Test of Mathematics Solution Subjective 73 - Coefficients of a Quadratic
This is a Test of Mathematics Solution Subjective 73 (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.
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Problem
Consider the equation $ {x^3 + Gx + H = 0} $, where G and H are complex numbers. Suppose that this equation has a pair of complex conjugate roots. Show that both G and H are real.
Solution
Let three roots of the equation $ {x^3 + Gx + H = 0} $ are
$ {\alpha, \beta, \gamma} $ [ Let $ {\alpha, \beta} $ are complex conjugates] .Now ${\alpha \beta \gamma} $ = - H ... (i)
$ {\alpha + \beta +\gamma} $ = 0 ... (ii)
$ {\alpha \beta + \beta \gamma + \gamma \alpha} $ = G ... (iii)
From (ii) we get $ {\alpha + \beta + \gamma}$
= 0 [ $ {\alpha, \beta} $ are complex conjugates so they are real]
${\Rightarrow} $ $ {\gamma} $ = real Now as $ {\gamma} $ = real
$ {\beta \gamma + \gamma \alpha} $ = $ {\gamma (\beta + \alpha)} $ =$ {real \times real} $ = $ {real} $ ... (iv) $ {\alpha, \beta} $ are complex conjugates so $ {\alpha \beta = real} $ ... (v) From (iv) & (v) we get
$ {\alpha \beta + \beta \gamma + \gamma \alpha}$ = $ {real + real = real} $ ${\Rightarrow}$ G = real [from (iii)] Now $ {\alpha, \beta} $ is real and $ {\alpha, \beta} $ is real so ${\alpha \beta \gamma} $ = real $ {\Rightarrow}$ H = real.
Test of Mathematics Solution Subjective 70 - Equal Roots

This is a Test of Mathematics Solution Subjective 70 (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
Suppose that all roots of the polynomial equation
$ {\displaystyle{x^4 - 4x^3 + ax^2 +bx + 1}} $ = 0 are positive real numbers.
Show that all the roots of the polynomial are equal.
Solution
$ {\displaystyle{x^4 - 4x^3 + ax^2 +bx + 1}} $ = 0
If the roots are $ {\displaystyle{\alpha}} $, $ {\displaystyle{\beta}} $, $ {\displaystyle{\gamma}} $ and $ {\displaystyle{\lambda}} $ .
then $ {\displaystyle{\alpha}} $, $ {\displaystyle{\beta}} $, $ {\displaystyle{\gamma}} $ and $ {\displaystyle{\lambda}} $ = 1
& $ {\displaystyle{\alpha}} $ + $ {\displaystyle{\beta}} $ + $ {\displaystyle{\gamma}} $ + $ {\displaystyle{\lambda}} $ = 4.
Now all of $ {\displaystyle{\alpha}} $, $ {\displaystyle{\beta}} $, $ {\displaystyle{\gamma}} $ and $ {\displaystyle{\lambda}} $ are positive so AM-GM inequality is applicable.
$ {\displaystyle{\frac{\alpha + \beta + \gamma + \lambda}{4}}}{\ge}$ $ {(\alpha\beta\lambda)^{\frac{1}{4}}}$
$ {\Rightarrow} $ $ {\frac{4}{4}} {\ge}$ $ {1^{\frac{1}{4}}}$
$ {\Rightarrow} $ 1 $ {\ge} $ 1
Now we know equality in AM-GM occours if all the numbers are equal.So $ {\displaystyle{\alpha}} $, $ {\displaystyle{\beta}} $, $ {\displaystyle{\gamma}} $ and $ {\displaystyle{\lambda}} $ are all equal.
Logarithms and Equations | AIME I, 2000 | Question 9
Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.
Logarithms and Equations - AIME I 2000
\(log_{10}(2000xy)-log_{10}xlog_{10}y=4\) and \(log_{10}(2yz)-(log_{10}y)(log_{10}z)=1\) and \(log_{10}(zx)-(log_{10}z)(log_{10}x)=0\) has two solutions \((x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})\) find \(y_{1}+y_{2}\).
- is 905
- is 25
- is 840
- cannot be determined from the given information
Key Concepts
Check the Answer
Answer: is 25.
AIME I, 2000, Question 9
Polynomials by Barbeau
Try with Hints
Rearranging equations we get \(-logxlogy+logx+logy-1=3-log2000\) and \(-logylogz+logy+logz-1=-log2\) and \(-logxlogz+logx+logz-1=-1\)
taking p, q, r as logx, logy and logz, \((p-1)(q-1)=log2\) and \((q-1)(r-1)=log2\) and \( (p-1)(r-1)=1\) which is first system of equations and multiplying the first three equations of the first system gives \((p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}\) gives \((p-1)(q-1)(r-1)=+-(log2)\) which is second equation
from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives \(y_{1}=20\),\(y_{2}=5\) then \(y_{1}+y_{2}=25\).
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Probability in Games | AIME I, 1999 | Question 13
Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.
Probability in Games - AIME I, 1999 Question 13
Forty teams play a tournament in which every team plays every other team exactly once. No ties occur,and each team has a 50% chance of winning any game it plays.the probability that no two team win the same number of games is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find \(log_{2}n\)
Key Concepts
Check the Answer
Answer: 742.
AIME, 1999 Q13
Course in Probability Theory by Kai Lai Chung .
Try with Hints
\({40 \choose 2}\)=780 pairings with \(2^{780}\) outcomes
no two teams win the same number of games=40! required probability =\(\frac{40!}{2^{780}}\)
the number of powers of 2 in 40!=[\(\frac{40}{2}\)]+[\(\frac{40}{4}\)]+[\(\frac{40}{8}\)]+[\(\frac{40}{16}\)]+[\(\frac{40}{32}\)]=20+10+5+2+1=38 then 780-38=742.
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Theory of Equations | AIME I, 2015 | Question 1
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Theory of Equations.
Theory of Equations - AIME I, 2015
The expressions A=\(1\times2+3\times4+5\times6+...+37\times38+39\)and B=\(1+2\times3+4\times5+...+36\times37+38\times39\) are obtained by writing multiplication and addition operators in an alternating pattern between successive integers.Find the positive difference between integers A and B.
- is 722
- is 250
- is 840
- cannot be determined from the given information
Key Concepts
Check the Answer
Answer: is 722.
AIME I, 2015, Question 1
Elementary Number Theory by Sierpinsky
Try with Hints
A = \((1\times2)+(3\times4)\)
\(+(5\times6)+...+(35\times36)+(37\times38)+39\)
B=\(1+(2\times3)+(4\times5)\)
\(+(6\times7)+...+(36\times37)+(38\times39)\)
B-A=\(-38+(2\times2)+(2\times4)\)
\(+(2\times6)+...+(2\times36)+(2\times38)\)
=722.
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Area of Equilateral Triangle | AIME I, 2015 | Question 4
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 from Geometry, based on Area of Equilateral Triangle (Question 4).
Area of Triangle - AIME I, 2015
Point B lies on line segment AC with AB =16 and BC =4. Points D and E lie on the same side of line AC forming equilateral triangle ABD and traingle BCE. Let M be the midpoint of AE, and N be the midpoint of CD. The area of triangle BMN is x. Find \(x^{2}\).
- is 107
- is 507
- is 840
- cannot be determined from the given information
Key Concepts
Check the Answer
Answer: is 507.
AIME, 2015, Question 4
Geometry Revisited by Coxeter
Try with Hints
Let A(0,0), B(16,0),C(20,0). let D and E be in first quadrant. then D =\((8,8\sqrt3)\), E=\((18,2\sqrt3\)).
M=\((9,\sqrt3)\), N=(\(14,4\sqrt3\)), where M and N are midpoints
since BM, BN, MN are all distance, BM=BN=MN=\(2\sqrt13\). Then, by area of equilateral triangle, x=\(13\sqrt3\) then\(x^{2}\)=507.
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Probability Problem | Combinatorics | AIME I, 2015 - Question 5
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Probability.
Probability Problem - AIME I, 2015
In a drawer Sandy has 5 pairs of socks, each pair a different color. on monday sandy selects two individual socks at random from the 10 socks in the drawer. On tuesday Sandy selects 2 of the remaining 8 socks at random and on wednesday two of the remaining 6 socks at random. The probability that wednesday is the first day Sandy selects matching socks is \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.
- is 107
- is 341
- is 840
- cannot be determined from the given information
Key Concepts
Check the Answer
Answer: is 341.
AIME, 2015, Question 5
Geometry Revisited by Coxeter
Try with Hints
Wednesday case - with restriction , select the pair on wednesday in \(5 \choose 1 \) ways
Tuesday case - four pair of socks out of which a pair on tuesday where a pair is not allowed where 4 pairs are left,the number of ways in which this can be done is \(8 \choose 2\) - 4
Monday case - a total of 6 socks and a pair not picked \(6 \choose 2\) -2
by multiplication and principle of combinatorics \(\frac{(5)({5\choose 2} -4)({6 \choose 2}-2)}{{10 \choose 2}{8 \choose 2}{6 \choose 2}}\)=\(\frac{26}{315}\). That is 341.
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Probability of tossing a coin | AIME I, 2009 | Question 3
Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.
Probability of tossing a coin - AIME I, 2009 Question 3
A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to \(\frac{1}{25}\) the probability of five heads and three tails. Let p=\(\frac{m}{n}\) where m and n are relatively prime positive integers. Find m+n.
Key Concepts
Check the Answer
Answer: 11.
AIME, 2009
Course in Probability Theory by Kai Lai Chung .
Try with Hints
here \(\frac{8!}{3!5!}p^{3}(1-p)^{5}\)=\(\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}\)
then \((1-p)^{2}\)=\(\frac{1}{25}p^{2}\) then 1-p=\(\frac{1}{5}p\)
then p=\(\frac{5}{6}\) then m+n=11
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Equations with number of variables | AIME I, 2009 | Question 14
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2009 based on Equations with a number of variables.
Equations with number of variables - AIME 2009
For t=1,2,3,4, define \(S^{t}=a^{t}_1+a^{t}_2+...+a^{t}_{350}\), where \(a_{i}\in\){1,2,3,4}. If \(S_{1}=513, S_{4}=4745\), find the minimum possible value for \(S_{2}\).
- is 905
- is 250
- is 840
- cannot be determined from the given information
Key Concepts
Check the Answer
Answer: is 905.
AIME, 2009, Question 14
Polynomials by Barbeau
Try with Hints
j=1,2,3,4, let \(m_{j}\) number of \(a_{i}\) s = j then \(m_{1}+m{2}+m{3}+m{4}=350\), \(S_{1}=m_{1}+2m_{2}+3m_{3}+4m_{4}=513\) \(S_{4}=m_{1}+2^{4}m_{2}+3^{4}m_{3}+4^{4}m_{4}=4745\)
Subtracting first from second, then first from third yields \(m_{2}+2m_{3}+3m_{4}=163,\) and \(15m_{2}+80m_{3}+255m_{4}=4395\) Now subtracting 15 times first from second gives \(50m_{3}+210m_{4}=1950\) or \(5m_{3}+21m_{4}=195\) Then \(m_{4}\) multiple of 5, \(m_{4}\) either 0 or 5
If \(m_{4}=0\) then \(m_{j}\) s (226,85,39,0) and if \(m_{4}\)=5 then \(m_{j}\) s (215,112,18,5) Then \(S_{2}=1^{2}(226)+2^{2}(85)+3^{2}(39)+4^{2}(0)=917\) and \(S_{2}=1^{2}(215)+2^{2}(112)+3^{2}(18)+4^{2}(5)=905\) Then min 905.
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Exponents and Equations | AIME I, 2010 Question 3
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Exponents and Equations.
Exponents and Equations - AIME 2010
Suppose that y=\(\frac{3x}{4}\) and \(x^{y}=y^{x}\). The quantity x+y can be expressed as a rational number \(\frac{r}{s}\) , where r and s are relatively prime positive integers. Find r+s.
.
- is 107
- is 529
- is 840
- cannot be determined from the given information
Key Concepts
Check the Answer
Answer: is 529.
AIME, 2010, Question 3.
Elementary Number Theory by Sierpinsky
Try with Hints
y=\(\frac{3x}{4}\) into \(x^{y}=y^{x}\) and \(x^{\frac{3x}{4}}\)=\((\frac{3x}{4})^{x}\) implies \(x^{\frac{3x}{4}}\)=\((\frac{3}{4})^{x}x^{x}\) implies \(x^{-x}{4}\)=\((\frac{3}{4})^{x}\) implies \(x^{\frac{-1}{4}}=\frac{3}{4}\) implies \(x=\frac{256}{81}\).
y=\(\frac{3x}{4}=\frac{192}{81}\).
x+y=\(\frac{448}{81}\) then 448+81=529.
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