ISI 2014 PEA Answer Key

[Anakin Skywalker]

CONTENTS DELETED

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[Jim_Moriarity]

How to do question number 27?

I'm getting the answer as B.
not D.
Please help me out with it.

[Jim_Moriarity]

Question 20. Please.
Thanks

[Ankita]

This is how i solved it:

kth term of AP: Tk=a+(k-1)d
Infinite sum of GP : a/(1-d)

a+(k-1)d = a/(1-d)
a-ad+(k-1)d-(k-1)d^2=a
d[(k-1)-a-(k-1)d]=0

(k-1)(1-d)=a

For d= 1/3    or    1/5      or    1/9
1-d  = 2/3    or    4/5      or    8/9 respectively
a  =   2/3(k-1) ; 4/5(k-1) ; 8/9(k-1) ------> all three give composite numbers no matter what the value of "k"
Only for 1-d = 1/2 => d=1/2
we can have "a" as prime

[Neha Aggarwal]

In question 14 won't the answer would be a since it is asking us head after even number of tosses so sample space should be H,TTH,TTTTH,.......... according to which prob will be 2/3

[Dr. Strange]

Even number of tosses include the last toss as well.
So favourable outcomes are TH,TTTH,TTTTTH and so on.

[Komal jain]

Please provide the solution for ques 15

[Dr. Strange]

 This is a binomial distribution with p(success)= 1/2 and n=10
Mean score = mean of this binomial distribution= np = 5
so P(SCORE =5)=p (NO. OF SUCCESS=5)= C(10,5)* (1/2)^5 * (1/2)^5  = 63/256

[varnika1880]

M not able to do question 13 plz help

[Dr. Strange]

eqn 1, x2>= x1-3

 x2=x1-3+ h1 .......(1)

eqn 2, x3>= x2+ 2

 x3=x2 + 2 + h2 = x1+h1+h2-1

eqn 3, x4>= x3-10

 x4=x1+h1+h2+h3-11   .......(2)

where h1,h2,h3 are non negative values



(2)- (1) gives   x4-x2 = h2 +h3 -8
since h2 and h3 are nonnegative

so x4-x2 >= -8

now from eqn 4

x4-x2<=alpha

so

-8<=x4-x2<=alpha

so alpha should be greater than -8 but all options are less than -8

hence option d none of these is the answer

[varnika1880]

Thanks a lot

[varnika1880]

Guys q 27 plz ?

[JMKeynes]

For q27 answer is (d).

If the function is of the form f(z)=z,
then from the given equation (let's call it it eq1): f(x+y)*f(x-y)=
􀀀(f(x) + f(y))^2 - 4*x^2*f(y),
we have (x+y)*(x-y) = (x+y)^2 - 4*x^2*y

Solving we get a quadratic equation 2*x^2-x-y=0.
Solving for x, we get,

 x=(1+(1+8y)^.5) or x=(1-(1+8y)^.5)..........(2)

So for a given value of y, eq1 will not hold for any x not satisfying (2). But as per the question eq1 should hold for all x,y that are real.
Contradiction!
So a function of the form f(z)=z cannot hold. Or, f(2)=2 cannot hold.

[Saunok]

Hi, can you post the solution to Q15? I can't find out the value of the sum

[Spirit1011]

in q14,the heads has to occur after even no. of tosses that means the series must be H+TTH+TTTTH+...