ISI-2009 MATH QUESTIONS

[jack]

Q.1- If f(x) is a real valued function such that
       2f(x)+3f(-x)=55-7x
for every xER, then f(3) equals           (E means belongs)
A) 40
B) 32
C) 26
D) 1

Q.2- If X=2^65 and Y=2^64+2^63+......+2^1+2^o, then
A) Y=X+2^64
B) X=Y
c)Y=X+1
D)X-1

Q.3- Two persons, A and B, make an appointment to meet at the train station between 4 P.m and 5 P.m. They agree that each is to wait not more than 15 minutes for the other. Assuming that each is independently equally likely to arrive at any point during the hour, find the probability that they meet.
A) 15/16
B) 7/16
C) 5/24
D) 22/175

[The Villain]

I'm getting quest 1 b
quest 3 b

[jack]

could you please give me the explaination?

[The Villain]

q1 ...replace x by -x in the function ..
then solve the 2 equations...u wilk get 32

q3...will upload the soln in sometime

[Granpa Simpson]

Answer to Q2 will be option d i.e. Y=X-1,
The solution can be obtained by expressing Y as a sum of finite G.P series with first term 2^0 (i.e. 1) and common ratio r= 2, and n= 65, this will give Y= 2^(65)-1= X-1

[Homer Simpson]

oh yeah, mixed up the formula! its definitely 2(d)

[The Villain]

Pls check out

[Dreyfus]

For q2.....y is a sum of gp with term being 1 nd r being 2....nd y turns out to (2^65)-1.....therefore Y=X-1

[jack]

@ ron
I am not able to solve the Q1 with your hint
so give me the whole solution
thank you

[L14]

Hi jack
2f(x)+3f(-x)=55-7x
Put x=-x
2f(-x)+3f(x)=55+7x

4f(x)+6f(-x)=2(55-7x)-----a
6f(-x)+9f(x)=3(55+7x)-----b
b-a=> 5f(x)=55+35x
f(x)=11+7x
f(3)=32

Pls can someone explain 3. Ron, the image is too small to be read.

[Homer Simpson]

can you explain 3?

[Granpa Simpson]

question 3 has to be solved using geometric probability concept...On x take 1 hr coordinate and y too take 1 hour as coordinates (this one 1 basically represents the distance from the origin on two axes, one hour is taken since the time span is 4:00 to 5:00 i.e 1 hr, if the time span would have been 4:00 to 6:00 then we would have taken 2 hrs as the respective distance from origin on both the axes, this would change with changing times), in this case however its 1 hr. Now a unit rectangle with area 1 can be formed. Also mark the waiting time in the graphical solution as (1/4), since each person waits 15 mins. do this on both the axes. Now a shaded region can be observed in the graph. the probability that they meet will be given by the P(of getting a point in the shaded area), remember here in every case the total sample space is 1. solving in the above manner ul get the answer. Also there's a direct formula for this type of problems which is given by, required probability = 1-{(1-w1)^2)/2}-{(1-w2)^2)/2}. also you can put w1=w2=(15/60)=1/4 for the above problem and solve, where w1 and w2 are waiting times.

[Homer Simpson]

 very well explained. thank you!

[Granpa Simpson]

Any sort of problem in probability involving variables as distance or time can be solved using this concept..:)

[jack]

thank you so much subhayu , ben10 and ron

[jack]

@ Subhayu
you can do V.K.Kapoor for probability

[rongmon]

9. Let f, g and h be real valued functions defined as follows: f(x) = x(1 − x), g(x) = x/2 and h(x) = min{f(x), g(x)} with 0 <= x <= 1. Then h is


A continuous and differentiable
B is differentiable but not continuous
C is continuous but not differentiable
D is neither continuous nor differentiable


27. If a probability density function of a random variable X is given by f(x) = kx(2 − x), 0<= x <= 2, then mean of X is
A 1/2
B 1
C 1/5
D 3/4


Thanks.

[hs]

for q27 its b
first calculate k using integral of pdf over o to 2 equal to 1
than calculate expectatin
 for questn 9 its c
h takes value x/2 from (0,1/2) and x(1-x) from (1/2,1) , so we have a kink at 1/2

[Granpa Simpson]

for q9 the answer will be c, because z(x)= g(x) for all 0<x<=1/2 and z(x)=f(x) for all 1/2<x<=1.

[rongmon]

Thanks guys