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\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 years, 2 months ago

No vote yet
1 vote

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Problem 1

If a,b,c \in R - {0}, such that both roots of the equation ax2+bx+c=0ax^2 + bx + c = 0 lies in (2,3), then both roots of the equation a(x+1)2+b(x21)+c(x1)2=0a{(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 years, 2 months ago

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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.

Prince Loomba - 3 years, 2 months ago

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You may post the next problem.

Rajdeep Dhingra - 3 years, 2 months ago

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For a function g(x),how can one conclude that roots lie between a and b just because g(a)*g(b) > 0 ? Couldn't the graph between a and b be both above/below the x-axis without intersecting and the condition would be still valid but no roots would exist between a and b?

Jacob Sony - 6 months, 3 weeks ago

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Problem 5

Consider a polynomial P(x) = x223569x+kx^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 years, 2 months ago

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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 years, 2 months ago

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Damn I edited my comment.

Who will post the next question?

Harsh Shrivastava - 3 years, 2 months ago

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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

Harsh Shrivastava - 3 years, 2 months ago

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Decide who will post the next one

Prince Loomba - 3 years, 2 months ago

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

Harsh Shrivastava - 3 years, 2 months 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 years, 2 months ago

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PROBLEM 7

sin(x)+sin2(x)=1\sin(x)+\sin^{2}(x)=1 and acos12(x)+bcos10(x)+ccos8(x)+dcos6(x)2=0a\cos^{12}(x)+b\cos^{10}(x)+c\cos^{8}(x)+d\cos^{6}(x)-2=0

Find the value of a+c+d2\frac {a+c+d}{2}

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

Type: Integer type

Prince Loomba - 3 years, 2 months ago

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SOLUTION 2 :

sinx=1(sinx)2=(cosx)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)32=0 a (sinx)^6 + b (sinx)^5 + c (sinx)^4 + d (sinx)^3 - 2 = 0

Now , take cube of equation sinx+(sinx)2=1 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

(sinx)6+3(sinx)5+3(sinx)4+(sinx)31=0 (sinx)^6 + 3 (sinx)^5 + 3 (sinx)^4 + (sinx)^3 - 1 = 0

2(sinx)6+6(sinx)5+6(sinx)4+2(sinx)32=0 2 (sinx)^6 + 6 (sinx)^5 + 6 (sinx)^4 + 2 (sinx)^3 - 2 = 0

Aniket Sanghi - 3 years, 2 months ago

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Thats the perfect example of elegance!

Prakhar Bindal - 3 years, 1 month ago

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@Prakhar Bindal Yeah!! Thanks bro! :)

Aniket Sanghi - 3 years, 1 month ago

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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

Prakhar Bindal - 3 years, 2 months ago

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

Aniket Sanghi - 3 years, 2 months 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 years, 2 months ago

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Answer is 2

sinx/cos3x=1/2(2sinxcosx/cosxcos3x)=1/2(sin2x/cosxcos3x)=1/2(tan3xtanx)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(tan9xtan3x)sin3x/cos9x=1/2*(tan9x-tan3x)

sin9x/cos27x=1/2(tan27xtan9x)sin9x/cos27x=1/2*(tan27x-tan9x)

Adding these 3 equations, we get P=1/2KP=1/2*K or K/P=2K/P=2

Prince Loomba - 3 years, 2 months ago

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Yeah right!

Prakhar Bindal - 3 years, 2 months ago

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@Prakhar Bindal Posted the next one

Prince Loomba - 3 years, 2 months 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 years, 2 months ago

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ff can be written as x=188tan(x).tan(x+1) \displaystyle \sum_{x=1}^{88} \tan(x) . \tan(x+1)

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

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

Summing this we get f=cot2(1)89f = \cot^2 (1) -89

So g=89g = 89

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

neelesh vij - 3 years, 2 months ago

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Yes you can post the next question. Its right AF

Prince Loomba - 3 years, 2 months ago

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Yeah neatly done bro! +1

Prakhar Bindal - 3 years, 1 month ago

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the answer is B right?

Rex Holmes - 3 years, 2 months ago

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

Prince Loomba - 3 years, 2 months 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 years, 2 months ago

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@Rex Holmes Wrong

Prince Loomba - 3 years, 2 months ago

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PROBLEM 15

Find the value of summation:

r=1k(3)r1(3n2r1)\displaystyle \sum_{r=1}^{k} (-3)^{r-1} \binom{3n}{2r-1}

where k=3n2k = \frac{3n}{2} and nn is an even positive integer

This is an old JEE question.

neelesh vij - 3 years, 2 months ago

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(1+x)3n=3nC0+3nC1x0.5+........ {(1 + \sqrt{x})}^{3n} = 3nC0 + 3nC1 {x}^{0.5} + ........

(1x)3n=3nC03nC1x0.5+........ {(1 - \sqrt{x})}^{3n} = 3nC0 - 3nC1 {x}^{0.5} + ........

Subtract the equations and put x=3 x = -3

You will get

(1+3i)3n(13i)3n23i=Required \frac {{ (1 + \sqrt{3} i )}^{3n} - {( 1 - \sqrt{3} i )}^{3n} }{2\sqrt{3}i } = Required equation equation

23n23i(cosπn+isinπncosπn+sinπn)=0 \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 years, 1 month ago

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Post problem 16

Prince Loomba - 3 years, 1 month ago

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@Prince Loomba If u have one , then post na ! I am not getting what to post

Aniket Sanghi - 3 years, 1 month ago

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Ans 0 , RIGHT?

Aniket Sanghi - 3 years, 1 month ago

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ya bro!

neelesh vij - 3 years, 1 month ago

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Problem 2

Find the minimum value of

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

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

TYPE: Integer type

Prince Loomba - 3 years, 2 months ago

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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 years, 2 months ago

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Problem 3

In a triangle ABC, asin3A+sin3B+sin3Cb-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 years, 2 months ago

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Is a=2 and b=2.

Archit Agrawal - 3 years, 2 months ago

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No wrong

Prince Loomba - 3 years, 2 months ago

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

Rajdeep Dhingra - 3 years, 2 months ago

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Minimum -2 and maximum 33/23\sqrt 3/2

Prince Loomba - 3 years, 2 months ago

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Please post solution also.

Rajdeep Dhingra - 3 years, 2 months 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 years, 2 months ago

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PROBLEM 4

If x2+ax+bx^{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 years, 2 months ago

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C)

Rajdeep Dhingra - 3 years, 2 months ago

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Please post solution and next question

Prince Loomba - 3 years, 2 months ago

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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 years, 2 months ago

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Ok you can post next question

Prince Loomba - 3 years, 2 months ago

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@Prince Loomba Sorry I am a little busy.So I cant.Please u do post

Kaustubh Miglani - 3 years, 2 months ago

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@Kaustubh Miglani No I wont. Its your turn.

Prince Loomba - 3 years, 2 months ago

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Should we also start a mechanics contest again?

@Rajdeep Dhingra @Prince Loomba

Harsh Shrivastava - 3 years, 2 months ago

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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 ,,, ;)

A Former Brilliant Member - 3 years, 2 months ago

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After mechanics, we will start NT :)

Harsh Shrivastava - 3 years, 2 months ago

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@Harsh Shrivastava Bro try Problem 7

Prince Loomba - 3 years, 2 months ago

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

Prince Loomba - 3 years, 2 months ago

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

Prakhar Bindal - 3 years, 2 months ago

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Problem 9

Consider a rational function of the form of

a0x0+a1x1+.....+amxmb0x0+b1x1+.....+bnxn\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 xx tends to \infty, the function tends to \infty.

It is given that 0m,n100\leq m,n \leq 10

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

Find SOD(c)SOD (c)

Note SODSOD denotes repeated sum of digits for example

SOD(189)=SOD(1+8+9)=SOD(18)=SOD(1+8)=SOD(9)=9SOD(189)=SOD (1+8+9)=SOD (18)=SOD (1+8)=SOD (9)=9.

Type: Integer type

Prince Loomba - 3 years, 2 months ago

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The given function will tend to \infty when m>n . So we have to find ordered pairs (m,n) in [0,10][0,10] where m>n which is equal to (112)=55\binom{11}{2} = 55 .

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

Aditya Chauhan - 3 years, 2 months ago

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Right. Are you posting next one or shall I?

Prince Loomba - 3 years, 2 months 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 years, 2 months ago

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Problem 10

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

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

Type: Integer type

Prince Loomba - 3 years, 2 months ago

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Is the answer Infinity?

Rex Holmes - 3 years, 2 months ago

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No and SOD stands for sum of digits, refer problem 9

Prince Loomba - 3 years, 2 months ago

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

Therefore, the sum is effective only till ii where i<3i < 3 .

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

Harsh Shrivastava - 3 years, 2 months ago

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Problem 12 IF aa and bb are the roots of equation x+1=cx(1cx)x+1=c x (1-cx) and c1,c2c_{1},c_{2} be two values of cc, determined from the equation ab+ba=π2\frac {a}{b}+\frac {b}{a}=\pi -2.

If c12c22+c22c12+2=k(π+1π1)2\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 kk.

Type: Integer type

Prince Loomba - 3 years, 2 months ago

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

Aniket Sanghi - 3 years, 2 months ago

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

Prince Loomba - 3 years, 2 months ago

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c2x2x(c1)+1=0c^{2}x^{2}-x(c-1)+1=0

ab+ba=a2+b2ab=(c1)22\dfrac{a}{b}+\dfrac{b}{a} = \dfrac{a^{2}+b^{2}}{ab} = (c-1)^{2}-2

(c1)22=π2(c-1)^{2}-2=\pi-2

c22c+1π=0c^{2}-2c+1-\pi = 0

c12c22+c22c12+2=(c12+c22)2c12c22\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}}

c12+c22=42(1π)=2π+2c_{1}^{2}+c_{2}^{2} = 4 - 2(1-\pi) = 2\pi + 2

c12c22+c22c12+2=(2π+2)2(π1)2\dfrac{c_{1}^{2}}{c_{2}^{2}}+\dfrac{c_{2}^{2}}{c_{1}^{2}} +2 = \dfrac{(2\pi+2)^{2}}{(\pi-1)^{2}}

k=4\boxed{k}=4

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

Aditya Chauhan - 3 years, 2 months ago

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Problem 13

Let1p1-1\leq p \leq 1

Show that the equation 4x33xp=04x^{3}-3x-p=0 has 11 unique root in the interval [1/2,1][1/2,1] and identify it. This is JEE problem with exact wordings.

Prince Loomba - 3 years, 2 months ago

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is the root 1

Rex Holmes - 3 years, 2 months ago

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No it is in terms of p

Prince Loomba - 3 years, 2 months ago

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oh then the answer must be p+1

Rex Holmes - 3 years, 2 months ago

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No and p+1 is out of the given range

Prince Loomba - 3 years, 2 months ago

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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]

Prakhar Bindal - 3 years, 2 months ago

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Exact answer....

Prince Loomba - 3 years, 2 months ago

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Note:Everybody please note that new topics have been added

Prince Loomba - 3 years, 2 months ago

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problem 16

If x={32n/83^{2n}/8} where {} denotes fractional part

Find value of sec18xsec^{-1}8x

Prince Loomba - 3 years, 1 month ago

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x=32n8=9n8=(8+1)n8=(n0)8n+(n1)8n1+....+(nn1)8+(nn)8=18=18 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}

sec1(1)=0sec^{-1}(1) = 0

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

Aditya Chauhan - 3 years, 1 month ago

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Right post next

Prince Loomba - 3 years, 1 month ago

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problem 17

Find the value of r=0nnCrr+1\displaystyle \sum_{r=0}^{n}{\frac {^{n}C_{r}}{r+1}}

Prince Loomba - 3 years, 1 month ago

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we know that

(1+x)n=r=0n(nr) \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 01(1+x)ndx=2n+11n+1 \displaystyle \int_{0}^{1} (1+x)^n dx = \dfrac{2^{n+1} -1}{n+1}

neelesh vij - 3 years, 1 month ago

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Post next question

Prince Loomba - 3 years, 1 month ago

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Bro! It is 2n+11n+1 \boxed{\frac{ {2}^{n+1} - 1 }{ n + 1}}

Aniket Sanghi - 3 years, 1 month ago

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@Aniket Sanghi Thanks for that I have edited it

neelesh vij - 3 years, 1 month 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 years, 1 month ago

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PROBLEM 18

Evaluate

r=1n1(k=1r1k)×(1)r1(nr) \displaystyle \sum_{r=1}^{n-1} \left( \sum_{k=1}^r \dfrac 1k \right) \times (-1)^{r-1} \binom{n}{r}

neelesh vij - 3 years, 1 month ago

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I Am attaching photos of my solution

Prakhar Bindal - 3 years, 1 month ago

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Are you sure that this series converges?

Prakhar Bindal - 3 years, 1 month ago

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Sorry for inconvenience, I have edited the question

neelesh vij - 3 years, 1 month ago

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

Prakhar Bindal - 3 years, 1 month ago

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I dont have any question so you can post next!

Prince Loomba - 3 years, 1 month ago

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Problem 19

How many numbers between 11 and 1000010000 ,both inclusive contain different digits when represented in the scientific notation?

Example: 3450=3.45×1033450=3.45×10^{3} is right and 4010=4.01×1034010=4.01×10^{3} is right but 4001=4.001×1034001=4.001×10^{3} is wrong as it contain 2 zeroes.

Prince Loomba - 3 years, 1 month ago

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

Aniket Sanghi - 3 years, 1 month ago

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

Prince Loomba - 3 years, 1 month 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 years, 1 month ago

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Allen test paper question.

Archit Agrawal - 3 years, 1 month ago

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Bro I know haha

Prince Loomba - 3 years, 1 month ago

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Ik that Bro

Prince Loomba - 3 years, 1 month ago

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Answer has to be in which form ? @Prince Loomba

Devang Agarwal - 3 years, 1 month ago

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(a/b)×pi+(c /d)arccot3

Prince Loomba - 3 years, 1 month ago

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Post next problem anybody.

Prince Loomba - 3 years, 1 month ago

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Problem 6

Let number of pairs of positive integers that satisfy the equation x3+y3=(x+y)2 x^3 +y^3 = (x+y)^2 be nn.

Find the value of n/3n/3.

Type-: Integer Type.

Harsh Shrivastava - 3 years, 2 months ago

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See the first line of the note

Prince Loomba - 3 years, 2 months ago

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Post other. Its out of the topics list

Prince Loomba - 3 years, 2 months ago

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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.

Harsh Shrivastava - 3 years, 2 months ago

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@Harsh Shrivastava Its infinite

Prince Loomba - 3 years, 2 months ago

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

Harsh Shrivastava - 3 years, 2 months ago

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Its infinite. x=0 y=any nonnegative integer

Prince Loomba - 3 years, 2 months ago

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Sorry correction done.

Harsh Shrivastava - 3 years, 2 months ago

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Please see typo has been corrected.

Harsh Shrivastava - 3 years, 2 months ago

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Now its 0

Prince Loomba - 3 years, 2 months ago

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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

Prince Loomba - 3 years, 2 months ago

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You committed a common mistake.

Answer is wrong.

Harsh Shrivastava - 3 years, 2 months 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 years, 2 months ago

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(2,2).

Harsh Shrivastava - 3 years, 2 months ago

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@Harsh Shrivastava 2^3+2^3=16 and 4^3=64. Wrong

Prince Loomba - 3 years, 2 months 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 years, 2 months ago

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@Harsh Shrivastava Try your luck again. I am right

Prince Loomba - 3 years, 2 months ago

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@Prince Loomba I have edited the question.

Harsh Shrivastava - 3 years, 2 months ago

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

Prince Loomba - 3 years, 2 months ago

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Prince Loomba - 3 years, 2 months ago

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Correct post the next question.

Harsh Shrivastava - 3 years, 2 months ago

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Problem 11

Given that the maximum value of i=1n(1+sin2xi)(1+cos2xi)\displaystyle \prod_{i=1}^{n}(1+\sin ^2 x_{i})(1 + \cos ^2 x_{i} ) is the minimum value of 9102y10+2410y10\dfrac{9^{10}}{2}y^{10} + \dfrac{2}{4^{10}y^{10}} .

Find the value of n.

Type : Integer Type.

Harsh Shrivastava - 3 years, 2 months ago

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Answer is 5

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

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

From these 2 we can conclude n=5n=5

Prince Loomba - 3 years, 2 months ago

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Your solution is wrong though answer is correct.

Harsh Shrivastava - 3 years, 2 months 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 years, 2 months ago

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

Harsh Shrivastava - 3 years, 2 months ago

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@Harsh Shrivastava Actually it's just observation that a product is maximum when all terms are as close as possible, differentiation can also be used!

Prince Loomba - 6 months, 1 week ago

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