value?
	(a) 5	(b) +10
	(c) –5	(d) –10
17.	In the above question, what will be the minimum value of the function?
	(a) –14	(b) 11
	(c) 86	(d) 0
18.	Find the maximum value of the function 1/(x² – 3x + 2).
	(a) 11/4	(b) 1/4
	(c) 0	(d) None of these
19.	Find the minimum value of the function f(x) = log₂(x² – 2x + 5).
	(a) –4	(b) 2
	(c) 4	(d) –2
20.	f(x) is any function and f ^–1(x) is known as inverse of f(x), then f ^–1(x) of f(x) = + 1 is
	(a) – 1	(b) x – 1
	(c)	(d)
*Directions for Questions 21 to 23:* Read the instructions below and solve.
	f(x) = f(x – 2) – f(x – 1), x is a natural number
	f(1) = 0, f(2) = 1
21.	The value of f(8) is
	(a) 0	(b) 13
	(c) –5	(d) –9
22.	The value of f(7) + f(4) is
	(a) 11	(b) –6
	(c) –12	(d) 12
23.	What will be the value of ?
	(a) –12	(b) –15
	(c) –14	(d) –13
24.	What will be the domain of the definition of the function f(x) = ^8–xC_5–xfor positive values of x?
	(a) {1, 2, 3}	(b) {1, 2, 3, 4}
	(c) {1, 2, 3, 4, 5}	(d) {1, 2, 3, 4, 5, 6, 7, 8}

Directions for Questions 25 to 38:

Mark a if f(–x) = f(x)

Mark b if f(–x) = – f(x)

Mark c if neither a nor b is true

Mark d if f(x) does not exist at at least one point of the domain.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

Directions for Questions 39 to 43: Define the following functions:

(i)a @ b =

(ii)a # b = a² – b²

(iii)(a ! b) =

39.	Find the value of {[(3@4)!(3#2)] @ [(4!3)@(2#3)]}.
	(a) –0.75	(b) –1
	(c) –1.5	(d) –2.25
40.	Find the value of (4#3)@(2!3).
	(a) 3.25	(b) 3.5
	(c) 6.5	(d) 7
41.	Which of the following has a value of 0.25 for a = 0 and b = 0.5?
	(a) a @ b	(b) a # b
	(c) Either a or b	(d) Cannot be determined
42.	Which of the following expressions has a value of 4 for a = 5 and b = 3?
	(a)	(b) (a!b)(a@b)
	(c)	(d) Both (b) and (c)
43.	If we define a$b as a³ – b³, then for integers a, b > 2 and a > b which of the following will always be true?
	(a) (a@b) > (a!b)	(b) (a@b) ≥ (a!b)
	(c) (a#b) < (a$b)	(d) Both a and c

Directions for Questions 44 to 48: Define the following functions:

	(a) (a M b) = a – b	(b) (a D b) = a + b
	(c) (a H b) = (ab)	(d) (a P b) = a/b
44.	Which of the following functions will represent a² – b²?
	(a) (a M b) H (a D b)	(b) (a H b) M (a P b)
	(c) (a D b)/(a M b)	(d) None of these
45.	Which of the following represents a²?
	(a) (a M b) H (a D b) + b²
	(b) (a H b) M (a P b) + b²
	(c)
	(d) Both (a) and (c)
46.	What is the value of (3M4H2D4P8M2)?
	(a) 6.5	(b) 6
	(c) –6.5	(d) None of these
47.	Which of the four functions defined has the maximum value?
	(a) (a M b)	(b) (a D b)
	(c) (a P b)	(d) Cannot be determined
48.	Which of the four functions defined has the minimum value?
	(a) (a M b)	(b) (a D b)
	(c) (a H b)	(d) Cannot be determined
49.	If 0 < a < 1 and 0 < b < 1 and a > b, which of the 4 expressions will take the highest value?
	(a) (a M b)	(b) (a D b)
	(c) (a P b)	(d) Cannot be determined
50.	If 0 < a <1 and 0 < b < 1 and if a < b, which of the following expressions will have the highest value?
	(a) (a M b)	(b) (a D b)
	(c) (a P b)	(d) Cannot be determined
51.	A function F(n) is defined as F(n – 1) = for all natural numbers ‘n’. If F(1) = 3, then what is the value of [F(1)] + [F(2)] +…+ [F(1000)]?

	(Here, [x] is equal to the greatest integer less than or equal to ‘x’)
	(a) 1001	(b) 1002
	(c) 3003	(d) None of these
52.	For the above question find the value of the expression: F(1) × F(2) × F(3) × F(4) ×….F(1000)
	(a) 2001	(b) 1999
	(c) 2004	(d) 1997
53.	A function f (x) is defined for all real values of x as f (x) = ax² + bx + c. If f (3) = f (–3) = 18, f (0) = 15, then what is the value of f (12)?
	(a) 63	(b) 159
	(c) 102	(d) None of these
54.	Two operations, for real numbers x and y, are defined as given below.
	(i) M (x q y) = (x + y)²
	(ii) f(x y y) = (x − y)²
	If M(x² q y²) = 361 and M(x² y y²) = 49, then what is the value of the square root of ((x²y²) + 3)?
	(a) ±81	(b) ±9
	(c) ±7	(d) ±11
55.	The function Y(m) = [m], where [m] represents the greatest integer less than or equal to m. Two real numbers x and y are such that Y(4x + 5) = 5y + 3 and Y(3y + 7) = x + 4, then find the value of x²× y².
	(a) 1	(b) 2
	(c) 4	(d) None of these
56.	A certain function always obeys the rule: If f (x.y) = f(x). f(y) where x and y are positive real numbers. A certain Mr. Mogambo found that the value of f (128) = 4, then find the value of the variable M = f (0.5). f (1). f (2). f(4). f (8). f (16). f (32). f (64). f (128). f (256)
	(a) 128	(b) 256
	(c) 512	(d) 1024
57.	x and y are non negative integers such that 4x + 6y = 20, and x² £ M/y^2/3for all values of x, y. What is the minimum value of M ?
	(a) 2^2/3	(b) 2^1/3
	(c) 2^4/3	(d) 4^2/3
58.	Let Y(x) = and q(x) = 3x² + 2. Find the value of q ( Y(–7)).
	(a) 12	(b) 14
	(c) 50	(d) 42
59.	If F(a + b) = F(a).F(b) ÷ 2, where F(b) π 0 and F(a)π 0, then what is the value of F(12b)?
	(a) (F(b))¹²	(b) F(b))¹²÷ 2
	(c) (F(b))¹²÷ 2¹²	(d) (F(b))¹²÷ 2¹¹
60.	A function a = q(b) is said to be reflexive if b = q(a). Which of the following is/are reflexive functions?
	(i)	(ii)
	(iii)
	(a) All of these are reflexive
	(b) Only (i) and (ii) are reflexive
	(c) Only (i) and (iii) are reflexive
	(d) None of these are reflexive.

LEVEL OF DIFFICULTY (II)

1.	Find the domain of the definition of the function y = 1/ (4 – x²)^1/2.
	(a) (–2, 2)
	(b) [–2, 2]
	(c) (–•, –2) » (2, •) excluding –2 and 2
	(d) (2, •)
2.	The domain of definition of the function
	y = + (x + 2)^1/2 is
	(a) (–3, –2)	(b) [0, 1)
	(c) [–2, 1]	(d) [–2, 1) excluding 0
3.	The domain of definition of y = [log₁₀ ]^1/2 is
	(a) [1, 4]	(b) [–4, –1]
	(c) [0, 5]	(d) [–1, 5]
4.	Which of the following functions is an odd function?
	(a) 2^–x.x	(b) 2^x^{– x.x.x.x}
	(c) Both (a) and (b)	(d) Neither (a) nor (b)
5.	The domain of definition of y = [1– \|x\| ]^1/2 is
	(a) [–1, 0]	(b) [0, 1]
	(c) (–1, 1)	(d) [–1, 1]
6.	The domain of definition of y = [3/(4 – x²)] + log₁₀ (x³ – x) is
	(a) (–1, 0) » (1, •)	(b) Not 2 or –2
	(c) (a) and (b) together	(d) None of these
7.	If f(t) = 2^t, then f(0), f(1), f(2) are in
	(a) AP	(b) HP
	(c) GP	(d) Cannot be said
8.	Centre of a circle x² + y² =16 is at (0, 0). What will be the new centre of the circle if it gets shifted 3 units down and 2 units left?
	(a) (2, 3)	(b) (–2, –3)
	(c) (–2, 3)	(d) (2, –3)
9.	If u(t) = 4t – 5, v(t) = t² and f(t) = 1/t, then the formula for u(f(v(t))) is
	(a)	(b)
	(c) – 5	(d) None of these
10.	If f(t) = , g(t) = t/4 and h(t) = 4t – 8, then the formula for g(f(h(t))) will be
	(a)	(b) 2 – 8
	(c)	(d)
11.	In the above question, find the value of h(g(f(t))).
	(a) – 8	(b)
	(c)	(d) None of these
12.	In question number 10, find the formula of f(h(g(t))).
	(a) – 8	(b)
	(c) 2 – 8	(d) None of these
13.	The values of x, for which the functions f(x) = x, g(x) = and h(x) = x²/x are identical, is
	(a) 0 £ x	(b) 0 < x
	(c) All real values	(d) All real values except 0
14.	Which of the following is an even function?
	(a) e^x	(b) e^–x
	(c) e^x + e^–x	(d)
15.	The graph of y = (x + 3)³ + 1 is the graph of y = x³ shifted
	(a) 3 units to the right and 1 unit down
	(b) 3 units to the left and 1 unit down
	(c) 3 units to the left and 1 unit up
	(d) 3 units to the right and 1 unit up
16.	If f(x) = 5x³ and g(x) = 3x⁵, then f(x).g(x) will be
	(a) Even function	(b) Odd function
	(c) Both	(d) None of these
17.	If f(x) = x² and g(x) = log_ex, then f(x) + g(x) will be
	(a) Even function	(b) Odd function
	(c) Both	(d) Neither (a) nor (b)
18.	If f(x) = x³ and g(x) = x²/5, then f(x) – g(x) will be
	(a) Odd function	(b) Even function
	(c) Neither (a) nor (b)	(d) Both
19.	f(x) is any function and f^–1(x) is known as inverse of f(x), then f^–1(x) of f(x) = 1/(x – 2) is
	(a) + 2	(b)
	(c) + 0.5	(d) None of these
20.	f(x) is any function and f^–1(x) is known as inverse of f(x), then f^–1(x) of f(x) = e^x is
	(a) –e^x	(b) e^–x
	(c) log_ex	(d) None of these
21.	f(x) is any function and f^–1(x) is known as inverse of f(x), then f^–1(x) of f(x) = x/(x – 1), x π 1 is
	(a) x/(1 + x)	(b)
	(c) x/(x – 1)	(d) –x/(x + 1)
22.	Which of the following functions will have a minimum value at x = –3?
	(a) f(x) = 2x³– 4x + 3	(b) f(x) = 4x⁴ – 3x + 5
	(c) f(x) = x⁶ – 2x – 6	(d) None of these

Directions for Questions 23 to 32:

Mark (a) if f(–x) = f(x)

Mark (b) if f(–x) = – f(x)

Mark (c) if neither (a) nor (b) is true

Mark (d) if f(x) does not exist at at least one point of the domain.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

Directions for Questions 33 to 36: If f(x) is represented by the graph below.

33.Which of the following will represent the function –f(x)?

34.Which of the following will represent the function –f(x) + 1?

35.Which of the following will represent the function f(x) – 1?

36.Which of the following will represent the function f(x) + 1?


(a)	(b)

(c)	(d)

Directions for Questions 37 to 40: Define the following functions:

f(x, y, z) = xy + yz + zx

g(x, y, z) = x²y + y²z + z²x and

h(x, y, z) = 3 xyz

Find the value of the following expressions:

37.	h[f(2, 3, 1), g(3, 4, 2), h(1/3, 1/3, 3)]
	(a) 0	(b) 23760
	(c) 2640	(d) None of these
38.	g[ f (1, 0, 0), g(0, 1, 0), h(1, 1, 1)]
	(a) 0	(b) 9
	(c)12	(d) None of these
39.	f[ f (1, 1, 1), g(1, 1, 1), h(1, 1, 1)]
	(a) 9	(b) 18
	(c)27	(d) None of these
40.	f(1, 2, 3) – g(1, 2, 3) + h(1, 2, 3)
	(a) –6	(b) 6
	(c)12	(d) 8
41.	If f(x) = 1/ g(x), then which of the following is correct?
	(a) f(f(g(f(x)))) = f(g(g(f(f(x)))))
	(b) f(g(g(f(f(x))))) = f(f(g(g(g(x)))))
	(c) g(g(f(f(g(f(x)))))) = f(f(g(g(f(g(x))))))
	(d) f(g(g(g(f(x))))) = g(g(f(f(f(x)))))
42.	If f(x) = 1/g(x), then the minimum value of f(x) + g(x), f(x) > 0 and g(x) > 0, will be
	(a) 0
	(b) 2
	(c) Depends upon the value of f(x) and g(x)
	(d) None of these

Directions for Questions 43 to 45:

If R(a/b) = Remainder when a is divided by b;

Q(a/b) = Quotient obtained when a is divided by b;

SQ(a) = Smallest integer just bigger than square root of a.

43.	If a = 12, b = 5, then find the value of SQ[R {(a + b)/b}].
	(a) 0	(b) 1
	(c) 2	(d) 3
44.	If a = 9, b = 7, then the value of Q [[SQ(ab)+b]/a] will be
	(a) 0	(b) 1
	(c) 2	(d) None of these
45.	If a =18, b = 2 and c = 7, then find the value of Q [{SQ(ab) + R(a/c)}/b].
	(a) 3	(b) 4
	(c) 5	(d) 6

Directions for Questions 46 to 48: Read the following passage and try to answer questions based on them.

[x] = Greatest integer less than or equal to x

{x} = Smallest integer greater than or equal to x.

46.	If x is not an integer, what is the value of ([x] – {x})?
	(a) 0	(b) 1
	(c) –1	(d) 2
47.	If x is not an integer, then ({x} + [x]) is
	(a) An even number
	(b) An odd integer
	(c) > 3x
	(d) < x
48.	What is the value of x if 5 < x < 6 and {x} + [x] = 2x?
	(a) 5.2	(b) 5.8
	(c) 5.5	(d) 5.76
49.	If f(t) = t² + 2 and g(t) = (1/t) + 2, then for t = 2, f [g(t)] – g[f(t)] = ?
	(a) 1.2	(b) 2.6
	(c) 4.34	(d) None of these
50.	Given f(t) = kt +1 and g(t) = 3t + 2. If fog = gof, find k.
	(a) 2	(b) 3
	(c) 5	(d) 4
51.	Let F(x) be a function such that F(x) F(x + 1) = – F(x – 1)F(x–2)F(x–3)F(x–4) for all x ≥ 0. Given the values of If F (83) = 81 and F(77) = 9, then the value of F(102) equals to
	(a)27	(b) 54
	(c) 729	(d) Data Insufficient
52.	Let f (x) = 121– x², g (x) = \| x – 8\| + \| x + 8 \| and h(x) = min (f (x), g (x)}. What is the number of integer values of x for which h(x) is equal to a positive integral value?
	(a) 17	(b) 19
	(c) 21	(d) 23
53.	If the function R(x) = max (x² – 8, 3x, 8), then what is the max value of R(x)?
	(a) 4	(b)
	(c) •	(d) 0
54.	If the function R(x) = min (x² – 8, 3x, 8), what is the max value of R(x)?
	(a)4	(b) 8
	(c) •	(d) None of these
55.	The minimum value of ax²+ bx + c is 7/8 at x = 5/4. Find the value of the expression at x = 5, if the value of the expression at x = 1 is 1.
	(a) 75	(b) 79
	(c) 121	(d) 129
56.	Find the range of the function f (x) = (x + 4)(5 – x)(x + 1).
	(a) [–2, 3]	(b) ( –•, 20]
	(c) (– •,+ •)	(d) [–20, •)
57.	The function f (x) is defined for positive integers and is defined as:
	f (x) = 6^x – 3, if x is a number in the form 2n.
	= 6^x + 4, if x is a number in the form 2n + 1.
	What is the remainder when f (1) + f (2) + f (3) + ... + f (1001) is divided by 2?
	(a) 1	(b) 0
	(c) –1	(d) None of the above
58.	p, q and r are three non-negative integers such that p + q + r = 10. The maximum value of pq + qr + pr + pqr is
	(a) ≥ 40 and < 50	(b) ≥ 50 and < 60
	(c) ≥ 60 and < 70	(d) ≥ 70 and < 80
59.	A function a(x) is defined for x as 3a(x) + 2a (2 – x) = (x + 3)². What is the value of [G (–5)] where [x] represents the greatest integer less than or equal to x?
	(a) 37	(b) –38
	(c) –37	(d) Cannot be determined
60.	For a positive integer x, f (x + 2) = 3 + f (x), when x is even and f (x + 2) = x + f (x), when x is odd. If f (1) = 6 and f (2) = 4, then find f ( f ( f ( f (1)))) + f ( f ( f ( f (2)))).
	(a) 1375	(b) 1425
	(c) 1275	(d) None of these

LEVEL OF DIFFICULTY (III)

1.	Find the domain of the definition of the function y = 1/(x – \| x \| )^1/2.
	(a) –• < x < •	(b) –• < x < 0
	(c) 0 < x < •	(d) No where
2.	Find the domain of the definition of the function y = (x – 1)^1/2 + 2(1 – x)^1/2 + (x² + 3)^1/2.
	(a) x = 0	(b) [1, •)
	(c) [–1, 1]	(d) 1
3.	Find the domain of the definition of the function y = log₁₀ [(x – 5)/(x² – 10x + 24)] – (x + 4)^1/2.
	(a) x > 6	(b) 4 < x < 5
	(c) Both a and b	(d) None of these
4.	Find the domain of the definition of the function y = [(x – 3)/(x + 3)]^1/2 + [(1 – x)/(1 + x)]^1/2.
	(a) x > 3	(b) x < –3
	(c) -3 £ x £ 3	(d) Nowhere
5.	Find the domain of the definition of the function y = (2x² + x + 1)^–3/4.
	(a) x ≥ 0	(b) All x except x = 0
	(c) –3 £ x £ 3	(d) Everywhere
6.	Find the domain of the definition of the function y = (x² – 2x – 3)^1/2 – 1/(–2 + 3x – x²)^1/2.
	(a) x > 0	(b) –1 < x < 0
	(c) x²	(d) None of these
7.	Find the domain of the definition of the function y = log₁₀ [1 – log₁₀(x² – 5x + 16)].
	(a) (2, 3]	(b) [2, 3)
	(c) [2, 3]	(d) None of these
8.	If f(t) = (t – 1)/(t + 1), then f(f(t)) will be equal to
	(a) 1/t	(b) –1/t
	(c) t	(d) –t
9.	If f(x) = e^x and g(x) = log_ex then value of fog will be
	(a) x	(b) 0
	(c) 1	(d) e
10.	In the above question, find the value of gof.
	(a) x	(b) 0
	(c) 1	(d) e
11.	The function y = 1/x shifted 1 unit down and 1 unit right is given by
	(a) y – 1 = 1/ (x + 1)	(b) y – 1 = 1/(x – 1)
	(c) y + 1 = 1/(x – 1)	(d) y + 1 = 1/(x + 1)
12.	Which of the following functions is an even function?
	(a) f(t) = (a^t + a^–t)/(a^t – a^–t)
	(b) f(t) = (a^t + 1)/(a^t – 1)
	(c) f(t) = t. (a^t – 1)/(a^t + 1)
	(d) None of these
13.	Which of the following functions is not an odd function?
	(a) f(t) = log₂(t + +1)
	(b) f(t) = (a^t + a^–t)/(a^t – a^–t)
	(c) f(t) = (a^t + 1)/(a^t – 1)
	(d) All of these
14.	Find fof if f(t) = t/(1 + t²)^1/2.
	(a) 1/(1+ 2t²)^1/2	(b) t/(1 + 2t²)^1/2
	(c) (1 + 2t²)	(d) None of these
15.	At what integral value of x will the function attain its maximum value?
	(a) 3	(b) 4
	(c) –3	(d) None of these
16.	Inverse of f(t) = (10^t – 10^–t)/(10^t + 10^–t) is
	(a) 1/2 log {(1 – t)/(1 + t)}
	(b) 0.5 log {(t – 1)/(t + 1)}
	(c) 1/2 log₁₀(2^t – 1)
	(d) None of these
17.	If f(x) = \|x – 2\| , then which of the following is always true?
	(a) f(x) = (f(x))²	(b) f(x ) = f(–x)
	(c) f(x) = x – 2	(d) None of these

Directions for Questions 18 to 20: Read the instructions below and solve:

f(x) = f(x – 2) – f(x – 1), x is a natural number

f(1) = 0, f(2) = 1

18.	The value of f(x) is negative for
	(a) All x > 2
	(b) All odd x(x > 2)
	(c) For all even x(x > 0)
	(d) f(x) is always positive
19.	The value of f[f(6)] is
	(a) 5	(b) – 1
	(c) – 3	(d) – 2
20.	The value of f(6) – f(8) is
	(a) f(4) + f(5)	(b) f(7)
	(c) – {f(7) + f(5)}	(d) – f(5)
21.	Which of the following is not an even function ?
	(a) f(x) = e^x + e^–x	(b) f(x) = e^x – e^–x
	(c) f(x) = e^2x + e^–2x	(d) None of these
22.	If f(x) is a function satisfying f(x). f(1/x) = f(x) + f(1/x) and f(4) = 65, what will be the value of f(6)?
	(a) 37	(b) 217
	(c) 64	(d) None of these

Directions for Questions 23 to 34:

Mark (a) if f(–x) = f(x),

Mark (b) if f(–x) = – f(x)

Mark (c) if neither (a) nor (b) is true

Mark (d) if f(x) does not exist at at least one point of the domain.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34.

Directions for Questions 35 to 40: Define the functions:

A(x, y, z) = Max (max (x, y), min (y, z) min (x, z))

B(x, y, z) = Max (max (x, y), min (y, z) max (x, z))

C(x, y, z) = Max (min (x, y), min (y, z) min (x, z))

D(x, y, z) = Min (max (x, y), max (y, z) max (x, z))

Max (x, y, z) = Maximum of x, y and z.

Min (x, y, z) = Minimum of x, y and z.

Assume that x, y and z are distinct integers.

35.	For what condition will A(x, y, z) be equal to Max (x, y, z)?
	(a) When x is maximum	(b) When y is maximum
	(c) When z is maximum	(d) Either (a) or (b)
36.	For what condition will B(x, y, z) be equal to Min (x, y, z)?
	(a) When y is minimum	(b) When z is minimum
	(c) Either (a) or (b)	(d) Never
37.	For what condition will A(x, y, z) not be equal to B (x, y, z)?
	(a) x > y > z	(b) y > z > x
	(c) z > y > x	(d) None of these
38.	Under what condition will C(x, y, z) be equal to B(x, y, z)?
	(a) x > y > z	(b) z > y > x
	(c) Both a and b	(d) Never
39.	Which of the following will always be true?
	(I) A(x, y, z) will always be greater than Min (x, y, z)
	(II) B (x, y, z) will always be lower than Max (x, y, z)
	(III) A(x, y, z) will never be greater than B(x, y, z)
	(a) I only	(b) III only
	(c) Both a and b	(d) All the three
40.	The highest value amongst the following will be
	(a) Max/Min	(b) A/B
	(c) C/D	(d) Cannot be determined

Directions for Questions 41 to 49: Suppose x and y are real numbers. Let f(x, y) = |x + y| F(f(x, y)) = –f(x, y) and G(f(x, y)) = – F(f(x, y))

41.	Which one of the following is true?
	(a) F(f(x, y)).G(f(x, y)) = –F(f(x, y)). G.(f(x, y))
	(b) F(f(x, y)).G(f(x, y)) < –F(f(x, y)). G.(f(x, y))
	(c) G(f(x, y)).f(x, y) = F(f(x, y)). (f(x, y)
	(d) G(f(x, y)).F(f(x, y)) = f(x, y). f(x, y)
42.	Which of the following has a² as the result?
	(a) F(f(a, –a)).G(f(a, –a))
	(b) –F(f(a, a).G(f(a, a))/4
	(c) F(f(a, a)).G(f(a, a))/2²
	(d) f(a, a).f(a, a)
43.	Find the value of the expression.
	...
	(a) 3/2	(b) 2/3
	(c) 1	(d) 2
44.	Which of the following is equal to
	?
	(a)
	(b)
	(c)
	(d) None of these
	Now if A(f(x, y) = f(x, y)
	B(f(x, y) = –f(x, y)
	C(f(x, y) = f(x, y)
	D(f(x, y) = –f(x, y) and similarly
	Z (f(x, y) = –f(x, y)
	Now, solve the following:
45.	Find the value of A(f(0, 1)) + B(f(1, 2)) + C(f(2, 3)) +…+ Z(f(25, 26)).
	(a) –50	(b) –52
	(c) –26	(d) None of these
46.	Which of the following is true?
	(i) A(f(0, 1)) < B(f(1, 2)) < C(f(2, 3)…
	(ii) A(f(0, 1)). B(f(1, 2)) > B(f(1, 2)).C(f(2, 3)) > C(f(2, 3)). D(f(3, 4))
	(iii) A(f(0, 0)) = B(f(0, 0)) = C(f(0, 0)) = … = Z(f(0, 0))
	(a) only (i) and (ii)	(b) only (ii) and (iii)
	(c) only (ii)	(d) only (i)
47.	If max (x, y, z) = maximum of x, y and z
	Min (x, y, z) = minimum of x, y and z
	f(x, y) = \|x + y\|
	F(f(x, y)) = –f(x, y)
	G(f(x, y) = – F(f(x, y))
	Then find the value of the following expression:
	Min (max [f(2, 3), F(f(3, 4)), G(f(4, 5))], min [f(1, 2), F(f(–1, 2)), G(f(1, –2))], max [f(–3 –4),
	f(–5 –1), G(f(– 4, –6))])
	(a) –1	(b) –7
	(c) –6	(d) –10
48.	Which of the following is the value of
	Max. [f(a, b), F(f(b, c), G(f(c, d))]
	For all a > b > c > d?
	(a) Anything but positive
	(b) Anything but negative
	(c) Negative or positive
	(d) Any real value
49.	If another function is defined as P(x, y) = which of the following is second lowest in value?
	(a) Value of P(x, y) for x = 2 and y = 1
	(b) Value of P(x, y) for x = 3 and y = 4
	(c) Value of P(x, y) for x = 3 and y = 5
	(d) Value of P(x, y) for x = 3 and y = 2
50.	If f(s) = (b^s + b^–s)/2, where b > 0. Find f(s + t) + f(s – t).
	(a) f(s) – f(t)	(b) 2 f(s).f(t)
	(c) 4 f(s).f(t)	(d) f(s) + f(t)

Questions 51 to 60 are all actual questions from the XAT exam.

51.	A0, A1, A2,...... is a sequence of numbers with
	A0 = 1, A1 = 3, and At = (t +1) At–1 – t At–2 = 2, 3, 4,....
	Conclusion I. A8 = 77
	Conclusion II. A10 = 121
	Conclusion III. A12 = 145
	(a)Using the given statement, only Conclusion I can be derived.
	(b)Using the given statement, only Conclusion II can be derived.
	(c)Using the given statement, only Conclusion III can be derived.
	(d)Using the given statement, Conclusion I, II and III can be derived.
	(e)Using the given statement, none of the three Conclusions I, II and III can be derived.
52.	A, B, C be real numbers satisfying A < B < C, A + B + C = 6 and AB + BC + CA = 9
	Conclusion I. 1 < B < 3
	Conclusion II. 2 < A < 3
	Conclusion III. 0 < C < 1
	(a)Using the given statement, only Conclusion I can be derived.
	(b)Using the given statement, only Conclusion II can be derived.
	(c)Using the given statement, only Conclusion III can be derived.
	(d)Using the given statement, Conclusion I, II and III can be derived.
	(e)Using the given statement, none of the three Conclusions I, II and III can be derived.
53.	If F (x, n) be the number of ways of distributing “x” toys to “n” children so that each child receives at the most 2 toys, then F(4, 3) = ___?
	(a) 2	(b) 6
	(c) 3	(d) 4
	(e) 5
54.	The figure below shows the graph of a function f (x). How many solutions does the equation f ( f (x)) = 15 have?
	(a) 5	(b) 6
	(c) 7	(d) 8
	(e) Cannot be determined from the given graph

55.	A sequence of positive integers is defined as A_n₊₁ = A²_n_{+ 1} for each n ≥ 0. What is the value of Greatest Common Divisor of A900 and A₁₀₀₀?
	I. A0 = 1	II. A₁ = 2
56.	A manufacturer produces two types of products— A and B, which are subjected to two types of operations, viz., grinding and polishing. Each unit of product A takes 2 hours of grinding and 3 hours of polishing whereas product B takes 3 hours of grinding and 2 hours of polishing. The manufacturer has 10 grinders and 15 polishers. Each grinder operates for 12 hours/day and each polisher 10 hours/day. The profit margin per unit of A and B are ` 5/- and ` 7/- respectively. If the manufacturer utilises all his resources for producing these two types of items, what is the maximum profit that the manufacturer can earn?
	(a) ` 280/-	(b) ` 294/-
	(c) ` 515/-	(d) ` 550/-
	(e) None of the above
57.	Consider a function f (x) = x⁴ + x³ + x² + x + 1, where x is a positive integer greater than 1. What will be the remainder if f (x⁵) is divided by f (x)?
	(a) 1	(b) 4
	(c) 5	(d) A monomial in x
	(e) A polynomial in x
58.	For all real numbers x, except x = 0 and x = 1, the function F is defined by F
	If 0 < a < 90° then F((cosec a)²) =
	(a) (sin a)²	(b) (cos a)²
	(c) (tan a)²	(d) (cot a)²
	(e) (sec a)²
59.	F(x) is a fourth order polynomial with integer coefficients and with no common factor. The roots of F(x) are –2, –1, 1, 2. If p is a prime number greater than 97, then the largest integer that divides F (p) for all values of p is:
	(a) 72	(b) 120
	(c) 240	(d) 360
	(e) None of the above.
60.	If x = (9 + 4)⁴⁸ = [x] + f , where [x] is defined as integral part of x and f is a fraction, then x(1 – f) equals–
	(a) 1
	(b) Less than 1
	(c) More than 1
	(d) Between 1 and 2
	(e) None of the above

ANSWER KEY

Level of Difficulty (I)

1. (b)	2. (d)	3. (a)	4. (c)
5. (a)	6. (d)	7. (b)	8. (a)
9. (c)	10. (b)	11. (d)	12. (b)
13. (d)	14. (a)	15. (a)	16. (c)
17. (a)	18. (d)	19. (b)	20. (c)
21. (b)	22. (b)	23. (a)	24. (c)
25. (a)	26. (d)	27. (c)	28. (c)
29. (c)	30. (d)	31. (a)	32. (a)
33. (a)	34. (a)	35. (b)	36. (c)
37. (d)	38. (d)	39. (c)	40. (a)
41. (a)	42. (d)	43. (d)	44. (a)
45. (a)	46. (c)	47. (d)	48. (d)
49. (d)	50. (d)	51. (b)	52. (a)
53. (a)	54. (b)	55. (d)	56. (d)
57. (c)	58. (b)	59. (d)	60. (c)
*Level of Difficulty (II)*
1. (a)	2. (d)	3. (a)	4. (d)
5. (d)	6. (c)	7. (c)	8. (b)
9. (c)	10. (c)	11. (a)	12. (b)
13. (b)	14. (c)	15. (c)	16. (a)
17. (d)	18. (c)	19. (a)	20. (c)
21. (c)	22. (d)	23. (b)	24. (a)
25. (d)	26. (d)	27. (c)	28. (a)
29. (c)	30. (d)	31. (b)	32. (d)
33. (b)	34. (a)	35. (c)	36. (d)
37. (c)	38. (a)	39. (c)	40. (b)
41. (c)	42. (b)	43. (c)	44. (b)
45. (c)	46. (c)	47. (b)	48. (c)
49. (d)	50. (a)	51. (a)	52. (c)
53. (c)	54. (b)	55. (b)	56. (c)
57. (a)	58. (c)	59. (c)	60. (a)
*Level of Difficulty (III)*
1. (d)	2. (d)	3. (c)	4. (d)
5. (d)	6. (d)	7. (d)	8. (b)
9. (a)	10. (a)	11. (c)	12. (c)
13. (a)	14. (b)	15. (a)	16. (b)
17. (d)	18. (b)	19. (c)	20. (b)
21. (b)	22. (b)	23. (a)	24. (b)
25. (b)	26. (d)	27. (b)	28. (a)
29. (c)	30. (a)	31. (d)	32. (a)
33. (d)	34. (a)	35. (d)	36. (d)
37. (c)	38. (d)	39. (c)	40. (d)
41. (b)	42. (b)	43. (c)	44. (b)
45. (c)	46. (b)	47. (a)	48. (b)
49. (b)	50. (b)	51. (e)	52. (a)
53. (b)	54. (e)	55. (d)	56. (b)
57. (c)	58. (b)	59. (d)	60. (a)

Solutions and Shortcuts

Level of Difficulty (I)

1.y = ÈxÈ will be defined for all values of x. From = – • to + •

Hence, option (b).

2.For y = to be defined, x should be non-negative. i.e. x ≥ 0.

3.Since the function contains a in it, x ≥ 0 would be the domain.

4.For (x – 2)^1/2 to be defined x ≥ 2.

For (8 – x)^1/2 to be defined x £ 8.

Thus, 2 £ x £ 8 would be the required domain.

5.(9 – x²) ≥ 0 fi –3 £ x £ 3.

6.The function would be defined for all values of x except where the denominator viz: x² – 4x + 3 becomes equal to zero.

The roots of x² – 4x + 3x = 0 being 1, 3, it follows that the domain of definition of the function would be all values of x except x = 1 and x = 3.

7.f(x) = x and g(x) = ()² would be identical if is defined.

Hence, x ≥ 0 would be the answer.

8.f(x) = x is defined for all values of x.

g(x) = x²/x also returns the same values as f(x) except at x = 0 where it is not defined.

Hence. option (a).

9.f(x) = fi f(3x) = .

Option (c) is correct.

10.7 f(x) = 7 e^x.

11.While log x² is defined for – • < x < • , 2 log x is only defined for 0 < x < •. Thus, the two functions are identical for 0 < x < •.

12.y – axis by definition.

13.Origin by definition.

14.x^–8 is even since f(x) = f(–x) in this case.

15.(x + 1)³ is not odd as f(x) π –f(–x).

16.dy/dx = 2x + 10 = 0 fi x = –5.

17.Required value = (–5)² + 10(–5) + 11

= 25 – 50 + 11 = –14.

18.Since the denominator x²– 3x + 2 has real roots, the maximum value would be infinity.

19.The minimum value of the function would occur at the minimum value of (x² – 2x + 5) as this quadratic function has imaginary roots.

For y = x² – 2x + 5

dy/dx = 2x – 2 = 0 fi x = 1

fi x² – 2x + 5 = 4.

Thus, minimum value of the argument of the log is 4.

So minimum value of the function is log₂ 4 = 2.

20.y = 1/x + 1

Hence, y – 1 = 1/x

fi x = 1/(y – 1)

Thus f^–1(x) = 1/(x – 1).

21–23.

f(1) = 0, f(2) = 1,

f(3) = f(1) – f(2) = –1

f(4) = f(2) – f(3) = 2

f(5) = f(3) – f(4) = –3

f(6) = f(4) – f(5) = 5

f(7) = f(5) – f(6) = –8

f(8) = f(6) – f(7) = 13

f(9) = f(7) – f(8) = –21

21.13

22.–8 + 2 = –6

23.0 + 1 – 1 + 2 – 3 + 5 – 8 + 13 – 21 = –12.

24.For any ⁿC_r, n should be positive and r ≥ 0.

Thus, for positive x, 5 – x ≥ 0

fi x = 1, 2, 3, 4, 5.

Directions for Questions 25 to 38: You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd.

Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things.

Either the function is returning two values for one value of x. (as in questions 26, 30, 37 and 38) or the function has a break in between (not seen in any of these questions).

We see even functions in: 25, 31, 32, 33 and 34, [Symmetry about the y axis].

We see odd functions in question 35.

While the figures in Questions 27, 28, 29 and 36 are neither odd nor even.

39.{[(3@4)! (3 #2)] @ [(4!3) @ (2 # 3)]}

{[(3.5) ! (5)] @ [ (0.5) @ (–5)]}

{[–0.75] @ [–2.25]} = –1.5.

40.(7) @ (–0.5) = 3.25.

41.0 @ 0.5 = 0.25. Thus, a

42.b = (1) (4) = 4.

C =

16/4 = 4

Hence, both (b) and (c).

43.(a) will always be true because (a + b)/2 would always be greater then (a – b)/2 for the given value range.

Further, a² – b² would always be less than a³ – b³. Thus, option (d) is correct.

44–48.

44.Option a = (a – b) (a + b) = a² – b²

45.Option a = (a² – b²) + b² = a².

46.3 – 4 × 2 + 4/8 – 2 = 3 – 8 + 0.5 – 2 = – 6.5

(using BODMAS rule)

47.The maximum would depend on the values of a and b. Thus, cannot be determined.

48.The minimum would depend on the values of a and b. Thus, cannot be determined.

49.Any of (a + b) or a/b could be greater and thus we cannot determine this.

50.Again (a + b) or a/b can both be greater than each other depending on the values we take for a and b.

E.g. for a = 0.9 and b = 0.91, a + b > a/b.

For a = 0.1 and b = 0.11, a + b < a/b

51.Given that F(n – 1) = , we can rewrite the expression as F(n) = (2F (n – 1) –1)/((F (n – 1)).

For n = 2: F(2) = .

The value of F(3) would come out as 7/5 and F(4) comes out as 9/7 and so on. What we realise is that for each value of n, after and including n = 2, the value of F(n) = .

This means that the greatest integral value of F(n) would always be 1 for n = 2 to n = 1000.

Thus, the value of the given expression would turn out to be:

3 + 1 × 999 = 1002. Option (b) is the correct answer.

52.From the solution to the previous question, we already know how the value of the given functions at n = 1, 2, 3 and so on would behave.

Thus, we can try to see what happens when we write down the first few terms of the expression:

F(1) × F(2) × F(3) × F(4) × …. F(1000)

= .

53.Since f (0) = 15, we get c =15.

Next, we have f (3) = f (–3) = 18. Using this information, we get:

9a + 3b + c = 9a – 3b + c Æ 3b = –3b

\ 6b = 0 Æ b = 0.

Also, since

f (3) = 9a + 2b + c = 18 Æ we get: 9a + 15 = 18 Æ a = 1/3

The quadratic function becomes f (x) = x²/3 + 15. f (12) = 144/3 + 15 = 63.

54.What you need to understand about M(x² q y²) is that it is the square of the sum of two squares. Since M(x² q y²) = 361, we get (x² + y²)² = 361, which means that the sum of the sqaures of x and y viz. x² + y² = 19. (Note it cannot be –19 as we are talking about the sum of two squares, which cannot be negative under any circumstance).

Also, from M(x²y y²) = 49,we get (x² – y²)² = 49, Æ (x² – y²) = ± 7

Based on these two values, we can solve for two distinct situations:

(a)When x² + y² = 19 and x²– y²= 7, we get x²= 13 and y² = 6

(b)When x² + y² = 19 and x²– y² = –7, we get x² = 6 and y² = 13

In both cases, we can see that the value of: ((x²y²) + 3) would come out as 13 × 6 + 3 = 81 and the square root of its value would turn out to ±9. Option (b) is correct.

55.The first thing you need to understand while solving this question is that, since [m] will always be integral, hence Y(4x + 5) will also be integral. Since Y(4x + 5) = 5y + 3, naturally, the value of 5y + 3 will also be integral. This means that y is an integer. By a similar logic, the value of x will also be an integer considering the second equation: Y(3y + 7) = x + 4.

Using, this logic we know that Y(4x + 5) = 4x + 5 (because, whenever m is an integer the value of [m] = m).

This leads us to two linear equations as follows:

4x + 5 = 5y +…(i)

3y + 7 = x + 4…(ii)

Solving simultaneously, we will get: x = –2 and y = –3. Thus, x²× y² = 4 × 9 = 36.

56.Since f (128) = 4, we can see that the product of f (256). f (0.5) = f (256 × 0.5) = f (128) = 4.

Similarly, the products f (1). f (128) = f (2). f (64)

= f (4). f (32) = f (8). f (16) = 4.

Thus, M = 4 × 4 × 4 × 4 × 4 = 1024.

Option (d) is the correct answer.

57.The only values of x and y that satisfy the equation 4x + 6y = 20 are x = 2 and y = 2 (since, x, y are non negative integers). This gives us: 4 ≤ M/2^2/3. M has to be greater than 2^4/3 for this expression to be satisfied. Option (c) is correct.

58.q ( Y(–7)) = q(–2) = 14. Option (b) is correct.

59.F(2b) = F(b + b) = F(b).F(b) ÷ 2 = (F(b))²÷ 2

Similarly, F(3b) = F(b + b + b) = F(b + b).F(b) ÷ 2 = {F(b)²÷ 2}.{F(b)} ÷ 2 = (F(b))³÷ 2²

Similarly, F (4b) = (F(b))⁴÷ 2³.

Hence, F(12b) = (F(b))¹²÷ 2¹¹. Option (d) is correct.

60.To test for a reflexive function as defined in the problem use the following steps:

Step 1: To start with, assume a value of ‘b’ and derive a value for ‘a’ using the given function.

Step 2: Then, insert the value you got for ‘a’ in the first step into the value of ‘b’ and get a new value of ‘a’. This value of ‘a’ should be equal to the first value of ‘b’ that you used in the first step. If this occurs the function would be reflexive. Else it is not reflexive.

Checking for the expression in (i) if we take b = 1, we get:

a = 8/1 = 8. Inserting, b = 8 in the function gives us a = 29/29 = 1. Hence, the function given in (i) is reflexive.

Similarly checking the other two functions, we get that the function in (ii) is not reflexive while the function in (iii) is reflexive.

Thus, Option (c) is the correct answer.

Level of Difficulty (II)

1.For the function to be defined 4 – x² > 0

This happens when –2 < x < 2.

Option (a) is correct.

2.For the function to be defined two things should happen

(a)(1 – x) > 0 fi x < 1 and

(b)(x + 2) ≥ 0 fi x ≥ – 2. Also x π 0

Thus, option (d) is correct.

≥ 1 fi 1 £ x£ 4.

4.Neither 2^–x◊x nor 2^{x–x◊x◊x◊x} is an odd function as for neither of them is f(x) = –f(–x).

5.1 – |x| should be non negative.

[–1, 1] would satisfy this.

6.4 – x² π 0 and (x³ – x) > 0 fi (–1, 0) » (1, •) but not 2 or –2.

7.f(0) = 1, f(1) = 2 and f(2) = 4

Hence, they are in G.P.

8.x would become – 2 and y = –3.

9.u(f(v(t))) = u (f(t²)) = u (1/t²) = – 5.

10.g(f(h(t))) = g (f(4t – 8)) = g

11.h(g(f(t)) = h (g()) = h (/4)

= – 8

12.f(h(g(t))) = f(h(t/4)) = f(t – 8) = .

13.All three functions would give the same values for x > 0. As g(x) is not defined for negative x, and h(x) is not defined for x = 0.

14.e^x + e^–x = e^–x + e^x

Hence, this is an even function.

15.(x + 3)³ would be shifted 3 units to the left and hence (x + 3)³ + 1 would shift 3 units to the left and 1 unit up. Option (c) is correct.

16.f(x) ◊ g(x) = 15x⁸ which is an even function. Thus, option (a) is correct.

17.(x²+ log_ex) would be neither odd nor even since it obeys neither of the rules for even function (f(x) = f(–x)) nor for odd functions (f(x) = –f(–x)).

18.(x³ – x²/5) = f(x) – g(x) is neither even nor odd.

19.y = 1/(x – 2) fi (x – 2) = 1/y fi x = 1/y + 2.

Hence, f ^–1(x) = 1/x + 2.

20.y = e^x

fi log_e y = x.

fi f ^–1(x) = log_e x.

21.y = x/(x – 1)

fi (x – 1)/x = 1/y

fi 1 – (1/x) = 1/y

fi 1/x = 1 – 1/y fi 1/x = (y – 1)/y

fi x = y/(y – 1)

Hence, f ^–1(x) = x/(x – 1).

22.If you differentiate each function with respect to x, and equate it to 0 you would see that for none of the three options will get you a value of x = –3 as its solution. Thus, option (d) viz. None of these is correct.

Directions for Questions 23 to 32: You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd.

Also, option (d) would occur if the function does not exist atleast one point of the domain. This means one of two things.

Either the function is returning two values for one value of x or the function has a break in between. This is seen in Questions 25, 26, 30 and 32.

We see even functions in Questions 24 and 28. [Symmetry about the y axis]. We see odd functions in Questions 23 and 31.

While the figures in Questions 27 and 29 are neither odd nor even.

Even fi 24, 28,

Odd 23, 31.

Neither 27, 29,

doesn’t exist: 25, 26, 30 and 32.

33.–f(x) would be the mirror image of the function, about the ‘x’ axis which is seen in option (b).

34.–f(x) + 1 would be mirror image about the x axis and then shifted up by 1. Option (a) satisfies this.

35.f(x) – 1 would shift down by 1 unit. Thus option (c) is correct.

36.f(x) + 1would shift by 1 unit. Thus, option (d) is correct.

37.The given function would become h[ 11, 80, 1] = 2640.

38.The given function would become g[ 0, 0, 3] = 0.

39.The given function would become f[3, 3, 3] = 27.

40.f(1, 2, 3) – g(1, 2, 3) + h (1, 2, 3) = 11 – 23 + 18 = 6.

41.The number of g’s and f’s should be equal on the LHS and RHS since both these functions are essentially inverse of each other.

Option (c) is correct.

42.The required minimum value would occur at f(x) = g(x) = 1.

43.SQ [R[(a + b)/b]] = SQ [R[17/5]] fi SQ [2] = 2.

44.Q [[SQ(63) + 7 ]/9] = Q [[ 8 + 7]/9] = Q [15/9] = 1.

45.Q [[SA (36) + R (18/7)]/2] = Q [(7 + 4)/2] = Q [11/2] = 5.

46.[x] – {x} = –1

47.[x] + {x} will always be odd as the values are consecutive integers.

48.At x = 5.5, the given equation can be seen to be satisfied as: 6 + 5 = 2 × 5.5 = 11.

49.f(g(t)) – g(f(t)) = f(2.5) – g(6) = 8.25 – 2.166 = 6.0833.

50.fog = f(3t + 2) = K (3t + 2) + 1

gof = g (kt + 1) = 3( kt + 1) + 2

K( 3t + 2) + 1 = 3 (kt + 1) + 2

fi 2 k + 1 = 5

fi k = 2.

51.When the value of x = 81 and 82 is substituted in the given expression, we get,

F (81) F (82) = – F (80) F (79) F(78) F(77)…(i)

F (82) F (83) = – F (81) F (80) F(79) F(78)…(ii)

On dividing (i) by (a), we get

fi F(81) × F(81) = 81 × 9

fi F(81) = 27

Option (a) is the correct answer.

52.In order to understand this question, you first need to develop your thought process about what the value of h(x) is in various cases. A little bit of trial and error would show you that the value of h (x) since it depends on the minimum of f (x) and g (x), would definitely be dependant on the value of f (x) once x becomes greater than 11 or less than –11. Also, the value of g(x) is fixed as an integer at 16. It can be observed that when f (x) > g(x) i.e. f(x) > 16, the values of h(x) would also be 16 and hence would be a positive integer.

With this thought when you look at the expression of f (x) = 121–x², you realise that the value of x can be –10, –9, –8, –7, …0, 1, 2, 3,….8, 9, 10, i.e., 21 values of x when h (x) = g(x) = 16. When we use x = 11 or x = –11, the vlaue of f (x) = min (0, 16) = 0 and is not a positive integral value.

Hence, the correct answer is Option (c).

53.Since, R(x) is the maximum amongst the three given functions, its value would always be equal to the highest amongst the three. It is easy to imagine that x²– 8 and 3x are increasing functions, therefore the value of the function is continuously increasing as you increase the value of x. Similarly x²– 8 would be increasing continuously as you go farther and farther down on the negative side of the x-axis. Hence, the maximum value of R(x) would be infinity. Option (c) is the correct answer.

54.In this case, the value of the function, is the minimum of the three values. If you visualise the graphs of the three functions (viz: y = x²– 8, y = 3x and y = 8) you realise that the function y = 3x (being a straight line) will keep going to negative infinity as you move to the left of zero on the negative side of the x-axis.

Hence, the minimum value of the function R(x) after a certain point (when x is negative) would get dictated by the value of 3x. This point will be the intersection of the line y = 3x and the function y = x²– 8 when x is negative.

The two intersection points of the line (3x) and the quadratic curve (x²– 8) would be got by equating 3x = x²– 8. Solving this equation tells us that the intersection points are:

R(x) would depend on the following structures based on the value of x:

(i)When x is smaller than , the value of the function R(x) would be given by the value of 3x.

(ii)When x is between and 4 the value of the function R(x) would be given by the value of x²– 8, since that would be the least amongst the three functions.

(iii)After x = 4, on the positive side of the x-axis, the value of the function would be defined by the third function viz: y = 8.

A close look at these three ranges would give you that amongst these three ranges, the third range would yield the highest value of R(x). Hence, the maximum possible value of R(x) = 8. Option (b) is correct.

55.The expression is 2x² – 5x + 4, and its value at x = 5 would be equal to 100 – 25 + 4 = 79. Option (b) is correct.

56.At x = 0, the value of the function is 20 and this value rejects the first option. Taking some higher values of x, we realise that on the positive side, the value of the function will become negative when we take x greater than 5 since the value of (5 – x) would be negative. Also, the value of f (x) would start tending to –•, as we take bigger values of x.

Similarly, on the negative side, when we take the value of x lower than –4, f (x) becomes positive and when we take it farther away from 0 on the negative side, the value of f (x) would continue tending to +•. Hence, Option (c) is the correct answer.

57.The remainder when 6 ^x+ 4 is divided by 2 would be 0 in every case (when x is odd)

Also, when x is even, we would get 6 ^x– 3 as an odd number. In every case the remainder would be 1 (when it is divided by 2.)

Between f (2), f (4), f (6),….f (1000) there are 500 instances when x is even. In each of these instances the remainder would be 0 and hence the remainder would be 0 (in total). Option (b) is correct.

58.The product of p, q and r will be maximum if p, q and r are as symmetrical as possible. Therefore, the possible combination is (4, 3, 3).

Hence, maximum value of pq + qr + pr + pqr = 4 × 3 + 4 × 3 + 3 × 3 + 4 × 3 × 3 = 69.

Hence, Option (c) is correct.

59.The equation given in the question is: 3a(x) + 2a (2–x) = (x + 3)²….(i)

Replacing x by (2–x) in the above equation, we get

3a(2–x) + 2a(x) = (5–x)²

Solving the above pairs of equation, we get

a(x) = (x²+ 38 – 23)/5

Thus, G(–5) = –188/5 = –37.6. The value of [–37.6] = –37. Hence, Option (c) is the correct answer.

60.The first thing you do in this question is to create the chain of values of f (x) for x = 1, 2, 3 and so on.

The chain of values would look something like this:

When x is odd			When x is even
f (1)	Value is given	6	f (2)	Value is given	4
f (3)	= 1 + f (1)	7	f (4)	= 3 + f (2)	7
f (5)	= 3 + f (3)	10	f (6)	= 3 + f (4)	10
f (7)	= 5 + f (5)	15	f (8)	= 3 + f (6)	13
f (9)	= 7 + f (7)	22	f (10)	= 3 + f (8)	16
f (11)	= 9 + f (9)	31

In order to evaluate the value of the embedded function represented by f ( f ( f ( f ( f ( f (1))))), we can use the above values and think as follows:

f ( f ( f ( f (1)))) = f ( f ( f (6))) = f ( f (10)) = f (16) = 25

Also, f ( f ( f ( f (2) = f ( f ( f (4))) = f ( f (7)) = f (15) = 55

Hence, the product of the two values is 25 × 55 =1375.

Option (a) is correct.

Level of Difficulty (III)

1.x – |x| is either negative for x < 0 or 0 for x ≥ 0. Thus, option (d) is correct.

2.The domain should simultaneously satisfy:

x – 1 ≥ 0, (1 – x) ≥ 0 and (x² + 3) ≥ 0.

Gives us: x ≥ 1 and x £ 1

The only value that satisfies these two simultaneously is x = 1.

3.For the function to exist, the argument of the logarithmic function should be positive. Also, (x + 4) ≥ 0 should be obeyed simultaneously.

For to be positive both numerator and denominator should have the same sign. Considering all this, we get:

4 < x < 5 and x > 6.

Option (c) is correct.

4.Both the brackets should be non-negative and neither (x + 3) nor (1+ x) should be 0.

For (x – 3)/(x + 3) to be non negative we have x>3 or x< – 3.

Also for (1– x)/(1+ x) to be non-negative –1 < x < 1. Since there is no interference in the two ranges, Option (d) would be correct.

8.f(f(t)) = f((t – 1)/(t + 1))

= =

= –2/2t = –1/t.

9.fog = f (log_ex) = e^logex = x.

10.gof = g(e^x) = log_e e^x= x.

11.Looking at the options, one unit right means x is replaced by (x – 1). Also, 1 unit down means –1 on the RHS.

Thus, (y + 1) = 1/(x – 1)

12.For none of these is f(t) = f(–t)

Thus, Option (d) is correct.

13.Option (b) is odd because:

= –1 ×

Similarly option (c) is also representing an odd function. The function in option (a) is not odd.

14.f (f(t)) = f [t/(1+ t²)^1/2] = t/(1 + 2t²)^1/2.

15.By trial and error it is clear that at x = 3, the value of the function is 19. At other values of ‘x’ the value of the function is less than 19.

17.Take different values of n to check each option. Each of Options (a), (b) and (c) can be ruled out. Hence, Option (d) is correct.

Solutions to 18 to 20:

f(1) = 0, f(2) = 1,

f(3) = f(1) – f(2) = –1

f(4) = f(2) – f(3) = 2

f(5) = f(3) – f(4) = –3

f(6) = f(4) – f(5) = 5

f(7) = f(5) – f(6) = –8

f(8) = f(6) – f(7) = 13

18.It can be seen that f(x) is positive wherever x is even and negative whenever x is odd.

19.f(f(6) = f(5) = –3.

20.f(6) – f(8) = 5 – 13 = –8 = f(7).

21.Option (b) is not even since e^x – e^–xπ e^–x– e^x.

22.We have f(x) ◊ f(1/x) = f(x) + f(1/x)

fi f(1/x) [f(x) – 1] = f(x)

For x = 4, we have f(1/4) [f(4) – 1] = f (4)

fi f(1/4) [64] = 65

fi f(1/4) = 65/64 = 1/64 + 1

This means f(x) = x³ + 1

For f(6) we have f(6) = 216 + 1 = 217.

Directions for Questions 23 to 34: You essentially have to mark (a) if it is an even function, mark (b) if it is an odd function, mark (c) if the function is neither even nor odd.

Also, Option (d) would occur if the function does not exist at, atleast one point of the domain. This means one of two things.

Either the function is returning two values for one value of x or the function has a break in between (as in questions 26, 31 and 33).

We see even functions in Questions 23, 28, 30, 32 and 34 [Symmetry about the y axis]. We see odd functions in Questions 24, 25 and 27.

While the figure in Question 29 is neither odd nor even.

Solutions to 35–40:

In order to solve this set of questions first analyse each of the functions:

A(x, y, z) = will always return the value of the highest between x and y.

B(x, y, z) will return the value of the maximum amongst x, y and z.

C(x, y, z) and D(x, y, z) would return the second highest values in all cases while max (x, y, z) and min(x, y, z) would return the maximum and minimum values amongst x, y, and z respectively.

35.When either x or y is maximum.

36.This would never happen.

37.When z is maximum, A and B would give different values. Thus, option (c) is correct.

38.Never.

39.I and III are always true.

40.We cannot determine this because it would depend on whether the integers x, y, and z are positive or negative.

Solutions to 41 to 49:

F(x, y) is always positive

F(f(x, y) is always negative

G(f(x, y) is always positive

41.F × G would always be negative while – F × G would always be positive.

Hence, Option (b) F × G < – F × G is correct.

42.Option (b) can be seen to give us 4a²/4 = a².

43.(5 – 1)/(1+ 3) = 4/4 = 1.

44.The given expression = (45 – 10)/(5 + 2) = 35/7 = 5.

Option (b) = 20/4 = 5.

Directions for Questions 45 to 49: Do the following analysis:

A(f(x, y)) is positive

B (f(x, y)) is negative

C(f(x, y)) is positive

D (f(x, y)) is positive

E(f(x, y)) is positive and so on.

45.1 – 3 + 5 – 7 + 9 – 11 + … – 51

= ( 1 + 5 + 9 + 13 + … 49) – (3 + 7 + 11 … + 51)

= –26

46.Verify each statement to see that (ii) and (iii) are true.

47.The given expression becomes:

Min ( max [5, – 7, 9], min [3, – 1 , 1], max [7, 6, 10])

= Min [ 9, – 1, 10]

= – 1.

48.The given expression becomes:

Max [Èa + bÈ, – Èb + cÈ, Èc + dÈ]

This would never be negative.

49.The respective values are:

–3/2, –7/12, –8/15, and –5/6.

Option (b) is second lowest.

50.Let s = 1, t = 2 and b = 3

Then, f(s + t) + f(s – t)

= f(3) + f(–1) = (3³ + 3^–3)/2 + (3^–1+ 3¹)/2

= [(27 + (1/27)]/2 + [3 + (1/3)]/2

= 730/54 + 10/6

= 820/54 = 410/27

Option (b) 2 f(s) × f(t) gives the same value.

51.This question is based on the logic of a chain function. Given the relationship

At = (t + 1) At – 1 – t At – 2

We can clearly see that the value of A2 would depend on the values of A₀ and A₁. Putting t = 2 in the expression, we get:

A₂ = 3A1 – 2A0 = 7; A₃ = 19; A₄ = 67 and A₅ = 307. Clearly, A6 onwards will be larger than 307 and hence none of the three conclusions are true. Option (e) is the correct answer.

52.In order to solve this question, we would need to check each of the value ranges given in the conclusions: Checking whether Conclusion I is possible

For B = 2, we get A + C = 4 (since A + B + C = 6). This transforms the second equation AB + BC + CA = 9 to:

2(A + C) + CA = 9 Æ CA = 1.

Solving CA = 1 and A + C = 4 we get: (4 – A)A = 1 Æ A² –4A + 1 = 0 Æ A = 2 + 3^1/2 and C = 2–3^1/2. Both these numbers are real and it satisfies A < B < C and hence, Conclusion I is true.

Checking Conclusion II: If we chose A = 2.5, the condition is not satisfied since we get the other two variables as (3.5 + 11.25^1/2) ÷ 2 ª 3.4 and (3.5 – 11.25^1/2) ÷ 2 ª 0.1. In this case, A is no longer the least value and hence Conclusion II is rejected.

Checking Conclusion III we can see that 0 < C < 1 cannot be possible since C being the largest of the three values has to be greater than 3 (the largest amongst A, B, and C would be greater than the average of A, B, C).

Option (a) is correct.

53.The number of ways of distributing n identical things to r people such that any person can get any number of things including 0 is always given by ⁿ^+r–1C_r_–1. In the case of F(4,3), the value of n = 4 and r = 3 and hence the total number of ways without any constraints would be given by ^4+3–1C_3–1 = ⁶C₂ = 15. However, out of these 15 ways of distributing the toys, we cannot count any way in which more than 2 toys are given to any one child. Hence, we need to reduce as follows:

The distribution of 4 toys as (3, 1, 0) amongst three children A, B and C can be done in 3! = 6 ways.

Also, the distribution of 4 toys as (4, 0, 0) amongst three children A, B and C can be done in 3 ways.

Hence, the value of F(4, 3) = 15 – 6 – 3 = 6.

Option (b) is correct.

54.f ( f (x) = 15 when f (x) = 4 or f (x) = 12 in the given function. The graph given in the figure becomes equal to 4 at 4 points and it becomes equal to 12 at 2 points in the figure. This gives us 6 points in the given figure when f(f (x) =15. However, the given function is continuous beyond the part of it which is shown between –10 and +13 in the figure. Hence, we do not know how many more solutions to f (f (x) = 15 would be there. Hence, Option (e) is the correct answer.

55.The given function is a chain function where the value of A_n₊₁ depends on the value A_n.

Thus for n = 0, A1 = A₀² + 1.

For n = 1, A₂ = A₁² + 1 and so on.

In such functions, if you know the value of the function at any one point, the value of the function can be calculated for any value till infinity.

Hence, Statement I is sufficient by itself to find the value of the GCD of A₉₀₀ and A₁₀₀₀.

So also, the Statement II is sufficient by itself to find the value of the GCD of A₉₀₀ and A₁₀₀₀.

Hence, Option (d) is correct.

56.This question can be solved by first putting up the information in the form of a table as follows:

	Product A	Product B	No of machines available	No of Hours/day per Machine.	Total Hrs. per day available for each activity
Grinding	2 hr	3 hr	10	12	120
Polishing	3 hr	2 hr	15	10	150
Profit	` 5	` 7

On the surface, the profit of Product B being higher, we can think about maximising the number of units of Product B. Grinding would be the constraint when we maximise Product B production and we can produce a maximum of 120 ÷ 3 = 40 units of Product B to get a profit of ` 280. The clue that this is not the correct answer comes from the fact that there is a lot of ‘polishing’ time left in this situation. In order to try to increase the profit we can check that if we reduce production of Product B and try to increase the production of Product A, does the profit go up?

When we reduce the production of Product B by 2 units, the production of Product A goes up by 3 units and the profit goes up by +1 (–2 × 7 + 3 × 5 gives a net effect of +1). In this case, the grinding time remains the same (as there is a reduction of 2 units × 3 hours/unit = 6 hours in grinding time due to the reduction in Product B’s production, but there is also a simultaneous increase of 6 hours in the use of the grinders in producing 3 units of Product A). Given that a reduction in the production of Product B, with a simultaneous maximum possible increase in the production of Product A, results in an increase in the profit, we would like to do this as much as possible. To think about it from this point this situation can be tabulated as under for better understanding:

	Product A Production (A)	Product B Production(B)	Grinding Machine Usage = 3A + 2B	Polishing Machine Usage = 2A + 3B	Time Left on Grinding Machine	Time Left on Polishing Machine	Profit = 7A+5B
Case 1	40	0	120	80	0	70	280
Case 2	38	3	120	85	0	65	281
Case 3	36	6	120	90	0	60	281
The limiting case would occur when we reduce the time left on the polishing machine to 0. That would happen in the following case:
Optimal case	12	42	120	150	0	0	294

Hence, the answer would be 294.

57.The value of f (x) as given is: f (x) = x⁴ + x³ + x² + x + 1 = 1 + x + x² + x³ + x⁴ + x⁵. This can be visualised as a geometric progression with 5 terms with the first term 1 and common ratio x. The sum of the GP = f (x) =

The value of f (x⁵) = x²⁰ + x¹⁵ + x¹⁰ + x⁵ + 1 and this can be rewritten as:

F(x⁵)= (x²⁰–1) + (x¹⁵–1) + (x¹⁰–1) + (x⁵–1) + 5. When this expression is divided by f (x) = we get each of the first four terms of the expression would be divisible by it, i.e. (x²⁰–1) would be divisible by f (x) = and would leave no remainder (because x²⁰–1 can be rewritten in the form (x⁵–1) × (x¹⁵ + x¹⁰ + x⁵+ 1) and when you divide this expression by we get the remainder as 0.)

A similar logic would also hold for the terms (x¹⁵–1), (x¹⁰–1) and (x⁵–1). The only term that would leave a remainder would be 5 when it is divided by

Also, for x ≥ 2 we can see the value of would be more than 5. Hence, the remainder would always be 5 and Option (c) is the correct answer.

58.Start by putting = (cosec a)² in the given expression

Now for 0 < a < 90°

= (cosec a)² fi x = = cos² a

Hence, Option (b) is correct.

59.Given that the roots of the equation F(x)= 0 are –2, –1, 1 and 2 respectively and the F(x) is a polynomial with the highest power of x as x⁴, we can create the value of

F(x) = (x + 2)(x + 1)(x –1)(x − 2)

Hence, F(p) = (P + 2)(P + 1)(P –1)(P − 2)

It is given to us that P is a prime number greater than 97. Hence, p would always be of the form 6n ± 1 where n is a natural number greater than or equal to 17.

Thus, we get two cases for F(p).

Case 1: If p = 6n + 1.

F(6n + 1) = (6n + 3)(6n + 2)(6n)(6n –1)

= 3(2n + 1)· 2(3n + 1)(6n)(6n –1)

= (36)(2n + 1)(3n + 1)(n)(6n –1)…(i)

If you try to look for divisibility of this expression by numbers given in the options for various values of n ≥ 17, we see that the for n = 17 and 18 both 360 divides the value of F(p). However at n = 19, none of the values in the four options divides 36 × 39 × 58 × 19 × 113. In this case however, at n = 19, 6n + 1 is not a prime number hence, this case is not to be considered. Whenever we put a value of n as a value greater than 17, such that 6n+1 becomes a prime number, we also see that the value of F(p) is divisible by 360. This divisibility by 360 happens since the expression (2n + 1)(3n + 1)(n)(6n –1) … is always divisible by 10 in all such cases. A similar logic can be worked out when we take p = gn–1. Hence, the Option (d) is the correct answer.

60.In order to solve this question, we start from the value of x = (9 + 4)⁴⁸.

Let the value of x(1 – f ) = xy. (We are assuming (1 – f ) = y, which means that y is between 0 to 1).

The value of x = (9 + 4)⁴⁸ can be rewritten as [⁴⁸C₀9⁴⁸+⁴⁸C₁9⁴⁷(4)+⁴⁸C₂9⁴⁶(4)²+…+ ⁴⁸C₄₇(9)(4)⁴⁷ + ⁴⁸C₄₈(4)⁴⁸] using the binary theorem.

In this value, it is going to be all the odd powers of the (4) which would account for the value of ‘f ’ in the value of x. Thus, for instance it can be seen that the terms ⁴⁸C₀9⁴⁸, ⁴⁸C₂9⁴⁶(4)², ….⁴⁸C₄₈(4)⁴⁸ would all be integers. It is only the terms:⁴⁸C₁9⁴⁷(4), ⁴⁸C₃9⁴⁵(4)³,… ⁴⁸C₄₇(9)(4)⁴⁷ which would give us the value of ‘f ’ in the value of x.

Hence, x(1–f) = x [1 – ⁴⁸C₁9⁴⁷(4) – ⁴⁸C₃9⁴⁵(4)³ –…– ⁴⁸C₄₇ (9)(4)⁴⁷]

In order to think further from this point, you would need the following thought. Let y = (9–4)⁴⁸.

Also, x + y ={⁴⁸C₀9⁴⁸ + ⁴⁸C₁9⁴⁷(4) + ⁴⁸C₂9⁴⁶(4)²+…+ ⁴⁸C₄₇(9)(4)⁴⁷ + ⁴⁸C₄₈(4)⁴⁸} + {⁴⁸C₀9⁴⁸– ⁴⁸C₁9⁴⁷ (4) + ⁴⁸C₂ 9⁴⁶ (4)² +…– ⁴⁸C₄₇(9) (4)⁴⁷ + ⁴⁸C₄₈(4)⁴⁸} = 2{⁴⁸C₀9⁴⁸ + ⁴⁸C₂9⁴⁶(4)²+…⁴⁸C₄₈(4)⁴⁸} – the bracket in this expression has only retained the even terms which are integral. Hence, the value of x+y is an integer.

Further, x+y = [x] + f and hence, if x+y is an integer, [x] + f + y would also be an integer. This automatically means that f+y must be an integer (as [x] is an integer).

Now, the value of y is between 0 to 1 and hence when we add the fractional part of x i.e. ‘f ’ to y, and we need to make it an integer, the only possible integer that f + y can be equal to is 1.

Thus, if f + y = 1 Æ y = (1 – f ).

In order to find the value of x(1 – f ) we can find the value of x × y.

Then, x(1 – f ) = x × y = (9 + 4)⁴⁸ × (9 – 4)⁴⁸ = (81 – 80)⁴⁸ = 1

x(1 – f) = 1

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