ISI-2009 MATH QUESTIONS

[Ridhika]

How did u guy do the following questions??
21. If (x − a)^2 + (y − b)^2 = c
then 1 + (dy/dx)^2
is independent of
A a
B b
C c
D Both b and c

26. The point on the curve y = x^2 − 1 which is nearest to the point (2, −0.5) is
A (1,0)
B (2,3)
C (0, −1)
D None of the above

28. Suppose X is the set of all integers greater than or equal to 8. Let f : X → <.
and f(x + y) = f(xy) for all x, y ≥ 4. If f(8) = 9, then f(9) =
A 8.
B 9.
C 64.
D 81.
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For 26 when i minimzed the distance formula [(x-2)^2+(x^2-0.5)^2]^(1/2)

 I got x=1/(2)^1/2 (one by root 2) and y = -0.5 .. is there a calculation mistake in my answer or is the approach wrong??

[Dreyfus]

For 21 ...find dy/dx ...then square it and add 1 to it...and then check which of a,b,c are not in the expression!

[Dreyfus]

For q28 f(9)=f(8+1)
f(8+1)=f(8*1)=f(8)
So f(8)=f(9)=9

[Granpa Simpson]

You can also simply tell that c will be eliminated because the (dy/dx) term will be independent of c but will contain a and b, hence (dy/dx)^2 and (1+(dy/dx)^2)...also you can do it in the way Vaibhav pointed out..thats the purest and safest way to solve, however to save time you can use this..!!!

[Ridhika]

Vaibhav I thought of this solution for 28 but the problem is that the question says x,y >=4.. when we take 8,1.. 1<4 .. so how does the condition hold ?

----- Reply message -----
From: "Vaibhav Garg [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: ISI-2009 MATH QUESTIONS
Date: Fri, May 9, 2014 11:35 PM

For q28 f(9)=f(8+1)
f(8+1)=f(8*1)=f(8)
So f(8)=f(9)=9

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[Ridhika]

Thanks for 21 guys ! I was making a random mistake :) 

----- Reply message -----
From: "subhayu [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: ISI-2009 MATH QUESTIONS
Date: Fri, May 9, 2014 11:38 PM

You can also simply tell that c will be eliminated because the (dy/dx) term will be independent of c but will contain a and b, hence (dy/dx)^2 and (1+(dy/dx)^2)...also you can do it in the way Vaibhav pointed out..thats the purest and safest way to solve, however to save time you can use this..!!!

---

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NAML