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JEE Algebra Contest

TOPICS for week 1+2.

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

Rules

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

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

  3. Please make a substantial comment.

  4. 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.

  5. If the one who solves the last problem does not post a new problem in 2 hours, the creator of the previous problem has the right to post another problem.

  6. It is NOT compulsory to post original problems. But make sure it has not been posted on Brilliant.

  7. If a diagram is involved in your problem please make sure it is drawn by a computer program.

  8. Format your solution in \(\LaTeX\), picture solution will be accepted but only picture will not be, i.e. you can use picture for diagram or something, but not for the complete solution. Also make sure your solution is detailed and make sure to proof all claims.

  9. Do not post problems on same topic frequently like 3 continuous inequality problem are not allowed. 2 are sufficient.

  10. For those who are on slack Please post in #general that new problem is up along with problem number and the link to contest (https://brilliant.org/discussions/thread/algebra-contest/?sort=new)

  11. Handwritten picture solution are accepted only when they are easily understandable!!

Note by Rajdeep Dhingra
3 months, 2 weeks ago

No vote yet
1 vote

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PROBLEM 15

Find 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. Neelesh Vij · 3 months, 2 weeks ago

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

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

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

You will get

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

\( \boxed {\frac{{2}^{3n}}{2\sqrt{3}i } ( cos\pi n + i sin\pi n - cos\pi n + sin \pi n ) = 0 }\) Aniket Sanghi · 3 months, 1 week ago

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@Aniket Sanghi Post problem 16 Prince Loomba · 3 months, 1 week ago

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@Prince Loomba If u have one , then post na ! I am not getting what to post Aniket Sanghi · 3 months, 1 week ago

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@Neelesh Vij Ans 0 , RIGHT? Aniket Sanghi · 3 months, 1 week ago

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@Aniket Sanghi ya bro! Neelesh Vij · 3 months, 1 week ago

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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 Prakhar Bindal · 3 months, 2 weeks ago

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@Prakhar Bindal \(f\) 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 \(\displaystyle h = \dfrac{\log(89)}{\log(9)} \) Neelesh Vij · 3 months, 2 weeks ago

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@Neelesh Vij Yeah neatly done bro! +1 Prakhar Bindal · 3 months, 1 week ago

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@Neelesh Vij Yes you can post the next question. Its right AF Prince Loomba · 3 months, 2 weeks ago

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@Prakhar Bindal the answer is B right? Rex Holmes · 3 months, 2 weeks ago

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@Rex Holmes He is gone. Tell complete answer then I will tell. Ik the answer but I cant post Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba the answer is B cause cot^2 doesn't give you a perfect square or a perfect cube so it must be B Rex Holmes · 3 months, 2 weeks ago

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@Rex Holmes Wrong Prince Loomba · 3 months, 2 weeks ago

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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 \(\frac {a+c+d}{2}\)

NOTE a,b,c,d are non zero even integers

Type: Integer type Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba 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

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

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

\( 2 (sinx)^6 + 6 (sinx)^5 + 6 (sinx)^4 + 2 (sinx)^3 - 2 = 0 \) Aniket Sanghi · 3 months, 2 weeks ago

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@Aniket Sanghi Thats the perfect example of elegance! Prakhar Bindal · 3 months, 1 week ago

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@Prakhar Bindal Yeah!! Thanks bro! :) Aniket Sanghi · 3 months, 1 week ago

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@Prince Loomba 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 Prakhar Bindal · 3 months, 2 weeks ago

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@Prakhar Bindal 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 ) . Aniket Sanghi · 3 months, 2 weeks ago

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Problem 20

A=(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| Prince Loomba · 3 months ago

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@Prince Loomba Answer has to be in which form ? @Prince Loomba Devang Agarwal · 3 months ago

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@Devang Agarwal Post next problem anybody. Prince Loomba · 3 months ago

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@Devang Agarwal (a/b)×pi+(c /d)arccot3 Prince Loomba · 3 months ago

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@Prince Loomba Allen test paper question. Archit Agrawal · 3 months ago

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@Archit Agrawal Ik that Bro Prince Loomba · 3 months ago

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@Archit Agrawal Bro I know haha Prince Loomba · 3 months ago

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Problem 19

How 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. Prince Loomba · 3 months, 1 week ago

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@Prince Loomba 1st make it clear , is number 4000 right? (Means can it be counted ?) Aniket Sanghi · 3 months ago

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@Aniket Sanghi Yes 4000 is right 4.0×10^3. I have given such example for 4010 bro Prince Loomba · 3 months ago

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PROBLEM 18

Evaluate

\( \displaystyle \sum_{r=1}^{n-1} \left( \sum_{k=1}^r \dfrac 1k \right) \times (-1)^{r-1} \binom{n}{r} \) Neelesh Vij · 3 months, 1 week ago

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@Neelesh Vij I Am attaching photos of my solution

Prakhar Bindal · 3 months, 1 week ago

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@Neelesh Vij I dont have any question so you can post next! Prince Loomba · 3 months, 1 week ago

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@Neelesh Vij Bro initially sigma should be from 1 to n not n-1 !! Prakhar Bindal · 3 months, 1 week ago

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@Neelesh Vij Are you sure that this series converges? Prakhar Bindal · 3 months, 1 week ago

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@Prakhar Bindal Sorry for inconvenience, I have edited the question Neelesh Vij · 3 months, 1 week ago

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problem 17

Find the value of \(\displaystyle \sum_{r=0}^{n}{\frac {^{n}C_{r}}{r+1}}\) Prince Loomba · 3 months, 1 week ago

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@Prince Loomba 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 \( \displaystyle \int_{0}^{1} (1+x)^n dx = \dfrac{2^{n+1} -1}{n+1} \) Neelesh Vij · 3 months, 1 week ago

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@Neelesh Vij Bro! It is \( \boxed{\frac{ {2}^{n+1} - 1 }{ n + 1}} \) Aniket Sanghi · 3 months, 1 week ago

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@Aniket Sanghi I saw the method and not the answer and thats why I thought it to be right! Prince Loomba · 3 months, 1 week ago

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@Aniket Sanghi Thanks for that I have edited it Neelesh Vij · 3 months, 1 week ago

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@Neelesh Vij Post next question Prince Loomba · 3 months, 1 week ago

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problem 16

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

Find value of \(sec^{-1}8x\) Prince Loomba · 3 months, 1 week ago

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

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

There are supposed to be { } around every fraction which i was not able to add. Aditya Chauhan · 3 months, 1 week ago

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@Aditya Chauhan Right post next Prince Loomba · 3 months, 1 week ago

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Note:Everybody please note that new topics have been added Prince Loomba · 3 months, 1 week ago

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Problem 13

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

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. Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba 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] Prakhar Bindal · 3 months, 2 weeks ago

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@Prakhar Bindal Exact answer.... Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba oh then the answer must be p+1 Rex Holmes · 3 months, 2 weeks ago

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@Rex Holmes No and p+1 is out of the given range Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba is the root 1 Rex Holmes · 3 months, 2 weeks ago

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@Rex Holmes No it is in terms of p Prince Loomba · 3 months, 2 weeks ago

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Problem 12 IF \(a\) 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 type Prince Loomba · 3 months, 2 weeks ago

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

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

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

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

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

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

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

\(\boxed{k}=4\)

Someone else may post the next question. @Prince Loomba @Aniket Sanghi Aditya Chauhan · 3 months, 2 weeks ago

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@Prince Loomba K = 4 ; I will edit sol after coming back from my class Aniket Sanghi · 3 months, 2 weeks ago

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@Aniket Sanghi 4 is right. Post solution and then the next question after coming. Prince Loomba · 3 months, 2 weeks ago

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Problem 11

Given 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. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava 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 \(n=5\) Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Justify your assumption why max occurs when all terms are equal? Harsh Shrivastava · 3 months, 2 weeks ago

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@Prince Loomba Your solution is wrong though answer is correct. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Whaat is wrong. Differentiate. 1 more thing, the min value is 2 times of that I have written. This impies that n is not integer Prince Loomba · 3 months, 2 weeks ago

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Problem 10

Lets 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 type Prince Loomba · 3 months, 2 weeks ago

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

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

Therefore answer = SOD(10000+200+4) = SOD(10204) = 7. Harsh Shrivastava · 3 months, 2 weeks ago

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@Prince Loomba Is the answer Infinity? Rex Holmes · 3 months, 2 weeks ago

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@Rex Holmes No and SOD stands for sum of digits, refer problem 9 Prince Loomba · 3 months, 2 weeks ago

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Problem 9

Consider 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 type Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba 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. Aditya Chauhan · 3 months, 2 weeks ago

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@Aditya Chauhan Right. Are you posting next one or shall I? Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba I don't have any good question to post at the moment so you may post the next one. Aditya Chauhan · 3 months, 2 weeks ago

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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) Prakhar Bindal · 3 months, 2 weeks ago

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@Prakhar Bindal 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 \(K/P=2\) Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Yeah right! Prakhar Bindal · 3 months, 2 weeks ago

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@Prakhar Bindal Posted the next one Prince Loomba · 3 months, 2 weeks ago

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Should we also start a mechanics contest again?

@Rajdeep Dhingra @Prince Loomba Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava 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 ,,, ;) Chinmay Sangawadekar · 3 months, 2 weeks ago

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@Chinmay Sangawadekar After mechanics, we will start NT :) Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Bro try Problem 7 Prince Loomba · 3 months, 2 weeks ago

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@Harsh Shrivastava Hey guys @Harsh Shrivastava @Chinmay Sangawadekar @Rajdeep Dhingra the problem is corrected now it was a bit flawed initially Prakhar Bindal · 3 months, 2 weeks ago

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@Harsh Shrivastava 3 contests are running now. Later. Now: geometry,algebra,jee algebra Prince Loomba · 3 months, 2 weeks ago

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Problem 6

Let 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. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava

Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Correct post the next question. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Aftr the calculations, answer is 1 and the 3 pairs are 2,2 , 1,2 , 2,1 Prince Loomba · 3 months, 2 weeks ago

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@Harsh Shrivastava Now its 0, 3xy (x+y)=0 implies x=0 or y=0 or x=-y. Any of the case is not possible Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba (2,2). Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Try your luck again. I am right Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba I have edited the question. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava 2^3+2^3=16 and 4^3=64. Wrong Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Bro I am really sorry.

I have mistyped the question.

I have edited the question one last time.

Sorry. Harsh Shrivastava · 3 months, 2 weeks ago

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@Prince Loomba You committed a common mistake.

Answer is wrong. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Then tell me one of the pairs so that I get my answer is incorrect Prince Loomba · 3 months, 2 weeks ago

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@Harsh Shrivastava Please see typo has been corrected. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Now its 0 Prince Loomba · 3 months, 2 weeks ago

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@Harsh Shrivastava Its infinite. x=0 y=any nonnegative integer Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Sorry correction done. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Post other. Its out of the topics list Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba If I tell the topic. the problem will become very easy Harsh Shrivastava · 3 months, 2 weeks ago

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

The problem is alright.

It is within topic. Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Its infinite Prince Loomba · 3 months, 2 weeks ago

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@Harsh Shrivastava See the first line of the note Prince Loomba · 3 months, 2 weeks ago

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Problem 5

Consider 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. Rajdeep Dhingra · 3 months, 2 weeks ago

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@Rajdeep Dhingra Decide who will post the next one Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba I have a good problem so can I post because I answered wrong but due to calculation mistake? Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava Yeah you can. I have posted so many. I was just writing same solution as you posted and then I got a chance to get this question correct Prince Loomba · 3 months, 2 weeks ago

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@Rajdeep Dhingra 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 Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Damn I edited my comment.

Who will post the next question? Harsh Shrivastava · 3 months, 2 weeks ago

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@Rajdeep Dhingra 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 Harsh Shrivastava · 3 months, 2 weeks ago

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PROBLEM 4

If \(x^{2}+ax+b\) 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 correct Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba 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 Kaustubh Miglani · 3 months, 2 weeks ago

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@Kaustubh Miglani Ok you can post next question Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Sorry I am a little busy.So I cant.Please u do post Kaustubh Miglani · 3 months, 2 weeks ago

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@Kaustubh Miglani No I wont. Its your turn. Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba C) Rajdeep Dhingra · 3 months, 2 weeks ago

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@Rajdeep Dhingra Please post solution and next question Prince Loomba · 3 months, 2 weeks ago

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Problem 3

In 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 type Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Minimum -2 and maximum \(3\sqrt 3/2\) Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Please post solution also. Rajdeep Dhingra · 3 months, 2 weeks ago

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@Rajdeep Dhingra Let y = sin3A + sin3B + sin3C for sin3A to be non positive

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. Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba It's more than 24 hours. Post the solution and the next problem. Rajdeep Dhingra · 3 months, 2 weeks ago

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@Prince Loomba Is a=2 and b=2. Archit Agrawal · 3 months, 2 weeks ago

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@Archit Agrawal No wrong Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba Is b=4? Kaustubh Miglani · 3 months, 2 weeks ago

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@Kaustubh Miglani No wrong b is not 4 Prince Loomba · 3 months, 2 weeks ago

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Problem 2

Find the minimum value of

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

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

TYPE: Integer type Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba 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 Prakhar Bindal · 3 months, 2 weeks ago

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Problem 1

If 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. Rajdeep Dhingra · 3 months, 2 weeks ago

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@Rajdeep Dhingra 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. Prince Loomba · 3 months, 2 weeks ago

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@Prince Loomba You may post the next problem. Rajdeep Dhingra · 3 months, 2 weeks ago

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