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 Rohit Gupta's messages in the community 1 2 3 4 5 6 GO Go to Page...
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 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]
Let us first  calculate  no. of numbers when exactly one of the digits is 5:

when 5 comes in first place:         no. of numbers = 9*9 = 81
(as second and third digit may be any of 0,1,2,3,4,6,7,8,9 (total 9 ways))

when 5 comes in second place:    no. of numbers = 8*9 = 72
(as first digit may be any of 1,2,3,4,6,7,8,9 (total 8 ways) and third digit may be any of 0,1,2,3,4,6,7,8,9 (total 9 ways))

when 5 comes in third place:        no. of numbers = 8*9 = 72
(as first digit may be any of 1,2,3,4,6,7,8,9 (total 8 ways) and second digit may be any of 0,1,2,3,4,6,7,8,9 (total 9 ways))

=>      total no. of numbers when exactly one of the digits is 5
= 81 + 72 + 72 = 225    ANS of (2)

Now number of numbers when exactly two of the digits are 5:
when 5 comes in 1st and 2nd place:  no. of numbers = 9
(as third digit may be any of 0,1,2,3,4,6,7,8,9 (total 9 ways))

when 5 comes in 2nd and 3rd place:    no. of numbers = 8
(as first digit may be any of 1,2,3,4,6,7,8,9 (total 8 ways))

when 5 comes in 3rd and 1st place:        no. of numbers = 9
(as second digit may be any of 0,1,2,3,4,6,7,8,9 (total 9 ways))

=>      total no. of numbers when exactly two of the digits are 5 = 26

and  when all 3 digits are 5:     number of numbers = 1

=> total no. of numbers when at least one of the digits is 5
= 225 + 26 + 1 = 252    ANS of (1)

 Discussion Forums -> This Post 9 points    (1    in 3 votes )   [?]
AJ1991!

i^i = (cos pi/2 + i sin pi/2)^i

= (e ^ ( i * pi/2 )) ^ i

= e ^  ( i * pi/2 * i)

= e^(-pi/2)   ANS
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
e^( - pi/2)
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
Krzme!

[x ][ 0] (sinx/x)^(1/x2 ) = [x ][ 0] (1 + (sinx/x - 1))^(1/x2 )

=
[x ][ 0] (1 + (sinx/x - 1))^[(1/(sinx/x - 1))*(sinx/x - 1)*(1/x2 )]

= e^[
[x ][ 0] (sinx/x - 1)*(1/x2 )]

(using
[n][ 0] (1 + n)^1/n here [x ][ 0] (sinx/x - 1) = 0)

= e^[[x ][ 0] ((sinx - x)/x)*(1/x2 )

= e^[[x ][ 0] ((x - x3/3! +  x5/5!.............. - x)/x)*(1/x2 )

= e^[[x ][ 0] ((- x3/3! +  x5/5!..............)/x3 )

= e^[[x ][ 0] (- 1/3! +  x2/5!..............

= e^[[x ][ 0] (- 1/6 + 0 + 0..............)/x3 )

= e^(-1/6)   ANS

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
Greatdreamss!

good attempt but complicated!  however keep it up!
 Discussion Forums -> This Post 35 points    (7    in 7 votes )   [?]
akash!

suppose A - event that a six came
A' - even that a six did not come
B - event that a man reports six
T - Man tells the truth   =>    P(T) = 2/3

we have to find P(A/B)

P(A/B) = P(AB)/P(B)

= P(A)P(B/A) / (P(BA) + P(BA'))

= P(A)P(B/A) / ( P(A)P(B/A) + P(A') P(B/A') )

= (1/6)(2/3) / ((1/6)(2/3) + (5/6)(1/3)*(1/5) )

= 2/3

Now here almost all the books have committed mistakes in bold, big part;

they put P(B/A') = 1/3 i.e. prob that person reported 6 given that 6 did not come
but it is wrong. 1/3 is just the prob that person tells a lie. Suppose if 3 came, then 1/3 is the probability that person says false that is he says any of these: 1,2,4,5,6. But we want him to say 6 only, so multiply by 1/5

prob that person reported 6, given that 6 did not come
= (prob that person lies) * (prob that person says 6 only among the 5 choices he has to tell a lie)
= 1/3 * 5/6

Thus correct answer of this ques is 2/3

Any how if u can understand then probability that it was a six, given that person reported six will be equal to the probability that person tells the truth, as only if he had told the truth, it would have been a six. Thus 2/3.

For any more clarifications write here!
 Discussion Forums -> This Post 30 points    (6    in 6 votes )   [?]
Mallika!

One diagonal is of course the line passing through (-3 ,5) and (-1,3)

=>    y - 3 = (-2/2) (x + 1)     =>    x + y - 2 = 0

Mid point of this diagonal is  ((-3-1)/2 , (5+3)/2)  =  (-2, 4)
slope of other diagonal will be -1/(-1) = 1

=> equation of other diagonal will be y - 4 =  x  + 2 => x - y  + 6 = 0

Suppose A is (-3 ,5), C is (-1,3) and O is (-2, 4)
AO = sqrt(2)
suppose B, D are other vertices of the square
if B is (h,k)   =>   (h,k) lies on   x - y  + 6 = 0        =>     h - k + 6 = 0    => k = h + 6

thus B is (h, h+6)
Now OB = sqrt(2)
=>      (h+2)^2 + (h+6-4)^2 = 2       =>      h = -3, -1      =>    k = 3, 5

Thus B is (-3, 3) and D is (-1, 5)

Eqn of AB:       x = -3
Eqn of BC:       y = 3
Eqn of CD:       x = -1
Eqn of DA:       y = 5
Eqn of Diagonal AC:
x + y - 2 = 0
Eqn of Diagonal BD:   x - y  + 6 = 0

Mallika!   ur answers are wrong, sorry

spideyunlimited!

at one place u r finding the equation of line as y -x1 = m(x - y1)

so ur one eqn of diagonal is wrong.

(ok..... u corrected it :)

 Discussion Forums -> This Post 10 points    (2    in 2 votes )   [?]
a,b,c,d and p are real numbers, and

(a^2 + b^2 + c^2)p^2 -  2(ab+bc+cd)p + (b^2+c^2+d^2) = 0

=> p is a root of (a^2 + b^2 + c^2)x^2 -  2(ab+bc+cd)x + (b^2+c^2+d^2) = 0

since p is real      =>      D >= 0 for the given equation

=>    4 (ab+bc+cd)^2 - 4 (a^2 + b^2 + c^2)(b^2+c^2+d^2) >= 0

=>    - (ac - b^2) - (bd - c)^2 - (ad - bc)^2 >= 0      (after simplifying)

=>    (ac - b^2) + (bd - c)^2 + (ad - bc)^2  <=  0

since a,b,c,d are real numbers, all terms above will be no negative.

=>    (ac - b^2) + (bd - c)^2 + (ad - bc)^2  =  0

=>    (ac - b^2) = 0, (bd - c)^2 = 0, (ad - bc)^2 = 0

=>     b/a = c/b,  c/b = d/c, ad = bc      => a,b,c,d are in G.P.   Done!

 Discussion Forums -> This Post 15 points    (3    in 3 votes )   [?]
hsbhatt!

Using triangle inequality,

|bz + c| <=   |bz| + |c|
=>         -
|bz + c| >= - |bz| -  |c|         (multiply with -1 both sides)
=>
|z2| - |bz + c| >= |z2| - |bz| -  |c|      ....... (i)  (add |z2| both sides)

Again u
sing triangle inequality,

|z2+bz+c| >= |z2| - |bz + c|

=>      |z2+bz+c| >= |z2| - |bz| -  |c|     (using (i), if a>b, b>c then a>c)

when |z| = 1

1 >= 1 - |b| - |c|   which is possible only when   |b| = |c| = 0  =>      b = c = 0

Done!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
sboosy!

(2 power any prime number) -1 gives one group  of prime numbers. e.g. 3, 7, 31 etc.

However, there is another group of prime numbers too which can't be obtained by
(2 power any prime number) -1. e.g.  13, 19 etc.

But if u want to prove that there is no largest prime number by using
(2 power any prime number) -1 form, you are, sorry to say that, wrong.

As in that case largest prime number is
232,582,657-1 (discovered in sept, 06, as far as i remember)

that means 232,582,657-1 is the largest prime number known of (2 power any prime number) -1 type.

Correct me IF i m somewhere wrong. I may be wrong as this topic is very very controversial among mathematicians. Who knows u will find a bigger prime number than the existing one and will be the next Ramanujan :)

But Yes, there is no largest prime number and that is possible becoz they may not be of (2 power any prime number) -1 type.

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
sorry for duplicate post, i saw the ques late, by the time i was typing, he might have posted it. well, a simple one. sorry again
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
take log, say   y = log (2000)^2000 = 2000 log 2000 = 2000(log 1000 + log 2)
= 2000(3 log 10 + log 2) = 2000(3 + 0.3010) = 6602

now log (2000)^2000 = 6602 => number of digits must have been 6603

when u take log of abcde it is 4.xyz i.e. (no. of digits - 1).something

 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
ananth_patri!

If you talk about slope, then you are using derivative only.

please remember that derivative is nothing  but the rate of change, or slope for a line.

if someone says derivative of a function is negative, that means slope of tangent is negative there.

however ur way is little diff from actual one..... not a big deal...... please try things by first principle when needed.
 Discussion Forums -> This Post 2 points    (0    in 1 votes )   [?]
Hummm.....

It feels good to prove something in a primitive way.

ok, Definition of a decreasing function comes as,

if f(x1) > f(x2) <=> x1 < x2 for all x1, x2 in domain of f(x), then we call f(x) a decreasing function.

here let us take any general x1, x2 such that x1 < x2

Now,  x1 < x2  <=> -x1 > -x2       (multiply by -1 both sides)

<=> - 3x1  > -3x2          (multiply by 3 both sides)
<=> 2 - 3x1 > 2 - 3x2      (add 2 both sides)
<=> f(x1) > f(x2)

<=> f(x) is a decreasing function.   Done!

( here  <=> is the equivalent sign.)

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
Please write here only if any queries!
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
man111

Karthik is  right! please post the question properly!
 Discussion Forums -> This Post 25 points    (5    in 5 votes )   [?]
harshatr!

we have,

f(x-y) = f(x)f(y) - f(a-x) f(a+y),       a is a constant

put x=y=0 => f(0) = f(0) f(0) - f(a) f(a) => f(a)^2 =1-1 =0 => f(a) = 0 (using f(0) = 1)

put x = 0    => f(-y) = f(0) f(y) - f(a) f(a+y) = 1.f(y) - 0.f(a+y) = f(y)

=> f(-y) = f(y)    => f(x) is an even function

put x = y = a    => f(0) = f(a)f(a) - f(0) f(2a)    => 1 = 0 - f(2a)       (using f(0) = 1
and  f(a) = 0)
=> f(2a) = -1

Now, put x = 2a, y = x    => f(2a - x) = f(2a) f(x) - f(-a) f(a+x)

= -1 . f(x) - f(a) f(a+x)      (using f(2a) = -1
and f(x) is even)
= - f(x) - 0

= - f(x)

Thus          f(2a - x) = - f(x)  that is (c) is true     Done!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
hsbhatt!

cos a + cos b + cos c = 2cos (a+b)/2 cos (a-b)/2 + 1 - 2(sin c/2)^2

= 1 + 2cos (pi-c)/2 cos (a-b)/2 - 2sin c/2. sin (pi-(a+b))/2

= 1 + 2sin c/2 (cos (a-b)/2 - cos (a+b)/2)

= 1 + 2 sin c/2 (2sin a/2 sin b/2)

= 1 + 4 sin a/2 sin b/2 sin c/2

sin a/2 sin b/2 sin c/2 > 0 since a,b,c < pi

=>
cos a + cos b + cos c > 1    Done!

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
You must be knowing that area of triangle = (1/2) * base * height

=>      = 0.5 * a * p     ('a' is one of the sides and 'p' is the corresponding height)

Now I wrote   T = 0.5 ap

I denoted  by ' T '. got it?
 Discussion Forums -> This Post 5 points    (1    in 1 votes )   [?]
No tobeiitian!, term with y^-3 will have k^6, you are having k^3.

What's your name btw, if you don't want me to call you tobeiitian :)
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