**TOPICS** for week 1+2.

Logarithms, Trigonometry(Ratios and Identies) , Quadratic Equations, Squence and series, Binomial Theorem

**Rules**

I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

Please make a substantial comment.

Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 24 hours. Then, you must post the solution and you have the right to post a new problem.

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## Comments

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TopNewestPROBLEM 15Find the value of summation:

\(\displaystyle \sum_{r=1}^{k} (-3)^{r-1} \binom{3n}{2r-1}\)

where \(k = \frac{3n}{2} \) and \(n\) is an even positive integer

This is an old JEE question.

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\( {(1 + \sqrt{x})}^{3n} = 3nC0 + 3nC1 {x}^{0.5} + ........ \)

\( {(1 - \sqrt{x})}^{3n} = 3nC0 - 3nC1 {x}^{0.5} + ........\)

Subtract the equations and put \( x = -3 \)

You will get

\( \frac {{ (1 + \sqrt{3} i )}^{3n} - {( 1 - \sqrt{3} i )}^{3n} }{2\sqrt{3}i } = Required \) \( equation \)

\( \boxed {\frac{{2}^{3n}}{2\sqrt{3}i } ( cos\pi n + i sin\pi n - cos\pi n + sin \pi n ) = 0 }\)

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Post problem 16

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Ans 0 , RIGHT?

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ya bro!

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PROBLEM 14

Define f= tan1tan2 + tan2tan3 ..............+tan88tan89

Define g = cot^2(1) -f

define h = log(g)/log(9) where all angles are measured in degrees

Then which of the following statements is true

A) g is prime number

B) g is composite number neither a perfect square nor a perfect cube

C) g is composite number and a perfect square

D) g is composite number and a perfect cube

E) h lies in (1,2)

F) h lies in (2,3)

G) h lies in (3,4)

H) h lies in (0,1)

One or more than one correct type

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\(f\) can be written as \( \displaystyle \sum_{x=1}^{88} \tan(x) . \tan(x+1) \)

As \( \displaystyle \tan( (x+1) -x) ) = \dfrac{ \tan(x+1) - \tan(x) }{1 + \tan(x) . \tan(x+1) } \)

Using this we get the value of \( \displaystyle \tan(x) . \tan(x+1) = \dfrac{ \tan(x+1) - \tan(x)}{\tan(1) } -1 \)

Summing this we get \(f = \cot^2 (1) -89\)

So \(g = 89\)

And \(\displaystyle h = \dfrac{\log(89)}{\log(9)} \)

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Yeah neatly done bro! +1

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Yes you can post the next question. Its right AF

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the answer is B right?

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He is gone. Tell complete answer then I will tell. Ik the answer but I cant post

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PROBLEM 8

Define

K = Tan(27x)-Tan(x)

and

P = sin(x)/cos(3x) +sin(3x)/cos(9x) + sin(9x)/cos(27x)

Assume That x lies in common domain of all the given functions.

Calculate the ratio K/P

Give 0 As your answer if you think that answer depends on x if you think answer is independent of x

And please solve it properly dont just put some value of x to get the ratio .

Type- Integer (0000 to 9999)

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Answer is 2

\(sinx/cos3x=1/2*(2 sinx cosx/cosx cos 3x)=1/2*(sin2x/cosx cos 3x)=1/2*(tan 3x-tanx)\)

Similarly \(sin3x/cos9x=1/2*(tan9x-tan3x)\)

\(sin9x/cos27x=1/2*(tan27x-tan9x)\)

Adding these 3 equations, we get \(P=1/2*K\) or \(K/P=2\)

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Yeah right!

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PROBLEM 7\(\sin(x)+\sin^{2}(x)=1\) and \(a\cos^{12}(x)+b\cos^{10}(x)+c\cos^{8}(x)+d\cos^{6}(x)-2=0\)

Find the value of \(\frac {a+c+d}{2}\)

NOTEa,b,c,d are non zero even integersType: Integer type

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SOLUTION 2 :

\( sinx = 1 - (sinx)^2 = (cosx)^2 \)

Now, putting this in the given eq gives

\( a (sinx)^6 + b (sinx)^5 + c (sinx)^4 + d (sinx)^3 - 2 = 0\)

Now , take cube of equation \( sinx + (sinx)^2 = 1 \) to get

\( (sinx)^6 + 3 (sinx)^5 + 3 (sinx)^4 + (sinx)^3 = 1 \)

\( (sinx)^6 + 3 (sinx)^5 + 3 (sinx)^4 + (sinx)^3 - 1 = 0\)

\( 2 (sinx)^6 + 6 (sinx)^5 + 6 (sinx)^4 + 2 (sinx)^3 - 2 = 0 \)

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Thats the perfect example of elegance!

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Answer to the problem is 5

Firstly Notice the fact that sinx = cos^2(x)

Putting in values we get

asin^6(x) +bsin^5(x)+csin^4(x)+dsin^3(x) = 2 1.....

Now sinx+sin^2(x) = 1

From this we will find values of successive powers of sinx which come to be

sin^2(x) = 1-sinx

sin^3(x) = 2sinx-1

sin^4(x) = 2-3sinx

sin^5(x) = 5sinx-3

sin^6(x) = 5-8sinx

Substituting these values in 1........

And then putting sinx = root 5 -1 /2

In 1..

We will get an expression in terms of a,b,c,d

Now a,b,c,d are integers , RHS is integer therefore irrational term on LHS should be Zero and Integral terms should be equal

Solving you will get

8a-5b+3c-2d = 0

and

18a-11b+7c-4d = 4

Eliminate b from both sides and solving you get a+c+d = 10

Actually the problem was flawed initially.

It took me and @Prince Loomba quite a lot of time to make the problem as it appears now.

Actually According to initial problem statement and answer that prince had only one quadruple was given as the answer. but then i proved that infinitely many quadruples exist. Then prince Told that a,b,c,d are integers . And then after quite a long discussion on slack the final agreement was done that a+c+d has a fixed value not a+b+c+d.

Although are many quadruples exist which satisfy that equation but value of a+c+d will be fixed for each quadruple

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Bro! You can solve this problem in much shorter way , just take cube of \( sinx + (sinx)^2 = 1 \) ( to get the eq you formed in sin by replacing cos^2 by sin ) .

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Problem 5Consider a polynomial P(x) = \(x^2 - 23569x + k\). It has 2 prime number roots. Find the value of k. (Give the answer as sum of digits like 57 = 5+7 = 12 = 1+2 = 3.)

Type:- Integer Type.Log in to reply

Answer is 1

Harsh is right but his answer is wrong

Since 23569 is an odd number,and we know that sum of 2 primes is always even if none of them is 2,therefore two roots of this equation are 2 and 23567.

Thus k = 47134

SOD(k) =19=1+9=10=1+0=1

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Damn I edited my comment.

Who will post the next question?

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Decide who will post the next one

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I have a good problem so can I post because I answered wrong but due to calculation mistake?

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Since 23569 is an odd number,and we know that sum of 2 primes is always even if none of them is 2,therefore two roots of this equation are 2 and 23567.

Thus k = 47134

SOD(k) = 1

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Problem 1If a,b,c \(\in\) R - {0}, such that both roots of the equation \(ax^2 + bx + c = 0\) lies in (2,3), then both roots of the equation \(a{(x+1)}^2 +b(x^2 -1) + c{(x-1)}^2 = 0\) always lies in

A) [1,4]

B)(2,3)

C)[1,3)

D)[1,2)

Type:- ONE OR MORE THAN ONE CORRECT.

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Let the first function be f (x). We know that f (2).f (3)>0 since the roots lie between (2,3) or (4a+2b+c) (9a+4b+c)>0

Let the second be g (x). We know that g (1)=4a and g (2)=9a+4b+c and g (3)=16a+8b+4c=4 (4a+2b+c)

From the above inequality, we figure out that g (3).g (2)>0. So the roots of equation always lie in (2,3) which is included in ABC.

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You may post the next problem.

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Problem 20A=(1/1) arccot(1/1)+(1/2) arccot (1/2)+(1/3) arccot (1/3) and B= (1) arccot (1)+(2)arccot(2)+(3)arccot(3)

Find the value of |B-A|

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Answer has to be in which form ? @Prince Loomba

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Post next problem anybody.

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(a/b)×pi+(c /d)arccot3

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Allen test paper question.

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Ik that Bro

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Bro I know haha

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Problem 19How many numbers between \(1\) and \(10000\) ,both inclusive contain different digits when represented in the scientific notation?

Example: \(3450=3.45×10^{3}\) is right and \(4010=4.01×10^{3}\) is right but \(4001=4.001×10^{3}\) is wrong as it contain 2 zeroes.

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1st make it clear , is number 4000 right? (Means can it be counted ?)

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Yes 4000 is right 4.0×10^3. I have given such example for 4010 bro

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PROBLEM 18Evaluate

\( \displaystyle \sum_{r=1}^{n-1} \left( \sum_{k=1}^r \dfrac 1k \right) \times (-1)^{r-1} \binom{n}{r} \)

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I Am attaching photos of my solution

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I dont have any question so you can post next!

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Bro initially sigma should be from 1 to n not n-1 !!

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Are you sure that this series converges?

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Sorry for inconvenience, I have edited the question

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problem 17Find the value of \(\displaystyle \sum_{r=0}^{n}{\frac {^{n}C_{r}}{r+1}}\)

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we know that

\( \displaystyle (1+x)^n = \sum_{r=0}^{n} \binom{n}{r} \)

Simply integrating it from 0 to 1 gives the required result

So answer is \( \displaystyle \int_{0}^{1} (1+x)^n dx = \dfrac{2^{n+1} -1}{n+1} \)

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Bro! It is \( \boxed{\frac{ {2}^{n+1} - 1 }{ n + 1}} \)

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Post next question

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problem 16If x={\(3^{2n}/8\)} where {} denotes fractional part

Find value of \(sec^{-1}8x\)

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\( x = \dfrac{3^{2n}}{8}= \dfrac{9^{n}}{8} = \dfrac{(8+1)^{n}}{8} = \dfrac{\binom{n}{0}8^{n}+\binom{n}{1}8^{n-1}+....+ \binom{n}{n-1}8+ \binom{n}{n}}{8} = {\dfrac{1}{8}}=\dfrac{1}{8}\)

\(sec^{-1}(1) = 0 \)

There are supposed to be { } around every fraction which i was not able to add.

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Right post next

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Note:Everybody please note that new topics have been addedLog in to reply

Problem 13Let\(-1\leq p \leq 1\)

Show that the equation \(4x^{3}-3x-p=0\) has \(1\) unique root in the interval \([1/2,1]\) and identify it. This is JEE problem with exact wordings.

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Looking at the range you should definitely think of sine and cosine functions as there range is also the same.

Then recall the trigonometric identity cos(3x) = 4cos^3(x)-3cos(x)

So we carefully substitute x = cos(y)

Now problem is over!.

The root will cos[arccos(p)/3]

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Exact answer....

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oh then the answer must be p+1

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No and p+1 is out of the given range

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is the root 1

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No it is in terms of p

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Problem 12IF \(a\) and \(b\) are the roots of equation \(x+1=c x (1-cx)\) and \(c_{1},c_{2}\) be two values of \(c\), determined from the equation \(\frac {a}{b}+\frac {b}{a}=\pi -2\).If \(\frac {c_{1}^{2}}{c_{2}^{2}}+\frac {c_{2}^{2}}{c_{1}^{2}}+2=k (\frac {\pi+1}{\pi-1})^{2}\), find the value of \(k\).

Type:Integer typeLog in to reply

\(c^{2}x^{2}-x(c-1)+1=0\)

\(\dfrac{a}{b}+\dfrac{b}{a} = \dfrac{a^{2}+b^{2}}{ab} = (c-1)^{2}-2 \)

\((c-1)^{2}-2=\pi-2\)

\(c^{2}-2c+1-\pi = 0\)

\(\dfrac{c_{1}^{2}}{c_{2}^{2}}+\dfrac{c_{2}^{2}}{c_{1}^{2}} +2 = \dfrac{(c_{1}^{2}+c_{2}^{2})^{2}}{c_1^{2}c_2^{2}}\)

\(c_{1}^{2}+c_{2}^{2} = 4 - 2(1-\pi) = 2\pi + 2\)

\(\dfrac{c_{1}^{2}}{c_{2}^{2}}+\dfrac{c_{2}^{2}}{c_{1}^{2}} +2 = \dfrac{(2\pi+2)^{2}}{(\pi-1)^{2}}\)

\(\boxed{k}=4\)

Someone else may post the next question. @Prince Loomba @Aniket Sanghi

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K = 4 ; I will edit sol after coming back from my class

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4 is right. Post solution and then the next question after coming.

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Problem 10Lets define a function named prince (x,n) such that prince (x,n)=[x]+[x/n]+...+[x/\(n^{\infty}\)]

Find the SOD (prince (10000,50)) without calculating each floor function value

Type:Integer typeLog in to reply

\(\lfloor \frac{a}{b}\rfloor = 0\) if b>a.

Therefore, the sum is effective only till \(i\) where \(i < 3 \).

Therefore answer = SOD(10000+200+4) = SOD(10204) = 7.

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Is the answer Infinity?

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No and SOD stands for sum of digits, refer

problem 9Log in to reply

Problem 9Consider a rational function of the form of

\(\frac {a_{0}x^{0}+a_{1}x^{1}+.....+a_{m}x^{m}}{b_{0}x^{0}+b_{1}x^{1}+.....+b_{n}x^{n}}\)

As \(x\) tends to \(\infty\), the function tends to \(\infty\).

It is given that \(0\leq m,n \leq 10\)

If the number of ordered pairs \((m,n)\) that satisfy the given conditions is \(c\),

Find \(SOD (c)\)

Note\(SOD\) denotes repeated sum of digits for example\(SOD(189)=SOD (1+8+9)=SOD (18)=SOD (1+8)=SOD (9)=9\).

Type: Integer typeLog in to reply

The given function will tend to \(\infty\) when m>n . So we have to find ordered pairs (m,n) in \([0,10]\) where m>n which is equal to \(\binom{11}{2} = 55 \).

SOD(55)=SOD(10)=1.

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Right. Are you posting next one or shall I?

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Should we also start a mechanics contest again?

@Rajdeep Dhingra @Prince Loomba

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As soon as this contest or any one of the other ends we will immediately start up with mech .... btw I also wanted to start up with Analytical and regular NT contest ,,, ;)

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After mechanics, we will start NT :)

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Hey guys @Harsh Shrivastava @Chinmay Sangawadekar @Rajdeep Dhingra the problem is corrected now it was a bit flawed initially

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3 contests are running now. Later. Now: geometry,algebra,jee algebra

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PROBLEM 4If \(x^{2}+ax+b\) is an integer for every integer x, then

A)a is always an integer but b need not be an integer

B)b is always an integer but a need not be an integer

C)a and b are always integers

D)none of these

Type: Single correctLog in to reply

Put x=0 Then we have b must be an integer Put x=1 Then we have 1+a+b is an innteger Since 1,b both are integers a is also integer

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Ok you can post next question

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C)

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Please post solution and next question

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Problem 3In a triangle ABC, \(-a\leq sin3A+sin3B+sin3C \leq b\). That is this can be minimum -a and maximum b.

Find the value of [a+2b]

Type: Integer typeLog in to reply

Minimum -2 and maximum \(3\sqrt 3/2\)

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Please post solution also.

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we have 2pi < 3A < 3pi

2pi/3< A <pi

since A + B + C = pi all of sin3A, sin3B, sin3C can’t be negative.

Let us take sin3A = – 1 or A = pi/2

sin3A = – 1, sin3B = – 1 and sin3C = 0 is possible So the minimum value is – 2.

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It's more than 24 hours. Post the solution and the next problem.

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Is a=2 and b=2.

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No wrong

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Is b=4?

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No wrong b is not 4

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Problem 2Find the minimum value of

|sinx+cosx+tanx+cotx+secx+cosecx|

If the answer is \(\theta\), find [\(\theta\)], where [.] Denotes floor function

TYPE: Integer typeLog in to reply

We make the careful substitution sinx+cosx = t

Converting all in terms of sinx and cosx and writing sinxcosx =(t^2-1)/2

and simplifying i got the expression as

y = t + 2/(t-1)

Just remember we need to find minimum value in (-root 2 , root 2) as range of sinx+cosx is that only

Differentiate this expression to get that function increases from (-infinite,1-root 2) and (1+root 2 , infinite) and decreases in the rest.

So for minimum value of modulus we check that value of t= 1-root 2 and -root 2 the smaller one will be the answer

On solving i got minimum value of modulus = 2root2 -1

And hence box theta = 1

I Dont have much time and good algebra problem so i give the right to @Prince Loomba to post the next

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Problem 11Given that the maximum value of \(\displaystyle \prod_{i=1}^{n}(1+\sin ^2 x_{i})(1 + \cos ^2 x_{i} )\) is the minimum value of \(\dfrac{9^{10}}{2}y^{10} + \dfrac{2}{4^{10}y^{10}} \).

Find the value of n.

Type :Integer Type.Log in to reply

Answer is 5

The product is max when all terms equal 3/2. Or we can write it as \((3/2)^{2n}\)

The min value of the second as calculate by AM-GM is \((3/2)^{10}\).

From these 2 we can conclude \(n=5\)

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Justify your assumption why max occurs when all terms are equal?

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Your solution is wrong though answer is correct.

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Problem 6Let number of pairs of positive integers that satisfy the equation \( x^3 +y^3 = (x+y)^2\) be \(n\).

Find the value of \(n/3\).

Type-: Integer Type.

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Correct post the next question.

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Aftr the calculations, answer is 1 and the 3 pairs are 2,2 , 1,2 , 2,1

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Now its 0, 3xy (x+y)=0 implies x=0 or y=0 or x=-y. Any of the case is not possible

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(2,2).

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I have mistyped the question.

I have edited the question one last time.

Sorry.

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You committed a common mistake.

Answer is wrong.

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Please see typo has been corrected.

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Now its 0

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Its infinite. x=0 y=any nonnegative integer

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Sorry correction done.

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Post other. Its out of the topics list

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If I tell the topic. the problem will become very easy

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Well I have read the note and I very well know what to post.

The problem is alright.

It is within topic.

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See the first line of the note

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