f(x+y)+f(x-y)=2f(x)f(y)

put y= -x

f(0)+f(2x)=2f(x)f(-x)  -----1

putx= -y

f(0)+f(-2y)=2f(-y)f(y) -----2

since function is variable immaterial

subtracting 1 from  2 we get

f(2x)-f(-2x)=0

which means f is even

now in 1 put x=0

which gives f(0)=either 0 or 1.

 

case1 (f(0)=0)

put y=0 in main equation

f(x)+f(x)=2f(x)f(0)

2f(x)[1-f(0)]=0

since f(0)=0

f(x)=0 thus it is a constant function

 

case2 (f(0)=1)

there r some possibilities where each value is different

but one immediate conclusion is f assuming a constant value

of 1

 

Another way:

I think (correct me if i am wrong)

that the fn given is such that it can assume some constant value k

then according to question

k+k=2k*k

which gives value of k as either 1 or 0

 

f(x) can be a constant function taking wither 0 or 1

Consider the three to A , B , C

assume they r standing in 1 row one behind the other C being the last

C can see the colours of A and B

 

now if they were both white he would have himself told the answer as blue on his head as there has to b atleast 1 blue

so it was some other combination

 

next come to B.assume he sees a white on A's head

now comes the point.If A is white and still C coudnt answer implies B was for

sure blue because otherwise Cwud have himself told the answer

 

so A could not have been white

all this goes on in his head

so he comes with the answer that he is wearing blue

 

 

consider the eqn of line to b

 

x+2/cosa =y+3/sina=r

 

x=rcosa-2

y=rsina-3

 

for r=r1 the pt lies on the 1st given line

 

so rcosa+3rsina+20

 

r=r2 it lies on second line

 

so rcosa+rsina=4

 

r1*r2=20

 

so     80/(3sina+cosa)(cosa+sina)=20

 

solving 2sin2a-cos2a=2

 

it satisfies for a=pi/4

 

hence the answer

 

 

the pattern goes like this

 

(6!/3!)-1 is divisible by7

 

but (8!/4!)-1 is not divisible by 9

 

but again (10!/5!)-1 is divisible by 11

 

then u might say that id denominator is odd it works

 

but even that is wrong

 

because (14!/7!)not divisible by 15

 

as far as i know the conditions for such a divisibility are as follows

 

1.denominator should be odd.eg 6!/3!

 

2.the number with u divide should be prime

 

since in the question asked we have 48! in the dr.(48 being even)

 

it is not divisible

 

Ya thats what i think.

 

sometimes even arihant can be wrong

substituting 4 in place of x we get p to b -17/4

 

now in the second quadratic since equal roots, discriminant =0

 

that is b ^2 = 4ac

 

so q= 289/64

f(x)=x+sinx

differentiating

we get 1+cosx

for a fn to have an inverse it should be strictly increasing or decreasing.

1+cosx>0

implies cosx>-1

since u didnt mention any restriction on interval

cosx can take -1 value as we all know.

 

so it is not always strictly increasing.

 

so no inverse exists

bhatt ji

 

i understand what u say.

i wrote that as the answer because when we add squares of numbers,

it could end with any of the numbers.

 

but when u consider only one number square

the answer is not same bcos squares of numbers do not end

with all possible numbers.

 

for example a number ending with 2 or 3 or 7 or 8 can never be a square.

 

anyway i understand ur point completely

 

A number when divided by seven can leave as remainder either one of

 

0,1,2,3,4,5,6

 

if u want the answer to b perfectly divisible by 7 it implies remainder=0

 

it is just a question of choosing 0 from the set 0,1,2,3,4,5,6

 

the probability of which is 1/7