# A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that $\tan {A}$ and $\tan {B}$ are the roots of the quadratic equation $x^{2}+px+q=0$, find the value of

$\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}$

Solution:

Let $\alpha=\tan{A}$ and $\beta=\tan{B}$ be the two roots of the given quadratic equation: $x^{2}+px+q=0$

By Viete’s relations between roots and coefficients:

$\alpha+\beta=\tan{A}+\tan{B}=-p$ and $\alpha \beta = \tan{A}\tan{B}=q$ but we also know that $\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}$

Now, let us call $E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)}$ which in turn is same as

$\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)$

We have already determined $\tan{(A+B)}$ in terms of p and q above.

Now, again note that $\sin^{2}{\theta}+\cos^{2}{\theta}=1$ which in turn gives us that $\tan^{2}{\theta}+1=\sec^{2}{\theta}$ so we get:

$\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}}$ so that

$\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}$

Hence, the given expression E becomes:

$(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q)$, which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.

# |log{xx_{1}}| + |log{xx_{2}}| + …+ |log{xx_{n}}| + |log{x/x_{1}}| + |log{x/x_{2}}| + …+|log{x/x_{n}}|= |log{x_{1}}+ log{x_{2}}+ ….+log{x_{n}}|

Solve the following :

Find all positive real numbers $x, x_{1}, x_{2}, \ldots, x_{n}$ such that

$|\log{xx_{1}}|+|\log{xx_{2}}| + \ldots + |\log{xx_{n}}| + |\log{\frac{x}{x_{1}}}| + |\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}|= |\log{x_{1}}+ \log{x_{2}}+\log{x_{3}}+ \ldots + \log{x_{n}}|$
…let us say this is given equality A

Solution:

Use the following inequality: $|a-b| \leq |a| + |b|$ with equality iff $ab \leq 0$

So, we observe that : $|\log{xx_{1}}|+|\log{\frac{x}{x_{1}}}| \geq |\log{xx_{1}}-\log{\frac{x}{x_{1}}}| = |\log{x_{1}^{2}}|=2 |\log{x_{1}}|$,

Hence, LHS of the given equality is greater than or equal to:

$2(|\log{x_{1}}|+|\log{x_{2}}|+|\log{x_{3}}|+ \ldots + |\log{x_{n}}|)$

Now, let us consider the RHS of the given equality A:

we have to use the following standard result:

$|\pm a_{} \pm a_{2} \pm a_{3} \ldots \pm a_{n}| \leq |a_{1}|+|a_{2}| + \ldots + |a_{n}|$

So, applying the above to RHS of A:

$|\log{x_{1}}+\log{x_{2}}+\ldots + \log{x_{n}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|$.

But, RHS is equal to LHS as given in A:

That is, $|\log{xx_{1}}|+|\log{xx_{2}}|+ \ldots + |\log{xx_{n}}| +|\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}|+ \ldots + |\log{\frac{x}{x_{n}}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|$

Now, just a few steps before we proved that LHS is also greater than or equal to : That is,

$|\log{xx_{1}}|+|\log{xx_{2}}|+\ldots + |\log{xx_{n}}|+ |\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}| \geq 2(|\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|)$

The above two inequalities are like the following: $x \leq y$ and $x \geq 2y$; so what is the conclusion? The first inequality means $x2y$ or $x=2y$; clearly it means the only valid solution is $x=2y$.

Using the above brief result, we have here:

$|\log{x_{1}}|+|\log{x_{2}}|+ \ldots +|\log{x_{n}}| =2(|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|)$

Hence, we get $|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|=0$, which in turn means that (by applying the definition of absolute value):

$|\log{x_{1}}|=|\log{x_{2}}|= \ldots =|\log{x_{n}}|$, which implies that $x_{1}=x_{2}= \ldots x_{n}=1$.

Substituting these values in the given logarithmic absolute value equation, we get:

$n \times |\log{x}|+ n \times |\log{x}|=0$, that is $2n \times |\log{x}|=0$, and as $n \neq 0$, this implies that $|\log{x}|=0$ which in turn means $x=1$ also.

# Axiomatic Method : A little explanation

I) Take an English-into-English dictionary (any other language will also do). Start with any word and note down any word occurring in its definition, as given in the dictionary. Take this new word and note down any word appearing in it until a vicious circle results. Prove that a vicious circle is unavoidable no matter which word one starts with , (Caution: the vicious circle may not always involve the original word).

For example, in geometry the word “point” is undefined. For example, in set theory, when we write or say : $a \in A$ ; the element “a” ‘belongs to’ “set A” —- the word “belong to” is not defined.

So, in all branches of math or physics especially, there are such “atomic” or “undefined” terms that one starts with.

After such terms come the “axioms” — statements which are assumed to be true; that is, statements whose proof is not sought.

The following are the axioms based on which equations are solved in algebra:

1. If to equals we add equals, we get equals.
2. If from equals we take equals, the remainders are equal.
3. If equals are multiplied by equals, the products are equal.
4. If equals are divided by equals (not zero), the quotients are equal.

More later,

Nalin Pithwa.

# Check your mathematical induction concepts

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If $n=1$, the result is evident.

Step 2: By the induction hypothesis the result is true when $n=k$; we must prove that it is correct when $n=k+1$. Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$. If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$; by induction hypothesis these numbers are all equal:

$a_{1}=a_{2}= \ldots = a_{k}$.

If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$; by the induction hypothesis these numbers are all equal:

$a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$.

It follows easily that all $k+1$ numbers in S are equal.

*************************************************************************************

Regards,

Nalin Pithwa

# Tutorial on Basic Set Theory and Functions: for PRMO, RMO and IITJEE Mains maths

I) Prove that every function can be represented as a sum of an even function and an odd function.

II)Let A, B, C be subsets of a set S. Prove the following statements and illustrate them with Venn Diagrams:

2a) The famous DeMorgan’s laws in their basic forms: $A^{'} \bigcup B^{'} = (A \bigcap B)^{'}$ and $A^{'} \bigcap B^{'} = (A \bigcup B)^{'}$. Assume that both sets A and B are subsets of Set S. In words, the first is: union of complements is the complement of intersection; the second is: intersection of two complements is the complement of the union of the two sets.

Sample Solution:

Let us say that we need to prove: $A^{'}\bigcap B^{'}=(A \bigcup B)^{'}$.

Proof: It must be shown that the two sets have the same elements; in other words, that each element of the set on LHS is an element of the set on RHS and vice-versa.

If $x \in A^{'} \bigcap B^{'}$, then $x \in A^{'}$ and $x \in B^{'}$. This means that $x \in S$, and $x \notin A$ and $x \notin B$. Since $x \notin A$ and $x \notin B$, hence $x \notin A \bigcup B$. Hence, $x \in (A \bigcup B)^{'}$.

Conversely, if $x \in (A \bigcup B)^{'}$, then $x \in S$  and $x \notin A \bigcup B$. Therefore, $x \notin A$ and $x \notin B$. Thus, $x \in A^{'}$ and $x \in Y^{'}$, so that $x \in A^{'} \bigcap B^{'}$. QED.

2b) $A \bigcap (B \bigcup C) = (A \bigcap B)\bigcup (A \bigcap C)$.

2c) $A \bigcup (B \bigcap C) = (A \bigcup B) \bigcap (A \bigcup C)$

III) Prove that if I and S are sets and if for each $i \in I$, we have $X_{i} \subset S$, then $(\bigcap_{i \in I} X_{i})^{'} = \bigcup_{i \in I}(X_{i})^{'}$.

Sample Solution:

It must be shown that each element of the set on the LHS is an element of the set on RHS, and vice-versa.

If $x \in (\bigcap_{i \in I} X_{i})^{'}$, then $x \in S$ and $x \notin \bigcap_{i \in I} X_{i}$. Therefore, $x \notin X_{i}$, for at least one $j \in I$. Thus, $x \in (X_{i})^{'}$, so that $x \in \bigcup_{i \in I}(X_{i})^{'}$.

Conversely, if $x \in \bigcup_{i \in I}(X_{i})^{i}$, then for some $j \in I$, we have $x \in (X_{i})^{'}$. Thus, $x \in S$ and $x \notin X_{i}$. Since $x \notin X_{i}$, we have $x \notin \bigcap_{i \in I}X_{i}$. Therefore, $x \in \bigcap_{i \in I}(X_{i})^{'}$. QED.

IV) If A, B and C are sets, show that :

4i) $(A-B)\bigcap C = (A \bigcap C)-B$

4ii) $(A \bigcup B) - (A \bigcap B)=(A-B) \bigcup (B-A)$

4iii) $A-(B-C)=(A-B)\bigcup (A \bigcap B \bigcap C)$

4iv) $(A-B) \times C = (A \times C) - (B \times C)$

V) Let I be a nonempty set and for each $i \in I$ let $X_{i}$ be a set. Prove that

5a) for any set B, we have : $B \bigcap \bigcup_{i \in I} X_{i} = \bigcup_{i \in I}(B \bigcap X_{i})$

5b) if each $X_{i}$ is a subset of a given set S, then $(\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'}$

VI) Prove that if $f: X \rightarrow Y$, $g: Y \rightarrow Z$, and $Z \rightarrow W$ are functions, then : $h \circ (g \circ f) = (h \circ g) \circ f$

VII) Let $f: X \rightarrow Y$ be a function, let A and B be subsets of X, and let C and D be subsets of Y. Prove that:

7i) $f(A \bigcup B) = f(A) \bigcup f(B)$; in words, image of union of two sets is the union of two images;

7ii) $f(A \bigcap B) \subset f(A) \bigcap f(B)$; in words, image of intersection of two sets is a subset of the intersection of the two images;

7iii) $f^{-1}(C \bigcup D) = f^{-1}(C) \bigcup f^{-1}(D)$; in words, the inverse image of the union of two sets is the union of the images of the two sets.

7iv) $f^{-1}(C \bigcap D)=f^{-1}(C) \bigcap f^{-1}(D)$; in words, the inverse image of intersection of two sets is intersection of the two inverse images.

7v) $f^{-1}(f(A)) \supset A$; in words, the inverse of the image of a set contains the set itself.

7vi) $f(f^{-1}(C)) \subset C$; in words, the image of an inverse image of a set is a subset of that set.

For questions 8 and 9, we can assume that the function f is $f: X \rightarrow Y$ and a set A lies in domain X and a set C lies in co-domain Y.

8) Prove that a function f is 1-1 if and only if $f^{-1}(f(A))=A$ for all $A \subset X$; in words, a function sends different inputs to different outputs iff a set in its domain is the same as the inverse of the image of that set itself.

9) Prove that a function f is onto if and only if $f(f^{-1}(C))=C$ for all $C \subset Y$; in words, the image of a domain is equal to whole co-domain (which is same as range) iff a set in its domain is the same as the image of the inverse image of that set.

Cheers,

Nalin Pithwa