@joyjitpal said:If f(x)*f(1/x) = f(x) + f(1/x) and f(3) = 28, then f(4) is:a. 63 b. 65 c. 17 d. None of these
is oa 65?
@adityaknsit said:...koi batado....
getting 65 with the formula n^3+1
ie f(4) = 4^3 +1 = 65 OA kya hai
ie f(4) = 4^3 +1 = 65 OA kya hai
@shattereddream said:getting 65 with the formula n^3+1 ie f(4) = 4^3 +1 = 65 OA kya hai
how did u arrive at n^3+1
@joyjitpal oa?....
@joyjitpal said:If f(x)*f(1/x) = f(x) + f(1/x) and f(3) = 28, then f(4) is:a. 63 b. 65 c. 17 d. None of these
If f(x)*f(1/x) = f(x) + f(1/x)
function is defined for x^n + 1 or X^n -1
after analysis f(3) = 28 = 27 +1
3 ^3 +1
so f(x) = x ^3 +1
f(4) = 4^3 +1
= 65
function is defined for x^n + 1 or X^n -1
after analysis f(3) = 28 = 27 +1
3 ^3 +1
so f(x) = x ^3 +1
f(4) = 4^3 +1
= 65
@krishh99 said:Hyderabad CAT 2013 aspirants can join herehttp://www.pagalguy.com/forums/cat-and-related-bschools/hyderabad-cat-2013-aspirants-t-84019/p-3594265
kindly post the correct link this thread is closed
@soumitrabengeri said:Here is oneThree diggers dug a ditch 324 m deep in 6 days working simultaneously. During one shift, the third digger digs as many more meters more than the second as the second digs more than the first. The third digger's work in 10 days is equal to the first digger's work in 14 days.How many meters does the first digger dig per shift?1) 15m2) 18m3) 21m4) 27m
Answer: 3) 21
If g(x)g(y) = g(x) + g(y) + g(xy) - 2 for all real x and y and g(2) = 5, then g(3) is:
a. 10 b. 24 c. 21 d. None of these
a. 10 b. 24 c. 21 d. None of these
g(2)g(1)=g(2)+g(1)+g(2)-2
from here u get g(1)=8/9
g(3)g(1)=g(3)+g(1)+g(3)-2 substitute the value of g(1) here u will get g(3)=10
@ajeetaryans said:If f(x)*f(1/x) = f(x) + f(1/x)function is defined for x^n + 1 or X^n -1after analysis f(3) = 28 = 27 +13 ^3 +1so f(x) = x ^3 +1f(4) = 4^3 +1 = 65
Bhai if this is the case X^n+1
then
we have If f(x)*f(1/x) = f(x) + f(1/x)
substituting the values
(28 * 27/28) != (28 + 27/28)
then
we have If f(x)*f(1/x) = f(x) + f(1/x)
substituting the values
(28 * 27/28) != (28 + 27/28)
g(2)g(1)=g(2)+g(1)+g(2)-2
from here u get g(1)=8/9
g(3)g(1)=g(3)+g(1)+g(3)-2 substitute the value of g(1) here u will get g(3)=10
@joyjitpal step 1: substitute x=y=1; youll get a quadratic equation in g(1), i,e you get two values.
step 2: substitue x=3 and y=1 solve the Quadratic eqauation for g(3)
Ans is '1' if iam not wrong
@joyjitpal said:If g(x)g(y) = g(x) + g(y) + g(xy) - 2 for all real x and y and g(2) = 5, then g(3) is:a. 10 b. 24 c. 21 d. None of these
x=2, y=1
5g(1)= 5+g(1)+5-2
g(1)=2
simliarly, g(0)=1
g(2)=5
function suiting this profile = (x)^2 +1
so, g(3)= 10? :O
Two Boxes A and B were filled with a mixture of rice and Dal - in A ratio of 5:3 and in B in the ratio of 7:3.What quantity must be taken from the first to form a mixture that shall contain 8kg of rice and 3 kg of dal ?
@dushyantagarwal said:Bhai if this is the case X^n+1then we have If f(x)*f(1/x) = f(x) + f(1/x)substituting the values(28 * 27/28) != (28 + 27/28)
We will put 4 here and not 28 i think its clear now
find a remainder of 2^133 divided by 133?
@IIM-A2013 said:Two Boxes A and B were filled with a mixture of rice and Dal - in A ratio of 5:3 and in B in the ratio of 7:3.What quantity must be taken from the first to form a mixture that shall contain 8kg of rice and 3 kg of dal ?
is it req. to mix anything from A with the content of B ????