Test of Mathematics Solution Subjective 116 - Angles in a Triangle

This is a Test of Mathematics Solution Subjective 116 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

If A, B, C are the angles of a triangle, then show that $ \displaystyle { \sin A + \sin B - \cos C \le \frac {3 \sqrt{3}}{2}}$

Solution:

$ \displaystyle { \sin A + \sin B - \cos C } $
$ \displaystyle { = \sin A + \sin B - \cos (\pi - (A+B)) } $
$ \displaystyle { = \sin A + \sin B + \sin (A+B) } $
$ \displaystyle { = 2\sin \frac{(A+B)}{2} \cos \frac{(A-B)}{2} + 2\sin \frac{(A+B)}{2} \cos \frac{(A+B)}{2} } $
$ \displaystyle { = 2\sin \frac{(A+B)}{2} \left( \cos \frac{(A-B)}{2} + \cos \frac{(A+B)}{2}\right ) } $
$ \displaystyle { = 2\sin \frac{(A+B)}{2} 2\cos \frac{A}{2} \cos \frac{B}{2} } $
$ \displaystyle { = 4\sin \frac{(\pi -C)}{2} \cos \frac{A}{2} \cos \frac{B}{2} } $
$ \displaystyle { = 4\sin \left(\frac{\pi}{2} - \frac{C}{2} \right) \cos \frac{A}{2} \cos \frac{B}{2} } $
$ \displaystyle { = 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} } $

We apply Jensen's Inequality and Arithmetic Mean - Geometric Mean inequality here. Since cosine function is concave in the interval $ [0, \frac{\pi}{2} ] $, we have
$ \displaystyle { \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \frac{\cos \frac{C}{2} + \cos \frac{A}{2} +\cos \frac{B}{2}}{3} \le \cos \left ( \frac{1}{3}\times \frac{A}{2} + \frac{1}{3}\times \frac{B}{2} + \frac{1}{3}\times \frac{C}{2} \right ) } $
This implies $ \displaystyle { \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \cos \left ( \frac{A+B+C}{6}\right ) } $
$ \displaystyle { \Rightarrow \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \cos \frac{\pi}{6} } $
$ \displaystyle { \Rightarrow \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \frac{\sqrt{3}}{2} } $
$ \displaystyle { \Rightarrow \cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \le \frac{3\sqrt{3}}{8} } $
$ \displaystyle { \Rightarrow 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \le \frac{3\sqrt{3}}{2} } $

Chatuspathi:

What is this topic: Properties of triangle
What are some of the associated concept: Jensen Inequality, Arithmetic Mean-Geometric Mean Inequality
Where can learn these topics: I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Inequality’ module.
Book Suggestions: 'Secrets in Inequality' by Pam Kim Hung

Test of Mathematics Solution Subjective 55 - Partition of a set of functions

This is a Test of Mathematics Solution Subjective 55 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

For a finite set A, let |A| denote the number of elements in set A.

(a) Let F be the set of all functions f:{1, 2, ... , n} --> {1, 2, ... , k} $ (n \ge 3, k \ge 2 ) $ satisfying $ f(i) \ne f(i+1) $ for every i, $ 1\le i \le n-1 $. Show that |F| = $ k(k-1)^{(n-1)}$.

Solution

The function f may take the element 1 in domain to one of the k numbers in codomain (k choices for 1). Since f(2) cannot equal f(1), we have (k-1) choices for 2 to go. Similarly f(3) cannot equal f(2) (but it can equal f(1)) so there are k-1 choices for 3 and so on. In total the number of functions possible equals $ k \times (k-1) \times (k-1) ... (n-1) \times = k (k-1)^{(n-1)} $.

(b) Let c(n,k) denote the number of functions in F such that $ f(n) \ne f(1) , n \ge 4$. Then show that $ c(n,k) = k(k-1)^{(n-1)} - c(n-1, k) $

Solution: We consider the following partition of the set of functions F:
(1) functions in which $ f(n) \ne f(1)$
(2) functions in which f(n) = f(1)
Note the above partition is well defined as they are mutually exclusive and exhaustive. Functions of type (1) are given by c(n, k) and functions of type two is given by c(n-1, k) because if f(n) = f(1), f(n-1) cannot equal f(1) as F contains functions for which $ f(i) \ne f(i+1) $
As functions of type 1 and type 2
makes up all the functions in F we have $ c(n,k) + c(n-1, k) = k(k-1)^{(n-1)} $.

(c) Using part (b) or otherwise prove $ c(n, k) = (k-1)^{(n-1)} + (-1)^n (k-1) $

Solution: We use induction on part (b). Suppose $ c(n, k) = (k-1)^{(n-1)} + (-1)^n (k-1) $. Then $ c(n+1, k) = k (k-1)^{(n)} - c(n, k) = k (k-1)^{(n)} - (k-1)^{(n-1)} + (-1)^n (k-1) $

Test of Mathematics Solution Subjective 50 -Dictionary Ranking

This is a Test of Mathematics Solution Subjective 50 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

All the permutation of the letters $a,b,c,d,e$ are written down and arranged in alphabetical order as in dictionary. Thus the arrangement $abcde$ is in first position and $abced$ is in second position. What is the position of the word $debac$?

Solution:

According to the arrangement in a dictionary, number of words:

i) starting with $a$ = 4!

ii) starting with $b$ = 4!

iii) starting with $c$ = 4!

iv) starting with $da$ = 3!

v) starting with $db$ = 3!

vi) starting with $dc$ = 3!

vii) starting with $dea$ = 2!

viii) starting with $deba$ = 1!

The last case gives us the word itself. So the position of the word will be = $3*4! + 3*3! + 2! +1$ = $93$

$debac$ is in the $93^{rd}$ position.

Test of Mathematics Solution Subjective 49 - Arrangement of Similar Items

This is a Test of Mathematics Solution Subjective 49 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem:

$x$ red balls, $y$ black balls,$z$ white balls are to be arranged in a row. Suppose that any two balls of the same color are indistinguishable. Given $x+y+z=30$ prove that number of possible arrangements is maximum when $x=y=z=10$.

Solution:

Given a total of $n$ objects out of which $r$ are of the same type, total number of arrangements possible is given by $\frac{n!}{r!}$. Therefore in this case the total number of arrangements is $\frac{30!}{x!y!z!}$.

Now we have to maximise $\frac{30!}{x!y!z!}$.

Thus we need to minimise ${x!y!z!}$. As the question claims that all have to be equal to $10$, let us consider that they are not all equal to $10$. So the cases that arise are:

Case 1: $x>10$ , $y=10$, $z<10$

Case 2: $x>10$, $y>10$, $z<10$

Case 3: $x>10$, $y<10$, $z<10$

These cases are considered without any loss of generality as any other possible case will just be another permutation of the given cases.

So as the question claims the ideal condition is $x!y!z!=(10!)^3$.

Now considering Case 1, as $x+y+z=30$, $x!y!z!=(10+a)!(10)!(10-a)!$ = $(10!)^3\frac{(10+1)(10+2)\cdots(10+a)}{(10-1)(10-2)\cdots(10-a)}> (10!)^3$. So this is not a possible option as we have to minimise $x!y!z!$.

Now taking Case 2, $x!y!z!=(10+a)!(10+b)!(10-a-b)!$ = $(10!)^3\frac{[(10+1)(10+2)\cdots(10+a)][(10+1)(10+2)\cdots(10+b)]}{(10-1)(10-2)\cdots(10-a-b)}> (10!)^3$. So again not a possibility.

Similarly it can be shown for Case 3 also, thus proving that any other value of $x,y$ and $z$ does not minimise $x!y!z!$.

Hence $x=y=z=10$ is the solution that maximises the number of possible arrangements.

Test of Mathematics Solution Subjective 48 - The Gifts Distribution

This is a Test of Mathematics Solution Subjective 48 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Find the different number of ways $5$ different gifts can be presented to $3$ children so that each child receives at least one gift.

Solution:

There are two possible ways in which the gifts can be distributed.

Case 1: They are distributed as $2,2,1$.

So first we choose the children who get $2$ gifts each in $^3C_2$ ways. Then we choose the gifts in $\frac{5!}{2!.2!}$ ways.

Thus total number of ways = $3.\frac{5!}{2!2!}= 90$ ways.

Case 2: They are distributed as $3,1,1$.

So first we choose the child who gets $3$ gifts in $^3C_1$ ways. Then we choose the gifts in $\frac{5!}{3!}$ ways.

Thus total number of ways = $3.\frac{5!}{3!}= 60$ ways.

Therefore total number of ways to distribute the gifts = $90+60$ = $150$ ways.

Test of Mathematics Solution Subjective 43-Integer Root

This is a Test of Mathematics Solution Subjective 43 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Show that the equation $ x^3 + 7x - 14(n^2 +1) = 0 $ has no integral root for any integer n.

Solution:

We note that $ 14(n^2 +1) - 7x = x^3 $ implies $ x^3 $ is divisible by 7. This implies x is divisible by 7 (as 7 is a prime number). Suppose x= 7x'. Hence we can rewrite the given equation as:

$ 7^3 x'^3 + 7 \times 7 x' - 14 (n^2 +1 ) = 0 $.

Cancelling out a 7 we have $ 7^2 {x'}^3 + 7{x'} = 2(n^2 +1) $. Since 7 divides left hand side, it must also divide the right hand side. Since 7 cannot divide 2, it must divide $ n^2 + 1 $ as 7 and 2 are coprime. Note that 7 cannot divide $ n^2 +1 $ as square of a number always gives remainder 0, 1, 4, 2 when divided by 7 and never 6. But if $ n^2 + 1 $ is divisible by 7 then $ n^2 $ must give remainder 6 when divided by 7. Hence contradiction.

Necessary Lemma: square of a number always gives remainder 0, 1, 4, 2 when divided by 7

$ n \equiv 0, \pm 1 , \pm 2 , \pm 3 \mod 7\Rightarrow n^2 \equiv 0, 1, 4, 9 (=2) \mod 7 $

Key Ideas: Modular Arithmetic

Test of Mathematics Solution Subjective 38 - When 30 divides a prime

Test of Mathematics Solution Subjective 38 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

Show that if a prime number p is divided by 30, the remainder is either prime or 1.

Discussion

Suppose p = 30Q + R

(here p is the prime, Q and R are quotient and remainders respectively when p is divided by 30).

For all primes less than 30, Q = 0 and R=p. So that satisfies the claim of this problem.

If p > 30, suppose the remainder R is composite (not a prime). Since R 6 and N > 6 then MN (=R) > 36 > 30. But R < 30. So both M and N cannot exceed 6. Suppose M < 6. Then M must be divisible by 2, 3, or 5. We consider the case when M is divisible by 2 (other cases are analogous).

Suppose M = 2M'

R = MN = 2M'N

p = 30Q + 2M'N

But then the right hand side is divisible by 2. Hence the left hand side is also divisible by 2. But that is not possible as p is a prime larger than 30.

Hence R cannot be composite. This implies R is either a prime or 1.

Test of Mathematics Solution Subjective 37 - The prime 13

Test of Mathematics Solution Subjective 37 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

Supposed p is a prime Number such that (p-1)/4 and (p+1)/2 are also primes. Show that p=13.

Discussion:

p is not 2 or 3 (otherwise (p-1)/4 would not be an integer).Hence p must be an odd prime. Also p-1 is divisible by 4

p = 4t + 1 (say)

(p-1)/4 = t

(p+1)/2 = 2t + 1

Hence t, 2t+1, 4t+1 are all primes.

If t = 3 then these numbers are 3, 7 and 13.

If t is not 3 then t must produce 1 or 2 remainders when divided by 3 (t is a prime, hence cannot be divisible by 3).

$ \displaystyle {t \equiv 1 \text{mod} 3 \Rightarrow 2t +1 \equiv 3 \equiv 0 \text {mod} 3} $

But 2t +1 is a prime. So it is impossible that 3 divides 2t +1. Hence t cannot be 1 mod 3.

Similarly $ \displaystyle {t \equiv 2 \text{mod} 3 \Rightarrow 4t +1 \equiv 9 \equiv 0 \text {mod} 3} $

But 4t +1 is a prime. So it is impossible that 3 divides 4t +1. Hence t cannot be 1 mod 3.

Therefore we can have no other t such that the given condition is satisfied. Hence t must be 3 and the prime must be 13.

Test of Mathematics Solution Subjective 36 - Invariance Principle

Test of Mathematics Solution Subjective 36 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

Let $ a_1 , a_2 , ... , a_n $ be n numbers such that each $ a_i $ is either 1 or -1. If $ a_1 a_2 a_3 a_4 + a_2 a_3 a_4 a_5 + ... + a_n a_1 a_2 a_3 = 0 $ then prove that 4 divides n.

Solution

Let us denote each product of four numbers as $ b_i $ . For example $ b_1 = a_1 a_2 a_3 a_4 $. Note that the last one begins with $ a_n $ indicating there are n $ b_i $'s.

We check the equation modulo 4 (that is in each step we change 'something' in the equation and check what happens to the remainder of the sum when divided by 4).In the first step we note that equation is 0 mod 4 (since 0 when divided by 4 gives 0 as the remainder).

Next we convert one of the a's which is -1 into +1. Note that each $ a_i $ appears in exactly 4 $ b_j $. Hence converting the one $ a_i $ from -1 to +1 will alter the values of 4 $ b_j $ Thus we may have five cases:

All the four b's were -1. Hence after conversion all of their values will change into +1. Therefore the value of the total sum will increase by +8 (as previously the four b's added up to -4 and now they add up to +4). Thus no change of the sum modulo 4.
Exactly three b's were -1 and the remaining was +1. Hence after conversion three of their values will change into +1 and the remaining one changes from +1 to -1. Therefore the value of the total sum will increase by +4 (as previously the four b's added up to -2 ((-1) + (-1) + (-1) + (+1)) and now they add up to +2). Thus no change of the whole sum modulo four.
Exactly two b's were -1 and the remaining two were +1. Hence after conversion two of their values will change into +1 and the remaining two will change from +1 to -1. Therefore the value of the total sum will increase by 0 (as previously the four b's added up to 0 ((-1) + (-1) + (+1) + (+1)) and now they add up to 0). Thus no change of the whole sum modulo four.
Exactly one b was -1 and the remaining were +1. Hence after conversion three of their values will change into -1 and the remaining one changes from -1 to +1. Therefore the value of the total sum will decrease by 4 (as previously the four b's added up to 2 ((+1) + (+1) + (+1) + (-1)) and now they add up to -2). Thus no change of the whole sum modulo four.
All the four b's were +1. Hence after conversion all of their values will change into -1. Therefore the value of the total sum will decrease by 8 (as previously the four b's added up to +4 and now they add up to -4). Thus no change of the sum modulo 4.

Thus we see that changing one $ a_i $ -1 to +1 does not change the value of the total sum modulo 4. Similarly we change all negative one's into positive one's (and in each step the sum remains invariant in modulo 4) to get n. Thus $ n \equiv 0 \mod 4 $ . This implies 4 divides n.

Key Idea

Modular arithmetic, invariance principle.

Test of Mathematics Solution Subjective 33 - Symmetrical Minima

Test of Mathematics Solution Subjective 33 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

Let $k$ be a fixed odd positive integer. Find the minimum value of $x^2 + y^2$, where $x,y$ are non-negative integers and $x+y=k$.

Solution

We have $y=k-x$. Therefore we get an equation in $x$ where $k$ is a constant, precisely $f(x) = x^2 + (k-x)^2$.

To minimise, we differentiate $f(x)$ w.r.t $x$.

So $f'(x) = 4x-2k = 0$ (for minimum $f(x)$)

That gives us $x=\frac{k}{2}$.

But the question tell us that $k$ is odd and $x$ is an integer. therefore we have to take the closest possible integer value to $\frac{k}{2}$, which is $\frac{k+1}{2}$ and $\frac{k-1}{2}$.

As already defined, taking $x$ to be any one of the above $y$ automatically takes the other value.

Therefore the minimum value of $x^2 + y^2$ is given by $(\frac{k+1}{2})^2 + (\frac{k-1}{2})^2$ $= \frac{k^2+1}{2}$.