3.Ans (E)
f(x) =-3x
f(a)= -3a--(1)
f(b)= -3b---(2)
f(a+b)= -3 (a+b)
(1) + (2)
f(a) + f (b) = -3a + (-3b) = -3 (a+b)
f (a+b) =f (a) + f (b)
2. ans (E)
(sqrt x^2)/ x = (+/- x)/x
Hello,
I recently did a CAT exam. I missed a few questions on Math and, unfortunately, there is no explanation for the solution. I hoped the wizards here could please help. I would love to hear how you would solve any of these problems
1. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is
A. b/n 2 and 10
B. b/n 10 and 20
C. b/n 20 and 30
D. b/n 30 and 40
E. Greater than 40
2. If x <> 0, then square root (x^2) / x =
A. -1
B. 0
C. 1
D. x
E. abs(x)/x
I thought it was c, but no dice
3. For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers a and b
A. f(x) = x^2
B. f(x) = x +1
C. f(x) = sqr (x)
D. f(x) = 2/x
E. f(x) = -3x
Thank you kindly
1. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is
A. b/n 2 and 10
B. b/n 10 and 20
C. b/n 20 and 30
D. b/n 30 and 40
E. Greater than 40
I think in this question b/n means between.
And the answer is E
Last edited by Md. Minuddin; 05-29-2009 at 11:09 AM.
Minuddin and everyone else,
Thanks for the responses. I understand 2 and 3, but question 1 is a mystery. The answer is E. The question is from the GMATPrep software.
Is there a way to view an explanation to the solution? The software tells you the answer, but does not explain the answer.
Hopefully someone can figure out why it is E
hmm.. b/n between? I didn't get it.
My answer is here,
h(n) = 2 x 4 x .....n
h(100) = 2 x 4 x ...100
h(100) + 1 = (2 x 4 x ...100 + 1) = 2^50 (1 x 2 x...50) +1
if we divide this any number below 50;
1 would be the remainder. So that the smallest prime factor would be greater than 50.
Ans (E)
Last edited by dissa; 05-29-2009 at 02:32 AM.
There are currently 1 users browsing this thread. (0 members and 1 guests)