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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)
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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
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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
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Greatdreamss!
good attempt but complicated! however keep it up!
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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(A  B)/P(B) = P(A)P(B/A) / (P(B A) + P(B A')) = 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!
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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 :)
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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!
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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 using 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!
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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.
Please see internet to confirm.
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.
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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
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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
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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.
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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.)
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Please write here only if any queries!
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 man111
Karthik is right! please post the question properly!
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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!
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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!
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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?
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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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Vriti Edustore
