341You have badly mangled up Q1 I think. Kindly edit
Q2) z = 1/2 + i/2
1..FIND THE MIN. VALUE of
A+3CA+2+C+4BA+B+2C-8CA+B+3C
(A,B,C.ε R+)
2..IF A={zεC:Z=X+Xi-1 FOR X ε R } AND MOD(Z)≤MOD(ω) FOR ω ε A THEN Z IS
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5 Answers
341
341I'll do that. But I wish you would edit your Q1.
Q2 - They have constituted a set A consisting of complex numbers such that Z = (X-1) + iX where X is a real number.
We are asked to find a number Z in this set of least modulus.
|Z|^2 = X^2 + (X-1)^2|Z|^2 = X^2 + (X-1)^2 = 2X^2 -2X+1 = 2\left(X - \frac{1}{2} \right)^2 + \frac{1}{2} \ge \frac{1}{2}
Minimum occurs when X = 1/2 = 1-X. That's how we get
Z = 1/2 + i/2
[Note, if you are familiar with inequalities, you can do this much faster. From Cauchy Schwarz, 2( X^2 + (1-X)^2 ) \ge (X+1-X)^2 = 1 \Rightarrow X^2 + (1-X)^2 \ge \frac{1}{2}
with equality occurring when X = 1-X which means X = 1/2
1for 1st q ...
did u meant this \frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c}.
if yes
check this http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147142
341Any tutor who puts in Chinese MO qns to coach IIT aspirants needs a change of job