ISI 2004 ME - I Answer Key

[Bankelal]

Hi n1,

Sorry for late response.

We can take f(x)=e^(-x)
hence f(x)*f'(x)<0
thus |f(x)| is a decreasing fn

[neha 1]

thanks :)

[rongmon]

What is the logic/ concept behind Q27.

Thanks.

[rongmon]

Q 16 and Q 28 as well.

Thank you.

[Ridhika]

Hey Rongmon..

For 15 basically you can arrange all the boys in 5! ways.. now the girls can be arranged as thus:
_ B1_ B2_ B3 _ B4_B5_

where the blanks represent places that can be occupied by girls.. So 4 girls can be arranged in 6P4 ways.. therefore ans is 6P4 x 5!

[Dreyfus]

For 27...the concept is of diagnolisation of a matrix using the eigen values and eigen vectors

[Jim_Moriarity]

Question No.12
 IN this case shouldn't it be A) Limit don't exist as in because at x<1 its limit value is -1 at x=1 its 0 and x>1 its 1.

[deepti]

I'm gttng c for question no. 30.. if we tk y=0 we'll get x=3 & -3 so its crossing x axis 2 times

[Jim_Moriarity]

You're right, deepti.

[deepti]

Also in question no. 3 can u plz tell wht i shud do aftr taking derivative of the function? ?

On May 9, 2015 11:52 AM, "Jim_Moriarity [via Discussion forum]" <[hidden email]> wrote:

You're right, deepti.

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[Jim_Moriarity]

tell the question.

[deepti]

3rd

On May 9, 2015 12:00 PM, "Jim_Moriarity [via Discussion forum]" <[hidden email]> wrote:

tell the question.

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[Jim_Moriarity]

The function changes its sign only once.
And the derivative is positive, increasing function.
Only one root.

[deepti]

Ok thnkq

On May 9, 2015 12:05 PM, "Jim_Moriarity [via Discussion forum]" <[hidden email]> wrote:

The function changes its sign only once.
And the derivative is positive, increasing function.
Only one root.

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