# Ratio and proportion: practice problems: set II: pRMO, preRMO or IITJEE foundation maths

Problem 1:

If $\frac{y+z}{pb+qc} = \frac{z+x}{pc+qa} = \frac{x+y}{pa+qb}$, then show that $\frac{2(x+y+z)}{a+b+c} = \frac{(b+c)x+(c+a)y+(a+b)z}{bc+ca+ab}$

Problem 2:

If $\frac{x}{a} = \frac{y}{b} = \frac{z}{b}$, show that $\frac{x^{3}+a^{3}}{x^{2}+a^{2}} +\frac{y^{3}+b^{3}}{y^{2}+b^{2}} + \frac{z^{3}+c^{3}}{z^{2}+c^{2}} = \frac{(x+y+z)^{3}+(a+b+c)^{3}}{(x+y+z)^{2}+(a+b+c)^{2}}$

Problem 3:

If $\frac{2y+2z-x}{a} = \frac{2z+2x-y}{b} = \frac{2x+2y-z}{c}$, show that $\frac{x}{2b+2c-a} = \frac{y}{2c+2a-b} = \frac{z}{2a+2b-c}$

Problem 4:

If $(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2}) = (ax+by+cz)^{2}$, prove that $x:a = y:b = z:c$

Problem 5:

If $l(my+nz-lx) = m(nz+lx-my) = n(lx+my-nz)$, prove that $\frac{y+z-x}{l} = \frac{z+x-y}{m} = \frac{x+y-z}{n}$

Problem 6:

Show that the eliminant of

$ax+cy+bz=0$

$cx+by+az=0$

$bx+ay+cz=0$

is $a^{3}+b^{3}+c^{3}-3abc=0$

Problem 7:

Eliminate x, y, z from the equations:

$ax+hy+gz=0$

$hx+by+fz=0$

$gx+fy+cz=0$.

This has significance in co-ordinate geometry. (related to conics).

Problem 8:

If $x=cy+bz$, $y=az+cx$, $z=bx+cy$, show that $\frac{x^{2}}{1-a^{2}} = \frac{y^{2}}{1-b^{2}} = \frac{z^{2}}{1-c^{2}}$.

Problem 9:

Given that $a(y+z)=x$, $b(z+x)=y$, $c(x+y)=z$, prove that $bc+ab+ca+2abc=1$

Problem 10:

Solve the following system of equations:

$3x-4y+7z=0$

$2x-y-2z=0$

$3x^{3}-y^{3}+z^{3}=18$

Problem 11:

Solve the following system of equations:

$x+y=z$

$3x-2y+17z=0$

$x^{3}+3y^{3}+2z^{3}=167$

Problem 12:

Solve the following system of equations:

$7yz+3zx=4xy$

$21yz-3zx=4xy$

$x+2y+3z=19$

Problem 13:

Solve the following system of equations:

$3x^{2}-2y^{2}+5z^{2}=0$

$7x^{2}-3y^{2}-15z^{2}=0$

$5x-4y+7z=0$

Problem 14:

If $\frac{l}{\sqrt{a}-\sqrt{b}} + \frac{m}{\sqrt{b}-\sqrt{c}} + \frac{n}{\sqrt{c}-\sqrt{a}} =0$,

and $\frac{l}{\sqrt{a}+\sqrt{b}} + \frac{m}{\sqrt{b}+\sqrt{c}} + \frac{n}{\sqrt{c}+\sqrt{c}} = 0$,

prove that $\frac{l}{(a-b)(c-\sqrt{ab})} = \frac{m}{(b-c)(a-\sqrt{ab})} = \frac{n}{(c-a)(b-\sqrt{ac})}$

Problem 15:

Solve the following system of equations:

$ax+by+cz=0$

$bcx+cay+abz=0$

$xyz+abc(a^{3}x+b^{3}y+c^{3}z)=0$

Cheers,

Nalin Pithwa

# Every function can be written as a sum of an even and an odd function

Let $f(x)$ be any well-defined function.

We want to express it as a sum of an even function and an odd function.

Let us define two other functions as follows:

$F(x) = \frac{f(x)+f(-x)}{2}$ and $G(x)=\frac{f(x)-f(-x)}{2}$.

Claim I: F(x) is an even function.

Proof I; Since by definition $F(x)= \frac{f(x)+f(-x)}{2}$, so $F(-x) = \frac{f(-x) +f(-(-x))}{2}=\frac{f(-x)+f(x)}{2} \Longrightarrow F(x) = F(-x)$ so that F(x) is indeed an even function.

Claim 2: G(x) is an odd function.

Proof 2: Since by definition $G(x) = \frac{f(x)-f(-x)}{2}$, so $G(-x) = \frac{f(-x)-f(-(-x))}{2} = \frac{f(-x)-f(x)}{2} = -\frac{f(x)-f(-x)}{2} = -G(-x) \Longrightarrow G(x) = -G(-x)$ so that G(x) is indeed an odd function.

Claim 3: $f(x)= F(x) + G(x)$

Proof 3: $F(x)+ G(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2} = \frac{f(x)}{2} + \frac{f(-x)}{2} + \frac{f(x)}{2} - \frac{f(-x)}{2} = f(x)$ indeed.

# Cyclic expressions, fractions: Pre RMO, PRMO, IITJEE foundation 2019

In order to solve the following tutorial sheet, it helps to solve/understand and then apply the following beautiful cyclic relations or identities:

(Note if these look new to you, then you need to check the truth of all them; if all are v v familiar to you, just go ahead and crack the tutorial sheet below):

Core Identities in Cyclic Expressions:
1) $(b-c)+(c-a)+(a-b)=0$
2) $a(b-c)+b(c-a)+c(a-b)=0$
3) $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=-(a-b)(b-c)(c-a)$
4) $bc(b-c)+ca(c-a)+ab(a-b)=-(a-b)(b-c)(c-a)$
5) $a(b^{2}-c^{2})+b(c^{2}-a^{2})+c(a^{2}-b^{2})=(a-b)(b-c)(c-a)$

Solve or simplify the following:

1) $\frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}$
2) $\frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}$
3) $\frac{a^{2}}{(a-b)(a-c)} + \frac{b^{2}}{(b-c)(b-a)} + \frac{c^{2}}{(c-a)(c-b)}$
4) $\frac{a^{3}}{(a-b)(a-c)} + \frac{b^{3}}{(b-c)(b-a)} + \frac{c^{3}}{(c-a)(c-b)}$
5) $\frac{a(b+c)}{(a-b)(c-a)} + \frac{b(a+c)}{(a-b)(b-c)} + \frac{a(a+b)}{(c-a)(b-c)}$
6) $\frac{1}{a(a-b)(a-c)} + \frac{1}{b(b-c)(b-a)} + \frac{1}{c(c-a)(c-b)}$
7) $\frac{bc}{a(a^{2}-b^{2})(a^{2}-c^{2})} + \frac{ca}{b(b^{2}-c^{2})(b^{2}-a^{2})} + \frac{ab}{c(c^{2}-a^{2})(c^{2}-b^{2})}$
8) $\frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$
9) $\frac{bc(a+d)}{(a-b)(a-c)} + \frac{ca(b+d)}{(b-c)(b-a)} + \frac{ab(c+d)}{(c-a)(c-b)}$
10) $\frac{1}{(a-b)(a-c)(x-a)} + \frac{1}{(b-c)(b-a)(x-b)} + \frac{1}{(c-a)(c-b)(x-c)}$
11) $\frac{a^{2}}{(a-b)(a-c)(x+a)} + \frac{b^{2}}{(b-c)(b-a)(x+b)} + \frac{c^{2}}{(c-a)(c-b)(x+c)}$
12) $a^{2}\frac{(a+b)(a+c)}{(a-b)(a-c)} + b^{2}\frac{(b+c)(b+a)}{(b-c)(b-a)} + c^{2}\frac{(c+a)(c+b)}{(c-a)(c-b)}$
13) $\frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}$
14) $\frac{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)+2(c-a)(a-b)(b-c)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}$
15) $\frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)}$
16) $\frac{a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3}}{(a-b)(b-c)(c-a)}$
17) $\frac{\frac{b-c}{a} + \frac{c-a}{b} + \frac{a-b}{c}}{\frac{1}{a}(\frac{1}{b^{2}}-\frac{1}{c^{2}})+\frac{1}{b}(\frac{1}{c^{2}}-\frac{1}{a^{2}})+\frac{1}{c}(\frac{1}{a^{2}}-\frac{1}{b^{2}})}$^
18) $\frac{a^{2}(\frac{1}{a^{2}}-\frac{1}{b^{2}})+b^{2}(\frac{1}{a^{2}}-\frac{1}{c^{2}})+c^{2}(\frac{1}{b^{2}}-\frac{1}{a^{2}})}{\frac{1}{bc}(\frac{1}{c}-\frac{1}{b})+\frac{1}{ca}(\frac{1}{a}-\frac{1}{c})+\frac{1}{ab}(\frac{1}{b}-\frac{1}{c})}$
19) $\frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-c)(b-a)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}$

More later,
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