quant-3

[wanderwanderer]

1.for which of the following function is f(a+b) = f(a) + f(b) :

(a) f(y) = x^2
(b) f(y) = x + 1
(c) f(y) = sqrt(x)
(d) 2/(sqrt(x))
(e) -3x

2.For every +ve integer n , the function f(n) is defined to be product of all even integers from 2 to n. If p is the smallest prime factor of h(100)+1 then p is :

(a) between 2 and 10
(b) between 10 and 20
(c) between 20 and 30
(d) between 30 and 40
(e) > 40

Data Sufficiency:
3.In xy-plane at what 2 points does the graph of y=(x+a)(x+b) intersects x-axis
(a)a+b = -1
(b)graph intersects y-axis at (0,-6)

4.If xyz > 0 is x>0
(1) xy > 0
(2) xz > 0


Thanks

[sommukh]

1. I am not sure if the function is expressed correctly or not . I think it should be F(x) = x^2 etc. In that case
f(x)=x^2=>f(a+b) = (a+b)^2 != a^2+b^2 =f(a)+f(b)
f(x)=x+1=>f(a+b) = a+b+1 != a+1+b+1 = f(a)+f(b)
f(x)=sqrt(x)=>f(a+b) =sqrt(a+b) != sqrt(a)+sqrt(b) =f(a)+f(b) ...etc

however for
f(x)=-3x => f(a+b) = -3(a+b) = -3(a) -3(b) =f(a)+f(b)

2. For every +ve integer n , the function f(n) is defined to be product of all even integers from 2 to n => f(100)+1 is not divisible by anything between 2 to 100 => P>100=> P>40

3. y=(x+a)(x+b) = > it will cut x axis at -a,0 or -b,0
a. It does not get us anywhere so insufficient (there are infinite number pairs satisfying this condition)
b. putting x=0 in the equation we get y=ab=> ab= -6 this alone also insufficient
However combinig both we can get specific values of a and b.
So IMO answer is C

[sommukh]

xyz>0 ....(A)

a. xy>0 => (x>0 && y>0) || (x<0 && y<0). In either case z must be > 0.
Insufficient ...
b. xz>0 => (x>0 && z>0) || (x<0 && z<0). In either case y must be > 0.
Insufficient ...

(a)+(b) => x>0, y>0, z>0 (Only possibility satisfying all requirements)

[llutsar]

The answer for 1. is E.

a.)i.) (a+b)^2 = a^2+2ab+b^2

ii.) a^2 +b^2
NOT EQUAL

b.)i.) (a+b) + 1
ii.) a+1+b+1= a+b+2
NOT EQUAL
c.)i.) sqrt(a+b)
ii.) sqrt(a) + sqrt(b)
NOT EQUAL
d.)i.) 2/sqrt(a+b)
ii.) 2/sqrt(a)+2/sqrt(b)
NOT EQUAL
e.)i.) -3(a+b) = -3a – 3b
ii.) -3a+(-3b)
EQUAL

[Dr_GMAT]

I agree...it is E.

[Crack_Mat]

Originally Posted by wanderwanderer

1.for which of the following function is f(a+b) = f(a) + f(b) :

(a) f(y) = x^2
(b) f(y) = x + 1
(c) f(y) = sqrt(x)
(d) 2/(sqrt(x))
(e) -3x

2.For every +ve integer n , the function f(n) is defined to be product of all even integers from 2 to n. If p is the smallest prime factor of h(100)+1 then p is :

(a) between 2 and 10
(b) between 10 and 20
(c) between 20 and 30
(d) between 30 and 40
(e) > 40

Data Sufficiency:
3.In xy-plane at what 2 points does the graph of y=(x+a)(x+b) intersects x-axis
(a)a+b = -1
(b)graph intersects y-axis at (0,-6)

4.If xyz > 0 is x>0
(1) xy > 0
(2) xz > 0


Thanks

answer:
1) E
2) ?
3) C
4) C

Is it that f(n) + 1 will not be divisible by any number in between 2 to n? I don't think so?
Can anybody throw some light on 2) please?

[bretania]

CrackMat, could you please explain 3?

I also don't know 2. But IMO, it will be E (more than 40).
My approach is not elegant and it took me a long time to come up with my observation so there should be a better approach to this problem. I've seen this problem somewhere as well.

Anyway.. for 2.
If you have up to 100, you will have 50 even numbers there.
2, 4, 6, 8, .... 100

Notice that each factor above is just a multiple of 2 of all positive numbers from 1 to 50.
1, 2, 3, 4, .... 50

Now, let's observe smaller numbers. Say we only have 2 even numbers.
2, 4
this corresponds to the following number times 2
1, 2 (*)

Notice then that the product of 2 and 4 + 1 = 9.
And the least prime number that can divide 9 is 3.. this is greater than any of the elements of your set (*).

We will notice this to be true for higher numbers actually.

Case A.
If you have a
product of 2, 4, 6, 8 and then add 1 to it (385)

The least prime that can divide the result above (is 5) will be greater than any of the following numbers: 1, 2, 3, 4

Case B.
If you have a product of 2, 4, and 6 and then add 1 to it (49), the resulting number is 49. And the least prime that can divide 49 (7) is greater than any of the ff number (1, 2 or 3).


So if you have up to F(100)+ 1,
we can conclude that the least prime (from our non-exhaustive observation) should be more than 50. And the choice nearest this approx is E.

[Crack_Mat]

Originally Posted by bretania

CrackMat, could you please explain 3?

I also don't know 2. But IMO, it will be E (more than 40).
My approach is not elegant and it took me a long time to come up with my observation so there should be a better approach to this problem. I've seen this problem somewhere as well.

Anyway.. for 2.
If you have up to 100, you will have 50 even numbers there.
2, 4, 6, 8, .... 100

Notice that each factor above is just a multiple of 2 of all positive numbers from 1 to 50.
1, 2, 3, 4, .... 50

Now, let's observe smaller numbers. Say we only have 2 even numbers.
2, 4
this corresponds to the following number times 2
1, 2 (*)

Notice then that the product of 2 and 4 + 1 = 9.
And the least prime number that can divide 9 is 3.. this is greater than any of the elements of your set (*).

We will notice this to be true for higher numbers actually.

Case A.
If you have a
product of 2, 4, 6, 8 and then add 1 to it (385)

The least prime that can divide the result above (is 5) will be greater than any of the following numbers: 1, 2, 3, 4

Case B.
If you have a product of 2, 4, and 6 and then add 1 to it (49), the resulting number is 49. And the least prime that can divide 49 (7) is greater than any of the ff number (1, 2 or 3).


So if you have up to F(100)+ 1,
we can conclude that the least prime (from our non-exhaustive observation) should be more than 50. And the choice nearest this approx is E.

Friend, y = (x+a)(x+b) , at what two point it intersects?
Let us analyse:
1) its a equation of parabola opening in +y direction!!
2) its roots are -a and -b!
This parabola may or may not have real roots , or in other words it can intersects x axis at two point, only if: roots are real and not equal!!

So if u remember for equation pX"2+qX+r = 0 , on solving, will give you two values of X or roots as:
X1 = -q/2p +0.5*sqrt ( (q/p)^2 - 4r/p )
and
X2 = -q/2p -0.5*sqrt ( (q/p)^2 - 4r/p )

for real root sqrt should be real.

Sum of the roots = -q/p;
product of roots = r/p;
------------This is the theory----------------
Now conditions:
1) a+b = -1 or -a -b = 1 or sum of the roots = 1 = -q/p;
2) ab = -6 or product of roots = -6 = r/p;

so ( (q/p)^2 - 4r/p ) is +ve hence real roots can be find and both conditions are needed!