JEE Algebra Contest

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

the answer is B right?

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.

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

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

Decide who will post the next one

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

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

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

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

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

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 $K/P=2$

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

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

Subtract the equations and put $x = -3$

You will get

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

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

Ans 0 , RIGHT?

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

Is a=2 and b=2.

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

Minimum -2 and maximum $3\sqrt 3/2$

C)

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

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 ,,, ;)

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

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

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.

Is the answer Infinity?

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

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

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

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

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

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

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

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

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

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

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

$\boxed{k}=4$

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

is the root 1

oh then the answer must be p+1

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]

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

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

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

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

I Am attaching photos of my solution

Are you sure that this series converges?

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

I dont have any question so you can post next!

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

Allen test paper question.

Answer has to be in which form ? @Prince Loomba

See the first line of the note

Post other. Its out of the topics list

Its infinite. x=0 y=any nonnegative integer

Please see typo has been corrected.

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

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

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 $n=5$