71Received this Question in my first test and Gone!!!!
Column I:
A) The no of solutions of the system of equations x+2y = 6 and |x-3| = y is m, then
B) If x and y are Integers and (x-8)(x-10) = 2y, the no. of solutions is n, then
C) The no. of Integral Solutions for the equation x+2y = 2xy is p, then
Column II:
P) m is the AM of n and p
Q) n is the GM of m and p
R) p is the HM of m and n
S) n = \frac{m^{p}+p^{m}}{mp}
T) m = \sqrt{n\sqrt{p\sqrt{n\sqrt{p}}}} .....∞
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8 Answers
11Hint for 1st sum : Graphs.
Hint for 2nd sum : take x=2ab.
Hint for 3rd sum : (x-1)(2y-1)=1
71Can you please Elaborate the second one.. Please!Please!
11Set x=2^a.b, so that the eqn at hand becomes (b-4)(b-5)=2^k.
Set (b-4)=c, so the eqn further becomes c^2-c-2^k=0.
Observe that c and (c-1) are co-prime, and their product is a kth power, so they must themselves be kth powers.
c(c-1)=m^k.n^k=(mn)^k=2^k.....From which it follows, c=2.
The remaining part, hopefully, can be completed.
11Complete the part, when both m and n are negative integers.
You will land up with 2 integral solutions (12,3) and (6,3). Thus p=2.
341B) Aliter:
We must have x-8 = 2^a; x-10 = 2^b or x-8 = -2^a; x-10 = -2^b
Taking the first set of equations, we obtain 2^a-2^b = 2
The only solution is a =2, b=1. Hence x =12, y=3
Taking the second set, we similarly obtain x=6, y=3
C) Aliter: Let 2y=z. Then x+z=xz. Its easy to see that we must have x|z and z|x. So x=±z.
Since z is even, the only solution is (0,0)
71THanks all! Wish I could do such sums mere by observation some day! :)
\int_{Maths}^{Infinity}{Knowledge} = theprophet,Soumik Sir