IIT JEE , AIEEE Forum - Mathematics , Algebra

iitaspirant1990

1.a new sequence is obtained from the sequence of the positive integers(1,2,3....) by deleting all the perfect squares.then find the 2003 term of the new sequence.

2.how many of the triangles whose verteces re chosen from the vertces of a rectangular block are acute.

a.0

b.6

c.8

d.24

3.If the 4th and the 13th term of a sequence(a+b)ⁿ is equal then find n.

4.the circle x²+y² =4 cuts the circle x²+ y²+2x+3y-5 inA and B.then find the equation of the circle on AB as diameter.

akhil_o

dreadful dreadful
edited

akhil_o

edited

Conjurer

3) 4th and 13th terms are equal so n = 3+12 = 15

Conjurer

4)
Subtracting both equations we get the common chord which is the diameter of the required circle. So equation of diameter is 2x +3y =1

Equate in first circle and get the points,you can find the center and the radius.

edison

Well done akhil.

please send other problems as separate posts.

iitaspirant1990

hey please tell me the very first two questions they are more important for me.

iitaspirant1990

another question
1.a fair coin is tossed 100 times .the probability of getting tails an odd number of times is ,
a.1/2
b.1/8
c.3/8
d.none of these.
2.two complex numbers a+ib and c+id are almost conjugate if |a-c| <= 1 and |b+d| <= 1.determine the maximum destance from origin of the almost conjugate of 2+3i.

nadeemoidu

1) in the first 2000 numbers , there are 44 squares. so when they are deleted , 2000 will become the 1956th term.
2025 is also a perfect square. So the 2000th term is 2045. 2003rd term is 2048.

2) a)0. which ever triangle u choose , there will be a right angle.

Frankly speaking , an expert should be able to do more than this.

sandeepramesh

3. write the general terms n equate coeffs :)

sandeepramesh

for the another question, note that we have

(1/2)^100(C0+C2+C4...C100) for even number of tails and also (1/2)^100(C1+C3+C5...C99) for odd number of tails so probability = 1/2 :)

hash_include

for the coin qsn:
100C1 (1/2)^100 + 100C3(1/2)^100.... 100C99 (1/2)^100
= (2^{100 -1})/2¹⁰⁰
= 1/2
ans (a)

nadeemoidu

answer for another question 2)
to maximise the distance from the origin , u need to maximise both the real and imaginary parts . the maximum value of real part is 3 and imaginary part is -4 . so the maximum distance is 5.

computer001

let it be x+iy |2-x|<=1 |3+y|<=1 max val of root(x^2 + y^2)
when |x| ,|y| max
so v have root(3^2 + (-4)^2) =5

computer001

sorry nadeem din notice

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