**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.

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.

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

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

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.

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

**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)Handwritten picture solution are accepted only when they are easily understandable!!

## 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. – Neelesh Vij · 3 months, 2 weeks ago

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

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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– Prince Loomba · 3 months, 1 week ago

Post problem 16Log in to reply

– Aniket Sanghi · 3 months, 1 week ago

If u have one , then post na ! I am not getting what to postLog in to reply

– Aniket Sanghi · 3 months, 1 week ago

Ans 0 , RIGHT?Log in to reply

– Neelesh Vij · 3 months, 1 week ago

ya bro!Log in to reply

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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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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– Prakhar Bindal · 3 months, 1 week ago

Yeah neatly done bro! +1Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Yes you can post the next question. Its right AFLog in to reply

– Rex Holmes · 3 months, 2 weeks ago

the answer is B right?Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

He is gone. Tell complete answer then I will tell. Ik the answer but I cant postLog in to reply

– Rex Holmes · 3 months, 2 weeks ago

the answer is B cause cot^2 doesn't give you a perfect square or a perfect cube so it must be BLog in to reply

– Prince Loomba · 3 months, 2 weeks ago

WrongLog in to reply

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 – Prince Loomba · 3 months, 2 weeks ago

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

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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– Prakhar Bindal · 3 months, 1 week ago

Thats the perfect example of elegance!Log in to reply

– Aniket Sanghi · 3 months, 1 week ago

Yeah!! Thanks bro! :)Log in to reply

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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– Aniket Sanghi · 3 months, 2 weeks ago

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 ) .Log in to reply

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| – Prince Loomba · 3 months ago

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@Prince Loomba – Devang Agarwal · 3 months ago

Answer has to be in which form ?Log in to reply

– Prince Loomba · 3 months ago

Post next problem anybody.Log in to reply

– Prince Loomba · 3 months ago

(a/b)×pi+(c /d)arccot3Log in to reply

– Archit Agrawal · 3 months ago

Allen test paper question.Log in to reply

– Prince Loomba · 3 months ago

Ik that BroLog in to reply

– Prince Loomba · 3 months ago

Bro I know hahaLog in to reply

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. – Prince Loomba · 3 months, 1 week ago

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– Aniket Sanghi · 3 months ago

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

– Prince Loomba · 3 months ago

Yes 4000 is right 4.0×10^3. I have given such example for 4010 broLog in to reply

PROBLEM 18Evaluate

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

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– Prince Loomba · 3 months, 1 week ago

I dont have any question so you can post next!Log in to reply

– Prakhar Bindal · 3 months, 1 week ago

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

– Prakhar Bindal · 3 months, 1 week ago

Are you sure that this series converges?Log in to reply

– Neelesh Vij · 3 months, 1 week ago

Sorry for inconvenience, I have edited the questionLog in to reply

problem 17Find 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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\( \displaystyle (1+x)^n = \sum_{r=0}^{n} \binom{n}{r} \)

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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– Aniket Sanghi · 3 months, 1 week ago

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

– Prince Loomba · 3 months, 1 week ago

I saw the method and not the answer and thats why I thought it to be right!Log in to reply

– Neelesh Vij · 3 months, 1 week ago

Thanks for that I have edited itLog in to reply

– Prince Loomba · 3 months, 1 week ago

Post next questionLog in to reply

problem 16If 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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\(sec^{-1}(1) = 0 \)

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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– Prince Loomba · 3 months, 1 week ago

Right post nextLog in to reply

Note:Everybody please note that new topics have been added – Prince Loomba · 3 months, 1 week agoLog 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. – Prince Loomba · 3 months, 2 weeks ago

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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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– Prince Loomba · 3 months, 2 weeks ago

Exact answer....Log in to reply

– Rex Holmes · 3 months, 2 weeks ago

oh then the answer must be p+1Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

No and p+1 is out of the given rangeLog in to reply

– Rex Holmes · 3 months, 2 weeks ago

is the root 1Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

No it is in terms of pLog in to reply

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 type – Prince Loomba · 3 months, 2 weeks agoLog in to reply

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

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

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

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

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

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

\(\boxed{k}=4\)

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

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– Aniket Sanghi · 3 months, 2 weeks ago

K = 4 ; I will edit sol after coming back from my classLog in to reply

– Prince Loomba · 3 months, 2 weeks ago

4 is right. Post solution and then the next question after coming.Log in to reply

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. – Harsh Shrivastava · 3 months, 2 weeks agoLog in to reply

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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– Harsh Shrivastava · 3 months, 2 weeks ago

Justify your assumption why max occurs when all terms are equal?Log in to reply

– Harsh Shrivastava · 3 months, 2 weeks ago

Your solution is wrong though answer is correct.Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Whaat is wrong. Differentiate. 1 more thing, the min value is 2 times of that I have written. This impies that n is not integerLog in to reply

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 type – Prince Loomba · 3 months, 2 weeks agoLog in to reply

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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– Rex Holmes · 3 months, 2 weeks ago

Is the answer Infinity?Log in to reply

problem 9– Prince Loomba · 3 months, 2 weeks agoLog 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 type – Prince Loomba · 3 months, 2 weeks agoLog in to reply

SOD(55)=SOD(10)=1. – Aditya Chauhan · 3 months, 2 weeks ago

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– Prince Loomba · 3 months, 2 weeks ago

Right. Are you posting next one or shall I?Log in to reply

– Aditya Chauhan · 3 months, 2 weeks ago

I don't have any good question to post at the moment so you may post the next one.Log in to reply

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

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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– Prakhar Bindal · 3 months, 2 weeks ago

Yeah right!Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Posted the next oneLog in to reply

Should we also start a mechanics contest again?

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

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– Chinmay Sangawadekar · 3 months, 2 weeks ago

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 ,,, ;)Log in to reply

– Harsh Shrivastava · 3 months, 2 weeks ago

After mechanics, we will start NT :)Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Bro try Problem 7Log in to reply

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

Hey guysLog in to reply

– Prince Loomba · 3 months, 2 weeks ago

3 contests are running now. Later. Now: geometry,algebra,jee algebraLog in to reply

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. – Harsh Shrivastava · 3 months, 2 weeks ago

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– Harsh Shrivastava · 3 months, 2 weeks ago

Correct post the next question.Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Aftr the calculations, answer is 1 and the 3 pairs are 2,2 , 1,2 , 2,1Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

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

– Harsh Shrivastava · 3 months, 2 weeks ago

(2,2).Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Try your luck again. I am rightLog in to reply

– Harsh Shrivastava · 3 months, 2 weeks ago

I have edited the question.Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

2^3+2^3=16 and 4^3=64. WrongLog in to reply

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

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– Prince Loomba · 3 months, 2 weeks ago

Then tell me one of the pairs so that I get my answer is incorrectLog in to reply

– Harsh Shrivastava · 3 months, 2 weeks ago

Please see typo has been corrected.Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Now its 0Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Its infinite. x=0 y=any nonnegative integerLog in to reply

– Harsh Shrivastava · 3 months, 2 weeks ago

Sorry correction done.Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Post other. Its out of the topics listLog in to reply

– Harsh Shrivastava · 3 months, 2 weeks ago

If I tell the topic. the problem will become very easyLog in to reply

The problem is alright.

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

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– Prince Loomba · 3 months, 2 weeks ago

Its infiniteLog in to reply

– Prince Loomba · 3 months, 2 weeks ago

See the first line of the noteLog in to reply

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. – Rajdeep Dhingra · 3 months, 2 weeks agoLog in to reply

– Prince Loomba · 3 months, 2 weeks ago

Decide who will post the next oneLog in to reply

– Harsh Shrivastava · 3 months, 2 weeks ago

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

– Prince Loomba · 3 months, 2 weeks ago

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 correctLog in to reply

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

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Thus k = 47134

SOD(k) = 1 – Harsh Shrivastava · 3 months, 2 weeks ago

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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 correct – Prince Loomba · 3 months, 2 weeks agoLog in to reply

– Kaustubh Miglani · 3 months, 2 weeks ago

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 integerLog in to reply

– Prince Loomba · 3 months, 2 weeks ago

Ok you can post next questionLog in to reply

– Kaustubh Miglani · 3 months, 2 weeks ago

Sorry I am a little busy.So I cant.Please u do postLog in to reply

– Prince Loomba · 3 months, 2 weeks ago

No I wont. Its your turn.Log in to reply

– Rajdeep Dhingra · 3 months, 2 weeks ago

C)Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

Please post solution and next questionLog in to reply

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 type – Prince Loomba · 3 months, 2 weeks agoLog in to reply

– Prince Loomba · 3 months, 2 weeks ago

Minimum -2 and maximum \(3\sqrt 3/2\)Log in to reply

– Rajdeep Dhingra · 3 months, 2 weeks ago

Please post solution also.Log in to reply

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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– Rajdeep Dhingra · 3 months, 2 weeks ago

It's more than 24 hours. Post the solution and the next problem.Log in to reply

– Archit Agrawal · 3 months, 2 weeks ago

Is a=2 and b=2.Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

No wrongLog in to reply

– Kaustubh Miglani · 3 months, 2 weeks ago

Is b=4?Log in to reply

– Prince Loomba · 3 months, 2 weeks ago

No wrong b is not 4Log in to reply

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 type – Prince Loomba · 3 months, 2 weeks agoLog in to reply

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 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. – Rajdeep Dhingra · 3 months, 2 weeks ago

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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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– Rajdeep Dhingra · 3 months, 2 weeks ago

You may post the next problem.Log in to reply