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I had given a detailed answer when u asked with respect to 92! question

19^92 = 19*19*19.....92 times
which is an odd number

any multiple of 92 is even because it contains a factor of 2 in it
so obviously 19^92 not divisible by 92

consider this way
1111(4 1's) divided by 270 Rem = 270
11111(5 1's) Rem = 0
111111(6 1's ) Rem= 1
(7 1's) Rem = 11
for 8 Rem = 111
for 9 Rem = 1111
for10Rem = 0
so similarly ....
for 15 Rem = 0
for 20 Rem = 0
.....
for 1110 Rem =0
so by pattern for 1111 Rem =1
that is 1111......(1111 times)
divided by 271 leaves 1 as remainder

I didnt read the first line my friends

so its my miss

anyway as far as i know mr feynmann is right

f(1)<0

f(2)>0

and by checking up we find only one root (real)

so i think there is only 1 real root

pls correct if wrong

a^2-b^2 (a-b)(a+b)is divisible by a-b

a^3-b^3 (a-b)(a^2+ab+b^2)is divisible by a-b

similarly a^4-b^4 is also divisible by a-b

it can be thus shown that a^n-b^n is always divisble by a-b for natural number n

thus a^1992-b^1992/a-b is an integer

similarly the other two

so int+int+int which is also an integer

sorry for being hasty
i didnt notice the square in the question
next time type the word square if u dont have special editor
anyway using same technique
answer is
-1/6*(1+1)
=-1/3

lim(x-0) [1/x-1/sinx]
=(sinx-x)/xsinx
consider sinx-x/x cubed
by applying l'hospital's rule we get value to b -1/6(when x tends to 0)
so given = (sinx-x)/x cubed * x square /sinx
so answer is 0

Could u retype the last question u have posted
there seem to be something missing in that

grouping sin square 12 with sin square 18 and sin square 21 with sin square 9
using the formula
sin(A+B)sin(A-B)=sin square (A)-sin square (B)
we get
1/2[sin12-sin6]
sin square 39 = (1- cos 78)/2
sin square 48 = (1- cos 96)/2
also cos 78= cos (90-12)=sin12
cos96=cos(90+6)= -sin 6
using both adding we get answer =1

what does the second question mean? is it sin square theta?

tan70-2cot40
tan70+2cot(180-40)
tan70+2cot140
tan70+2(cot^2(70)-1)/2cot70
tan70+cot70-tan70
=cot70

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

Let the point B on AB b (h1,k1)
Let the point C on AC b (h2,k2)

we form 2 equations respectively bcos they lie on the given lines

not only that since midpt is 5,6
h1+h2=10
k1+k2=12

2 equations 2 unknowns

hence the answer

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