CAT Question Bank  Functions

@SoumyaRanjanPatra How you got this value ? Answer is 720 + 7^2
f(x) = (x + 1)^2 for x = [0, 5]
Also, as f(x) has a degree six with coefficient of x^6 as 1, we can write
f(x) = x (x  1) (x 2) (x  3) (x  4) (x  5) + (x + 1)^2
so f(6) = 6 * 5 * 4 * 3 * 2 * 1 + 7^2
f(6) = 720 + 7^2

Method 1 :
2f(x) + f(1 – x) = x^2
2f(5) + f(4) = 25 (for x = 5)  (1)
2f(4) + f(5) = 16 (for x =  4)  (2)
Solve for f(5)  (1) * 2  (2)
3f(5) = 50  16 = 34
f(5) = 34/3Method 2 : [Solution by @vikas_saini ]
2f(x) + f(1  x) = x^2
2f(1  x) + f(x) = (1x)^2
3f(x) = 2x^2  (1x)^2
x = 5
3 f(5) = 34
f(5) = 34/3

f(1) = 4
8a = 4
a = 1/2
So, f(3) = 64a = 64 x 1/2 = 32

@rowdyrathore :Ans I believe is 359

@rowdyrathore : Ans to this question is D

@rowdyrathore: This was already asked in Q2. Anyways answer is 34/3

@rowdyrathore : 2009(4!)+ 5  correct me if I am wrong

@rowdyrathore : Kindly provide the solution to this.

@rowdyrathore
,
,
.
is it 36


,
.
.359

@rowdyrathore 704 ?.......................

1+2+4+7+11+16........................and so on till 30th term

f(1)=f(3) = f(5) = and so on
so f(511) =3
@SujanChhetri

@RowdyRathore F(x) = 3X  1.

@RowdyRathore 2009(5!) + 11.

@RowdyRathore It has to be 36 only, since f(1) = 1 and f(1/2) = 8, and there is no possibility of F(1/3) being negative.

F(x) = x^4360x^2+400
F(x) = (x^2+20)^2  400x^2
F(x) = (x^220x+20)(x^2+20x+20) = Prime Number
Since the value of second factor is always positive and greater than 1 therefore, 1st factor has
to be equals to 1.
x^220x+20 = 1
x^220x+19= 0
On solving,
x = 1, 19.
Putting back the values in the original equation we get f(x) = 41 and f(x) = 761.So, the sum will be 41+761 = 802