Category Archives: ISI Kolkatta Entrance Exam

Solutions to Birthday Problems: IITJEE Advanced Mathematics

In the following problems, each year is assumed to be consisting of 365 days (no leap year):

Problem 1:

What is the least number of people in a room such that it is more likely than not that at least two people will share the same birthday?

Solution 1:

The probability of the second person having a different birthday from the first person is \frac{364}{365}. The probability of the first three persons having different birthdays is \frac{364}{365} \times \frac{363}{365}. In this way, the probability of all n persons in a room having different birthdays is P(n) = \frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+1}{365}. For the value of n, when P(n) falls just below 1/2 is the least number of people in a room when the probability of at least two people having the same birthday becomes greater than one half (that is, more likely than not). Now, one can make the following table:

\begin{tabular}{|c|c|}\hline    N & P(n) \\ \hline    2 & 364/365 \\ \hline    3 & 0.9918 \\ \hline    4 & 0.9836 \\ \hline    5 & 0.9729 \\ \hline    6 & 0.9595 \\ \hline    7 & 0.9438 \\ \hline    8 & 0.9257 \\ \hline    9 & 0.9054 \\ \hline    10 & 0.8830 \\ \hline    11 & 0.8589 \\ \hline    12 & 0.8330 \\ \hline    13 & 0.8056 \\ \hline    14 & 0.7769 \\ \hline    15 & 0.7471 \\ \hline    16 & 0.7164 \\ \hline    17 & 0.6850 \\ \hline    18 &0.6531 \\ \hline    19 & 0.6209 \\ \hline    20 & 0.5886 \\ \hline    21 & 0.5563 \\ \hline    22 & 0.5258 \\ \hline    23 & 0.4956 \\ \hline    \end{tabular}

Thus, the answer is 23. One may say that during a football match with one referee, it is more likely than not that at least two people on the field have the same birthday! ūüôā ūüôā ūüôā

Problem 2:

You are in a conference. What is the least number of people in the conference (besides you) such that it is more likely than not that there is at least another person having the same birthday as yours?

Solution 2:

The probability of the first person having a different birthday from yours is \frac{364}{365}. Similarly, the probability of the first two persons not having the same birthday as yours is \frac{(364)^{2}}{(365)^{2}}. Thus, the probability of n persons not  having the same birthday as yours is \frac{(364)^{n}}{(365)^{n}}. When this value falls below 0.5, then it becomes more likely than not that at least another person has the same birthday as yours. So, the least value of n is obtained from (\frac{364}{365})^{n}<\frac{1}{2}. Taking log of both sides, we solve to get n>252.65. So, the least number of people required is 253.

Problem 3:

A theatre owner announces that the first person in the queue having the same birthday as the one who has already purchased a ticket will be given a free entry. Where (which position in the queue) should one stand to maximize the chance of earning a free entry?

Solution 3:

For the nth person to earn a free entry, first (n-1) persons must have different birthdays and the nth person must have the same birthday as that of one of these previous (n-1) persons. The probability of such an event can we written as

P(n) = [\frac{364}{365} \times \frac{363}{365} \times \frac{362}{365} \times \ldots \frac{365-n+2}{365}] \times \frac{n-1}{365}

For a maximum, we need P(n) > P(n+1). Alternatively, \frac{P(n)}{P(n+1)} >1. Using this expression for P(n), we get \frac{365}{365-n} \times \frac{n-1}{n} >1. Or, n^{2}-n-365>0. For positive n, this inequality is satisfied first for some n between 19 and 20. So, the best place in the queue to get a free entry is the 20th position.

More later,

Nalin Pithwa.

Can anyone have fun with infinite series?

Below is list of finitely many puzzles¬†on infinite series to keep you a bit busy¬†!! ūüôā Note that these puzzles do have an academic flavour, especially concepts of convergence and divergence of an infinite series.

Puzzle 1: A grandmother’s¬†vrat¬†(fast) requires her to keep odd number of lamps of finite capacity lit in a temple at any time during 6pm to 6am the next morning. Each oil-filled lamp lasts 1 hour and it burns oil at a constant rate. She is not allowed to light any lamp after 6pm but she can light any number of lamps before 6pm and transfer oil from some to the others throughout the night while keeping odd number of lamps lit all the time. How many fully-filled oil lamps does she need to complete her¬†vrat?

Puzzle 2: Two number theorists, bored in a chemistry lab, played a game with a large flask containing 2 liters of a colourful chemical solution and an ultra-accurate pipette. The game was that they would take turns to recall a prime number p such that p+2 is also a prime number. Then, the first number theorist would pipette out \frac{1}{p} litres of chemical and the second \frac{1}{(p+2)} litres. How many times do they have to play this game to empty the flask completely?

Puzzle 3: How farthest from the edge of a table can a deck of playing cards be stably overhung if the cards are stacked on top of one another? And, how many of them will be overhanging completely away from the edge of the table?

Puzzle 4: Imagine a tank that can be filled with infinite taps and can be emptied with infinite drains. The taps, turned on alone, can fill the empty tank to its full capacity in 1 hour, 3 hours, 5 hours, 7 hours and so on. Likewise, the drains opened alone, can drain a full tank in 2 hours, 4 hours, 6 hours, and so on. Assume that the taps and drains are sequentially arranged in the ascending order of their filling and emptying durations.

Now, starting with an empty tank, plumber A alternately turns on a tap for 1 hour and opens the drain for 1 hour, all operations done one at a time in a sequence. His sequence, by using t_{i} for i^{th} tap and d_{j} for j^{th} drain, can be written as follows: \{ t_{1}, d_{1}, t_{2}, d_{2}, \ldots\}_{A}.

When he finishes his operation, mathematically, after using all the infinite taps and drains, he notes that the tank is filled to a certain fraction, say, n_{A}<1.

Then, plumber B turns one tap on for 1 hour and then opens two drains for 1 hour each and repeats his sequence: \{ (t_{1},d_{1},d_{2}), (t_{2},d_{3},d_{4}), (t_{3},d_{4},d_{5}) \ldots \}_{B}.

At the end of his (B’s) operation, he finds that the tank is filled to a fraction that is exactly half of what plumber A had filled, that is, 0.5n_{A}.

How is this possible even though both have turned on all taps for 1 hour and opened all drains for 1 hour, although in different sequences?

I hope u do have fun!!

-Nalin Pithwa.

Lagrange’s Mean Value Theorem and Cauchy’s Generalized Mean Value Theorem

Lagrange’s Mean Value Theorem:

If a function f(x) is continuous on the interval [a,b] and differentiable at all interior points of the interval, there will be, within [a,b], at least one point c, a<c<b, such that f(b)-f(a)=f^{'}(c)(b-a).

Cauchy’s Generalized Mean Value Theorem:

If f(x) and phi(x) are two functions continuous on an interval [a,b] and differentiable within it, and phi(x) does not vanish anywhere inside the interval, there will be, in [a,b], a point x=c, a<c<b, such that \frac{f(b)-f(a)}{phi(b)-phi(a)} = \frac{f^{'}(c)}{phi^{'}(c)}.

Some questions based on the above:

Problem 1:

Form Lagrange’s formula for the function y=\sin(x) on the interval [x_{1},x_{2}].

Problem 2:

Verify the truth of Lagrange’s formula for the function y=2x-x^{2} on the interval [0,1].

Problem 3:

Applying Lagrange’s theorem, prove the inequalities: (i) e^{x} \geq 1+x (ii) \ln (1+x) <x, for x>0. (iii) b^{n}-a^{n}<ab^{n-1}(b-a) for b>a. (iv) \arctan(x) <x.

Problem 4:

Write the Cauchy formula for the functions f(x)=x^{2}, phi(x)=x^{3} on the interval [1,2] and find c.

More churnings with calculus later!

Nalin Pithwa.

 

 

Some Applications of Derivatives — Part II

Derivatives in Economics.

Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals.

In a manufacturing operation, the cost of production c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost (c) with respect to a level of production (x), so it is dc/dx.

For example, let c(x) represent the dollars needed needed to produce x tons of steel in one week. It costs more to produce x+h units, and the cost difference, divided by h, is the average increase in cost per ton per week:

\frac{c(x+h)-c(x)}{h}= average increase in cost/ton/wk to produce the next h tons of steel

The limit of this ratio as h \rightarrow 0 is the marginal cost of producing more steel when the current production level is x tons.

\frac{dc}{dx}=\lim_{h \rightarrow 0} \frac{c(x+h)-c(x)}{h}= marginal cost of production

Sometimes, the marginal cost of production is loosely defined to be the extra cost of producing one unit:

\frac{\triangle {c}}{\triangle {x}}=\frac{c(x+1)-c(x)}{1}

which is approximately the value of dc/dx at x. To see why this is an acceptable approximation, observe that if the slope  of c does not change quickly near x, then the difference quotient will be close to its limit, the derivative dc/dx, even if \triangle {x}=1. In practice, the approximation works best for large values of x.

Example: Marginal Cost

Suppose it costs c(x)=x^{3}-6x^{2}+15x  dollars to produce x radiators when 8 to 30 radiators are produced. Your shop currently produces 10 radiators a day. About how much extra cost will it cost to produce one more radiator a day?

Example : Marginal tax rate

To get some feel for the language of marginal rates, consider marginal tax rates. If your marginal income tax rate is 28% and your income increases by USD 1000, you can expect to have to pay an extra USD 280 in income taxes. This does not mean that you pay 28 percent of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes I with respect to income is dT/dI = 0.28. You will pay USD 0.28 out of every extra dollar you earn in taxes. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increase.

Example: Marginal revenue:

If r(x) = x^{3}-3x^{2}+12x gives the dollar revenue from selling x thousand candy bars, 5<= x<=20, the marginal revenue when x thousand are sold is

r^{'}(x) = \frac{d}{dx}(x^{3}-3x^{2}+12x)=3x^{2}-6x+12.

As with marginal cost, the marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 thousand candy bars a week, you can expect your revenue to increase by about r^{'}(10) = 3(100) -6(10) +12=252 USD, if you increase sales to 11 thousand bars a week.

Choosing functions to illustrate economics.

In case, you are wondering why economists use polynomials of low degree to illustrate complicated phenomena like cost and revenue, here is the rationale: while formulae for real phenomena are rarely available in any given instance, the theory of  economics can still provide valuable guidance. the functions about which theory speaks can often be illustrated with low degree polynomials on relevant intervals. Cubic polynomials provide a good balance between being easy to work with and being complicated enough to illustrate important points.

Ref: Calculus and Analytic Geometry by G B Thomas.

More later,

Nalin Pithwa

 

Could a one-sided limit not exist ?

Here is basic concept of limit :

Cyclic Fractions for IITJEE foundation maths

Consider the expression

\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}

Here, in finding the LCM of the denominators, it must be observed that there are not six different compound factors to be considered; for, three of them differ from the other three only in sign.

Thus,

(a-c)  =  -(c-a)

(b-a) = -(a-b)

(c-b) = -(b-c)

Hence, replacing the second factor in each denominator by its equivalent, we may write the expression in the form

-\frac{1}{(a-b)(c-b)}-\frac{1}{(b-c)(a-b)}-\frac{1}{(c-a)(b-c)} call this expression 1

Now, the LCM is (b-c)(c-a)(a-b)

and the expression is \frac{-(b-c)-(c-a)-(a-b)}{(b-c)(c-a)(a-b)}=0.,

Some Remarks:

There is a peculiarity in the arrangement of this example, which is desirable to notice. In the expression 1, the letters occur in what is known as cyclic order; that is, b follows a, a follows c, c follows b. Thus, if a, b, c are arranged round the circumference of a circle, if we may start from any letter and move round in the direction of  the arrows, the other letters follow in cyclic  order; namely, abc, bca, cab.

The observance of this principle is especially important in a large class of examples in which the differences of three letters are involved. Thus, we are observing cyclic order when we write b-c, c-a, a-b, whereas we are violating order by the use of arrangements such as b-c, a-c, a-b, etc. It will always be found that the work is rendered shorter and easier by following cyclic order from the beginning, and adhering to it throughout the question.

Homework:

(1) Find the value of \frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}

2) Find the value of \frac{b}{(a-b)(a-c)} + \frac{c}{(b-c)(b-a)} + \frac{a}{(c-a)(c-b)}

3) Find the value of \frac{z}{(x-y)(x-z)} + \frac{x}{(y-z)(y-x)} + \frac{y}{(z-x)(z-y)}

4) Find the value of \frac{y+z}{(x-y)(x-z)} + \frac{z+x}{(y-z)(y-x)} + \frac{x+y}{(z-x)(z-y)}

5) Find the value of \frac{b-c}{(a-b)(a-c)} + \frac{c-a}{(b-c)(b-a)} + \frac{a-b}{(c-a)(c-b)}

More later,

Nalin Pithwa

Solution of Triangles (Ambiguous Cases) : IIT JEE Maths

The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is, the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the solution of triangles.

  • If the three sides a, b, ¬†and c are given, angle A is obtained from \tan{(A/2)}= \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} or \cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2bc}. B and C can be obtained in a similar way.
  • If two sides b and c and the included angle A are given, then \tan{\frac{B-C}{2}}=\frac{b-c}{b+c}\cot{(A/2)} gives \frac{B-C}{2}. Also, \frac{B-C}{2}=90 \deg - \frac{A}{2}, so that B and C can be evaluated. The third side is given by a=b \frac{\sin{A}}{\sin{B}}, or, a^{2}=b^{2}+c^{2}-2bc \cos{A}.
  • If two sides b and c and the angle B (opposite to side b) are given, then \sin{C}=\frac{c}{b}\sin{B}. And, A=180\deg -(B+C), and a=\frac{b \sin{A}}{\sin{B}} give the remaining elements.

By applying the cosine rule, we have:

\cos{B}=\frac{a^{2}+c^{2} - b^{2}}{2ac}, or if we manipulate this, we get

a^{2}-(2c\cos{B})a+(c^{2}-b^{2})=0

or, a=c \cos{B} \pm \sqrt{b^{2}-(c\sin{B})^{2}}

This equation leads to the following cases:

Case I:

If b<c\sin{B}, no such triangle is possible.

Case II:

Let b=c\sin{B}. There are further following two cases:

Sub-case II a:

B is an obtuse angle, that is, \cos{B} is negative. There exists no such triangle.

Sub-case II b:

B is an acute angle, that is, \cos {B} is positive. There exists only one such triangle.

Case III:

Let b >c \sin{B}. There are following two cases further here also:

Sub-case IIIa:

B is an acute angle, that is, \cos {B} is positive. In this case, two values of a will exist if and only if c\cos{B} > \sqrt{b^{2}-(c \sin{B})^{2}} or, c>b, which means two such triangles are possible. If c<b, only one such triangle is possible.

Sub-case IIIb:

B is an obtuse angle, that is, \cos{B} is negative. In this case, triangle will exist if and only if \sqrt{b^{2}-(c \sin{B})^{2}} > c |\cos{B}| \Longrightarrow b > c. So, in this case, only one such triangle is possible. If b <c, there exists no such triangle.

Note:

If one side a and angles B and C are given, then A=180 \deg -(B+C), and b=a \frac{\sin{B}}{\sin{A}} and c=a\frac{\sin{C}}{\sin{A}}.

If the three angles A, B and C are given, we can only find the ratios of the three sides a, b, and c by using the sine rule(since there are infinite number of similar triangles possible).

More theory later,

Nalin Pithwa

 

Pappus’s theorem

Problem:

Given a point on the circumference of a cyclic quadrilateral, prove that the product of the distances from the point to any pair of opposite sides or to the diagonals are equal.

Proof:

Let a, b, c, d be the coordinates of the vertices A, B, C, D of the quadrilateral and consider the complex plane with origin at the circumcenter of ABCD. Without loss of generality, assume that the circumradius equals 1.

The equation of line AB is

\left | \begin{array}{ccc}    a & \overline{a} & 1 \\    b & \overline{b} & 1 \\    z & \overline{z} & 1 \end{array} \right | = 0.

This is equivalent to z(\overline{a}-\overline{b})-\overline{z}(a-b)=\overline{a}b-a\overline{b}, that is,

z+ab\overline{z}=a+b

Let point M_{1} be the foot of the perpendicular from a point M on the circumcircle to the line AB. If m is the coordinate of the point M, then

z_{M_{1}}=\frac{m-ab\overline{m}+a+b}{2}

and

d(M, AB)=|m-m_{1}|=|m-\frac{m-ab\overline{m}+a+b}{2}|=|\frac{(m-a)(m-b)}{2m}| since m \overline{m} = 1.

Likewise,

d(M, BC)=|\frac{(m-b)(m-c)}{2m}|, d(M, CD)=|\frac{(m-c)(m-d)}{2m}|

d(M, DA)=|\frac{(m-d)(m-a)}{2m}|, d(M, AC)=|\frac{(m-a)(m-c)}{2m}|

and d(M, BD)=|\frac{(m-b)(m-d)}{2m}|

Thus,

d(M, AB).d(M, CD)=d(M, BC).d(M, DA)=d(M, AC).d(M, BD) as claimed.

QED.

More later,

Nalin Pithwa

PS: The above example indicates how easy it is prove many fascinating theorems of pure plane geometry using the tools and techniques of complex numbers.

 

ISI or Pre-RMO practice problems — I

Problem #1.

A man started from home at 14:30 hours and drove to a village, arriving there when the clock indicated 15:15 hours. After staying for 25 min. he drove back by a different route of length 5/4 times the first route at a rate twice as fast, reaching home at 16:00 hours. As compared to the clock at home, the village clock is

(a) 10 min slow

(b) 5 min slow

(c) 5 min fast

(d) 20 min fast

Problem #2.

If \frac{a+b}{b+c}=\frac{c+d}{d+a}, then

(a) a=c

(b) either a=c or a+b+c+d=0

(c) a+b+c+d=0

(d) a=c and b=d.

Problem #3.

In an election, 10% of the voters on the voters list did not cast their votes and 60 voters cast their ballot papers blank. There were only two candidates. The winner was supported by 47% of all voters in the list and he got 308 votes more than his rival. The number of voters on the list was

(A) 3600

(B) 6200

(C) 4575

(D) 6028

I hope to post more such questions every week,

Nalin Pithwa

Maxima and Minima using calculus

Problem:

The vertices of an (n+1)-gon lie  on the sides of a regular n-gon and divide its perimeter into  parts of equal length. How should one construct the (n+1)- gon so that its area is :

(a) maximum

(b) minimum

Hint only:

[One of the golden rule of solving problems in math/physics is to draw diagrams, as had benn emphasized by the maverick American physics Nobel Laureate, Richard Feynman. He expounded this technique even in software development. So, in the present problem, first draw several diagrams.]

There exists a side B_{1}B_{2} of the (n+1) -gon that lies entirely on a side A_{1}A_{2} of the n-gon. Let b=B_{1}B_{2} and b=A_{1}A_{2}. Show that b=\frac{n}{n+1}a. Then, for x=A_{1}B_{1}, we have 0 \leq x \leq \frac{n}{n+1} and the area S of the (n+1) -gon is given by

S(x)=\frac{\sin{\phi}}{2}\Sigma_{i=1}^{n}(\frac{i-1}{n+1}a+x)(\frac{n-i+1}{n+1}a-x)

where \phi=\angle{A_{1}A_{2}A_{3}}. Thus, S(x) is a quadratic function of x. Show that S(x) is a minimal when x=0 or x=\frac{a}{n+1} and S(x) is maximal when x=\frac{a}{2(n+1)}.

Let me know if you have any trouble when you attempt it,

-Nalin Pithwa