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.

 

You and your research or you and your studies for competitive math exams

You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;

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