ISI 2015 PEA Answer Key - Discuss

[onionknight]

Yes, the answer is B. Lim x->0+ cos{sqrt x} = Lim x->0+ cos (sqrt x) =1 since x belongs to the interval and (0,1) and in this interval {x}=x

[deepti]

hlp me with question no. 2,3,8. also the ans of question no. 23 is c and for 25th its b

[onionknight]

Standard deviation won't change. The answer for 23 is B.

2.) Can you draw a continuous function (i.e. without lifting the tip of your pen) ranging from 0 to 1 without passing 1/2? I don't think you can. So that makes it D.

3.) f(x)  =  -2x+6 X<2
               f(2) =2
                 2       2<X<4

right derivative= lim h-> 0 (f(2+h)-f(2)) /h = 0
left derivative = lim h->0   (f(2-h)-f(2)) /h = -2

8.) Firstly, the question itself says x>=0 so rule out all negative numbers. for sqrt(x2-1) to be defined x2-1 must be non negative which means x must lie in  [1, infinity) . For log(sqrt(1-x)) to  be defined  1-x must be positive which means x must lie in [0,1). The intersection of these two sets is null set

[deepti]

I'm sending u the sol that i hv done fr this ans.

On Apr 30, 2015 7:02 AM, "onionknight [via Discussion forum]" <[hidden email]> wrote:

Standard deviation won't change. The answer for 23 is B.

2.) Can you draw a continuous function (i.e. without lifting the tip of your pen) ranging from 0 to 1 without passing 1/2? I don't think you can. So that makes it D.

3.) f(x)  =  -2x+6 X<2
               f(2) =2
                 2       2<X<4

right derivative= lim h-> 0 (f(2+h)-f(2)) /h = 0
left derivative = lim h->0   (f(2-h)-f(2)) /h = -2

8.) Firstly, the question itself says x>=0 so rule out all negative numbers. for sqrt(x2-1) to be defined x2-1 must be non negative which means x must lie in  [1, infinity) . For log(sqrt(1-x)) to  be defined  1-x must be positive which means x must lie in [0,1). The intersection of these two sets is null set

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[deepti]

And standard deviation is dependent of the chnges in origin that mean it'll chng if we add or subtract anything..it'll probably chng it's value

On Apr 30, 2015 8:08 AM, [hidden email] wrote:

I'm sending u the sol that i hv done fr this ans.

On Apr 30, 2015 7:02 AM, "onionknight [via Discussion forum]" <[hidden email]> wrote:

Standard deviation won't change. The answer for 23 is B.

2.) Can you draw a continuous function (i.e. without lifting the tip of your pen) ranging from 0 to 1 without passing 1/2? I don't think you can. So that makes it D.

3.) f(x)  =  -2x+6 X<2
               f(2) =2
                 2       2<X<4

right derivative= lim h-> 0 (f(2+h)-f(2)) /h = 0
left derivative = lim h->0   (f(2-h)-f(2)) /h = -2

8.) Firstly, the question itself says x>=0 so rule out all negative numbers. for sqrt(x2-1) to be defined x2-1 must be non negative which means x must lie in  [1, infinity) . For log(sqrt(1-x)) to  be defined  1-x must be positive which means x must lie in [0,1). The intersection of these two sets is null set

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[onionknight]

Yeah, this is exactly what I've done.

[onionknight]

I'm pretty certain it won't. Google it if you want.

[kk]

how to do ques number 7?

[L]

Easiest way is to just put x+2y and x-2y in the given options and check for which one you get xy.  

[kk]

YES..BUT HOW TO SOLVE THESE LINEAR TRANSFORMATIONS IN A PROPER WAY?

[deepti]

I hv checked it in s.c gupta also..sd will chng if we add or subtract anything

On Apr 30, 2015 8:37 AM, "onionknight [via Discussion forum]" <[hidden email]> wrote:

Yeah, this is exactly what I've done.

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[onionknight]

Adding the same constant 'a' to all the observations increases the value of mean by
a. However, adding the same constant 'a' to all the observations does not change the value
of s. That’s because adding a constant 'a' to all data values shifts the location of the data but
does not affect its spread.
For more proof, see the second page: http://faculty.uncfsu.edu/dwallace/lesson%206.pdf

[onionknight]

Put u= x+2y and V=x-2y. Solve for x and y in terms of u and v. You get x=u+v /2 y=u-v /4 which means xy= u2-v2 /8 . Just replace u and v by x and y

[deepti]

Yes u r ryt.. thankq for clearing this misconception of mine..plz hlp with 26th question

On Apr 30, 2015 3:35 PM, "onionknight [via Discussion forum]" <[hidden email]> wrote:

Put u= x+2y and V=x-2y. Solve for x and y in terms of u and v. You get x=u+v /2 y=u-v /4 which means xy= u2-v2 /8 . Just replace u and v by x and y

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[deepti]

If u go by rolle's theoram in this question then also it's d. Bcz the 3rd condition of rolle's theoram is nt satisfying in question no. 14th

[kk]

i think the answer of question number 24 will be c..if u find µ2 we find it is 4..

[deepti]

Plz hlp with the solution of this question 24th

[deepti]

Ans. Of question no. 27th should be c bcz here we have without replacement case. If u have assumed that white ball is drawn in the frst draw then in the next draw the no. Of white balls will reduce to w-1. So ans. Shud be c only

On Apr 30, 2015 5:38 PM, "deepti [via Discussion forum]" <[hidden email]> wrote:

Plz hlp with the solution of this question 24th

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[L]

Q. 24) E(X) = 2
          E(X^2) = 4
          E(X- mean (X) ) ^2  = 0

kk, how you are getting variance equal to 4?

[kk]

sorry its 0..it was a calculation mistake..its 0