@Brooklyn said:x+y property3 by any chance????
the answer to the maxima question is 2.
finally we lost T20 match to NZ by 1 run...
@sampras said:calculation mistake ... its coming as ... 6(10-3rt5)/55 ... is it correct... ????
bhai 2 is the answer.
1/(sin^2 y + 3 sin y cos y + 5 cos^2 y)
= 1/ (sin^2 y + cons^2 y + 3sin y cos y + 4 cos^2 y)
= 1/ (1+3sin y cos y + 4 cos^2 y)
For this to be max. denominator should be minimum.
so we now have to find out min (1+3sin y cos y + 4 cos^2 y)
Now cos^2 y = (1+cos 2y)/2 and sin y cos y = (sin 2y)/2
on putting these
we get in the deno as -
3 + 2 cos 2y + 3/2 sin 2y
now take 2y as x
so we get 3+2cos x + 3/2 sinx
now on differentiating - we get
-2 sin x + 3/2 cos x = 0
on solving we get - tan x = 3/4
This is possible when -
sin x = 3/5 and cos x = 4/5
or sin x = -3/5 and cos x = -4/5
for minima we take nagative value and solve. we get deno = 1/2
so the max value of the question will be 2
hope ye samajh me aaye
kisi aur ko koi aur tarika aata hai to pls batao
1/(sin^2 y + 3 sin y cos y + 5 cos^2 y)
= 1/ (sin^2 y + cons^2 y + 3sin y cos y + 4 cos^2 y)
= 1/ (1+3sin y cos y + 4 cos^2 y)
For this to be max. denominator should be minimum.
so we now have to find out min (1+3sin y cos y + 4 cos^2 y)
Now cos^2 y = (1+cos 2y)/2 and sin y cos y = (sin 2y)/2
on putting these
we get in the deno as -
3 + 2 cos 2y + 3/2 sin 2y
now take 2y as x
so we get 3+2cos x + 3/2 sinx
now on differentiating - we get
-2 sin x + 3/2 cos x = 0
on solving we get - tan x = 3/4
This is possible when -
sin x = 3/5 and cos x = 4/5
or sin x = -3/5 and cos x = -4/5
for minima we take nagative value and solve. we get deno = 1/2
so the max value of the question will be 2
hope ye samajh me aaye
kisi aur ko koi aur tarika aata hai to pls batao
@rahulshaitan said:the answer to the maxima question is 2.
still not getting max value 2...bhai .... rechecked my method at post #2361, can't find error...
find out the remainder when 8^201 divided by 9. approach plz
@sampras said: finally we lost T20 match to NZ by 1 run...
koi na yr practice thi
..........practice makes a man perfect
..........practice makes a man perfect
@rahulshaitan
hmmmmm so we can't arrive at the ans to maxima qn without differentiation... your comments on my method where did I go wrong... if at all...
@JS-M said:koi na yr practice thi..........practice makes a man perfect
ha yaar lets hope for best... though m not a great admirer of the game ... but T20 is new charm... fatafat results...
@rahulshaitan : i used same logic!! bt didnt simply d eqn n got 3 so nythng wrong in dat??
look:
we hav to minimize denominator:
sin^2 + 3siny cos y + 5 cos^y
from now on
c= cos y
s = sin y
2sc + 3c^2 - 3s^2 -10cs
3c^2-3s^2-8cs
solving we get soln as 3, -1/3
ab iske if we put it in soln : we get 1
so max value 1
am i doin smthng wrong ??
@sampras said: @rahulshaitanhmmmmm so we can't arrive at the ans to maxima qn without differentiation... your comments on my method where did I go wrong... if at all...
bhai check kiya aapka method bhi but kuch galat to usme bhi nahi laga bus ek he dikkat lag rahi hai wo ye ki is it necessary ki square term should be zero?
coz square ke bahar bhi sin cos hai. agar waha sin cos n to shayad ye laga sakte the but as sinre outside braclso so we cannot apply that sqaure should be equal to 0 method.
But this is my take ho sakta hme galat hu. Someone who is good at this please comment.
coz square ke bahar bhi sin cos hai. agar waha sin cos n to shayad ye laga sakte the but as sinre outside braclso so we cannot apply that sqaure should be equal to 0 method.
But this is my take ho sakta hme galat hu. Someone who is good at this please comment.
@Brooklyn said: @rahulshaitan : i used same logic!! bt didnt simply d eqn n got 3 so nythng wrong in dat??look:we hav to minimize denominator:sin^2 + 3siny cos y + 5 cos^yfrom now onc= cos ys = sin y2sc + 3c^2 - 3s^2 -10cs3c^2-3s^2-8cssolving we get soln as 3, -1/3ab iske if we put it in soln : we get 1so max value 1am i doin smthng wrong ??
bhai ye kaha se aaya?
2sc + 3c^2 - 3s^2 -10cs
I guess ye sin^2 + 3siny cos y + 5 cos^y ke equal nahi hai. If i m wrong please elaborate.
2sc + 3c^2 - 3s^2 -10cs
I guess ye sin^2 + 3siny cos y + 5 cos^y ke equal nahi hai. If i m wrong please elaborate.
@rahulshaitan said:bhai check kiya aapka method bhi but kuch galat to usme bhi nahi laga bus ek he dikkat lag rahi hai wo ye ki is it necessary ki square term should be zero? coz square ke bahar bhi sin cos hai. agar waha sin cos n to shayad ye laga sakte the but as sinre outside braclso so we cannot apply that sqaure should be equal to 0 method.But this is my take ho sakta hme galat hu. Someone who is good at this please comment.
thanks for reviewing my sol patiently, bro... what i feel since all terms in the denominator are positive... so only possibility that the whole expn attains min value is to make sq term as zero... anyway waiting for someone for rescue....from differentiation.... naam se hi darr lagta hai...
@sandeepv882 said: find out the remainder when 8^201 divided by 9. approach plz
bhai just go through this for remainders -
PS - Sorry for this long post but ye lambi he thi.
Jab MBA ke liye prepare kar raha tha tab banaye the ye notes.
Hope it helps you guys.
Remainders
in most of the major competitive exams questions on remainders are frequently asked. This post will clear the basics about solving remainder problems.
To clear the basic concept about remainders consider the following example-
Consider two numbers a and b which when divided by 8 leave remainder 3 and 4 respectively. Now think what will be the remainder when each of the following is divided by 8 :
1). a + b
2). a - b
3). a x b
4). a2
The solution to these questions is as follows-
Ans 1). 'a' can be written in the form 8x + 3 and 'b' can be written in the form 8y + 4. Now a + b = (8x + 3) + (8y + 4) = 8(x + y) + 7.
On dividing this by 8 we get a remainder 7 (which is equal to 3 + 4).
Ans 2). Similarly a - b = (8x + 3) - (8y + 4) = 8(x - y) – 1 = 8(x – y) -1 + 8 – 8
= 8( x – y – 1) + 7 (which is basically 3 – 4 = -1 or 7)
REMEMBER: In remainder problems if the remainder comes out to be negative then we have to add the lowest multiple of the divisor so that we get a positive number less than the divisor. So, in the above problem though the remainder was -1 but to make it positive we added 8 and thus the remainder obtained is 7.
Ans 3). Now,
a x b = (8x + 3)(8y + 4) = 64xy + 32x + 24y + 12
= 64xy + 32x + 24y + 8 + 4
=8(8xy + 4x + 3y + 1) + 4
Thus, on dividing this by 8 we get the remainder of 4 (which is nothing but remainder obtained when 3 x 4 is divided by 8)
Ans 4). Now, a2 = (8x + 3)(8x + 3)
=64x2 + 24x + 24x + 9
=8(8x2 + 6x + 1) + 1
So on dividing by 8 we get a remainder 1 which is nothing but the remainder left after dividing 3 x 3 by 8 (i.e. 1)
From these examples it would certainly be clear that whatever is the operation that is performed on the divisors, the same operation has to be performed on the remainders of respective divisors to get the remainder.
SOLVED EXAMPLES
Example 1
What is the remainder when 41564 is divided by 3?
Answer: (41564)/3
=(4 x 4 x 4 x …..1564 times)/4
=(1 x 1 x 1 x….1564 times)/3
Hence the remainder is 1.
Example 2
What is the remainder when 2128 is divided by 5?
Answer: (2128)/5
=(464)/5
=(-1)64/5
=1/5
Hence we get a remainder 1.
Example 3
What is the remainder when the product 1998 × 1999 × 2000 is divided by 7?
Answer: the remainders when 1998, 1999, and 2000 are divided by 7 are 3, 4, and 5 respectively. Hence the final remainder is the remainder when the product 3 × 4 × 5 = 60 is divided by 7.
So, the reminder is 4.
Example 4
What is the remainder when 2 2004 is divided by 7?
Answer: 2 2004 is again a product (2 × 2 × 2... (2004 times)). Since 2 is a number less than 7 we try to convert the product into product of numbers higher than 7. Notice that 8 = 2 × 2 × 2. Therefore we convert the product in the following manner-
2 2004 = 8 668 = 8 × 8 × 8... (668 times).
The remainder when 8 is divided by 7 is 1.
Hence the remainder when 8 668 is divided by 7 is the remainder obtained when the product 1 × 1 × 1... is divided by 7
Hence the remainder is 1.
Example 5
What is the remainder when 360 is divided by 10?
Answer: now 360 can be written as 930 and when this is divided by 10 we get (-1)30. Hence the remainder obtained is 1.
Example 6
What is the remainder when 2525 is divided by 9?
Answer: Now, 25 25 = (18 + 7) 25 = (18 + 7)(18 + 7)...25 times = 18K + 7 25
Hence remainder when 25 25 is divided by 9 is the remainder when 7 25 is divided by 9.
Now 7 25 = 7 3 × 7 3 × 7 3 .. (8 times) × 7 = 343 × 343 × 343... (8 times) × 7.
The remainder when 343 is divided by 9 is 1 and the remainder when 7 is divided by 9 is 7.
Hence the remainder when 7 25 is divided by 9 is the remainder we obtain when the product 1 × 1 × 1... (8 times) × 7 is divided by 9. The remainder is 7 in this case. Hence the remainder when 25 25 is divided by 9 is 7.
IMPORTANT METHODS AND THEOREMS TO SOLVE REMAINDER PROBLEMS
A). REMAINDER THEOREM
Remainder theorem states that when f(x), a polynomial function in x is divided by (x – a), then the remainder is f(a).
Example 1
What is the remainder when 8x4 + 6x3 + 4x2 + 2x + 9 is divided by (x - 1)?
Answer: Here f(x) = 8x4 + 6x3 + 4x2 + 2x + 9
Now from remainder theorem we first equate (x – 1) to zero and then proceed.
So, (x – 1) = 0 or x = 1.
Now from remainder theorem,
Remainder is given by f(1) = 8 + 6 + 4 + 2 + 9
Hence the remainder is 29
Example 2
What is the remainder when x 3 + 2x 2 + 5x + 3 is divided by x + 1?
Answer: The remainder when the expression is divided by (x - (-1)) will be f(-1).
Remainder = (-1) 3 + 2(-1) 2 + 5(-1) + 3 = -1
Example 3
What is the remainder when 290 is divided by 15?
Answer: since the numerator is in the powers of 2, so our first step will be to write the denominator i.e. 15 in powers of 2.
So, we get 15 = 24 – 1
Now the numerator can be rewritten in powers of 2 as ( 24 )22 x 22.
Applying remainder theorem we get the remainder as 1 x 22 = 4.
Thus, 4 is the required remainder.
Example 4
If 2x 3 -3x 2 + 4x + c is divisible by x - 1, find the value of c.
Answer: Since the expression is divisible by x - 1, the remainder f(1) should be equal to zero.
Or 2 - 3 + 4 + c = 0, or c = -3.
B). Euler's theorem
If M and N are the two numbers coprime to each other(i.e. their HCF is 1) and N = paqbrc….
Then Remainder [(MØ(N))/N] = 1, where Ø(N) = N [1 – (1/p)] x [1 – (1/q)] x [1 – (1/r)] x …… and is known as Euler's Totient function.
Ø(N) is also the number of numbers less than N and co-prime to N.
COROLLARY
FERMAT'S LITTLE THEOREM
If n in the above Euler theorem is a prime number, then Ø(N) = N[1 – (1/N)] = N – 1. Therefore, if M and N are co-prime to each other and N is a prime number then the remainder is
Remainder [(M(N-1)/N] = 1.
SOLVED EXAMPLES
Example 1
What is the remainder when 1761 is divided by 77?
Answer: 17 and 77 are co-prime to each other, therefore we can apply Euler's theorem here.
Now, 77 = 7 x 11
Ø(77) = 77 x [1 – (1/7)] x [1 – (1/11)] = 60.
Therefore, Remainder [(1761)/77] = Remainder [(1760 x 17)/ 77] = 17.
Example 2
What is the remainder when 533 is divided by 17?
Answer:
Now, 5 and 17 are co-prime and 17 is prime so we can apply Fermat's little theorem.
Ø(17) = 16
So, Remainder [(533)/17] = Remainder [(516x2 x 5)/17]
= Remainder [(1 x 5)/17] = 5.
Hence the remainder is 5.
C). Wilson's Theorem
If p is a prime number, ( p – 1)! + 1 is divisible by p.
EXAMPLE 1
Find the remainder when 18! Is divided by 19.
Answer:
18! = (18! + 1) -1 = (18! + 1) + 18 - 19
Every term except 18 is divisible by 19 in the above expression. Hence the remainder = the remainder obtained when 18 is divided by 19 = 18.
So, the remainder is 16.
D). Fermat's Theorem
If p is a prime number and N is co-prime to p, then Np – N is divisible by p.
Example1
What is the remainder when 165 – 16 is divided by 5?
Answer: now, as 5 is prime and 16 and 5 are co-prime so on applying Fermat's Theorem that (165 – 16) is completely divisible by 5. So, the remainder is 0.
E). Chinese remainder theorem
If a number N = a x b, where a and b are co-prime to each other and M is a number such that Remainder [M/a] = r1 and Remainder[M/b] = r2 then Remainder[M/N] = ar2x + br1y, where ax + by = 1.
Example
What is the remainder when 3101 is divided by 77?
Answer:
Now, 77 = 7 x 11
By Fermat's little theorem, Remainder[36/7] = 1 and Remainder[310/11] = 1.
Remainder[3101/7] = Remainder[396 x 35/7] = Remainder[(36)16 x 35/7] = Remainder[1 x 35/ 7] = 5 = r1.
Remainder[3101/11] = Remainder[3100 x 3/11] = Remainder[(310)10 x 31/7] = Remainder[1 x 3/ 11] = 3 = r2.
Now we will find x and y such that 7x + 11y = 1. By observation we can find out, x = -3 and y = 2.
Now we can say that Remainder[3101/77] = 7 x 3 x -3 + 11 x 5 x 2 = 47.
F).When the divisor is 9, 99, 999, etc.
When the remainder is to be solved and the divisor is 9, 99, 999, etc then
Suppose,
Remainder[abcdefghij/99] is to be solved then
Here the number is 99(2 digits) so break the original number in groups of 2 starting from right and add them and then divide this sum by 99 and the remainder so obtained is equal to the remainder which would be obtained if the original number were to be divided by 99. Thus, we get –
Remainder[abcdefghij/99] = Remainder[(ij + gh + ef + cd + ab) /99]
REMEMBER:
1).If the number were to be divided by 999 then we would have grouped the digits in triplets and this concept can further be extended to 9999(grouping of 4 digits), 99999( grouping of 5 digits), and so on.
2). This concept can be extended to 3, 33, 333, 3333, etc.
G). If p is a prime number greater than 5 then-
Remainder[(p – 1) times any digit/p] = 0
Example
What is the remainder when 111111 is divided by 7?
Answer: 7 is a prime number and 111111 has 6 (i.e. 7 – 1) digits so the remainder is 0.
PS - Sorry for this long post but ye lambi he thi.
Jab MBA ke liye prepare kar raha tha tab banaye the ye notes.
Hope it helps you guys.
@sandeepv882 said: find out the remainder when 8^201 divided by 9. approach plz
8 rem
( Binomial Theorem)
@sandeepv882 said: find out the remainder when 8^201 divided by 9. approach plz
its -1 as ... (9-1)^201/9 will give remainder (-1)^201
@sampras said:thanks for reviewing my sol patiently, bro... what i feel since all terms in the denominator are positive... so only possibility that the whole expn attains min value is to make sq term as zero... anyway waiting for someone for rescue....from differentiation.... naam se hi darr lagta hai...
haha. Me bhi google kar raha hu ispar agar kuch milta hu to batata hu .....
bhai waise differentiation itni muskil nahi hai sikh lo it will take 1 hour at a max.
bhai waise differentiation itni muskil nahi hai sikh lo it will take 1 hour at a max.