# Multiple type of questions...

• March 31st 2009, 05:35 AM
siddscool19
Multiple type of questions...
I have attached an image of the questions I want the answer for. I tried a lot but couldn't get. Basically the problem is I have entrance exam on 5th april and this is the sample paper. Since our syllabus changed therefore many of the questions we can't do.
Hope people can help here.

Here are my queries:

In Q.27. I want to know how to do it the answer is c and d. (I have the answer key with me.)

In Q.28. is the c) part correct? If yes how?

In Q29. how is the d) part correct?

In Q30. I don't know how to do this question at all never saw this question before. If anyone can explain this please surely do.

In Q32,33,34 How to do them also...

I know these are many questions but even if someone can solve any of them it would be really a great help :).
I know they should be to their respective threads but it would have taken a lot of time so I apologize for that.
• March 31st 2009, 07:04 AM
HallsofIvy
Quote:

Originally Posted by siddscool19
I have attached an image of the questions I want the answer for. I tried a lot but couldn't get. Basically the problem is I have entrance exam on 5th april and this is the sample paper. Since our syllabus changed therefore many of the questions we can't do.
Hope people can help here.

Here are my queries:

In Q.27. I want to know how to do it the answer is c and d. (I have the answer key with me.)

$p(x)= (x^2- 3x+ 2)(x^2- 7x+ a)$, $q(x)= (x^2- 8x+ 15)(x^2- 6x+ b)$ and you are told that the H.C.F (highest common factor) is (x-2)(x-4). The fact that that is a common factor means that x-2 and x-4 must be a factor of both p and q. We have $2^2- 3(2)+ 2= 4- 6+ 2= 0$ so x- 2 is a factor of $x^2- 3x+ 2$. Now look at x- 4. $4^2- 3(4)+ 2= 16- 12+ 2= 4$ so x- 4 is not a factor of $x^2- 3x+ 2$. $4^2- 7(4)+ a= 16-28+ a= a- 12$. In order that x-4 be a factor, a must equal 12.
Similarly, for $q(x)= (x^2- 8x+ 15)(x^2- 6x+ b)$. $2^2- 8(2)+ 15= 4- 16+ 15= 3$, not 0, so x- 2 must be a factor of $x^2- 6x+ b$. $2^2- 6(2)+ b= b- 8$ so b must be equal to 8.
You should also check that $4^2- 6(4)+ 8= 0$ so that x- 4 also is a factor of $x^2- 6x+ 8$

Quote:

In Q.28. is the c) part correct? If yes how?
You could do that using a calculator. Or, for c, use the fact that if $y= log_{0.4}(3)$ then $3= 0.4^y$ and if $x= log_4(3)$ then 3= 4^x[/tex]. Since 4 is larger than .4, it will take a smaller power to equal 3.

Quote:

In Q29. how is the d) part correct?
When i is any number between 0 and 45, 90-i is between 45 and 90. tan(90-i)= cot(i)= 1/tan(i). So ln(tan(90-i))= ln(1/tan(i))= -ln(tan(i)). Every number i, for i between 0 and 45, is canceled by i between 45 and 90.

Quote:

In Q30. I don't know how to do this question at all never saw this question before. If anyone can explain this please surely do.
You can only do problems you have seen before? Surely not!
The shaded part is all points that are not in A and not in B. "not in A" means "in the complement of A" and "not in B" means "in the complement of B". Finally, "in X and in Y" is "in X intersect Y" so this is "complement of A intersect complement of B".

Quote:

In Q32,33,34 How to do them also...

I know these are many questions but even if someone can solve any of them it would be really a great help :).
I know they should be to their respective threads but it would have taken a lot of time so I apologize for that.
• April 1st 2009, 05:28 AM
siddscool19
Can anyone else also solve these questions?
I really need help with these entrance exam on 5 April.