JEE-Advanced Maths Contest '16

Hello, guys!

This contest is the first JEE contest on Brilliant. The objective of this game is to get well prepared for JEE-Advanced 2016 preparation having fun as well. There is so much to learn from each other, and the contest is an approach to accomplish the same. Each Brilliant user is invited to participate in the contest under the following rules:

  1. I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
  2. You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.
  3. Please make a substantial comment.
  4. Make sure you know how to solve your problem before posting it. In case there is no one can answer it within 24 hours, then you must post the solution, and you have a right to post another problem.
  5. If the one who solves the last problem does not post his/her problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.
  6. The scope of questions is only what is covered under JEE-Advanced syllabus. If you are unaware of the syllabus cover, You can check out the information brochure.
  7. You can use tricks/short methods/specific cases to post the solution of a problem only if they are logically valid.
  8. In the case of any argument or disagreement among the solvers, the decision/judgement is passed to me.
  9. DO NOT ask the answer to the problem. Just post your detailed solution solution along with the answer. It will be highly helpful to gain confidence in your problem solving and rock in JEE.
  10. Proof problems are not allowed.
  11. You can post a problem only from Maths section.

    • Please write the detailed solutions to the problems.

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :

[Post your solution here]

PROBLEM xxx (number of problem) :

[Post your problem here]

Please try to post problems from all the spheres of JEE syllabus and share this note so that maximum users come to know about this contest and participate in it. (>‿◠)✌

Note by Sandeep Bhardwaj
3 years, 6 months ago

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Here is the first problem for the inauguration of the contest. It's an easy problem. All the best!

Problem 1:

In a sequence of independent trials, the probability of success in one trial is 14\frac 14. Find the probability that the second success takes place on or after the fourth trial.

Ritu Roy has solved this problem at the first place and provided the solution. Thanks!

Sandeep Bhardwaj - 3 years, 6 months ago

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Solution of problem 1

The required probability is

=13C.14.(34)2.14+14C.14.(34)3.14+15C.14.(34)4.14+...=9256.24=2732=_{ 1 }^{ 3 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 2 }.\frac { 1 }{ 4 } \quad +\quad _{ 1 }^{ 4 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 3 }.\frac { 1 }{ 4 } \quad +\quad _{ 1 }^{ 5 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 4 }.\frac { 1 }{ 4 } \quad +\quad ...\infty \\ =\frac { 9 }{ 256 } .24\\ =\boxed { \frac { 27 }{ 32 } }

Ritu Roy - 3 years, 6 months ago

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How did we get that 24? I used the reverse probability method 1 - 3C2.1/4.1/4.3/4 - 3C3.1/4.1/4.1/4 = 27/32

Shivam Mahajan - 3 years, 6 months ago

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Problem 10 :

In ΔABC\Delta \text{ABC}, AB=tan1(sin(2))\text{AB} = \tan^{-1}\left(\sin \left(\sqrt{2} \right) \right) and tan(A2)=ln(π)tan(B2)\tan \left(\dfrac{\text{A}}{2} \right) = \ln(\pi) \tan \left(\dfrac{\text{B}}{2} \right)
Then the vertex C\text{C} lies on

a) Ellipse

b) Parabola

c) Hyperbola

d) Straight Line

Ishan Singh - 3 years, 6 months ago

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@Aditya Kumar

I see that you've posted the next problem of contest i.e. Problem 11, but there is no solution of 10th problem by you. Can you clarify on this?

Sandeep Bhardwaj - 3 years, 6 months ago

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I had posted the solution! Some careless mod has deleted it! I can't believe it.

Aditya Kumar - 3 years, 6 months ago

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@Aditya Kumar Ooops, sorry for that! I don't know who did that.

Can you please post that again?

Sandeep Bhardwaj - 3 years, 6 months ago

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@Aditya Kumar @Aditya Kumar ,your solution was incorrect.

Saarthak Marathe - 3 years, 6 months ago

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Problem 3:

Let ff be a twice differentiable function, such that f(x)=f(x)f''(x)=-f(x) and f(x)=g(x)f'(x)=g(x) , h(x)=[f(x)]2+[g(x)]2h(x)=[f(x)]^2+[g(x)]^2. Given that h(5)=11h(5)=11, evaluate h(10)h(10) .

Vighnesh Shenoy has solved this problem at the first place and provided the solution. Thanks!

Nihar Mahajan - 3 years, 6 months ago

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Problem 51

If (a+ω)1+(b+ω)1+(c+ω)1+(d+ω)1=2ω1(a+\omega)^{-1}+(b+\omega)^{-1}+(c+\omega)^{-1}+(d+\omega)^{-1}=2\omega^{-1} and (a+ω)1+(b+ω)1+(c+ω)1+(d+ω)1=2ω1(a+\omega')^{-1}+(b+\omega')^{-1}+(c+\omega')^{-1}+(d+\omega')^{-1}=2\omega'^{-1} ; where ω\omega and ω\omega' are the complex cube roots of unity, then what is the value of (a+1)1+(b+1)1+(c+1)1+(d+1)1(a+1)^{-1}+(b+1)^{-1}+(c+1)^{-1}+(d+1)^{-1} ?

Anant Badal - 1 year, 11 months ago

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Problem 4:

If nN n \in N , evaluate the value of

(n0)+(n4)+(n8)+, \dbinom{n}{0} + \dbinom{n}{4} + \dbinom{n}{8} + \ldots ,

where (nr)=n!r!(nr)! \dbinom{n}{r} =\dfrac{n!}{r!(n-r)!} .

Akul Agrawal is the first person to solve this problem and provide the solution. Thanks!

A Former Brilliant Member - 3 years, 6 months ago

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2n2+1cosnπ4+2n4\frac { { 2 }^{ \frac { n }{ 2 } +1 }\cos { \frac { n\pi }{ 4 } } +{ 2 }^{ n } }{ 4 }

put i, -i,1, -1 in expansion of (1+x)^n and add

Akul Agrawal - 3 years, 6 months ago

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(1+x)n=r=0n(nr)xr \displaystyle (1+x)^n = \sum_{r=0}^{n} \dbinom{n}{r} x^r

1m+(1)m+im+(i)m=4,m=4k, 1^{m} + (-1)^{m} + i^{m} + (-i)^{m} = 4 , m = 4k, else it is equal to 0 0

Putting x=1,1,i,i \displaystyle x=1, -1, i, -i and adding, we have,

r=0(n4r)=(1+i)n+(1i)n+2n4\displaystyle \sum_{r=0}^{\infty} \dbinom{n}{4r} = \dfrac{(1+i)^n + (1-i)^n +2^n}{4}

Samuel Jones - 3 years, 6 months ago

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Problem 25 :
If p p is an odd prime number, then (2+5)p2p+1 \lfloor ( 2 + \sqrt{5})^{p} \rfloor - 2^{p+1} is always divisible by :

1)2p 1) 2p
2)3p 2) 3p
3)p+1 3) p+1
4)5p 4) 5p

Vighnesh Shenoy has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!

Sarvesh Nalawade - 3 years, 6 months ago

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(2+5)p+(25)p(2+5)p(2+5)p+(25)p+1 (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \le (2+\sqrt{5})^{p} \le (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} + 1
Now,
(2+5)p+(25)pI (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \in I
(2+5)p+(25)p+1I (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} + 1 \in I
(2+5)p=(2+5)p+(25)p \therefore \lfloor (2+\sqrt{5})^{p} \rfloor = (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p}
S(p)=(2+5)p2p+1=(2+5)p+(25)p2p+1 S(p) = \lfloor (2+\sqrt{5})^{p} \rfloor - 2^{p+1} = (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} - 2^{p+1}
S=2r=0p12(p2r)2p2r5r2p+1 \therefore S =2\displaystyle \sum_{r=0}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r} - 2^{p+1}
S=2p+1+2r=1p12(p2r)2p2r5r2p+1=2r=1p12(p2r)2p2r5r \therefore S = 2^{p+1} + \displaystyle 2\sum_{r=1}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r} - 2^{p+1} = \displaystyle 2\sum_{r=1}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r}
Clearly, S(p) S(p) is divisible by 2 2 and 5 5
S(p)=2r=1p12p2r(p12r1)2p2r5r S(p) = 2\displaystyle \sum_{r=1}^{\frac{p-1}{2}}\frac{p}{2r}\dbinom{p-1}{2r-1}2^{p-2r}5^{r}
Now since p is a prime number, gcd(r,p)=1,r[1,p1] gcd(r,p) = 1 , r \in \left[1,p-1\right]

Therefore, S(p) S(p) is divisible by p.
Since S(p) S(p) is divisible by 2,p,5 2 , p ,5 it is divisible by 2p,5p 2p, 5p

S(5)=1300 S(5) = 1300
S(5) S(5) is not divisible by 15=35 15 = 3\cdot5 and is also not divisible by 6=5+1 6 = 5 + 1

Thus,
S(p) S(p) is not always divisible by p+1 p + 1 and 3p 3p

The correct options are 1 1 and 4 4

A Former Brilliant Member - 3 years, 6 months ago

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Problem 36:

Let aa be a positive real number such that a3=6(a+1)a^3 = 6(a + 1) then, find the nature of the roots of x2+ax+a26=0x^2 + ax + a^2 - 6 = 0.

Both Mayank Chaturvedi and Vighnesh Shenoy have provided complete solution this problem. Thanks!

Saarthak Marathe - 3 years, 6 months ago

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A graphical approach.

We have to see the nature of discriminant

a2(4a224)or8a2Now,graphofa36a6=f(a)takestwoturnsata=2and2,whichshowsa(rootoffunction)ispositiveandatleast2rememberf(x)iscontinuous.ata=8thef(a)isnegative.Sorootoff(a)>8.Henceforequationa36a6=0,a>8;a2>8Wehave8a2<0.sotherootsarenonreal{ a }^{ 2 }-(4{ a }^{ 2 }-24)\\ or\quad 8-{ a }^{ 2 }\\ Now,\quad graph\quad of\quad { a }^{ 3 }-6a-6=f(a)\quad \\ takes\quad two\quad turns\quad at\quad a\quad =\sqrt { 2 } \quad and\quad -\sqrt { 2 } ,\quad \\ which\quad shows\quad a(root\quad of\quad function)\quad is\quad positive\quad and\quad at\quad least\quad \sqrt { 2 } \\remember\quad f(x)\quad is\quad continuous.\\ at\quad a=\sqrt { 8 } \quad the\quad f(a)\quad is\quad negative.\\ So\quad root\quad of\quad f(a)>\sqrt { 8 } .\quad \\ Hence\quad for\quad equation\quad { a }^{ 3 }-6a-6=0,\\ a>\sqrt { 8 } ;{ a }^{ 2 }>8\\ We\quad have\quad 8-{ a }^{ 2 }<0.\quad so\\the \quad roots\quad are\quad non\quad real

Mayank Chaturvedi - 3 years, 6 months ago

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Δ=(a)24(a26)=243a2=3(8a2) \Delta = (a)^{2} -4(a^{2}-6) = 24-3a^{2} = 3(8-a^{2})
Consider,
f(a)=a36a6 f(a) = a^{3} -6a - 6

f(a)=3a26=3(a22) f'(a) = 3a^{2} - 6 = 3(a^{2}-2)
For 0a2 0 \le a \le \sqrt{2} , f(a) is decreasing, increasing for a2 a \ge \sqrt{2}
f(0)=6<0 f(0) = - 6 < 0
f(2)=22626<0 f(\sqrt{2}) = 2\sqrt{2} -6\sqrt{2} - 6 < 0
f(8)=88686=286<0 f(\sqrt{8}) = 8\sqrt{8}-6\sqrt{8} - 6 = 2\sqrt{8} - 6 < 0

Thus, the positive root of f(a) f(a) is 8 \ge \sqrt{8}
a8 a \ge \sqrt{8}

Δ<0 \Delta < 0
Thus, the roots are not real.

A Former Brilliant Member - 3 years, 6 months ago

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Just completed solving the question and here you post the solution! But I think proof questions are not allowed, isn't it?

Miraj Shah - 3 years, 6 months ago

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@Miraj Shah Oh yes.

Mayank Chaturvedi - 3 years, 6 months ago

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@Miraj Shah I am changing the wording of the question ,as suggested by @Vighnesh Shenoy. @Vighnesh Shenoy and @Mayank Chaturvedi,please make changes in your solutions accordingly.

Saarthak Marathe - 3 years, 6 months ago

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@Miraj Shah I guess he should have asked if the roots were real or complex.

A Former Brilliant Member - 3 years, 6 months ago

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@A Former Brilliant Member Mayank should post the new question.

A Former Brilliant Member - 3 years, 6 months ago

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@A Former Brilliant Member OK. @Mayank Chaturvedi,please post the next question

Saarthak Marathe - 3 years, 6 months ago

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A non-calculus approach.

a36a6=0{a}^{3}-6a-6=0

Let a=b+2/b a=b+2/b

Therefore,(b+2/b)36(b+2/b)6=0 { \left( b+2/b \right) }^{ 3 }-6(b+2/b)-6=0

Simplifying we get that, b66b3+8=0 {b}^{6}-6{b}^{3}+8=0

Therefore, b3=4 {b}^3=4 or 22

Substitute these values to get aa.

That time we see that only one real solution of aa occurs which is, a=21/3+22/3a={2}^{1/3}+{2}^{2/3}

We see that, a26=6/a {a}^{2}-6=6/a

Substituting this value in x2+ax+a26=0{x}^{2}+ax+{a}^{2}-6=0 we get that,

ax2+a2x+6=0a{x}^{2}+{a}^{2}x+6=0

Assume that the roots of these quadratic equation are real, Then using formula for roots for quadratic equations,

x=a±a2242 x=\dfrac { -a\pm \sqrt { { a }^{ 2 }-24 } }{ 2 }

Then substituting the acquired value of aa in this equation we get that xx is a complex number. Hence, our assumption was wrong.

Saarthak Marathe - 3 years, 6 months ago

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Wondering, how did you thought of that a=b+(2/b).

Mayank Chaturvedi - 3 years, 6 months ago

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@Mayank Chaturvedi One way of solving cubic equation of the type ax3+bx+c=0ax^3+bx+c=0 is to take x=d+y/dx=d+y/d and manipulate the value of yy to get a solvable 6th6th degree equation in dd

Saarthak Marathe - 3 years, 6 months ago

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@Saarthak Marathe By solvable, do you mean quadratic type equations with higher degrees?

Mayank Chaturvedi - 3 years, 6 months ago

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@Mayank Chaturvedi yes

Saarthak Marathe - 3 years, 6 months ago

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@Saarthak Marathe Great!!!

Mayank Chaturvedi - 3 years, 6 months ago

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Problem 8: Find the locus of centers of the circles which touch the two circles

x2+y2=a2{x}^2+{y}^2={a}^2 and x2+y2=4ax{x}^2+{y}^2=4ax

externally.

Akul Agrawal has solved this problem in the first place and posted the solution. Thanks!

Rohit Ner - 3 years, 6 months ago

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Difference of distance of center of touching circle from the two given circles is constant. Hence equation will be hyperbola with (0,0) and (2a,0) as focii. Vertex is (a,0). difference between distance of center from focii is 2a-a=a.

Hence locus is

(xa)2(a2)2y2(3a2)2=1 \frac { { \left( x-a \right) }^{ 2 } }{ { \left( \frac { a }{ 2 } \right) }^{ 2 } } -\frac { { y }^{ 2 } }{ { \left( \sqrt { 3 } \frac { a }{ 2 } \right) }^{ 2 } } =1

Akul Agrawal - 3 years, 6 months ago

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Problem 41

tanx.tan2x.tan3x dx\displaystyle \int tanx. tan2x. tan3x\text{ }\,dx

Saarthak Marathe has posted the solution to the problem.Thanks!

Miraj Shah - 3 years, 5 months ago

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tan3x=tan2x+tanx1tanx.tan2xtanx.tan2x.tan3x=tan3xtanxtan2x\displaystyle tan3x=\frac{tan2x+tanx}{1-tanx.tan2x}\\ tanx.tan2x.tan3x=tan3x-tanx-tan2x

Now integrating we get,

tanx.tan2x.tan3x.dx=lnsec3x3lnsec2x2lnsecx+constant\displaystyle \int tanx.tan2x.tan3x.dx=\frac{ln|sec3x|}{3}-\frac{ln|sec2x|}{2}-ln|secx|+constant

Saarthak Marathe - 3 years, 5 months ago

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Correct!Post the next question.

Miraj Shah - 3 years, 5 months ago

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Problem 2:

Which real values of kk exist such that the following system of equations has no solution? {(k+1)x+8y=4kkx+(k+3)y=3k1 \begin{cases} (k + 1)x + 8y = 4k \\ kx + (k + 3)y = 3k - 1 \end{cases}

Nihar Mahajan has solved this problem at the first place and has posted the solution. Thanks!

Ritu Roy - 3 years, 6 months ago

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For the equations to have no solution, they should be parallel lines, hence their x-y coefficients are in proportion but not in proportion when ratio of the constant terms is included.

k+1k=8k+34k3k1\dfrac{k+1}{k} = \dfrac{8}{k+3}\neq \dfrac{4k}{3k-1}

First solving the quadratic from k+1k=8k+3\dfrac{k+1}{k} = \dfrac{8}{k+3} we have k=1,3k=1,3. However when k=3k=3 are substituted in 4k3k1\dfrac{4k}{3k-1} the ratio does not remain consistent. Hence, k=3k=3 is the solution and number of solutions is 11.

(Feel free to correct me)

Nihar Mahajan - 3 years, 6 months ago

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Problem 5:

r=17tan2(rπ16)=?\displaystyle \sum _{ r=1 }^{ 7 }{ { \tan }^{ 2 }\left( \frac { r\pi }{ 16 } \right) } =?

Samuel Jones is the first person to solve this problem and provide the solution. Thanks!

Akul Agrawal - 3 years, 6 months ago

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The sum can be written as 8+r=17sec2(rπ16)\displaystyle -8 +\sum_{r=1}^{7} \sec^2 \left(\dfrac{r \pi}{16}\right)

=8+r=171cos2(rπ16)\displaystyle =-8 + \sum_{r=1}^{7} \dfrac{1}{ \cos^2 \left(\dfrac{r \pi}{16}\right)}

Consider the 15th \displaystyle 15^{\text{th}} Chebyshev Polynomials of the Second kind

U15(x)=215k=115(xcos(rπn))\displaystyle U_{15} (x) = 2^{15}\prod_{k=1}^{15} \left(x-\cos \left(\dfrac{r \pi}{n} \right)\right)

Taking logarithm and differentiating two times at x=0x=0, we have

S=8+43=35\displaystyle S = -8 + 43 =\boxed{35}

This way, we can generalize to nn terms also.

Samuel Jones - 3 years, 6 months ago

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Problem 6:

Evaluate the sum of the non - real roots of the equation

x4+x35x212x6=0.x^4+x^3-5x^2-12x-6=0.

Aditya Kumar solved the problem at the first place and provided the solution. Thanks!

Samuel Jones - 3 years, 6 months ago

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The equation can be written as (x22x2)(x2+3x+3)=0\left( { x }^{ 2 }-2x-2 \right) \left( { x }^{ 2 }+3x+3 \right) =0

The non- real solutions are: x=12(3i+3)x=\frac { -1 }{ 2 } \left( \sqrt { 3 } i+3 \right) and x=12(3i3)x=\frac { 1 }{ 2 } \left( \sqrt { 3 } i-3 \right) \quad

Hence the sum is: 3-3

Aditya Kumar - 3 years, 6 months ago

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Just for enlightening, I'm posting my solution (again).

Let x=t1\displaystyle x = t-1 the equation converts to

t43t32t23t+1=0\displaystyle t^4-3t^3-2t^2-3t+1 = 0

    t23t23t+1t2=0\displaystyle \implies t^2 - 3t -2 -\dfrac{3}{t} + \dfrac{1}{t^2} = 0

Now, let y=t+1t\displaystyle y = t + \dfrac{1}{t}, we have,

y23y4=0\displaystyle y^2 -3y - 4 = 0

    (y4)(y+1)=0\displaystyle \implies (y-4)(y+1) = 0

    (t24t+1)(t2t+1)=0(y=t+1t)\displaystyle \implies (t^2-4t+1)(t^2-t+1) = 0 \quad \left( \because y = t + \dfrac{1}{t} \right)

    (x22x2)(x2+3x+3)=0(t=x+1) \displaystyle \implies (x^2 - 2x -2)( x^2 + 3x + 3) = 0 \quad \left( \because t = x+1 \right)

Only the second factor gives non - real roots since it's discriminant is negative, therefore sum is 3\boxed{-3}

Samuel Jones - 3 years, 6 months ago

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The equation factors out as,

(x22x2)(x2+3x+3)=0 ({x}^{2}-2x-2)({x}^{2}+3x+3)=0

Now solve x22x2=0{x}^{2}-2x-2=0 and x2+3x+3=0{x}^{2}+3x+3=0 .

We see that the discriminant of x2+3x+3=0{x}^{2}+3x+3=0 is negative.

Therefore the sum of the non-real roots are -3

Saarthak Marathe - 3 years, 6 months ago

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Problem 7:

If (1+x)n=C0+C1x+C2x2+...+Cnxn{ \left( 1+x \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+...+{ C }_{ n }{ x }^{ n }.

Then find a closed of: C02+2C12+3C22+...+(n+1)Cn2{ C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }+...+\left( n+1 \right) { C }_{ n }^{ 2 }

Rohit Ner has solved this problem at the first place and posted the solution. Thanks!

Aditya Kumar - 3 years, 6 months ago

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x(1+x)n=C0x+C1x2+C2x3++Cnxn+1x{(1+x)}^{n}=C_0 x+C_1 {x}^2 +C_2 {x}^3+\cdots +C_n {x}^{n+1}

Differentiating both sides, (1+x)n1(nx+1+x)=C0+2C1x+3C2x2++(n+1)Cnxn{(1+x)}^{n-1}(nx+1+x)=C_0 +2C_1 x +3C_2 {x}^2 +\cdots + (n+1)C_n {x}^{n}

Also (1+1x)n=C0+C1x+C2x2++Cnxn{\left(1+\dfrac{1}{x}\right)}^{n}=C_0 + \dfrac{C_1}{x} + \dfrac{C_2}{{x}^2} +\cdots +\dfrac{C_n}{{x}^{n}}

The given series is sum of all constant terms obtained in the expansion of

(1+x)n1((n+1)x+1).(1+1x)n{(1+x)}^{n-1}((n+1)x+1).{\left(1+\dfrac{1}{x}\right)}^{n}

which is same as coefficient of xn{x}^{n} in (1+x)2n1{(1+x)}^{2n-1} i.e. (2n1n)\binom{2n-1}{n}

plus coefficient of xn1{x}^{n-1} in (n+1)(1+x)2n1(n+1){(1+x)}^{2n-1} i.e. (n+1)(2n1n1)(n+1)\binom{2n-1}{n-1}

So the closed form is (2n1n)+(n+1)(2n1n1)\binom{2n-1}{n}+(n+1)\binom{2n-1}{n-1}

Rohit Ner - 3 years, 6 months ago

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Problem 9:

(On behalf of Akul Agarwal)

If f(x)=x+01t(x+t)f(t).dt\displaystyle f(x)=x+\int _{ 0 }^{ 1 }{ t(x+t)f(t).dt } , then find the value of the definite integral,

01f(x)dx\int _{ 0 }^{ 1 }{ f(x)dx }

Ishan Singh has provided the complete solution to the problem. Thanks!

Saarthak Marathe - 3 years, 6 months ago

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f(x)=x(1+01tf(t) dt)+01t2f(t) dt \displaystyle f(x) = x \left(1+\int_{0}^{1} t f(t) \ \mathrm{d}t \right) +\int_{0}^{1} t^2 f(t) \ \mathrm{d}t

=Ax+B\displaystyle = Ax +B (say)

    f(x)=x(1+01t(At+B) dt)+01t2(At+B) dt\displaystyle \implies f(x) = x\left(1+\int_{0}^{1} t (At+B) \ \mathrm{d}t \right) + \int_{0}^{1} t^2 (At+B) \ \mathrm{d}t

=x(1+A3+B2)+A4+B3\displaystyle = x \left(1+\frac{A}{3}+\frac{B}{2} \right) + \frac{A}{4} + \frac{B}{3}

    Ax+B=x(1+A3+B2)+A4+B3\displaystyle \implies Ax+B = x \left(1+\frac{A}{3}+\frac{B}{2} \right) + \frac{A}{4} + \frac{B}{3}

Comparing coefficients, we have,

A=6523\displaystyle A = \frac{65}{23}

    01f(t)dt=A1=4223\displaystyle \implies \int_{0}^{1} f(t) dt = A - 1 = \boxed{\dfrac{42}{23}}

Ishan Singh - 3 years, 6 months ago

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Problem 11:

If the medians of a ΔABC\Delta ABC make angles α,β,γ\alpha ,\beta ,\gamma with each other, then find the value of:cot(α)+cot(β)+cot(γ)+cot(A)+cot(B)+cot(C)\cot { \left( \alpha \right) } +\cot { \left( \beta \right) } +\cot { \left( \gamma \right) } +\cot { \left( A \right) } +\cot { \left( B \right) } +\cot { \left( C \right) }

Consider α,β,γ\alpha ,\beta ,\gamma to be acute.

Saarthak Marathe is the first person to solve this problem and provide the solution. Thanks!

Aditya Kumar - 3 years, 6 months ago

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The answer is 00.

Saarthak Marathe - 3 years, 6 months ago

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Problem 12:

Evaluate:

limx0{sinxx}1{tanxx}\large \displaystyle \lim_{ x\rightarrow 0 }{ \left\{ \frac { sinx }{ x } \right\} }^{ \frac { 1 }{ \left\{ \frac { tanx }{ x } \right\} } }

where,{.} \{ . \} is fractional part function.

Rishabh Cool has provided a complete solution to this problem. Thanks!

Saarthak Marathe - 3 years, 6 months ago

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Problem 13:

f2(n+1)f(n)+2f(n+1)=f(n) f^{ 2 }\left( n+1 \right) f\left( n \right) +2f\left( n+1 \right) =f\left( n \right)

Given f(0)=1f\left( 0 \right) =1

Find limn2nf(n)\lim_{n\to\infty} { 2 }^{ n }f\left( n \right)

Ashu Dablo has solved this problem at the first place and has posted the solution. Thanks!

Akul Agrawal - 3 years, 6 months ago

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@Akul Agrawal Simplify the given equation to the form:

f(n) f\left(n \right) = 2(f(n+1))1f2(n+1) \frac {2(f\left (n+1 \right))} { 1 - f^{2}\left (n+1 \right)}

Now we say that for f(n)=tanθf\left(n \right) = \tan \theta , f(n+1)=tanθ2f\left(n+1 \right)=\tan \frac{\theta} {2}

As f(1) =1, from above relation, we can say that f(n)=tanπ2(n+1)f\left(n \right)=\tan \frac{\pi}{2^{(n+1)}} which for n >,>0-> \infty , -> 0

multiply divide by tanπ2(n+1)tan \frac{\pi}{2^{(n+1)}}

now for x>0x->0 tanxx=1\frac {\tan x}{x} =1

So answer is π4 \frac{\pi}{4}

Sorry for my latex!

Ashu Dablo - 3 years, 6 months ago

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The equation becomes inconsistent for n=0 . Please check .

Keshav Tiwari - 3 years, 6 months ago

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Problem 14 :

If the sum of the first n terms of an AP is cn2 cn^2, then find the sum of the cubes of these n terms.

Saakshi Singh has posted the solution first. Thanks!

Ashu Dablo - 3 years, 6 months ago

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Using sum formula of AP, we get 2cn=2a+(n1)d2cn= 2a + (n-1)d

2cna=a+(n1)d(2cna)3=(a+(n1)d)3. \Rightarrow 2cn - a = a + (n-1)d \\ \Rightarrow (2cn - a)^3 = ( a + (n-1)d)^3 .

We have to find n=1n(a+(n1)d)3=n=1n(2cna)3\displaystyle \sum_{n=1}^{ n} \left( a + (n-1)d \right)^3 =\displaystyle \sum_{n=1}^{ n}(2cn-a)^3 . Nos putting n=1n=1, we get a=ca=c.

So we have sum of the cubes of nn terms

=c3n=1n(2n1)3= c^3 \cdot \displaystyle \sum_{ n=1}^{ n} (2n-1)^3

=c3(n2(2n21)). = c^3 \left( n^2(2n^2 - 1) \right) .

Saakshi Singh - 3 years, 6 months ago

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I've cleaned up your solution for now. But it's very easy to type an equation using latex which you can learn from this note. @Saakshi Singh

Sandeep Bhardwaj - 3 years, 6 months ago

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This one is a fairly easy one:

Problem 16:

The maximum value of the function f(x)=2x315x2+36x48f(x) = 2x^{3} - 15 x^{2} + 36x -48

for {xx2+209x x | x^{2} + 20 \leq 9x}

Vignesh Shenoy has posted the correct solution. Thanks!

Ashu Dablo - 3 years, 6 months ago

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x29x+200 x^{2} - 9x + 20 \le 0

(x4)(x5)0x[4,5] (x-4)(x-5) \le 0 \rightarrow x \in \text{[}4,5\text{]}

f(x)=2x315x2+36x48 f(x) = 2x^{3} - 15x^{2} + 36x - 48

Differentiating,

f(x)=6x230x+36 f(x) = 6x^{2} - 30x + 36

f(x)=6(x25x+6) f'(x) = 6(x^{2}-5x+6)

f(x)=6(x2)(x3) f'(x) = 6(x-2)(x-3)

For x[4,5] x \in \text{[}4,5\text{]} , f(x)>0f(x) f'(x) > 0 \rightarrow f(x) is increasing.

f(x)max=f(5)=7 f(x)_{max} = f(5) = 7

A Former Brilliant Member - 3 years, 6 months ago

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Problem 18:

Consider the ellipse :
x225+y216=1 \dfrac{x^{2}}{25} + \dfrac{y^{2}}{16} = 1

Let L be the length of perpendicular drawn from the origin to any normal of the ellipse. Find the maximum value of L.

Bonus : Solve it without calculus and generalize for any ellipse

Aditya Kumar has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!

Prakhar Bindal - 3 years, 6 months ago

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Solution to Problem 18:

Consider the point (5cos(θ),4sin(θ))(5\cos(\theta),4\sin(\theta)).

The equation of normal would be: 5xsec(θ)4ycsc(θ)5x\sec(\theta)-4y\csc(\theta)

Hence, distance d=925sec2(θ)+16csc2(θ)d=\left| \frac { 9 }{ \sqrt { 25\sec ^{ 2 }{ \left( \theta \right) } +16\csc ^{ 2 }{ \left( \theta \right) } } } \right|

To maximise dd, we have to minimise the denominator. By minimising the denominator(did that by calculus), the denominator comes out to be 9.

Hence dmax=1\boxed{d_{max}=1}

Aditya Kumar - 3 years, 6 months ago

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Absolutely Correct.

Generalisation for any ellipse x^2 / a^2 + y^2/b^2 = 1

Maximum distance of normal from origin = Modulus (b-a)

Also you could have avoided calculus by using AM Gm in last step by writing sec and cosec in terms of tan and cot

Prakhar Bindal - 3 years, 6 months ago

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Problem 26 :
Let na=aaantimes ^{n}a = \underbrace{a^{a^{\cdot^{\cdot^{a}}}}}_{n \text{times}}
f(x)=r=1x(rr) f(x) = \displaystyle \sum_{r=1}^{x} (^{r}r)

Find the last digit of f(15) f(15)

Mayank Chaturvedi has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks!

A Former Brilliant Member - 3 years, 6 months ago

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The last digit is 8. Short methods are always welcome. I went to find out last digit of individual terms. Yes!. This looks lengthy but when we note down last terms, last digits of 1,2,4,5,6,10,11,14,15 are easy to find. For rest terms, dividing exponents by 4, using modular arithmetic would do.

Mayank Chaturvedi - 3 years, 6 months ago

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Correct.

A Former Brilliant Member - 3 years, 6 months ago

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Problem 39:

Solve

x3dydx=y3+y2y2x2\displaystyle {x}^{3}\frac{dy}{dx}={y}^{3}+{y}^{2}\sqrt{{y}^{2}-{x}^{2}}

Saarthak Marathe - 3 years, 6 months ago

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The answer is xy=c(yy2x2)xy=c(y-\sqrt{y^{2}-x^{2}}) The above equation is a homogeneous equation and can be written of the form dydx=yx3+yx2y2x2\frac{dy}{dx}={\frac{y}{x}}^{3}+{\frac{y}{x}}^{2}*\sqrt{y^{2}-x^{2}} Making the substitution y=vxy=vx and we see that dydx=(v+dvdxx)\dfrac{dy}{dx}=(v+\dfrac{dv}{dx}*x). On simplification we get dvv(v21+vv21)=dxx \frac{dv}{v(v^{2}-1+v\sqrt{v^{2}-1})}=\frac{dx}{x}. Taking v21\sqrt{v^{2}-1} and then Multiplying and dividing by vv21v-\sqrt{v^{2}-1} we have dvv21dvv=dxx\frac{dv}{\sqrt{v^{2}-1}} -\frac{dv}{v}=\frac{dx}{x}. On integrating both sides we have lnvv21v=lnx+c\ln{\frac{v-\sqrt{v^{2}-1}} {v}}=\ln{x}+c. This on substituting back vv gives us the above answer.

Vignesh S - 3 years, 6 months ago

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Problem 43 :

Find the remainder when 323232 32^{32^{32}} is divided by 7.
Mayank Chaturvedi has solved this problem with the correct solution.

A Former Brilliant Member - 3 years, 5 months ago

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323(mod7)331(mod7)32310+2(3)310+2(1)99(mod7)323232(9)310+2(1)814(mod7)32\equiv -3 (mod 7)\\ {3}^{3}\equiv -1 (mod 7)\\ {32}^{3*10+2}\equiv {(-3)}^{3*10+2} \equiv (1)*9 \equiv 9 (mod 7)\\ {32}^{32^{32}}\equiv {(9)}^{3*10+2} \equiv (1)*81 \equiv 4 (mod 7)\\

Hence the answer is 4\boxed 4

Saarthak Marathe - 3 years, 5 months ago

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The answer is right, however you used
323232(3232)329324(mod7) 32^{32^{32}} \equiv (32^{32})^{32} \equiv 9^{32} \equiv 4 \pmod{7}

A Former Brilliant Member - 3 years, 5 months ago

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@A Former Brilliant Member Yes i agree with you. Have a check @Saarthak Marathe the answer is a coincidence this time.

Let me try once.

Reference-euler's theorem

φ(7)=6andgcd(32,7)=13261mod(7)............me1Nowwefind3232mod63232232444mod(6)......me2Usingme1andme2323232326x+4324mod(7)32444256=4mod(7)So4istheremainder:)\varphi (7)=6\quad and\quad gcd(32,7)=1\\ { 32 }^{ 6 }\equiv 1mod(7)............me-1\\ Now\quad we\quad find\quad { 32 }^{ 32 }mod\quad 6\\ { 32 }^{ 32 }\equiv { 2 }^{ 32 }\equiv 4^{ 4 }\equiv 4mod(6)......me-2\\ Using\quad me-1\quad and\quad me-2\\ { 32 }^{ { 32 }^{ 32 } }\equiv { 32 }^{ 6x+4 }\equiv { 32 }^{ 4 }mod(7)\\ { 32 }^{ 4 }\equiv 4^{ 4 }\equiv 256=4mod(7)\\ So\quad 4\quad is\quad the\quad remainder\quad :)

Mayank Chaturvedi - 3 years, 5 months ago

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@Mayank Chaturvedi Perfect.

A Former Brilliant Member - 3 years, 5 months ago

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Problem 44:

Find the locus of point of intersection of tangents to an ellipse x2a2+y2b2=1 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at two points,whose eccentric angles differ by a constant angle α\alpha .

Kunal Verma has provided a complete solution to this problem.Thanks!

Saarthak Marathe - 3 years, 5 months ago

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Point of intersection of tangents at points who's eccentric angles are i i and j j :-

x= a×cosi+j2cosij2 a \times \frac{\cos \frac{i+j}{2}}{\cos \frac{i-j}{2}} and y= b×sini+j2cosij2 b \times \frac{\sin \frac{i+j}{2}}{\cos \frac{i-j}{2}}

Thus x2a2 +y2b2 =sec2ij2\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{i-j}{2}

x2a2 +y2b2 =sec2α2\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{\alpha }{2}

Kunal Verma - 3 years, 5 months ago

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Correct! But prove your 1st statement in your solution and post the next question

Saarthak Marathe - 3 years, 5 months ago

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@Saarthak Marathe The thread should clear out older comments. It took ages to put up that solution. Won't live to see the proof of that uploaded. I'll just mention here that it was obtained by solving the parametric equations of tangents.

Kunal Verma - 3 years, 5 months ago

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Problem 45

The coefficient of xn6x^{n-6} in the expansion:

n!× [ x (n0) +(n1)(n0) ] [x2 (n1) +(n2)(n1) ] .... [xn (nn1) +(nn)(nn1) ] n! \times \ [ \ x \ - \frac{\binom{n}{0} \ + \binom{n}{1}}{\binom{n}{0}} \ ] \ [ \frac{x}{2} \ - \frac{\binom{n}{1} \ + \binom{n}{2}}{\binom{n}{1}} \ ] \ .... \ [ \frac{x}{n} \ - \frac{\binom{n}{n-1} \ + \binom{n}{n}}{\binom{n}{n-1}} \ ]