Indian Statistical Institute Entrance Exam

Functions — “s’wat” Math is about !! :-)

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Reference: Nordic Mathematical Contest 1987, R. Todev:

Question:

Let f be a function, defined for natural numbers, that is strictly increasing, such that values of the function are also natural numbers and which satisfies the conditions f(2)=a>2 and f(mn)=f(m)f(n) for all natural numbers m and n. Define the smallest possible value of a.

Solution:

Since, f(n)=n^{2} is a function satisfying the conditions of the problem, the smallest possible a is at most 4. Assume that a=3. It is easy to prove by induction that f(n^{k})={f(n)}^{k} for all k \geq 1. So, taking into account that f is strictly increasing, we get

{f(3)}^{4}=f(3^{4})=f(81)>f(64)=f(2^{6})={f(2)}^{6}=3^{6}=27^{2}>25^{2}=5^{4}

as well as {f(3)}^{8}=f(3^{8})=f(6561)<f(8192)=f(2^{13})={f(2)}^{13}=3^{13}<6^{8}.

So, we arrive at 5<f(3)<6. But, this is not possible, since f(3) is an integer. So, a=4.

Cheers,

Nalin Pithwa.

Algebra : max and min: RMO/INMO problem solving practice

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Question 1:

If x^{2}+y^{2}=c^{2}, find the least value of \frac{1}{x^{2}} + \frac{1}{y^{2}}.

Solution 1:

Let z^{'}=\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{y^{2}+x^{2}}{x^{2}y^{2}} = \frac{c^{2}}{x^{2}y^{2}}

z^{'} will be minimum when \frac{x^{2}y^{2}}{c^{2}} will be minimum.

Now, let z=\frac{x^{2}y^{2}}{c^{2}}=\frac{1}{c^{2}}(x^{2})(y^{2})….call this equation I.

Hence, z will be maximum when x^{2}y^{2} is maximum but (x^{2})(y^{2}) is the product of two factors whose sum is x^{2}+y^{2}=c^{2}.

Hence, x^{2}y^{2} will be maximum when both these factors are equal, that is, when

\frac{x^{2}}{1}=\frac{y^{2}}{1}=\frac{x^{2}+y^{2}}{1}=\frac{c^{2}}{1}. From equation I, maximum value of z=\frac{c^{2}}{4}. Hence, the least value of \frac{1}{x^{2}} + \frac{1}{y^{2}}=\frac{4}{c^{2}}.

Some basics related to maximum and minimum:

Basic 1:

Let a and b be two positive quantities, S their sum and P their product; then, from the identity:

4ab=(a+b)^{2}-(a-b)^{2}, we have

4P=S^{2}-(a-b)^{2} and S^{2}=4P+(a-b)^{2}.

Hence, if S is given, P is greatest when a=b; and if P is given, S is least when a=b. That is, if the sum of two positive quantities is given, their product is greatest when they are equal; and, if the product of two positive quantities is given, their sum is least when they are equal.

Basic 2:

To find the greatest value of a product the sum of whose factors is constant.

Solution 2:

Let there be n factors a,b,c,\ldots, k, and suppose that their sum is constant and equal to s.

Consider the product abc\ldots k, and suppose that a and b are any two unequal factors. If we replace the two unequal factors a and b by the two equal factors \frac{a+b}{2}, \frac{a+b}{2}, the product is increased, while the sum remains unaltered; hence, so long as the product contains two unequal factors it can be increased without altering the sum of the factors; therefore, the product is greatest when all the factors are equal. In this case, the value of each of the n factors is \frac{s}{m}, and the greatest value of the product is (\frac{s}{n})^{n}, or (\frac{a+b+c+\ldots+k}{n})^{n}.

Corollary to Basic 2:

If a, b, c, \ldots k are unequal, (\frac{a+b+c+\ldots+k}{n})^{2}>abc\ldots k;

that is, \frac{a+b+c+\ldots +k}{n} > (\frac{a+b+c+\ldots + k}{n})^{\frac{1}{n}}.

By an extension of the meaning of the terms arithmetic mean and geometric mean, this result is usually stated as follows: the arithmetic mean of any number of positive quantities is greater than the geometric mean.

Basic 3:

To find the greatest value of a^{m}b^{n}c^{p}\ldots when a+b+c+\ldots is constant; m,n, p, ….being positive integers.

Solution to Basic 3:

Since m,n,p, …are constants, the expression a^{m}b^{n}c^{p}\ldots will be greatest when (\frac{a}{m})^{m}(\frac{b}{n})^{n}(\frac{c}{p})^{p}\ldots is greatest. But, this last expression is the product of m+n+p+\ldots factors whose sum is m(\frac{a}{m})+n(\frac{b}{n})+p(\frac{c}{p})+\ldots, or a+b+c+\ldots, and therefor constant. Hence, a^{m}b^{n}c^{p}\ldots will be greatest when the factors \frac{a}{m}, \frac{b}{n}, \frac{c}{p}, ldots are all equal, that is, when

\frac{a}{m} = \frac{b}{n} = \frac{c}{p} = \ldots = \frac{a+b+c+\ldots}{m+n+p+\ldots}

Thus, the greatest value is m^{m}n^{n}p^{p}\ldots (\frac{a+b+c+\ldots}{m+n+p+\ldots})^{m+n+p+\ldots}.

Some examples using the above techniques:

Example 1:

Show that (1^{r}+2^{r}+3^{r}+\ldots+n^{r})>n^{n}(n!)^{r} where r is any real number.

Solution 1:

Since \frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n}>(1^{r}.2^{r}.3^{r}\ldots n^{r})^{\frac{1}{n}}

Hence, (\frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n})^{n}>1^{r}.2^{r}.3^{r} \ldots n^{r}

that is, >(n!)^{r}, which is the desired result.

Example 2:

Find the greatest value of (a+x)^{3}(a-x)^{4} for any real value of x numerically less than a.

Solution 2:

The given expression is greatest when (\frac{a+x}{3})^{3}(\frac{a-x}{4})^{4} is greatest; but, the sum of the factors of this expression is 3(\frac{a+x}{3})+4(\frac{a-x}{4}), that is, 2a; hence, (a+x)^{3}(a-x)^{4} is greatest when \frac{a+x}{3}=\frac{a-x}{4}, that is, x=-\frac{a}{7}. Thus, the greatest value is \frac{6^{3}8^{4}}{7^{7}}a^{r}.

Some remarks/observations:

The determination of maximum and minimum values may often be more simply effected by the solution of a quadratic equation than by the foregoing methods. For example:

Question:

Divide an odd integer into two integral parts whose product is a maximum.

Answer:

Let an odd integer be represented as 2n+1; the two parts by x and 2n+1-x; and the product by y; then (2n+1)x-x^{2}=y; hence,

2x=(2n+1)\pm \sqrt{(2n+1)^{2}-4y}

but the quantity under the radical sign must be positive, and therefore y cannot be greater than \frac{1}{4}(2n+1)^{2}, or, n^{2}+n+\frac{1}{4}; and since y is integral its greatest value must be n^{2}+n; in which case x=n+1, or n; thus, the two parts are n and n+1.

Sometimes we may use the following method:

Find the minimum value of \frac{(a+x)(b+x)}{c+x}.

Solution:

Put c+x=y; then the expression =\frac{(a-c+y)(b-c+y)}{y}=\frac{(a-c)(b-c)}{y}+y+a-c+b-c

which in turn equals

(\frac{\sqrt{(a-c)(b-c)}}{\sqrt{y}}-\sqrt{y})^{2}+a-c+b-c+2\sqrt{(a-c)(b-c)}.

Hence, the expression is a a minimum when the square term is zero; that is when y=\sqrt{(a-c)(b-c)}.

Thus, the minimum value is a-c+b-c+2\sqrt{(a-c)(b-c)}, and the corresponding value of x is \sqrt{(a-c)(b-c)}-c.

Problems for Practice:

  1. Find the greatest value of x in order that 7x^{2}+11 may be greater than x^{3}+17x.
  2. Find the minimum value of x^{2}-12x+40, and the maximum value of 24x-8-9x^{2}.
  3. Show that (n!)^{2}>n^{n} and 2.4.6.\ldots 2n<(n+1)^{n}.
  4. Find the maximum value of (7-x)^{4}(2+x)^{5} when x lies between 7 and -2.
  5. Find the minimum value of \frac{(5+x)(2+x)}{1+x}.

More later,

Nalin Pithwa.

Algebra question: RMO/INMO problem-solving practice

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Question:

If \alpha, \beta, \gamma be the roots of the cubic equation ax^{3}+3bx^{2}+3cx+d=0. Prove that the equation in y whose roots are \frac{\beta\gamma-\alpha^{2}}{\beta+\gamma-2\alpha} + \frac{\gamma\alpha-\beta^{2}}{\gamma+\\alpha-2\beta} + \frac{\alpha\beta-\gamma^{2}}{\alpha+\beta-2\gamma} is obtained by the transformation axy+b(x+y)+c=0. Hence, form the equation with above roots.

Solution:

Given that \alpha, \beta, \gamma are the roots of the equation:

ax^{3}+3bx^{2}+3cx+d=0…call this equation I.

By relationships between roots and co-efficients, (Viete’s relations), we get

\alpha+\beta+\gamma=-\frac{3b}{a} and \alpha\beta+\beta\gamma+\gamma\alpha=\frac{3c}{a}, and \alpha\beta\gamma=-\frac{d}{a}

Now, \gamma=\frac{\beta\gamma-\alpha^{2}}{\beta+\gamma-2\alpha}=\frac{\frac{\alpha\beta\gamma}{\alpha}-\alpha^{2}}{(\alpha+\beta+\gamma)-3\alpha}=\frac{-\frac{d}{a\alpha}-\alpha^{2}}{-\frac{3b}{a}-3\alpha}=\frac{d+a\alpha^{3}}{3\alpha(b+a\alpha)}, that is,

3xy(b+ax)=d+ax^{3}, or ax^{3}-3ayx^{2}-3byx+d=0…call this equation II.

Subtracting Equation II from Equation I, we get

3(b+ay)x^{2}+3(c+by)x=0

(b+ay)x+c+by=0 since x \neq 0

axy+b(x+y)+c=0 which is the required transformation.

Now, (ay+b)x=-(by+c), that is, x=-\frac{by+c}{ay+b}

Putting this value of x in Equation I, we get

-a(\frac{by+c}{ay+b})^{3}+3b(\frac{by+c}{ay+b})^{2}-3c(\frac{by+c}{ay+b})+d=0, that is,

a(by+c)^{3}-3b(by+c)^{2}(ay+b)+3c(by+c)(ay+b)^{2}-d(ay+b)^{3}=0, which is the required equation.

Cheers,

Nalin Pithwa.

Uses of mathematical induction to prove inequalities

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Some classic examples are presented below to illustrate the use of mathematical induction to prove inequalities:

Example 1:

Prove the inequality n<2^{n} for all positive integers n.

Proof 1:

Let P(n) be the proposition that n<2^{n}.

Basis step:

P(1) is true, because 1<2^{1}=2. This completes the basis step.

Inductive step:

We first assume the inductive hypothesis that P(k) is true for the positive integer k. That is, the inductive hypothesis P(k) is the statement that k<2^{k}. To complete the inductive step, we need to show that if P(k) is true, then P(k+1), which is the statement that k+1<2^{k+1} is also true. That is, we need to show that if k<2^{k}, then k+1<2^{k+1}. To show that this conditional statement is true for the positive integer k, we first add 1 to both sides of k<2^{k} and then note that 1\leq 2^{k}. This tells us that

k+1<2^{k}+1 \leq 2^{k} + 2^{k} = 2.2^{k}=2^{k+1}.

This shows that P(k+1) is true; namely, that k+1<2^{k+1}, based on the assumption that P(k) is true. The induction step is complete.

Therefore, because we have completed both the basis step and the inductive step, by the principle of mathematical induction we have shown that n<2^{n} is true for all positive integers n. QED.

Example 2:

Prove that 2^{n}<n! for every positive integer n with n \geq 4. (Note that this inequality is false for n=1, 2, 3).

Proof 2:

Let P(n) be the proposition that 2^{n}<n!

Basis Step:

To prove the inequality for n \geq 4 requires that the basis step be P(4). Note that P(4) is true because 2^{4} =16<24=4!.

Inductive Step:

For the inductive step, we assume that P(k) is true for the positive integer k with k \geq 4. That is, we assume that 2^{k}<k! with k \geq 4. We must show that under this hypothesis, P(k+1) is also also true. That is, we must show that if 2^{k}<k! for the positive integer k \geq 4, then 2^{k+1}<(k+1)!. We have

2^{k+1}=2.2^{k} (by definition of exposition)

which <2.k! (by the inductive hypothesis)

which <(k+1).k! because 2<k+1

which equals (k+1)! (by definition of factorial function).

This shows that P(k+1) is true when P(k) is true. This completes the inductive step of the proof.

We have completed the basis step and the inductive step. Hence, by mathematical induction P(k) is true for all integers n with n \geq 4. That is, we have proved that 2^{n}<n! is true for all integers n with n \geq 4. QED.

An important inequality for the sum of the reciprocals of a set of positive integers will be proved in the example below:

Example 3:

An inequality for Harmonic Numbers

The harmonic numbers H_{j}, j=1,2,3, \ldots are defined by H_{j}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{j}.

For instance, H_{4}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}. Use mathematical induction to show that H_{2*{n}} \geq 1 + \frac{n}{2}, whenever n is a nonnegative integer.

Proof 3:

To carry out the proof, let P(n) be the proposition that H_{2^{n}} \geq 1 + \frac{n}{2}.

Basis Step:

P(0) is true because H_{2^{0}}=H_{1}=1 \geq 1 + \frac{0}{2}.

Inductive Step:

The inductive hypothesis is the statement that P(1) is true, that is, H_{2^{k}} \geq 1+\frac{k}{2}, where k is a nonnegative integer. We must show that if P(k) is true, then P(k+1), which states that H_{2^{k+1}} \geq 1 + \frac{k+1}{2}, is also true. So, assuming the inductive hypothesis, it follows that

H_{2^{k+1}}=1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{2^{k}}+\frac{1}{2^{k}+1}+\ldots + \frac{1}{2^{k+1}} (by the definition of harmonic number)

which equals H_{2^{n}}+\frac{1}{2^{k}+1} + \ldots + \frac{1}{2^{k+1}} (by the definition of the 2^{k}th harmonic number)

\geq (1+\frac{k}{2})+\frac{1}{2^{k}+1}+\ldots + \frac{1}{2^{k+1}} (by the inductive hypothesis)

\geq (1+\frac{k}{2}) + 2^{k}\frac{1}{2^{k+1}} (because there are 2^{k} terms each greater than or equal to \frac{1}{2^{k+1}})

\geq (1+\frac{k}{2})+\frac{1}{2} (canceling a common factor of 2^{k} in second term), which in turn, equals 1+ \frac{k+1}{2}.

This establishes the inductive step of the proof.

We have completed the basis step and the inductive step. Thus, by mathematical induction P(n) is true for all nonnegative integers. That is, the inequality H_{2^{n}} \geq 1 + \frac{n}{2} for the harmonic number is valid for all non-negative integers n. QED.

Remark:

The inequality established here shows that the harmonic series 1+\frac{1}{2}+\frac{1}{3}+\ldots \frac{1}{n}+\ldots is a divergent infinite series. This is an important example in the study of infinite series.

More later,

Nalin Pithwa

Reference: Discrete Mathematics and its Applications by Kenneth H. Rosen, Seventh Edition.

Heights and Distances — II — problems for RMO and IITJEE maths

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This is one of the prime utilities of trigonometry —- to calculate heights and distances, called as surveying in civil engineering.

1. From the extremities of a horizontal base line AB, whose length is 1 km, the bearings of the foot C of a tower are observed and it is found that \angle {CAB} is 56 degrees and 23 minutes and \angle{CBA} is 47 degrees and 15 minutes, and the elevation of the tower from A is 9 degrees and 25 minutes, find the height of the tower.

2. A man in a balloon observes that the angle of depression of an object on the ground bearing due north is 33 degrees; the balloon drifts 3 km due west and the angle of depression is now found to be 21 degrees. Find the height of the balloon.

3. A tower PN stands on level ground. A base AB is measured at right angles to AN, the points A, B, and N being in the same horizontal plane, and the angles PAN and PBN are found to be \alpha and \beta respectively. Prove that the height of the tower is

AB\frac{\sin {\alpha}\sin{\beta}}{\sqrt{\sin{(\alpha-\beta)}\sin{(\alpha+\beta)}}}

If AB is 100m, \alpha = 70 degrees and \beta = 50 degrees, calculate the height.

4. At each end of a horizontal base of length 2a, it is found that the angular height of a certain peak is \theta and that at the middle point it is \phi. Prove that the vertical height of the peak is

\frac{a \sin {\theta}\sin{\phi}}{\sqrt{\sin{(\phi+\theta)}\sin{(\phi-\theta)}}}

5. To find the distance from A to P, a distance AB of 1 km. is measured for a convenient direction. At A the angle PAB is found to be 41 degrees 18 min and at B, the angle PBA is found to be 114 degrees and 38 min. What is the required distance to the nearest metre?

More later,

Nalin Pithwa

Heights and Distances — I — problems for RMO and IITJEE Maths

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1. A man observes that at a point due south of a certain tower its angle of elevation is 60 degrees. He then walks 100 meters due west on a horizontal plane and finds that the angle of elevation is 30 degrees. Find the height of the tower and his original distance from it.

2. At the foot of a mountain the elevation of its summit is found to be 45 degrees; after ascending 1000 m towards the mountain up a slope of 30 degree inclination, the elevation is found to be 60 degrees. Find the height of the mountain.

3. A square tower stands upon a horizontal plane, from which three of its upper corners are visible, their angular elevations are respectively 45 degrees, 60 degrees and 45 degrees. Show  that the height of the tower is to the breadth of one of its sides as \sqrt{6}(\sqrt{5}+1) to 4.

4. A lighthouse, facing north, sends out a fan-shaped beam of light extending from north-east to north-west. An observer on a steamer, sailing due west, first sees the light when is 5 km, away from the lighthouse and continues to see it for 30\sqrt{2} minutes. What is the speed of the steamer?

5. A man stands at a point X on the bank XY of a river with straight and parallel banks, and observes that the line joining X to a point Z on the opposite bank makes an angle of 30 degrees with XY. He then goes along the bank a distance of 200 meters to Y and finds that the angle ZYX is 60 degrees. Find the breadth of the river.

6. A man, walking due north, observes that the elevation of a balloon, which is due east of him and is sailing toward the north-west, is then 60 degrees; after he has walked 400 meters the balloon is vertically over his head; find its height supposing it to have always remained the same.

The above are some tricky trig problems ! What is the main trick ? I will suggest a very simple hint: draw as good diagrams as possible!

Nalin Pithwa