Category Archives: Pre-RMO

Set Theory Primer : Some basic thinking and problem solving

Reference: AMS, Student Mathematical Library: Basic Set Theory by A. Shen, et al. Chapter 1. Section 1.

Problem 1:

Consider the oldest mathematician amongst chess players and the oldest chess player amongst mathematicians. Could they be two different people?

Problem 2:

The same question for the best mathematician amongst chess players and the best chess player amongst mathematicians.

Problem 3:

One tenth of mathematicians are chess players, and one sixth of chess players are mathematicians. Which group (mathematicians or chess players) is bigger? What is the ratio of sizes of these two groups?

Problem 4:

Do there exist sets A, B and C such that A \bigcap B \neq \phi, A \bigcap C = \phi and (A\bigcap B)-C = \phi ?

Problem 5:

Which of the following formulas are true for arbitrary sets A, B and C:

i) (A \bigcap B) \bigcup C = (A \bigcup C) \bigcap (B \bigcup C)

ii) (A \bigcup B) \bigcap C = (A \bigcap C) \bigcup (B \bigcap C)

iii) (A \bigcup B) - C = (A-C)\bigcup B

iv) (A \bigcap B) - C = (A - C) \bigcap B

v) A - (B \bigcup C) = (A-B) \bigcap (A-C)

vi) A - (B \bigcap C) = (A-B) \bigcup (A-C)

Problem 6:

Give formal proofs of all valid formulas from the preceding problem. (Your proof should go like this : “We have to prove that the left hand side equals the right hand side. Let x be any element of the left hand side set. Then, ….Therefore, x belongs to the right hand side set. On the other hand, let…”)

Please give counterexamples to the formulas which are not true.

Problem 7:

Prove that the symmetric difference is associative:

A \triangle (B \triangle C) = (A \triangle B) \triangle C for any sets A, B and C. Hint: Addition modulo two is associative.

Problem 8:

Prove that:

(A_{1}\bigcap A_{2}\bigcap \ldots A_{n}) \triangle (B_{1} \bigcap B_{2} \bigcap \ldots B_{n}) = (A_{1} \triangle B_{1}) \bigcup (A_{2} \triangle B_{2}) \ldots (A_{n} \triangle B_{n}) for arbitrary sets A_{1}, A_{2}, \ldots, A_{n}, B_{1}, B_{2}, \ldots, B_{n}.

Problem 9:

Consider an inequality whose left hand side and right hand side contain set variables and operations \bigcap, \bigcup and -. Prove that if this equality is false for some sets, then it is false for some sets that contain at most one element.

Problem 10:

How many different expressions can be formed from set variables A and B by using union, intersection and set difference? (Variables and operations can be used more than once. Two expressions are considered identical if they assume the same value for each set of values of the variables involved.) Solve the same problem for three sets and for n sets. (Answer: In the general case, 2^{2^{n}-1})

Problem 11:

Solve the same problem if only \bigcup and \bigcap are allowed. For n=2 and n-3, this problem is easy to solve; however, no general formula for any n is known. This problem is also called “counting monotone Boolean functions in n variables”.)

Problem 12:

How many subsets does an n-element subset have?

Problem 13:

Assume that A consists of n elements and B \subset A consists of k elements. Find the number of different sets C such that B \subset C \subset A.

Problem 14:

A set U contains 2n elements. We select k subsets of A in such a way that none of them is a subset of another one. What is the maximum possible value of k? (Hint: Maximal k is achieved when all subsets have n elements. Indeed, imagine the following process: We start with an empty set and add random elements one by one until we get U. At most one selected set can appear in this process. On the other hand, the expected number of selected sets that appear during this process can be computed using the linearity of expectation. Take into account that the probability to come across some set Z \subset U is minimal when Z contains n elements, since all the sets of a given size are equiprobable.)

Your comments/solutions are welcome.

Regards,

Nalin Pithwa.

Purva building, 5A
Flat 06
Near Dimple Arcade, Thakur Complex
Mumbai, Maharastra 400101
India.

IITJEE Foundation Maths: Tutorial Problems II

  1. A, B, C start from the same place at the rates a, a+b, a+2b kilometres per hour respectively. B starts n hours after A, how long after B must C start in order that they may overtake A at the same instant, and how far will they then have walked?
  2. Find the distance between two towns when by increasing the speed 7 kilometres per hour a train can perform the journey in 1 hour less, and by reducing the speed 5 kilometres per hour can perform the journey in 1 hour more.
  3. A person buys a certain quantity of land. If he had bought 7 hectares more each hectare would have cost Rs 80 less; and if each hectare had cost Rs. 360 more, he would have obtained 15 hectares less, how much did he pay for the land?
  4. A can walk half a kilometre per hour faster than B; and three quarters of a kilometre per hour faster than C. To walk a certain distance C takes three-quarters of an hour more than B, and two hours more than A; find their rates of walking per hour.
  5. A person spends Rs. 15 in buying goods; if each kg had cost 25 paise more he would have got 5 kg less, but if each kg had cost 15 paise less, he would have received 5 kg more; what weight did he buy?
  6. Five silver coins weight 125 gm and are worth Rs. 6. Ten bronze coins weigh 500 gm and are worth 80 paise. A number of silver and bronze coins which are worth Rs. 134 weigh 11 kg and 250 gm. How many coins of each kind are there?
  7. A and B are playing for money; in the first game, A loses one half of his money, but in the second he wins one-quarter of what B then has. When they cease playing, A has won Rs. 6 and B has still Rs. 14.50 more than A; with what amounts did they begin?
  8. A, B, C each spend the same amount in buying different qualities of the same commodity. B pays 36 paise per kg less than A and obtains 750 gm more; C pays 60 paise per kg more than A and obtains one kg less; how much does each spend?

IITJEE Foundation practice or training problem sheet: I

  1. If the numerator of a fraction is increased by 5, it reduces to \frac{2}{3}, and if the denominator is increased by 9, it reduces to \frac{1}{3}. Find the fraction.
  2. Find a fraction such that it reduces to \frac{3}{5} if 7 is subtracted from its denominator, and reduces to \frac{3}{8} on subtracting 3 from its numerator.
  3. If unity is taken from the denominator of a fraction, it reduces to \frac{1}{2}; if 3 is added to the numerator it reduces to \frac{4}{7}, find the required fraction.
  4. Find a fraction which becomes \frac{3}{4} on adding 5 to the numerator and subtracting 1 from its denominator; and, reduces to \frac{1}{3} on subtracting 4 from the numerator and adding 7 to the denominator.
  5. If 9 is added to the numerator a certain fraction will be increased by \frac{1}{3}; if 6 is taken from the denominator the fraction reduces to \frac{2}{3}; find the required fraction.
  6. At what time between 9 and 10 o’clock are the hands of a watch together?
  7. When are the hands of a clock 8 minutes apart between the hours at 5 and 6 ?
  8. At what time between 10 and 11 o’clock is the hour hand six minutes ahead of the minute hand?
  9. At what time between 1 and 2 o’clock are the hands of a watch in the same straight line?
  10. At what times between 12 and 1 o’clock are the hands of a watch at right angles?
  11. A person buys 20 m of cloth and 25 m of canvas for Rs. 22.50. By selling the cloth at a gain of 15 per cent, and the canvas at a gain of 20 per cent, he clears Rs. 3.75. Find the price of each per metre.
  12. A dealer spends Rs. 6950 in buying horses at Rs, 250/- each and cows at Rs. 200/- each; through disease, he loses 20 percent of the horses and 25 % of the cows. By selling the animals at the price he gave for them, he receives Rs. 5400/-. Find how many of each kind he bought.
  13. The population of a certain district is 53000, of whom 835 can neither read nor write. These consists of 2 %, of all the males and 3 % of all the females; find the number of males and females.
  14. Two persons C and D start simultaneously from two places a kilometre apart, and walk to meet each other; if C walks p kilometres per hour, and D one kilometre per hour faster than C, how far will D have walked when they meet?
  15. A can walk a kilometres per hour faster than B; supposing that he gives B a start of c kilometres, and that B walks a kilometres per hour, how far will A have walked when he overtakes B?

Cheers,

Nalin Pithwa

Two cute problems in HP : IITJEE Foundations\Mains, pre RMO

Problem 1: 

If a^{2}, b^{2}, c^{2} are in AP, show that b+c, c+a, a+b are in HP.

Proof 1:

Note that a straight forward proof is not so easy.

Below is a nice clever solution:

By adding ab+bc+ca to each term, we see that:

a^{2}+ab+ac+bc, b^{2}+ab+ac+bc, c^{2}+ab+ac+bc are in AP.

that is, (a+b)(a+c), (b+c)(b+a), (c+a)(c+b) are in AP.

Dividing each term by (a+b)(b+c)(c+a).

\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} are in AP.

that is, b+c, c+a, a+b are in HP.

QED.

Problem 2:

If the p^{th}, q^{th}, r^{th}, s^{th} terms of an AP are in GP, show that p-q, q-r, r-s are in GP.

Proof 2:

Once again a straight forward proof is not at all easy.

Below is a “bingo” sort of proof ūüôā

With the usual notation, we have

\frac{a+(p-1)d}{a+(q-1)d} = \frac{a+(q-1)d}{a+(r-1)d} = \frac{a+(r-1)d}{a+(s-1)d}

Hence, each of the ratios is equal to

\frac{(a+(p-1)d)-(a+(q-1)d)}{(a+(q-1)d)-(a+(r-1)d)} = \frac{(a+(q-1)d)-(a+(r-1)d)}{(a+(r-1)d)-(a+(s-1)d)}

which in turn is equal  to \frac{p-q}{q-r} = \frac{q-r}{r-s}

Hence, p-q, q-r, r-s are in GP.

Cheers,

Nalin Pithwa

Binomial Theorem Tutorial problems I: IITJEE mains practice

I. Expand up to 5 terms the following expressions:

  1. (1+x)^{\frac{1}{2}}
  2. (1+x)^{\frac{7}{2}}
  3. (1-x)^{\frac{2}{5}}
  4. (1+x^{2})^{-2}
  5. (1-3x)^{\frac{1}{3}}
  6. (1-3x)^{\frac{-1}{2}}
  7. (1+2x)^{-\frac{1}{2}}
  8. (1+\frac{x}{3})^{-2}
  9. (1+\frac{2x}{3})^{\frac{3}{2}}
  10. (1+\frac{1}{2}a)^{-4}
  11. (2+x)^{-2}
  12. (9+2x)^{\frac{1}{2}}
  13. (8+12a)^{\frac{3}{2}}
  14. (9-6x)^{-\frac{3}{2}}
  15. (4a-8x)^{-\frac{1}{2}}

II. Write down and simplify:

  1. The 8th term of (1+2x)^{-\frac{1}{2}}
  2. The 11th term of (1-2x^{3})^{\frac{11}{2}}
  3. The 16th term of (1+3a^{2})^{\frac{16}{3}}
  4. The 6th term of (3a-2b)^{-1}
  5. The (r+1)^{th} term of (1-x)^{-2}
  6. The (r+1)^{th} term of (1-x)^{-4}
  7. The (r+1)^{th} term of (1+x)^{\frac{1}{2}}
  8. The (r+1)^{th} term of (1+x)^{\frac{11}{3}}
  9. The 14th term of (2^{10}-2^{7}x)^{\frac{13}{2}}
  10. The 7th term of (3^{8}+6^{4}x)^{\frac{11}{4}}

Regards,

Nalin Pithwa

Theory of Quadratic Equations: Part III: Tutorial practice problems: IITJEE Mains and preRMO

Problem 1:

Find the condition that a quadratic function of x and y may be resolved into two linear factors. For instance, a general form of such a function would be : ax^{2}+2hxy+by^{2}+2gx+2fy+c.

Problem 2:

Find the condition that the equations ax^{2}+bx+c=0 and a^{'}x^{2}+b^{'}x+c^{'}=0 may have a common root.

Using the above result, find the condition that the two quadratic functions ax^{2}+bxy+cy^{2} and a^{'}x^{2}+b^{'}xy+c^{'}y^{2} may have a common linear factor.

Problem 3:

For what values of m will the expression y^{2}+2xy+2x+my-3 be capable of resolution into two rational factors?

Problem 4:

Find the values of m which will make 2x^{2}+mxy+3y^{2}-5y-2 equivalent to the product of two linear factors.

Problem 5:

Show that the expression A(x^{2}-y^{2})-xy(B-C) always admits of two real linear factors.

Problem 6:

If the equations x^{2}+px+q=0 and x^{2}+p^{'}x+q^{'}=0 have a common root, show that it must be equal to \frac{pq^{'}-p^{'}q}{q-q^{'}} or \frac{q-q^{'}}{p^{'}-p}.

Problem 7:

Find the condition that the expression lx^{2}+mxy+ny^{2} and l^{'}x^{2}+m^{'}xy+n^{'}y^{2} may have a common linear factor.

Problem 8:

If the expression 3x^{2}+2Pxy+2y^{2}+2ax-4y+1 can be resolved into linear factors, prove that P must be be one of the roots of the equation P^{2}+4aP+2a^{2}+6=0.

Problem 9:

Find the condition that the expressions ax^{2}+2hxy+by^{2} and a^{'}x^{2}+2h^{'}xy+b^{'}y^{2} may be respectively divisible by factors of the form y-mx and my+x.

Problem 10:

Prove that the equation x^{2}-3xy+2y^{2}-2x-3y-35=0 for every real value of x, there is a real value of y, and for every real value of y, there is a real value of x.

Problem 11:

If x and y are two real quantities connected by the equation 9x^{2}+2xy+y^{2}-92x-20y+244=0, then will x lie between 3 and 6, and y between 1 and 10.

Problem 11:

If (ax^{2}+bx+c)y+a^{'}x^{2}+b^{'}x+c^{'}=0, find the condition that x may be a rational function of y.

More later,

Regards,

Nalin Pithwa.

Theory of Quadratic Equations: part II: tutorial problems: IITJEE Mains, preRMO

Problem 1:

If x is a real number, prove that the rational function \frac{x^{2}+2x-11}{2(x-3)} can have all numerical values except such as lie between 2 and 6. In other words, find the range of this rational function. (the domain of this rational function is all real numbers except x=3 quite obviously.

Problem 2:

For all real values of x, prove that the quadratic function y=f(x)=ax^{2}+bx+c has the same sign as a, except when the roots of the quadratic equation ax^{2}+bx+c=0 are real and unequal, and x has a value lying between them. This is a very useful famous classic result. 

Remarks:

a) From your proof, you can conclude the following also: The expression ax^{2}+bx+c will always have the same sign, whatever real value x may have, provided that b^{2}-4ac is negative or zero; and if this condition is satisfied, the expression is positive, or negative accordingly as a is positive or negative.

b) From your proof, and using the above conclusion, you can also conclude the following: Conversely, in order that the expression ax^{2}+bx+c may be always positive, b^{2}-4ac must be negative or zero; and, a must be positive; and, in order that ax^{2}+bx+c may be always negative, b^{2}-4ac must be negative or zero, and a must be negative.

Further Remarks:

Please note that the function y=f(x)=ax^{2}+bx+c, where a, b, c \in \Re and a \neq 0 is a parabola. The roots of this y=f(x)=0 are the points where the parabola cuts the y axis. Can you find the vertex of this parabola? Compare the graph of the elementary parabola y=x^{2}, with the graph of y=ax^{2} where a \neq 0 and further with the graph of the general parabola y=ax^{2}+bx+c. Note you will just have to convert the expression ax^{2}+bx+c to a perfect square form.

Problem 3:

Find the limits between which a must lie in order that the rational function \frac{ax^{2}-7x+5}{5x^{2}-7x+a} may be real, if x is real.

Problem 4:

Determine the limits between which n must lie in order that the equation 2ax(ax+nc)+(n^{2}-2)c^{2}=0 may have real roots.

Problem 5:

If x be real, prove that \frac{x}{x^{2}-5x+9} must lie between 1 and -\frac{1}{11}.

Problem 6:

Prove that the range of the rational function y=f(x)=\frac{x^{2}-x+1}{x^{2}+x+1} lies between 3 and \frac{1}{3} for all real values of x.

Problem 7:

If x \in \Re, Prove that the rational function y=f(x)=\frac{x^{2}+34x-71}{x^{2}+2x-7} can have no value between 5 and 9. In other words, prove that the range of the function is (x <5)\bigcup(x>9).

Problem 8:

Find the equation whose roots are \frac{\sqrt{a}}{\sqrt{a} \pm \sqrt(a-b)}.

Problem 9:

If \alpha, \beta are roots of the quadratic equation x^{2}-px+q=0, find the value of (a) \alpha^{2}(\alpha^{2}\beta^{-1}-\beta)+\beta^{2}(\beta^{2}\alpha^{-1}-\alpha) (b) (\alpha-p)^{-4}+(\beta-p)^{-4}.

Problem 10:

If the roots of lx^{2}+mx+n=0 be in the ratio p:q, prove that \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0

Problem 11:

If x be real, the expression \frac{(x+m)^{2}-4mn}{2(x-n)} admits of all values except such as those that lie between 2n and 2m.

Problem 12:

If the roots of the equation ax^{2}+2bx+c=0 are \alpha and \beta, and those of the equation Ax^{2}+2Bx+C=0 be \alpha+\delta and \beta+\delta, prove that \frac{b^{2}-ac}{a^{2}} = \frac{B^{2}-AC}{A^{2}}.

Problem 13:

Prove that the rational function y=f(x)=\frac{px^{2}+3x-4}{p+3x-4x^{2}} will be capable of all values when x is real, provided that p has any real value between 1 and 7. That is, under the conditions on p, we have to show that the given rational function has as its range the full real numbers. (Of course, the domain is real except those values of x for which the denominator is zero).

Problem 14:

Find the greatest value of \frac{x+2}{2x^{2}+3x+6} for any real value of x. (Remarks: this is maxima-minima problem which can be solved with algebra only, calculus is not needed). 

Problem 15:

Show that if x is real, the expression (x^{2}-bc)(2x-b-c)^{-1} has no real value between b and a.

Problem 16:

If the roots of ax^{2}+bx+c=0 be possible (real) and different, then the roots of (a+c)(ax^{2}+2bx+c)=2(ac-b^{2})(x^{2}+1) will not be real, and vice-versa. Prove this.

Problem 17:

Prove that the rational function y=f(x)=\frac{(ax-b)(dx-c)}{(bx-a)(cx-a)} will be capable of all real values when x is real, if a^{2}-b^{2} and c^{2}-a^{2} have the same sign.

Cheers,

Nalin Pithwa

Set theory, relations, functions: preliminaries: Part V

Types of functions: (please plot as many functions as possible from the list below; as suggested in an earlier blog, please use a TI graphing calculator or GeoGebra freeware graphing software): 

  1. Constant function: A function f:\Re \longrightarrow \Re given by f(x)=k, where k \in \Re is a constant. It is a horizontal line on the XY-plane.
  2. Identity function: A function f: \Re \longrightarrow \Re given by f(x)=x. It maps a real value x back to itself. It is a straight line passing through origin at an angle 45 degrees to the positive X axis.
  3. One-one or injective function: If different inputs give rise to different outputs, the function is said to be injective or one-one. That is, if f: A \longrightarrow B, where set A is domain and set B is co-domain, if further, x_{1}, x_{2} \in A such that x_{1} \neq x_{2}, then it follows that f(x_{1}) \neq f(x_{2}). Sometimes, to prove that a function is injective, we can prove the conrapositive statement of the definition also; that is, y_{1}=y_{2} where y_{1}, y_{2} \in codomain \hspace{0.1in} range, then it follows that x_{1}=x_{2}. It might be easier to prove the contrapositive. It would be illuminating to construct your own pictorial examples of such a function. 
  4. Onto or surjective: If a function is given by f: X \longrightarrow Y such that f(X)=Y, that is, the images of all the elements of the domain is full of set Y. In other words, in such a case, the range is equal to co-domain. it would be illuminating to construct your own pictorial examples of  such a function.
  5. Bijective function or one-one onto correspondence: A function which is both one-one and onto is called a bijective function. (It is both injective and surjective). Only a bijective function will have a well-defined inverse function. Think why! This is the reason why inverse circular functions (that is, inverse trigonometric functions have their domains restricted to so-called principal values). 
  6. Polynomial function: A function of the form f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\ldots + a_{n}x^{n}, where n is zero or positive integer only and a_{i} \in \Re is called a polynomial with real coefficients. Example. f(x)=ax^{2}=bx+c, where a \neq 0, a, b, c \in \Re is called a quadratic function in x. (this is a general parabola).
  7. Rational function: The function of the type \frac{f(x)}{g(x)}, where g(x) \neq 0, where f(x) and g(x) are polynomial functions of x, defined in a domain, is called a rational function. Such a function can have asymptotes, a term we define later. Example, y=f(x)=\frac{1}{x}, which is a hyperbola with asymptotes X and Y axes.
  8. Absolute value function: Let f: \Re \longrightarrow \Re be given by f(x)=|x|=x when x \geq 0 and f(x)=-x, when x<0 for any x \in \Re. Note that |x|=\sqrt{x^{2}} since the radical sign indicates positive root of a quantity by convention.
  9. Signum function: Let f: \Re \longrightarrow \Re where f(x)=1, when x>0 and f(x)=0 when x=0 and f(x)=-1 when x<0. Such a function is called the signum function. (If you can, discuss the continuity and differentiability of the signum function). Clearly, the domain of this function  is full \Re whereas the range is \{ -1,0,1\}.
  10. In part III of the blog series, we have already defined the floor function and the ceiling function. Further properties of these functions are summarized (and some with proofs in the following wikipedia links): (once again, if you can, discuss the continuity and differentiablity of the floor and ceiling functions): https://en.wikipedia.org/wiki/Floor_and_ceiling_functions
  11. Exponential function: A function f: \Re \longrightarrow \Re^{+} given by f(x)=a^{x} where a>0 is called an exponential function. An exponential function is bijective and its inverse is the natural logarithmic function. (the logarithmic function is difficult to define, though; we will consider the details later). PS: Quiz: Which function has a faster growth rate — exponential or a power function ? Consider various parameters.
  12. Logarithmic function: Let a be a positive real number with a \neq 1. If a^{y}=x, where x \in \Re, then y is called the logarithm of x with base a and we write it as y=\ln{x}. (By the way, the logarithmic function is used in the very much loved mp3 music :-))

Regards,

Nalin Pithwa

Rules for Inequalities

If a, b and c are real numbers, then

  1. a < b \Longrightarrow a + c< b + c
  2. a < b \Longrightarrow a - c < b - c
  3. a < b \hspace{0.1in} and \hspace{0.1in}c > 0 \Longrightarrow ac < bc
  4. a < b \hspace{0.1in} and \hspace{0.1in}c < 0 \Longrightarrow bc < ac special case: a < b \Longrightarrow -b < -a
  5. a > 0 \Longrightarrow \frac{1}{a} > 0
  6. If a and b are both positive or both negative, then a < b \Longrightarrow \frac{1}{b} < \frac{1}{a}.

Remarks:

Notice the rules for multiplying an inequality by a number: Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign.

Regards,

Nalin Pithwa.

Set Theory, Relations, Functions Preliminaries: Part III

FUNCTIONS:

Functions as a special kind of relation:

Let us first consider an example where set A is a set of men, and B is a set of positive real numbers. Let us say f is a relation from A to B given by : f = \{ (x,y) : x \in A, y \hspace{0.1in} is \hspace{0.1in} the \hspace{0.1in} weight \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} person \hspace{0.1in} x \}

Hence, f “relates” every man in set A to his weight in set B. That is,

i) Every man has some weight associated with him in set B. (ii) That weight is unique. That is, a person cannot have more than one weight (at a given time, of course) !! ūüôā This, of course, does not mean that two different persons, say P and Q may not have the same weight 100 kg ( the same element of set B). The only thing it means is that any one person, say P will have one and only one weight (100kg) at the time instant of measurement and not more than one weights (which would be crazy) at a time instant it is measured !!

Definition I (a function defined as a relation):

A function f from a set A (called domain) to a set B (called codomain) is a relation that associates or “pairs up” every element of domain A with a unique element of codomain B. (Note that whereas a relation from a set A to a set B is just a subset of the cartesian product A \times B).

Some remarks: The above definition is also motivated by an example of a function as a relation. On the other hand, another definition of a function can be motivated as follows:

We know that the boiling point of water depends on the height of water above sea level. We also know that the simple interest on a deposit in a bank depends on the duration of deposit held in the bank. In these and several such examples, one quantity, say y, depends on another quantity “x”.

Symbol: f: A \longrightarrow B; if x \in A, y \in B, then we also denote: f: x \longmapsto y; we also write y=f(x), read as “y is f of x”.

Here, y is called image of x under f and x is called the preimage of y under f.

Definition: Range: The set of all images in B is called the range of f. That is, Range = \{ f(x): x \in A\}

Note: (i) Every function is a relation but every relation need not be a function. (Homework quiz: find illustrative examples for the same) (ii) If the domain and codomain are not specified, they are assumed to be the set of real numbers.

In calculus, we often want to refer to a generic function without having any particular formula in mind. Leonhard Euler invented a symbolic way to say “y is a function of x” by writing

y = f(x) (“y equals f of x”)

In this equation, the symbol f represents the function. The letter x, called the independent variable, represents an input value from the domain of f, and y, the dependent variable, represents the corresponding output value f(x) in the range of f. Here is the formal definition of function: (definition 2):

A function from a set D to a set \Re is a rule that assigns a unique element f(x) in \Re to each element x in D.

In this definition, D=D(f) (read “D of f”) is the domain of the function f and \Re is the range (or codomain containing the range of f).

Think of a function f as a kind of machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain. In our scope, we will usually define functions in one of two ways:

a) by giving a formula such as y=x^{2} that uses a dependent variable y to denote the value of the function, or

b) by giving a formula such as f(x)=x^{2} that defines a function symbol f to name the function.

NOTE: there can be well-defined functions which do not have any formula at all; for example, let f(x) = 0 when x \in Q and f(x)=1, when x \in Q^{'}.

Strictly speaking, we should call the function f and not f(x) as the latter denotes the value of the function at the point x. However, as is common usage, we will often refer to the function as f(x) in order to name the variable on which f depends.

It is sometimes convenient to use a single letter to denote both a function and the dependent variable. For instance, we might say that the area A of a circle of radius r is given by the function : A(r)=\pi r^{2}.

Evaluation:

As we said earlier, most of the functions in our scope will be¬†real-valued function¬†of a¬†real variable,¬†functions whose domains and ranges are sets of real numbers. We evaluate such functions by susbtituting particular values from the domain into the function’s defining rule to calculate the corresponding values in the range.

Example 1:

The volume V of a ball (solid sphere) r is given by the function: V(r)=\frac{4}{3}\pi  r^{3}.

The volume of a ball of radius 3 meters is : V(3)=\frac{4}{3}\pi (3)^{3}=36 \pi m^{3}.

Example 2:

Suppose that the function F is defined for all real numbers t by the formula: F(t)=2(t-1)+3.

Evaluate F at the output values 0, 2, x+2, and F(2).

Solution 2:

In each case, we substitute the given input value for t into the formula for F:

F(0)=2(0-1)+3=-2+3=1

F(2)=2(2-1)+3=2+3=5

F(x+2)=2(x+2-1)+3=2x+3

F(F(2))=F(5)=2(5-1)+3=11

The Domain Convention

When we define a function y=f(x) with a formula and the domain is not stated explicitly, the domain is assumed to be the largest set of x-values for which the formula gives real x-values. This is the function’s so-called¬†natural domain.¬†If we want the domain to be restricted in some way, we must say so.

The domain of the function y=x^{2} is understood to be the entire set of real numbers. The formula gives a real value y-value for every real number x. If we want to restrict the domain to values of x greater than or equal to 2, we must write ” y=x^{2}” for x \geq 2.

Changing the domain to which we apply a formula usually changes the range as well. The range of y=x^{2} is [0, \infty). The  range of y=x^{2} where x \geq 2 is the set of all numbers obtained by squaring numbers greater than or equal to 2. In symbols, the range is \{ x^{2}: x \geq 2\} or \{ y: y \geq 4\} or [4,\infty)

Example 3:

Function : y = \sqrt{1-x^{2}}; domain [-1,1]; Range (y) is [0,1]

Function: y=\frac{1}{x}; domain (-\infty,0) \bigcup (0,\infty); Range (y) is (-\infty,0)\bigcup (0,\infty)

Function: y=\sqrt{x}; domain (0,\infty) and range (y) is (0,\infty)

Function y = \sqrt{4-x}, domain (-\infty,,4], and range (y) is [0, \infty)

Graphs of functions:

The graph of a function f is the graph of the equation y=f(x). It consists of the points in the Cartesian plane whose co-ordinates (x,y) are input-output pairs for f.

Not every curve you draw is the graph of a function. A function f can have only one value f(x) for each x in its domain so no vertical line can intersect the graph of a function more than once. Thus, a circle cannot be the graph of a function since some vertical line intersect the circle twice. If a is in the domain of a function f, then the vertical line x=a will intersect the graph of f in the single point (a, f(a)).

Example 4:¬†Graph the function y=x^{2} over the interval [-2.2]. (homework).Thinking further:¬†so plotting the above graph requires a table of x and y values; but how do we connect the points ? Should we connect two points by a straight line, smooth line, zig-zag line ??? How do we know for sure what the graph looks like between the points we plot? The answer lies in calculus, as we will see in later chapter. There will be a marvelous mathematical tool called the derivative to find a curve’s shape between plotted points. Meanwhile, we will have to settle for plotting points and connecting them as best as we can.¬†

PS: (1) you can use GeoGebra, a beautiful freeware for plotting various graphs, and more stuff https://www.geogebra.org/ (2) If you wish, you can use a TI-graphing calculator. This is a nice investment for many other things like number theory also. See for example,

https://www.amazon.in/Texas-Instruments-Nspire-Graphing-Calculator/dp/B004NBZAYS/ref=sr_1_2?crid=3JSHJUOZMDMUS&keywords=ti+nspire+cx&qid=1569334614&s=electronics&sprefix=TI+%2Caps%2C267&sr=1-2

Meanwhile, you need to be extremely familiar with graphs of following functions; plot and check on your own:

y=x^{3}, y=x^{2/3}, y=\sqrt{x}, y=\sqrt[3]{x}, y=\frac{1}{x}, y=\frac{1}{x^{2}}, y=mx, where m \in Z, y=x^{3/2}

Sums, Differences, Products and Quotients

Like numbers, functions can be added, subtracted, multiplied and divided (except where the the denominator is zero) to produce new functions. If f and g are functions, then for every x that belongs to the domains of BOTH f and g, we define functions: f+g, f-g, fg by the formulas:

(f+g)(x)=f(x)+g(x),

(f-g)(x)=f(x)-g(x)

(fg)(x)=f(x)g(x)

At any point D(f) \bigcap D(g) at which g(x) \neq 0, we can also define the function f/g by the formula:

(\frac{f}{g})(x)=\frac{f(x)}{g(x)}, where g(x) \neq 0

Functions can also be multiplied by constants. If c is a real number, then the function cf is defined for all x in the domain of f by (cf)(x)=cf(x)

Example 5:

Function f, formula y=\sqrt{x}, domain [0,\infty)

Function g, formula g(x)=\sqrt{(1-x)}, domain (-\infty, 1]

Function 3g, formula 3g(x)=3\sqrt{(1-x)}, domain (-\infty, 1]

Function f+g, formula (f+g)(x)=\sqrt{x}+\sqrt{(1-x)}, domain [0,1]=D(f) \bigcap D(g)

Function f-g, formula (f-g)(x)=\sqrt{x}-\sqrt{(1-x)}, domain [0.1]

Function g-f, formula (g-f)(x)=\sqrt{(1-x)}-\sqrt{x}, domain [0,1]

Function f . g, formula (f . g)(x)=f(x)g(x) = \sqrt{x(1-x)}, domain [0,1]

Function \frac{f}{g}, formula \frac{f}{g}(x)=\frac{f(x)}{g(x)}=\sqrt{\frac{x}{1-x}}, domain is [0,1)

Function \frac{g}{f}(x) = \frac{g(x)}{f(x)}=\sqrt{\frac{1-x}{x}}, domain (0,1]

Composite Functions:

Composition is another method for combining functions.

Definition:

If f and g are functions, the¬†composite¬†function f \circ g (f “circle” g) is defined by (f \circ g)(x)=f(g(x)). The domain of f \circ g consists of the numbers x in the domain of g for which g(x) lies in the domain of f.

The definition says that two functions can be composed when the image of the first lies in the domain of the second. To (f \circ g)(x) we first find g(x) and second find f(g(x)).

Clearly, in general, (f \circ g)(x) \neq (g \circ f)(x). That is, composition of functions is not commutative.

Example 6:

If f(x)=\sqrt{x} and g(x)=x+1, find (a) (f \circ g)(x) (b) (g \circ f)(x) (c) (f \circ f)(x) (d) (g \circ g)(x)

Solution 6:

a) (f \circ g)(x) = f(g(x))=\sqrt{g(x)}=\sqrt{x+1}, domain is [-1, \infty)

b) (g \circ f)(x)=g(f(x))=f(x)+1=\sqrt{x}+1, domain is [0, \infty)

c) (f \circ f)(x)=f(f(x))=\sqrt{f(x)}=\sqrt{\sqrt{x}}=x^{\frac{1}{4}}, domain is [0, \infty)

d) (g \circ g)(x)=g(g(x))=g(x)+1=(x+1)+1=x+2, domain is \Re or (-\infty, \infty)

Even functions and odd functions:

A function f(x) is said to be even if f(x)=f(-x). That is, the function possesses symmetry about the y-axis. Example, y=f(x)=x^{2}.

A function f(x) is said to be odd if f(x)=-f(-x). That is, the function possesses symmetry about the origin. Example y=f(x)=x^{3}.

Any function can be expressed as a sum of an even function and an odd function.

A function could be neither even nor odd.

Note that a function like y^{2}=x possesses symmetry about the x-axis !!

Piecewise Defined Functions:

Sometimes a function uses different formulas or formulae over different parts of its domain. One such example is the absolute value function:

y=f(x) = |x|=x, when x \geq 0 and y=-x, when x<0.

Example 7:

The function f(x)=-x, when x<0, y=f(x)=x^{2}, when 0 \leq x \leq 1, and f(x)=1, when x>1.

Example 8:

The greatest integer function:

The function whose value at any number x is the greatest integer less than or equal to x is called the greatest integer function or the integer floor function. It is denoted by \lfloor x \rfloor.

Observe that \lfloor 2.4 \rfloor =2; \lfloor 1.4 \rfloor =1; \lfloor 0 \rfloor =0; \lfloor -1.2 \rfloor =-2; \lfloor 2 \rfloor =2; \lfloor 0.2 \rfloor =0\lfloor -0.3 \rfloor =-1; \lfloor -2 \rfloor =-2.

Example 9:

The least integer function:

The function whose value at any number x is the smallest integer greater than or equal to x is called the least integer function or the integer ceiling function. It is denoted by \lceil x \rceil. For positive values of x, this function might represent, for example, the cost of parking x hours in a parking lot which charges USD 1 for each hour or part of an hour.

Cheers,

Nalin Pithwa