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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
8 months, 1 week 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 \(\frac 14\). 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 · 8 months, 1 week ago

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@Sandeep Bhardwaj Solution of problem 1

The required probability is

\(=_{ 1 }^{ 3 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 2 }.\frac { 1 }{ 4 } \quad +\quad _{ 1 }^{ 4 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 3 }.\frac { 1 }{ 4 } \quad +\quad _{ 1 }^{ 5 }{ C }.\frac { 1 }{ 4 } { .\left( \frac { 3 }{ 4 } \right) }^{ 4 }.\frac { 1 }{ 4 } \quad +\quad ...\infty \\ =\frac { 9 }{ 256 } .24\\ =\boxed { \frac { 27 }{ 32 } } \) Ritu Roy · 8 months, 1 week ago

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@Ritu Roy 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 · 7 months, 2 weeks ago

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Problem 10 :

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

a) Ellipse

b) Parabola

c) Hyperbola

d) Straight Line Ishan Singh · 8 months ago

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@Ishan Singh @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 · 7 months, 3 weeks ago

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@Sandeep Bhardwaj I had posted the solution! Some careless mod has deleted it! I can't believe it. Aditya Kumar · 7 months, 3 weeks ago

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@Aditya Kumar @Aditya Kumar ,your solution was incorrect. Saarthak Marathe · 7 months, 3 weeks 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 · 7 months, 3 weeks ago

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@Ishan Singh \[\dfrac{\tan(A/2)+\tan(B/2)}{\tan(A/2)-\tan(B/2)}=\dfrac{\ln \pi+1}{\ln \pi-1}=T~(Let)\]

\[\implies \dfrac{\sin(\frac{A+B}{2})}{\sin(\frac{A-B}{2})}=T\]

\[\implies \sin C=T(\sin A-\sin B)\] \[\implies a-b=\dfrac{c}{T}=\text{Constant}<c\] Here \(a,b,c\) denote sides opposite to angles \(A,B,C\) respectively.. A hyperbola..... Is it? Rishabh Cool · 8 months ago

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@Rishabh Cool @Ishan Singh

Please reply to Rishabh's comment. Sandeep Bhardwaj · 7 months, 3 weeks ago

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@Sandeep Bhardwaj Rishabh's solution is correct (I had stated it earlier). Aditya's solution was incorrect. Ishan Singh · 7 months, 3 weeks ago

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Problem 45

The coefficient of \(x^{n-6} \) in the expansion:

\( n! \times \ [ \ x \ - \frac{\binom{n}{0} \ + \binom{n}{1}}{\binom{n}{0}} \ ] \ [ \frac{x}{2} \ - \frac{\binom{n}{1} \ + \binom{n}{2}}{\binom{n}{1}} \ ] \ .... \ [ \frac{x}{n} \ - \frac{\binom{n}{n-1} \ + \binom{n}{n}}{\binom{n}{n-1}} \ ] \)

is equal to \( \binom{x}{y} \times \ z^y \)

Find x,y,z if all are integers( x, y and z can be in terms of \( n \) )

Sarvesh Nalawade has provided a complete solution to the problem. Kunal Verma · 7 months, 2 weeks ago

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@Kunal Verma Let S = \( n! \prod_{i=1}^{n} ( \frac{x}{i} - \frac{{n \choose i-1} + {n \choose i}}{{n \choose i-1}}) \)

\( Therefore, S= n! \prod_{i=1}^{n} ( \frac{x}{i} - \frac{{n+1 \choose i}}{{n \choose i-1}} ) \)

\( S = n! \prod_{i=1}^{n} ( \frac{x}{i} - \frac{n+1}{i} ) \)

\( Therefore, S = (x-n-1)^{n} \) .

Co-efficient of \(x^{n-6} = {n \choose 6} (n+1)^{6} \)

Therefore x=n , y=6, z= n+1 Sarvesh Nalawade · 7 months, 2 weeks ago

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@Sarvesh Nalawade You are required to find the individual values of \(x \) , \( y \) and \( z \), not the sum ,although it is correct. You may post the next problem after posting the solution. Kunal Verma · 7 months, 2 weeks ago

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@Kunal Verma Don't use the variable x for two purposes in the same question. Vighnesh Shenoy · 7 months, 2 weeks ago

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Problem 41

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

Saarthak Marathe has posted the solution to the problem.Thanks! Miraj Shah · 7 months, 2 weeks ago

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@Miraj Shah \(\displaystyle tan3x=\frac{tan2x+tanx}{1-tanx.tan2x}\\ tanx.tan2x.tan3x=tan3x-tanx-tan2x\)

Now integrating we get,

\(\displaystyle \int tanx.tan2x.tan3x.dx=\frac{ln|sec3x|}{3}-\frac{ln|sec2x|}{2}-ln|secx|+constant \) Saarthak Marathe · 7 months, 2 weeks ago

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@Saarthak Marathe Correct!Post the next question. Miraj Shah · 7 months, 2 weeks ago

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Problem 36:

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

Both Mayank Chaturvedi and Vighnesh Shenoy have provided complete solution this problem. Thanks! Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe A graphical approach.

We have to see the nature of discriminant

\({ a }^{ 2 }-(4{ a }^{ 2 }-24)\\ or\quad 8-{ a }^{ 2 }\\ Now,\quad graph\quad of\quad { a }^{ 3 }-6a-6=f(a)\quad \\ takes\quad two\quad turns\quad at\quad a\quad =\sqrt { 2 } \quad and\quad -\sqrt { 2 } ,\quad \\ which\quad shows\quad a(root\quad of\quad function)\quad is\quad positive\quad and\quad at\quad least\quad \sqrt { 2 } \\remember\quad f(x)\quad is\quad continuous.\\ at\quad a=\sqrt { 8 } \quad the\quad f(a)\quad is\quad negative.\\ So\quad root\quad of\quad f(a)>\sqrt { 8 } .\quad \\ Hence\quad for\quad equation\quad { a }^{ 3 }-6a-6=0,\\ a>\sqrt { 8 } ;{ a }^{ 2 }>8\\ We\quad have\quad 8-{ a }^{ 2 }<0.\quad so\\the \quad roots\quad are\quad non\quad real \) Mayank Chaturvedi · 7 months, 3 weeks ago

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@Saarthak Marathe A non-calculus approach.

\({a}^{3}-6a-6=0\)

Let \( a=b+2/b \)

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

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

Therefore, \( {b}^3=4\) or \(2\)

Substitute these values to get \(a\).

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

We see that, \( {a}^{2}-6=6/a\)

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

\(a{x}^{2}+{a}^{2}x+6=0 \)

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

\( x=\dfrac { -a\pm \sqrt { { a }^{ 2 }-24 } }{ 2 } \)

Then substituting the acquired value of \(a\) in this equation we get that \(x\) is a complex number. Hence, our assumption was wrong. Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe Wondering, how did you thought of that a=b+(2/b). Mayank Chaturvedi · 7 months, 3 weeks ago

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@Mayank Chaturvedi One way of solving cubic equation of the type \(ax^3+bx+c=0\) is to take \(x=d+y/d\) and manipulate the value of \(y\) to get a solvable \(6th\) degree equation in \(d\) Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe By solvable, do you mean quadratic type equations with higher degrees? Mayank Chaturvedi · 7 months, 3 weeks ago

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@Mayank Chaturvedi yes Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe Great!!! Mayank Chaturvedi · 7 months, 3 weeks ago

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@Saarthak Marathe \( \Delta = (a)^{2} -4(a^{2}-6) = 24-3a^{2} = 3(8-a^{2}) \)
Consider,
\( f(a) = a^{3} -6a - 6 \)

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

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

\( \Delta < 0 \)
Thus, the roots are not real. Vighnesh Shenoy · 7 months, 3 weeks ago

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@Vighnesh Shenoy Just completed solving the question and here you post the solution! But I think proof questions are not allowed, isn't it? Miraj Shah · 7 months, 3 weeks 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 · 7 months, 3 weeks ago

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@Miraj Shah Oh yes. Mayank Chaturvedi · 7 months, 3 weeks ago

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@Miraj Shah I guess he should have asked if the roots were real or complex. Vighnesh Shenoy · 7 months, 3 weeks ago

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@Vighnesh Shenoy Mayank should post the new question. Vighnesh Shenoy · 7 months, 3 weeks ago

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@Vighnesh Shenoy OK. @Mayank Chaturvedi,please post the next question Saarthak Marathe · 7 months, 3 weeks ago

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

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

Vighnesh Shenoy has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Sarvesh Nalawade · 7 months, 4 weeks ago

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@Sarvesh Nalawade \( (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \le (2+\sqrt{5})^{p} \le (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} + 1 \)
Now,
\( (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \in I \)
\( (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} + 1 \in I \)
\( \therefore \lfloor (2+\sqrt{5})^{p} \rfloor = (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} \)
\( S(p) = \lfloor (2+\sqrt{5})^{p} \rfloor - 2^{p+1} = (2+\sqrt{5})^{p} + (2-\sqrt{5})^{p} - 2^{p+1} \)
\( \therefore S =2\displaystyle \sum_{r=0}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r} - 2^{p+1} \)
\( \therefore S = 2^{p+1} + \displaystyle 2\sum_{r=1}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r} - 2^{p+1} = \displaystyle 2\sum_{r=1}^{\frac{p-1}{2}}\dbinom{p}{2r}2^{p-2r}5^{r}\)
Clearly, \( S(p) \) is divisible by \( 2 \) and \( 5 \)
\( S(p) = 2\displaystyle \sum_{r=1}^{\frac{p-1}{2}}\frac{p}{2r}\dbinom{p-1}{2r-1}2^{p-2r}5^{r} \)
Now since p is a prime number, \( gcd(r,p) = 1 , r \in \left[1,p-1\right]\)

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

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

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

The correct options are \( 1 \) and \( 4 \) Vighnesh Shenoy · 7 months, 4 weeks ago

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

\({x}^2+{y}^2={a}^2\) and \({x}^2+{y}^2=4ax\)

externally.

Akul Agrawal has solved this problem in the first place and posted the solution. Thanks! Rohit Ner · 8 months ago

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@Rohit Ner 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

\[ \frac { { \left( x-a \right) }^{ 2 } }{ { \left( \frac { a }{ 2 } \right) }^{ 2 } } -\frac { { y }^{ 2 } }{ { \left( \sqrt { 3 } \frac { a }{ 2 } \right) }^{ 2 } } =1 \] Akul Agrawal · 8 months ago

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Problem 4:

If \( n \in N \), evaluate the value of

\[ \dbinom{n}{0} + \dbinom{n}{4} + \dbinom{n}{8} + \ldots ,\]

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

Akul Agrawal is the first person to solve this problem and provide the solution. Thanks! Vighnesh Shenoy · 8 months, 1 week ago

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@Vighnesh Shenoy \( \displaystyle (1+x)^n = \sum_{r=0}^{n} \dbinom{n}{r} x^r\)

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

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

\(\displaystyle \sum_{r=0}^{\infty} \dbinom{n}{4r} = \dfrac{(1+i)^n + (1-i)^n +2^n}{4}\) Samuel Jones · 8 months, 1 week ago

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@Vighnesh Shenoy \(\frac { { 2 }^{ \frac { n }{ 2 } +1 }\cos { \frac { n\pi }{ 4 } } +{ 2 }^{ n } }{ 4 } \)

put i, -i,1, -1 in expansion of (1+x)^n and add Akul Agrawal · 8 months, 1 week ago

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Problem 3:

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

Vighnesh Shenoy has solved this problem at the first place and provided the solution. Thanks! Nihar Mahajan · 8 months, 1 week ago

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@Nihar Mahajan \( f''(x) = -f(x) \)

\( f'(x) = g(x) \)

\( f''(x) = g'(x) = -f(x) \)

\( h(x) = (f(x))^{2} + (g(x))^{2} \)
Differentiate,
\( h'(x) = 2 \cdot \left( f(x)f'(x) + g(x)g'(x) \right) \) \( h'(x) = 2 \cdot \left( f(x)g(x) + g(x)(-f(x)) \right) = 0 \)
Thus, \( h(x) \) is a constant.
\( h(x) = h(5) = 11 \) for all x. Vighnesh Shenoy · 8 months, 1 week ago

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Problem 2:

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

Nihar Mahajan has solved this problem at the first place and has posted the solution. Thanks! Ritu Roy · 8 months, 1 week ago

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@Ritu Roy 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.

\[\dfrac{k+1}{k} = \dfrac{8}{k+3}\neq \dfrac{4k}{3k-1}\]

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

(Feel free to correct me) Nihar Mahajan · 8 months, 1 week ago

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Problem 47 : In a \( \Delta ABC \) the ratio of side \(BC \) and \( AC \) to the circumradius is \( 2 \) and \( \dfrac{3}{2} \) respectively. If the ratio of length of angle bisectors of angle \( B \) to length of angle bisector of angle \( C \) is given by \( \dfrac{\alpha(\sqrt{\alpha}-1)}{\beta\sqrt{\gamma}} \) find
\( \dfrac{\alpha + \beta + \gamma}{3} \)

Note :
\( \alpha, \beta, \gamma \) are positive integers with \( gcd(\alpha, \beta ) = gcd(\alpha,\gamma) = gcd( \beta, \gamma = 1 \) Vighnesh Shenoy · 7 months, 2 weeks ago

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@Vighnesh Shenoy Firstly Using Sine Rule we get,

\( \frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =2R \)

Hence, we get \( A = \frac{\pi}{2} and \sin{B} = \frac{3}{4} \)

Let \( BC = 4k , AC = 3k , AB = \sqrt{7}k \)

Let BD and CE be the respective angle bisectors.

\( \frac { AD }{ CD } =\frac { BA }{ BC } \quad and\quad \frac { AE }{ EB } =\frac { AC }{ BC } \)

\( AD=\frac { 3\sqrt { 7 } }{ 4+\sqrt { 7 } }k \quad and\quad AE=\frac { 3 }{ \sqrt { 7 } }k \)

Then using Pythagoras Theorem , we get values of BD and CE as :

\(BD=\frac { 4\sqrt { 7 } }{ 1+\sqrt { 7 } } k\quad and\quad CE=\frac { 6\sqrt { 2 } }{\sqrt{ 7 }} k\quad \)

Their ratio will be : \( \dfrac{7(\sqrt{7} - 1 )}{9\sqrt2} \)

Hence, \( \alpha =7,\quad \beta =9,\quad \gamma =2 \)

Therefore, final answer is 6 Sarvesh Nalawade · 7 months, 2 weeks ago

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@Sarvesh Nalawade Getting the same thing @Sarvesh Nalawade. I think you should go ahead and post the solution Miraj Shah · 7 months, 2 weeks ago

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@Sarvesh Nalawade Correct. Vighnesh Shenoy · 7 months, 2 weeks ago

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Problem 46

If \( M = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \) is an orthogonal matrix with real entries , what is the minimum value of abc ? Sarvesh Nalawade · 7 months, 2 weeks ago

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@Sarvesh Nalawade For an orthogonal matrix M,
\( MM^{T} = I \) where I is the identity matrix.
Multiplying the matrices on LHS , and comparing with the identity matrix we get,
\( a^{2} + b^{2} + c^{2} = 1 \)
\( ab + bc + ca = 0 \)
From these two equations we get,
\( a + b + c = \pm 1 \)
\( f(x) = x^{3} - (a+b+c)x^{2} + (ab+bc+ca)x - d \) where d = abc )

Thus, a,b,c are roots of \( f(x) = 0 \)
\( f(x) = x^{3} - (a+b+c)x^{2} - d \)
Let \( a + b + c = p \)
\( f(x) =x^{3} - px^{2} - d \)
Differentiating,
\( f'(x) = 3x^{2} - 2px \)
The roots of \( f'(x) = 0 \) are,
\( x = 0, \dfrac{2p}{3} \)
For the equation to have 3 real roots, the first turn of the graph should be above x-axis, and second must be below x-axis.

Consider case of \( p =1 \)
The roots are \( x = 0 , \dfrac{2}{3} \)

\( \therefore f(0)f\left(\dfrac{2}{3}\right) \le 0 \)
\( \left(-d\right)\left(\dfrac{-4}{27}-d\right) \le 0 \)
\( \left(d\right)\left(d+\dfrac{4}{27}\right) \le 0 \)
\( d \in \left[\dfrac{-4}{27},0 \right] \)
When \( p = -1 \)
The roots of \( f'(x) = 0 \) are \( x = 0 , \dfrac{-2}{3} \)
\( f(0)f\left(\dfrac{-2}{3}\right) \le 0 \)
\( \left(-d\right)\left(\dfrac{4}{27}-d\right)\le 0 \)
\( d \in \left[0,\dfrac{4}{27}\right] \)
Range of d \( \left[\dfrac{-4}{27}, \dfrac{4}{27} \right] \)
The mininum value occurs when \( a,b,c \) are a permutation of \( \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{-1}{3} \).
The maximum value occurs when \( a,b,c \) are a permutation of \( \dfrac{-2}{3} , \dfrac{-2}{3} , \dfrac{1}{3} \) Vighnesh Shenoy · 7 months, 2 weeks ago

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@Vighnesh Shenoy Great! Please post the next problem. @Vighnesh Shenoy Sandeep Bhardwaj · 7 months, 2 weeks ago

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@Sarvesh Nalawade An alternate solution to @Vighnesh Shenoy 's already well written and beautiful solution:

\(a^2+b^2+c^2=1...(1) \\ab+bc+ca=0...(2)\)

Let

\(a=cos\alpha\\b=cos\beta\\c=cos\gamma\)

From \((1)\) we can write \(cos^2\alpha+cos^2\beta=sin^2\gamma...(3)\)

Now, using \((1)\) and \((2)\) we can write

\(cos\alpha+cos\beta+cos\gamma=\pm 1\\ cos\alpha+cos\beta=\pm 1-cos\gamma\)

Squaring both sides we get

\(cos^2\alpha+cos^2\beta+2cos\alpha .cos\beta=1+cos^2\gamma \mp 2cos\gamma\)

From \((3)\) we can write \(cos\alpha .cos\beta=cos^2\gamma \mp cos\gamma\)

Now multiplying \(cos\gamma\) on both sides we get:

\(cos\alpha .cos\beta .cos\gamma=cos^3\gamma\mp cos^2\gamma\)

Critical points of the expression on the right-hand side of the above equation

(1.) \(\{0,\frac{2}{3}\}\) when \(cos\alpha .cos\beta .cos\gamma=cos^3\gamma- cos^2\gamma\)

(2.) \(\{0,\frac{-2}{3}\}\) when \(cos\alpha .cos\beta .cos\gamma=cos^3\gamma+cos^2\gamma\)

On checking these values, minima occurs at \(cos\gamma = \frac{2}{3}\) and the minimum value is \(\frac{-4}{27}\), and maximum occur maxima occurs at \(cos\gamma = \frac{-2}{3}\) and the maximum value is \(\frac{4}{27}\) Miraj Shah · 7 months, 2 weeks ago

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Problem 39:

Solve

\(\displaystyle {x}^{3}\frac{dy}{dx}={y}^{3}+{y}^{2}\sqrt{{y}^{2}-{x}^{2}} \) Saarthak Marathe · 7 months, 2 weeks ago

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@Saarthak Marathe The answer is \(xy=c(y-\sqrt{y^{2}-x^{2}})\) The above equation is a homogeneous equation and can be written of the form \[\frac{dy}{dx}={\frac{y}{x}}^{3}+{\frac{y}{x}}^{2}*\sqrt{y^{2}-x^{2}}\] Making the substitution \(y=vx\) and we see that \(\dfrac{dy}{dx}=(v+\dfrac{dv}{dx}*x)\). On simplification we get \[ \frac{dv}{v(v^{2}-1+v\sqrt{v^{2}-1})}=\frac{dx}{x}\]. Taking \(\sqrt{v^{2}-1}\) and then Multiplying and dividing by \(v-\sqrt{v^{2}-1}\) we have \[\frac{dv}{\sqrt{v^{2}-1}} -\frac{dv}{v}=\frac{dx}{x}\]. On integrating both sides we have \[\ln{\frac{v-\sqrt{v^{2}-1}} {v}}=\ln{x}+c\]. This on substituting back \(v\) gives us the above answer. Vignesh S · 7 months, 2 weeks ago

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\(PROBLEM \quad 29\):

Time for an easy question.

Find the range of \(\beta\) such that \( (0,2\beta-1) \) lies on or inside the triangles formed by the lines. \( y +3x +2 = 0 \)
\( 3y- 2x- 5=0 \)
\( 4y+x-14=0 \) Mayank Chaturvedi · 7 months, 3 weeks ago

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@Mayank Chaturvedi Simply by drawing the graph I got \( \frac{4}{3} \leq \beta \leq \frac{9}{4} \)
. The triangle intersects Y-axis at points \( \left(0,\frac{5}{3}\right) \) and \( \left(0,\frac{7}{2}\right) \)
\( \therefore \frac{5}{3} \leq 2\beta-1 \leq \frac{7}{2} \)

\( \therefore \frac{4}{3} \leq \beta \leq \frac{9}{4} \)

Sarvesh Nalawade · 7 months, 3 weeks ago

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@Sarvesh Nalawade Great! Can you please add the graph for the sake of solution? Thanks! Sandeep Bhardwaj · 7 months, 3 weeks ago

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@Sarvesh Nalawade Correct Mayank Chaturvedi · 7 months, 3 weeks ago

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

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

Mayank Chaturvedi has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Vighnesh Shenoy · 7 months, 4 weeks ago

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@Vighnesh Shenoy 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 · 7 months, 4 weeks ago

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@Mayank Chaturvedi Correct. Vighnesh Shenoy · 7 months, 4 weeks ago

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Problem 17:

\( z \) is a complex number in the complex plane such that \(\Im(z) \ne 0 \).
If \( \dfrac{z^{2} + z + 1}{z^{2}-z+1} \in \Re \) find the value of \( 10|z| \)

Details :
\( \Im(z) \) denotes imaginary part of \( z \)
\( \Re \) denotes real numbers.

Prakhar Bindal has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Vighnesh Shenoy · 8 months ago

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Comment deleted 8 months ago

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@Prakhar Bindal i did a bit of bashing. I used the fact that conjugate a purely real number is equal to the number itself.

Just take conjugate of Given Expression and equate to original one.

Cross multiply and cancel like terms . Finally we will obtai

a = z b = conjugate of z

(ab-1)(a-b) = 0

So either a= b which is not possible as already given in problem that a is not purely real .

Hence mod(a) = 1 Prakhar Bindal · 8 months ago

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@Vighnesh Shenoy Just to show what I was going at.
My method :
\( \dfrac{z^{2}+z+1}{z^{2}-z+1} \in \Re \)
\( 1 +2\dfrac{z}{z^{2}-z+1} \in \Re \)
\( 1 + 2\dfrac{1}{z+\dfrac{1}{z} - 1} \in \Re \)
\( \therefore z + \dfrac{1}{z} \in \Re \)

Since the conjugate of a real number is equal to itself,
\( z + \dfrac{1}{z} = \overline{z} + \dfrac{1}{\overline{z}} \)
\( z - \overline{z} = \dfrac{1}{\overline{z}} - \dfrac{1}{z} \)
\( z - \overline{z} = \dfrac{z - \overline{z}}{z\overline{z}} \)
Since, \( z \) is not purely real, \( z \ne \overline{z} \)
\( \therefore z\overline{z} = 1 \)
\( |z|^{2} = 1 \rightarrow |z| = 1 \)
\( 10|z| = 10 \) Vighnesh Shenoy · 8 months ago

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This one is a fairly easy one:

Problem 16:

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

for {\( x | x^{2} + 20 \leq 9x\)}

Vignesh Shenoy has posted the correct solution. Thanks! Ashu Dablo · 8 months ago

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@Ashu Dablo \( x^{2} - 9x + 20 \le 0 \)

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

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

Differentiating,

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

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

\( f'(x) = 6(x-2)(x-3) \)

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

\( f(x)_{max} = f(5) = 7 \) Vighnesh Shenoy · 8 months ago

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Problem 14 :

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

Saakshi Singh has posted the solution first. Thanks! Ashu Dablo · 8 months ago

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@Ashu Dablo Using sum formula of AP, we get \[2cn= 2a + (n-1)d\]

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

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

So we have sum of the cubes of \(n\) terms

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

\[ = c^3 \left( n^2(2n^2 - 1) \right) .\] Saakshi Singh · 8 months ago

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@Saakshi Singh 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 · 8 months ago

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Problem 13:

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

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

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

Ashu Dablo has solved this problem at the first place and has posted the solution. Thanks! Akul Agrawal · 8 months ago

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@Akul Agrawal @Akul Agrawal Simplify the given equation to the form:

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

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

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

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

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

So answer is \( \frac{\pi}{4} \)

Sorry for my latex! Ashu Dablo · 8 months ago

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@Akul Agrawal The equation becomes inconsistent for n=0 . Please check . Keshav Tiwari · 8 months ago

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Problem 12:

Evaluate:

\[\large \displaystyle \lim_{ x\rightarrow 0 }{ \left\{ \frac { sinx }{ x } \right\} }^{ \frac { 1 }{ \left\{ \frac { tanx }{ x } \right\} } } \]

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

Rishabh Cool has provided a complete solution to this problem. Thanks! Saarthak Marathe · 8 months ago

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Comment deleted 8 months ago

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@Akul Agrawal Just to expand.... \(\{\dfrac{\sin x}{x}\}=\dfrac{\sin x}{x}\) and \(\{\dfrac{\tan x}{x}\}=\dfrac{\tan x}{x}-1\) as \(\sin x<x<\tan x\).

The form is \(1^{\infty}\). Therefore write it as:

\[ e^{\displaystyle\lim_{x\to 0}(\frac{\sin x}{x}-1)\times \frac{1}{\frac{\tan x}{x}-1}}\]

\[= e^{\displaystyle \lim_{x\to 0}\dfrac{\sin x-x}{\tan x-x}}\]

Apply Lhospital to get \(e^{-1/2}\). Rishabh Cool · 8 months ago

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Problem 11:

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

Consider \(\alpha ,\beta ,\gamma \) to be acute.

Saarthak Marathe is the first person to solve this problem and provide the solution. Thanks! Aditya Kumar · 8 months ago

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@Aditya Kumar The answer is \(0\).

Saarthak Marathe · 8 months ago

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Problem 7:

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

Then find a closed of: \[{ C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }+...+\left( n+1 \right) { C }_{ n }^{ 2 }\]

Rohit Ner has solved this problem at the first place and posted the solution. Thanks! Aditya Kumar · 8 months, 1 week ago

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@Aditya Kumar \(x{(1+x)}^{n}=C_0 x+C_1 {x}^2 +C_2 {x}^3+\cdots +C_n {x}^{n+1} \)

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

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

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

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

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

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

So the closed form is \(\binom{2n-1}{n}+(n+1)\binom{2n-1}{n-1}\) Rohit Ner · 8 months, 1 week ago

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Problem 6:

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

\[x^4+x^3-5x^2-12x-6=0.\]

Aditya Kumar solved the problem at the first place and provided the solution. Thanks! Samuel Jones · 8 months, 1 week ago

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

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

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

\(\displaystyle \implies t^2 - 3t -2 -\dfrac{3}{t} + \dfrac{1}{t^2} = 0 \)

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

\(\displaystyle y^2 -3y - 4 = 0\)

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

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

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

Only the second factor gives non - real roots since it's discriminant is negative, therefore sum is \(\boxed{-3}\) Samuel Jones · 8 months ago

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@Samuel Jones The equation can be written as \[\left( { x }^{ 2 }-2x-2 \right) \left( { x }^{ 2 }+3x+3 \right) =0\]

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

Hence the sum is: \(-3\) Aditya Kumar · 8 months, 1 week ago

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@Samuel Jones The equation factors out as,

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

Now solve \({x}^{2}-2x-2=0\) and \({x}^{2}+3x+3=0 \).

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

Therefore the sum of the non-real roots are -3 Saarthak Marathe · 8 months, 1 week ago

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Problem 5:

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

Samuel Jones is the first person to solve this problem and provide the solution. Thanks! Akul Agrawal · 8 months, 1 week ago

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

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

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

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

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

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

This way, we can generalize to \(n\) terms also. Samuel Jones · 8 months, 1 week ago

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Last problem for this page:Problem No. '50'

A cevian \(AQ\) of a equilateral \(\Delta\) \(ABC\) is extended to meet circumcircle at \(P\). If \(PB=50\) and \(PC=45\), find \(PQ\) upto 3 decimal places.

Deeparaj Bhat has provided the answer Vignesh S · 7 months, 2 weeks ago

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@Vignesh S Great! This is the last problem of this contest on this note. The solution poster will post the next problem on the continued part of the contest: JEE-Advanced Maths Contest (Continued).

Thanks! Sandeep Bhardwaj · 7 months, 2 weeks ago

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@Vignesh S Solution by Deeparaj Bhat:

Extend \(CP\) to \(D\) such that \(\Delta BDP\) is equilateral. Then, \[ \begin{align*}\angle BCD&=\angle QCP \\\angle QPC&=\angle ABC\quad \left(\because \text{ angles in the same segment are equal }\right)\\&=60^{\circ}\\&=\angle BDP\\\implies \Delta DCB &\sim \Delta PCQ\\\implies \frac{ DB }{PQ}&=\frac{ DC}{PC }\\&=1+\frac{PB}{PC}\quad \left( \because DC=DP+PC=BP+PC \right)\\\implies \frac{1}{PC}+\frac{1}{PB}&=\frac{1}{PQ}\quad \left( \because BP=DB \right)\end{align*} \]

Substituting the given values the answer comes out to be 23.684. Sandeep Bhardwaj · 7 months, 2 weeks ago

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@Vignesh S 23.684. I'll post the soln in some time... Deeparaj Bhat · 7 months, 2 weeks ago

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@Deeparaj Bhat Waiting for the solution. Sandeep Bhardwaj · 7 months, 2 weeks ago

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@Sandeep Bhardwaj Please provide the complete solution. Vignesh S · 7 months, 2 weeks ago

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Problem 48 : If \( \displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ \left( a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x } \right) ^{ 2 } } } =\quad \frac { \pi \left( a+b \right) }{ p{ \left( ab \right) }^{ q } } \) ;

where p and q are positive rational numbers and ab>0 , then find the value of pq

Vighnesh Shenoy has provided a complete solution to this problem. Sarvesh Nalawade · 7 months, 2 weeks ago

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@Sarvesh Nalawade \( I = \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))} \)
Put, \( \tan(x) = t \)

\(I = \displaystyle \int_{0}^{\infty} \dfrac{dt}{a+bt^{2}} = \dfrac{\pi}{2\sqrt{ab}} \)
\( \therefore \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))} =\dfrac{\pi}{2}(ab)^{\frac{-1}{2}} \)

For a continuous and differentiable function,
\(\displaystyle \dfrac{d}{dy} \int_{a}^{b} f(x,y)dx = \int_{a}^{b} \dfrac{\partial}{\partial y} f(x,y) dx \)

Differentiating with respect to a,

\( \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{\partial}{\partial a} \dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))} = \dfrac{\partial}{\partial a} \left( \dfrac{\pi}{2}(ab)^{\frac{-1}{2}}\right) \)

\( \therefore \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{-\cos^{2}dx}{(a\cos^{2}(x) + b\sin^{2}(x))^{2}} = \dfrac{-\pi}{4}b^{\frac{-1}{2}}a^{\frac{-3}{2}} \)

Differentiate with respect to b,

\( \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{\partial}{\partial b} \dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))} = \dfrac{\partial}{\partial b} \left( \dfrac{\pi}{2}(ab)^{\frac{-1}{2}}\right) \)

\( \therefore \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{-\sin^{2}dx}{(a\cos^{2}(x) + b\sin^{2}(x))^{2}} = \dfrac{-\pi}{4}b^{\frac{-3}{2}}a^{\frac{-1}{2}} \)

Adding both the integrals,

\( \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))^{2}} = \dfrac{\pi}{4\sqrt{ab}}\left(\dfrac{1}{a} + \dfrac{1}{b} \right) \)

\( \therefore \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))^{2}} = \dfrac{\pi(a+b)}{4(ab)^{\frac{3}{2}}} \)

\( p = 4 , q = \dfrac{3}{2} \)
\( pq = 6 \) Vighnesh Shenoy · 7 months, 2 weeks ago

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Problem 44:

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

Kunal Verma has provided a complete solution to this problem.Thanks! Saarthak Marathe · 7 months, 2 weeks ago

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@Saarthak Marathe Point of intersection of tangents at points who's eccentric angles are \( i \) and \( j \) :-

x= \( a \times \frac{\cos \frac{i+j}{2}}{\cos \frac{i-j}{2}} \) and y= \( b \times \frac{\sin \frac{i+j}{2}}{\cos \frac{i-j}{2}} \)

Thus \(\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{i-j}{2} \)

\(\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{\alpha }{2} \) Kunal Verma · 7 months, 2 weeks ago

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@Kunal Verma Correct! But prove your 1st statement in your solution and post the next question Saarthak Marathe · 7 months, 2 weeks 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 · 7 months, 2 weeks ago

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Problem 43 :

Find the remainder when \( 32^{32^{32}} \) is divided by 7.
Mayank Chaturvedi has solved this problem with the correct solution. Vighnesh Shenoy · 7 months, 2 weeks ago

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@Vighnesh Shenoy \(32\equiv -3 (mod 7)\\ {3}^{3}\equiv -1 (mod 7)\\ {32}^{3*10+2}\equiv {(-3)}^{3*10+2} \equiv (1)*9 \equiv 9 (mod 7)\\ {32}^{32^{32}}\equiv {(9)}^{3*10+2} \equiv (1)*81 \equiv 4 (mod 7)\\ \)

Hence the answer is \(\boxed 4\) Saarthak Marathe · 7 months, 2 weeks ago

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@Saarthak Marathe The answer is right, however you used
\( 32^{32^{32}} \equiv (32^{32})^{32} \equiv 9^{32} \equiv 4 \pmod{7} \) Vighnesh Shenoy · 7 months, 2 weeks ago

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@Vighnesh Shenoy 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

\(\varphi (7)=6\quad and\quad gcd(32,7)=1\\ { 32 }^{ 6 }\equiv 1mod(7)............me-1\\ Now\quad we\quad find\quad { 32 }^{ 32 }mod\quad 6\\ { 32 }^{ 32 }\equiv { 2 }^{ 32 }\equiv 4^{ 4 }\equiv 4mod(6)......me-2\\ Using\quad me-1\quad and\quad me-2\\ { 32 }^{ { 32 }^{ 32 } }\equiv { 32 }^{ 6x+4 }\equiv { 32 }^{ 4 }mod(7)\\ { 32 }^{ 4 }\equiv 4^{ 4 }\equiv 256=4mod(7)\\ So\quad 4\quad is\quad the\quad remainder\quad :)\) Mayank Chaturvedi · 7 months, 2 weeks ago

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@Mayank Chaturvedi Perfect. Vighnesh Shenoy · 7 months, 2 weeks ago

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Comment deleted 7 months ago

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@Saarthak Marathe That is an incorrect way of evaluation of power towers.
\( a^{b^{c}} \ne (a^{b})^{c} \) Vighnesh Shenoy · 7 months, 2 weeks ago

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Problem 42:

Evaluate

\(\displaystyle \lim_{n \rightarrow \infty} \left[\frac{{n}^{1/2}}{{n}^{3/2}}+\frac{{n}^{1/2}}{{(n+3)}^{3/2}}+\frac{{n}^{1/2}}{{(n+6)}^{3/2}}+........+\frac{{n}^{1/2}}{{(n+3(n-1))}^{3/2}}\right] \)

Vighnesh Shenoy has provided solution to this question. Thanks! Saarthak Marathe · 7 months, 2 weeks ago

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@Saarthak Marathe This problem can be easily solved by using the concept of definite integral as the sum of limits, which is applied as

\( \displaystyle \lim_{n \to \infty} \sum_{k=0}^{n-1} \dfrac{n^{\frac{1}{2}}}{(n+3k)^{\frac{3}{2}}} = \displaystyle \int_{0}^{1} \dfrac{1}{(1+3x)^{\frac{3}{2}}}dx = \left[ \dfrac{2}{3(\sqrt{1+3x})}\right]_{1}^{0} = \dfrac{2}{3} - \dfrac{2}{6} = \dfrac{2}{6} = \dfrac{1}{3} \) Vighnesh Shenoy · 7 months, 2 weeks ago

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@Vighnesh Shenoy Great! It would be awesome if you can make your solution more detailed. You can post the next problem. Thanks! Sandeep Bhardwaj · 7 months, 2 weeks ago

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Problem no. 40.

An easy question:Find the following \[\dbinom{n}{0}+3\dbinom{n}{1}+5\dbinom{n}{2}+........+(2n+1)\dbinom{n}{n}\] Miraj Shah has posted the solution to the problem. Thanks! Vignesh S · 7 months, 2 weeks ago

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@Vignesh S The above question can be written in the following manner:

\(\displaystyle \sum_{r=0}^{n} (2r+1)\dbinom{n}{r}\) \( =2\displaystyle\sum_{r=0}^{n}r\dbinom{n}{r} +\displaystyle\sum_{r=0}^{n}\dbinom{n}{r} = n2^n +2^n= \boxed{(n+1)2^n}\) Miraj Shah · 7 months, 2 weeks ago

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@Miraj Shah Please post the next question Vignesh S · 7 months, 2 weeks ago

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Problem 38:

Find the number of ways to go from \( (0,0) \) to \( (8,7) \) in a rectangle formed by vertices \( (0,0) , (8,0) , (0,7), (8,7) \). The person can only move from \( (i,j) \) to \( (i+1,j) \) OR \((i,j+1)\) OR \((i+1,j+1)\) in one step. Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe Let the person travel \(d\) diagonals in one trip. So the remaining \(15-2d\) sides,he travels up or right, which can be done in \(\dbinom{15-2d}{8-d}\) ways. The \(d\) diagonals can be placed in \(\dbinom{15-d}{d}\) ways.

So the total number of ways=\(\displaystyle \sum_{d=0}^{7} \dbinom{15-2d}{8-d}\dbinom{15-d}{d} \).

This summation turns out as \(\boxed {108545}\) Saarthak Marathe · 7 months, 2 weeks ago

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@Saarthak Marathe @Saarthak Marathe

No one has solved the problem in the time limit. Can you please post the solution and the next problem?

Thanks! Sandeep Bhardwaj · 7 months, 2 weeks ago

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PROBLEM 37

In how many ways 12 different books can be distributed among 5 children so that 2 get three books each, and 3 get two books each.

Saarthak Marathe and Miraj Shah had solved this question Mayank Chaturvedi · 7 months, 3 weeks ago

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@Mayank Chaturvedi The number of ways distributing \(n\) distinct things into \(r\) groups and arranging them among \(r\) people such that \(s_1, s_2,s_3,...s_r\) denotes the number of things in the respective groups is :

\(\large \frac{(n!)( r!)}{(s_1!)( s_2!) ...(s_r!)}\)

Note in the above formula \(s_1\neq s_2\neq s_3\neq ... s_r\)

For the given question \(n=12\); \(r=5\); \(s_1=s_2=3\) and \(s_3=s_4=s_5=2\).

Now, the above formula can be used but, with a slight modification. Since groups are only identified by the number of objects it has, therefore there is no difference between \(s_1\) and \(s_2\) and there is no difference between \(s_3, s_4\) and \(s_5\). Hence the number of cases will reduce by a factor of \(2!\) with respect to \(s_1\) and \(s_2\) and by \(3!\) with respect to \(s_3, s_4\) and \(s_5\). Hence, in totality the required answer is:

\(\large \frac{12!\times 5!}{(3!)^2(2!)^3(2!)(3!)} = 16632000\)

Is the above answer correct? @Mayank Chaturvedi Miraj Shah · 7 months, 3 weeks ago

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@Mayank Chaturvedi \(\dbinom{12}{6}\dbinom{6}{3}\dbinom{6}{2}\dbinom{4}{2}\dbinom{5}{3}=16632000 \)

First choose 2 people of 5 which get 3 books each. Select 6 books(3*2=6) out of 12 books for these two 2 people. Then distribution of these 6 books by letting the first person choose 3 books out of 6 and the rest goes to 2nd person. Now for the remaining 3 people,there are 6 books. 1st person selects 2 books out of 6 books and 2nd person chooses 2 out of 4 books and the remaining goes to the 3rd person. I hope it is clear now. @Mayank Chaturvedi Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe As you asked, your solution is Crystal clear But it would be interesting if we can find out an explanation for solution by @Miraj Shah.Which I feel is as good approach as yours. Mayank Chaturvedi · 7 months, 3 weeks ago

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@Saarthak Marathe Can you please tell what is the numerical figure you are getting after solving your expression? Miraj Shah · 7 months, 3 weeks ago

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@Miraj Shah 16632000 Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe OK! Thanks! Waiting for the next question from you... Miraj Shah · 7 months, 3 weeks ago

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Problem 35:

It can be proved that the areas \(S_0,S_1,S_2,S_3,...\) bounded by the \(x-axis\) and the half waves of the curve \(y=e^{-\alpha x}sin\beta x\), \(x\ge 0\) form a geometric progression. Find the common ratio of this geometric progression. Miraj Shah · 7 months, 3 weeks ago

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@Miraj Shah Answer: \({e}^{-\frac{\alpha\pi}{\beta}}\)

First I wrote the general integral for \({S}_{j}\) and then divided by \({S}_{j-1}\)

Is it correct? @Miraj Shah Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe The answer is correct! Can you post a clearer picture of your working if possible?

Thanks! Miraj Shah · 7 months, 3 weeks ago

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@Miraj Shah The problem is my mobile phone does not have a good camera. I'll tell u what I did. First I wrote the general integral for \({S}_{j}\) and then divided by \({S}_{j-1}\) Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe The method is correct! Just add this statement to your solution so that others can get an idea as to what was your approach! Good solution anyways! Miraj Shah · 7 months, 3 weeks ago

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Problem 34:

Solve:

\(\displaystyle ({x}^{2}+y)\frac{dy}{dx}=6x \) Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe We can write the question as follows:

\((x^2+y)dy=3\times 2xdx\)

Now let \(x^2=\lambda\)

\((\lambda+y)=3\large \frac{d\lambda}{dy}\)

We'll do another substitution here, let \(\lambda+y=t\)

Therefore,

\(t+3=3\large \frac{dt}{dy}\)

Now after rearranging the terms and integrating we get:

\(3ln|x^2+y+3| = y+c\)

Is this correct? @Saarthak Marathe Miraj Shah · 7 months, 3 weeks ago

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@Miraj Shah It is incorrect. You missed a number in your final answer Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe Got it! it should be \(x^2+y+3\) right? Miraj Shah · 7 months, 3 weeks ago

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@Miraj Shah Yes correct!! Post the next question Saarthak Marathe · 7 months, 3 weeks ago

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Problem 33:

Two persons \(A\) and \(B\) agree to meet at a place between \(11\) to \(12\) noon. The first one to arrive waits for \(20\) minutes and then leaves. If the time of their arrival be independent and random, what is the probability that \(A\) and \(B\) shall meet? Miraj Shah · 7 months, 3 weeks ago

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@Miraj Shah Just an alternate solution:

Let the time of arrival of \(A\) be denoted along the \(x-axis\) and that of \(B\) along the \(y-axis\).

Therefore, the required event is denoted graphically be the area enclosed in \(|x-y|\le 20\). Now the ratio of the area of the event to the area of sample space(\(=3600\)) gives the answer \(\frac{5}{9}\) Miraj Shah · 7 months, 3 weeks ago

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@Miraj Shah Probability of meeting If the first person comes between 0-40 minutes=\(2/3*1/3\)=\(2/9\)

Probability of meeting if first person comes between 40-60 minutes=\(1/3\)

Therefore,total probability=\(5/9\)

Is this correct? @Miraj Shah Saarthak Marathe · 7 months, 3 weeks ago

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@Saarthak Marathe Ya! The answer is correct. Miraj Shah · 7 months, 3 weeks ago

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Problem 32:

The area bounded by the curve \(y=x-{x}^2\) and the line \(y=mx\) equals \(\large\frac{9}{2}\).

Find the sum of all possible values of \(m\). Rohit Ner · 7 months, 3 weeks ago

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@Rohit Ner Let the points of intersection between \(y=x-x^2\) and \(y'=mx\) be \(x_1\) and \(x_2\). Therefore \(x_1, x_2\) are the roots of the equation \(x^2 +(m-1)x=0\)...\((1)\)

Given:

\(\displaystyle \int_{x_1}^{x_2} y-y'\, dx =\frac{9}{2}\)

\(\displaystyle \int_{x_1}^{x_2} x(1-m)-x^3\, dx=\frac{9}{2}\)

\(\large \frac{(1-m)(x_2-x_1)(x_2+x_1)}{2}-\frac{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)}{3}=\frac{9}{2}\)

Now using equation \((1)\) we can write \(x_2+x_1=1-m\); \(x_1x_2=0\) and \(x_2-x_1=|1-m|\)

Therefore finally,

\(m=-2\) or \(m=4\)

Therefore, the required answer is \(\boxed{2}\)

Is this the answer? Miraj Shah · 7 months, 3 weeks ago

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@Miraj Shah Correct. :). Waiting for the next problem. Rohit Ner · 7 months, 3 weeks ago

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Problem 31 :

Given that,
\( x^{2} + y^{2} + z^{2} = R^{2} \)

Let,
\( P = axy + byz \)
If \( P_{max} = R^{2}f(a,b) \)
Find \( f(3,4) \) Vighnesh Shenoy · 7 months, 3 weeks ago

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@Vighnesh Shenoy Let \(x=R\sin\alpha\sin\beta,y=R\cos\beta,z=R\cos\alpha\sin\beta\)

\(\begin{align} P&=y\left(ax+bz\right)\\&=\dfrac{{R}^2}{2}\sin 2\beta \left(a\sin\alpha+b\cos\alpha\right)\\{P}_{max}&=\dfrac{{R}^2}{2}\sqrt{{a}^2+{b}^2}\\\Rightarrow f(a,b)&=\dfrac{\sqrt{{a}^2+{b}^2}}{2}\\f(3,4)&=\frac{5}{2}\end{align}\) Rohit Ner · 7 months, 3 weeks ago

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@Rohit Ner Correct. Vighnesh Shenoy · 7 months, 3 weeks ago

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@Vighnesh Shenoy I got the answer 5/2. Applying cauchy-schwarz inequality

\(\sqrt { ({ a }^{ 2 }+b^{ 2 })(x^{ 2 }{ y }^{ 2 }+y^{ 2 }z^{ 2 }) } \ge (axy+byz)\\ \sqrt { ({ a }^{ 2 }+b^{ 2 }){ y }^{ 2 }(x^{ 2 }+z^{ 2 }) } \ge (axy+byz)\\ \sqrt { ({ a }^{ 2 }+b^{ 2 }){ y }^{ 2 }(R^{ 2 }-y^{ 2 }) } \ge (axy+byz)\\ \frac { { R }^{ 2 } }{ 2 } \sqrt { ({ a }^{ 2 }+b^{ 2 }) } \ge (axy+byz)\) Mayank Chaturvedi · 7 months, 3 weeks ago

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@Mayank Chaturvedi You guys decide who is going to post next. Vighnesh Shenoy · 7 months, 3 weeks ago

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Problem 30 :

Let P(x) be a polynomial of degree 11 such that : \( P(x) = \dfrac{1}{x+1} , 0 \le x \le 11 \)

Find the value of \( P(12) \) . Sarvesh Nalawade · 7 months, 3 weeks ago

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@Sarvesh Nalawade \( P(x)(x+1) = 1 \)
\( P(x)(x+1) - 1 = g(x) \)
\( g(x) \) has roots \( 0,1,2,3,4\ldots 11 \)
\( \therefore g(x) = ax(x-1)(x-2)\ldots(x-11) \)

\( P(x)(x+1)-1 = ax(x-1)(x-2)\ldots(x-11) \)
Substituting \( x = - 1 \),
\(-1 = a(-1)(-2)\ldots(-12) \)
\( -1 = 12!a \)
\( a = -\dfrac{1}{12!} \)
\( P(12)(13) - 1 = -\dfrac{1}{12!}\times 12! \)
\( P(12)(13) - 1 = -1 \)
\( P(12) = 0 \) Vighnesh Shenoy · 7 months, 3 weeks ago

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@Vighnesh Shenoy That's correct !! Sarvesh Nalawade · 7 months, 3 weeks ago

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Problem 28: (On behalf of Vighnesh Shenoy)

If \(\vec a\) and \(\vec b\) are two vectors such that \(|\vec{a}|=1, |\vec b|=4\) and \( \vec a \cdot \vec b =2\), then find the angle between \(\vec b\) and \(\vec c\) given that \(\vec c=\left( 2 \vec a \times \vec b \right) - 3 \vec b\).

Mayank Chaturvedi has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Sandeep Bhardwaj · 7 months, 3 weeks ago

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@Sandeep Bhardwaj Sir please extend this page by making a new note as it is becoming slower to load the contents and rendering the latex. Rohit Ner · 7 months, 3 weeks ago

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@Rohit Ner Yes sir @Sandeep Bhardwaj. The note is lagging too much and taking too much time to load. Saarthak Marathe · 7 months, 3 weeks ago

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@Sandeep Bhardwaj Earlier i wrote an answer telling angle between \(\vec c \) and -\(\vec b\) so you too take care of it.

Check this:

i have got answer 150 degrees

\(\vec { a } .\vec { b } =|a||b|cos\theta \\ \theta =\frac { \pi }{ 3 } \\ \vec { v } =2\vec { a } X \vec { b } .So\vec { |v| } =4\sqrt { 3 } \quad perpendicular\quad to\quad \vec { b } \\ \vec { c } =\vec { v } -3\vec { b } ,which\quad is\quad at\quad { 30 }^{ \circ }to\quad -\vec { b } \quad So\quad { 150 }^{ \circ } \quad angle \quad between \quad \vec c \quad\ and \quad \vec b \)

Note:\(\vec c \) is resultant of \(-3\vec b\) and \(\vec v\), which are perpendicular. We have |3b|=12 and |v|=4\(\sqrt 3\).So tan\(\theta\)=4\(\sqrt 3\)/12. \(\theta\)=30. Angle between \(\vec c \quad and \quad \vec 3b \quad is\quad same \quad as\quad angle\) between \(\vec c \quad and \quad \vec b\)

All credits to @Vighnesh Shenoy, Mayank Chaturvedi · 7 months, 3 weeks ago

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@Mayank Chaturvedi Great! Can you please post the next problem? @Mayank Chaturvedi Sandeep Bhardwaj · 7 months, 3 weeks ago

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PROBLEM 27:

\(\csc ^{ -1 }{ \sqrt { 5 } } +\csc ^{ -1 }{ \sqrt { 65 } } +\csc ^{ -1 }{ \sqrt { 325 } } +\ldots =\frac { n\pi }{ 32 } \)

If you were given a chance to visit n (its value to be found from above expression) towns, in any way you want irrespective of order. I tell you that you can do so in \({ A }^{ B }\) ways.A & B are naturals.Find minimum value of A+B.

\( A\quad 10\\ B\quad 8\\ C\quad 6\\ D\quad 4\).

Vighnesh Shenoy has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Mayank Chaturvedi · 7 months, 4 weeks ago

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@Mayank Chaturvedi \( S = \csc^{-1}(\sqrt{5}) + \csc^{-1}(\sqrt{65}) + \csc^{-1}(\sqrt{325}) \ldots \)
\( S = \sin^{-1}\left(\dfrac{1}{\sqrt{5}}\right) + \sin^{-1}\left(\dfrac{1}{\sqrt{65}}\right) + \sin^{-1}\left(\dfrac{1}{\sqrt{325}}\right) + \ldots \)

\(S = \tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{8} + \tan^{-1}\dfrac{1}{18} + \ldots \)
\( S = \displaystyle \sum_{k=1}^{\infty} \tan^{-1}\dfrac{1}{2k^{2}} = \sum_{k=1}^{\infty} \tan^{-1}(2k+1) - \tan^{-1}(2k-1) = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4} \)
\( \therefore n = 8 \)

I can visit 0 towns, 1 town, 2 towns, and so forth till 8 towns irrespective of order.
Number of ways I can do this = \( \displaystyle \sum_{r=0}^{8}\dbinom{8}{r} = 2^{8} \)
\( 2^{8} = 4^{4} = 16^{2} \)
Minimum value of A + B = 8.

Next time you give a summation, please give more terms. I am still not sure that the series I used is correct. Vighnesh Shenoy · 7 months, 4 weeks ago

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@Vighnesh Shenoy Correct . And yes i will take care of more terms in the series next time. Mayank Chaturvedi · 7 months, 3 weeks ago

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@Vighnesh Shenoy I think you mean \(\tan ^{ -1 }{ \frac { 1 }{ 2 } } \) in the third line, and not \(\tan ^{ -1 }{ \frac { 1 }{ 4 } } \) Abhineet Nayyar · 7 months, 4 weeks ago

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Time for an easy question.

Problem. 24 :
Let \( A_{1}, A_{2}, A_{3}, A_{4} \) be the areas of four faces of a tetrahedron, and \( h_{1}, h_{2}, h_{3}, h_{4} \) be the corresponding altitudes. Give that the volume of tetrahedron is 5 cubic units . Find the minimum value of the expression
\( \dfrac{(A_{1} + A_{2} + A_{3} + A_{4})(h_{1}+h_{2}+h_{3}+h_{4})}{5!} \)

Sarvesh Nalawade has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Vighnesh Shenoy · 7 months, 4 weeks ago

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@Vighnesh Shenoy Volume of tetrahedron = \( \dfrac{A \cdot h}{3} \)
\( \therefore 15 = A_{1} \cdot h_{1} = A_{2} \cdot h_{2} = A_{3} \cdot h_{3} = A_{4} \cdot h_{4} \)

We have to minimize :

\( S = \dfrac{15(A_{1}+A_{2}+A_{3}+A_{4})\left(\dfrac{1}{A_{1}} + \dfrac{1}{A_{2}} + \dfrac{1}{A_{3}} + \dfrac{1}{A_{4}}\right)}{5!} \)
Using AM GM ineqaulity :
\( \displaystyle \sum_{i=1}^{4}A_{i} \geq 4\left({A_{1} \cdot A_{2} \cdot A_{3} \cdot A_{4}}\right)^{\frac{1}{4}} \)

\( \displaystyle \sum_{i=1}^{4} \dfrac{1}{A_{i}} \geq 4\left(\dfrac{1}{A_{1} \cdot A_{2}\cdot A_{3}\cdot A_{4}}\right)^{\frac{1}{4}} \)
Multiplying these two we get :
\( \displaystyle \sum_{i=1}^{4}A_{i} \times \displaystyle \sum_{i=1}^{4} \dfrac{1}{A_{i}} \geq 16 \)

Hence,
\( S_{min} = \dfrac{15 \times 16}{5!} = 2 \) Sarvesh Nalawade · 7 months, 4 weeks ago

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@Sarvesh Nalawade Correct!
I used the Cauchy-Schwartz inequality instead of the AM-GM.
\( \left(\sqrt{A_{1}}^{2}+\sqrt{A_{2}}^{2}+\sqrt{A_{3}}^{2}+\sqrt{A_{4}}^{2}\right)\left(\dfrac{1}{\sqrt{A_{1}}^{2}} + \dfrac{1}{\sqrt{A_{2}}^{2}} + \dfrac{1}{\sqrt{A_{3}}^{2}} + \dfrac{1}{\sqrt{A_{4}}^{2}}\right) \geq \left(\sqrt{A_{1}}\dfrac{1}{\sqrt{A_{1}}} + \sqrt{A_{2}}\dfrac{1}{\sqrt{A_{2}}} + \sqrt{A_{3}}\dfrac{1}{\sqrt{A_{3}}} + \sqrt{A_{4}}\dfrac{1}{\sqrt{A_{4}}}\right)^{2} \)
Equality holds when :
\( A_{1} = A_{2} =A_{3} = A_{4} \) Vighnesh Shenoy · 7 months, 4 weeks ago

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Problem 23:

Find the closed form of \[\int _{ 0 }^{ \infty }{ \left\lfloor \frac { n }{ { e }^{ x } } \right\rfloor dx } \]

Statutory warning: Do not relate this to Gamma Function.

Vighnesh Shenoy has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Aditya Kumar · 8 months ago

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@Aditya Kumar \( I = \displaystyle \int_{0}^{\infty} \left \lfloor \dfrac{n}{e^{x}} \right \rfloor dx \)
\( e^{-x} = t \rightarrow dx = \dfrac{-dt}{t} \)
\( I = \displaystyle \int_{0}^{1} \left \lfloor nt \right \rfloor \dfrac{dt}{t} \)
\( I = \displaystyle \sum_{k=1}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} \dfrac{k}{t} dt \)
\( I = \displaystyle \sum_{k=1}^{n-1} k\log\left(\dfrac{k+1}{k}\right) \)
\( S = \displaystyle \sum_{k=1}^{n-1} \left[(k+1)\log(k+1)-k\log(k) - \log(k+1)\right]\)
\( S = n\log(n) - log(n!)\) Vighnesh Shenoy · 8 months ago

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PROBLEM 22

Let \(a(n)=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1} {2^n-1} \).. Then:

A: \(a(100) \leq 100 \)

B: \(a(100)>100 \)

C: \(a(200)\leq 100 \)

D: \(a(200)> 100 \)

For sake of clarification \(\displaystyle a\left( n \right) =\sum _{ r=1 }^{ { 2 }^{ n }-1 }{ \frac { 1 }{ r } } \)

Aditya Kumar has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Samarth Agarwal · 8 months ago

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@Samarth Agarwal Solution to problem 22:

Clearly \(a(n)\leq n\). Hence, \(a(100)\leq100\)

Now, for \(a(200)\), we need to do the following steps: \[a(n)>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{2^n}+...+\frac{1}{2^n}\right)\]

This is an AGP.

Hence, \(a(200)>\left(1-\frac{1}{2^200}\right)+100>100\).

Therefore options A and D are correct. Aditya Kumar · 8 months ago

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Problem 21

Find the equation of the director circle to the circle circumscribing the quadrilateral formed by the lines in order 2x+3y-2 = 0 , x+y =0 , 2x+5y-1 = 0 , 5x+27y-1 = 0

Samarth Agarwal has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Prakhar Bindal · 8 months ago

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@Prakhar Bindal The equation is \( {(x-11/6)}^{2}+{(y-25/6)}^{2}=349/9 \). The steps are same as that of Samarth's solution Saarthak Marathe · 8 months ago

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@Prakhar Bindal Circle curcumscribing quad is l1l2+(lambda)l3l4=0 lambda can be found by cofficient of xy=0

Here lambda came out to be - 0.5

So circle is (x-11/6)^2+(y-25/6)^2=698/36 and directer circle is (x-11/6)^2+(y-25/6)^2=698/18

Is this correct.? Samarth Agarwal · 8 months ago

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Problem 20:

A series of chords of a parabola \({y}^{2}=4ax\) are drawn so that their projections on the straight line,which is inclined at an angle \(\alpha\) to the axis,are of constant length \(c\). Find the locus of the midpoints of these chords.

Prakhar Bindal has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Saarthak Marathe · 8 months ago

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@Saarthak Marathe Assume end point of chord in parametric form (at^2 , 2at) and (as^2,2as)

Let The equation of line be y = xtan(alpha)+b

Now find foot of perpendicular from end points onto the line. Using distance formula between them and equate it to c .

Let midpoint be (h,k)

2h = a(t^2 + s^2)

2k = 2a(t+s)

From above two relation find value of t-s and t+s as they will come in expression of distance formula.

Substitute these into that expression and on simplifying further and replacing h and k by x and y.

We obtain locus as

p = alpha (inclination of line)

(y^2 - 4ax) (ycosp+2asinp)^2 + (ac)^2 = 0

Alternate : Find out Length of chord using distance of formula. Then find angle between chord and line by using standard results. And finally take component of length of chord along line and equate to c . That would reduce some of the calculation work .

Credits @Samarth Agarwal for alternate way Prakhar Bindal · 8 months ago

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@Prakhar Bindal I think it should be -(ac)^2 Rohit Ner · 8 months ago

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@Rohit Ner @Prakhar Bindal

Do you agree with Rohit Ner? If so, then please edit your solution accordingly. And please use latex into your solution to make it understandable. If you need guidance on latex, you can check out this one. Thanks! Sandeep Bhardwaj · 8 months ago

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@Sandeep Bhardwaj No sir saarthak who posted the problem confirmed that the answer is correct Prakhar Bindal · 8 months ago

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Problem 19:

If \[{ \left( 1-{ x }^{ 3 } \right) }^{ n }=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r }{ \left( 1-x \right) }^{ 3n-2r } } \] then find \(a_r\).

Saarthak Marathe has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Aditya Kumar · 8 months ago

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@Aditya Kumar Write \( {(1-{x}^{3})}^{n} = {(3x(1-x)+{(1-x)}^{3})}^{n}=\sum _{ r=0 }^{ n }{ { 3 }^{ r } } \binom{ n }{ r }{ (x(1-x)) }^{ r }{ (1-x) }^{ 3n-3r } \) Saarthak Marathe · 8 months ago

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@Aditya Kumar Answer is \({3}^{r}\cdot \dbinom{n}{r} \) Saarthak Marathe · 8 months ago

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Problem 18:

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

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 · 8 months ago

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@Prakhar Bindal Solution to Problem 18:

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

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

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

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

Hence \(\boxed{d_{max}=1}\) Aditya Kumar · 8 months ago

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@Aditya Kumar 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 · 8 months ago

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Problem 15:

If N is the number of ways in which 3 distinct numbers can be selected from the set {3, 3^2 ,3^3 ,...... ,3^10 } , so that they form a GP, then find the value of N/5.

Ashu Dablo has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! Saakshi Singh · 8 months ago

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@Saakshi Singh @Saakshi Singh

The problem is similar to having to get number of possible Ap's from the set of {1,2,3..10}

So number of possible Ap's is (8+6+4+2)=20

8 for common dif =1 (1,2,3) , (2,3,4) ........(8,9,10)

6 for common dif =2 (1,3,5) , (2,4,6) ........(6,8,10)

4 for common dif =3 (1,4,7) , (2,5,8) ........(4,7,10)

2 for common dif = 4 (1,5,9) and (2,6,10)

So the answer to your question is 4 Ashu Dablo · 8 months ago

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@Ashu Dablo the number of possibilities in that order is 8+6+4+2 but the the set containing these in reverse order can also be in gp ie(4,7,10) &(10,7,4) similar is the case with all other elements so the So number of possible GP's is 2(8+6+4+2)=220=40 so N/5 =8 this was also my solution Ashwin Kumar · 8 months ago

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@Ashwin Kumar We have to SELECT the numbers, not arrange them in order, which is why I said the answer is 4. Ashu Dablo · 8 months ago

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@Saakshi Singh My answer is 8 (NoteThe following is a table) Difference in power. No terms. Set(of powers) 1. 8. (123,234,...89 10) 2. 6. (135,246....68 10) 3. 4. (147....47 10) 4. 2. (159,2710) Total = 20 But r can be 3^{1,2,3} or 3^{-1,-2,-3} so N = 2x20 = 40 So N/5 =8 Ashwin Kumar · 8 months ago

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Problem 9:

(On behalf of Akul Agarwal)

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

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

Ishan Singh has provided the complete solution to the problem. Thanks! Saarthak Marathe · 8 months ago

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@Saarthak Marathe \( \displaystyle f(x) = x \left(1+\int_{0}^{1} t f(t) \ \mathrm{d}t \right) +\int_{0}^{1} t^2 f(t) \ \mathrm{d}t \)

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

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

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

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

Comparing coefficients, we have,

\(\displaystyle A = \frac{65}{23} \)

\(\displaystyle \implies \int_{0}^{1} f(t) dt = A - 1 = \boxed{\dfrac{42}{23}} \) Ishan Singh · 8 months ago

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Problem 49 :
\( a,b,c,d \) are real numbers such that,
\( a + 2b + 3c + 4d = 15 \)
Find the minimum value of
\( 9^{a} + 2\cdot9^{b} + 3\cdot9^{c} + 4\cdot 9^{d} \)

Vignesh S has posted the solution to this problem. Feel free to add new approaches. Thanks! Vighnesh Shenoy · 7 months, 2 weeks ago

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@Vighnesh Shenoy The answer is 270 Let \[f(x)=9^{a}+2\cdot 9^{b}+3\cdot 9^{c}+4\cdot 9^{d}\] and \[g(x)=a+2b+3c+4d\]. Using Lagrange Multipliers we have \[\dfrac{\partial{f(a,b,c,d)}}{\partial x}=k \times \dfrac{\partial{g(a,b,c,d)}}{\partial x}\] where \(x=a,b,c,d\) \[\dfrac{\partial{(9^{a}+2\cdot 9^{b}+3 \cdot 9^{c}+4 \cdot 9^{d})}}{\partial x}=k \dfrac{\partial (a+2b+3c+4d)}{\partial x}\] \[\ln(9) \cdot 9^{a}=k \times1\] Similarly doing for b,c,d we have \[2\ln(9)\cdot 9^{b}=k \times 2\] \[3\ln(9)\cdot9^{c}=k \times 3\] \[4\ln(9) \cdot 9^{d}=k \times4\] \[\implies a=b=c=d\]. So substituting in \(g\) the above condition we get \[10a=15\implies a=1.5\] Therefore \[\inf{f(x)}=27×(1+2+3+4)=270\] Vignesh S · 7 months, 2 weeks ago

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@Vignesh S the given expression (S) whose minimum value is to be calculated can be written as

S = 9^a + 9^b + 9^b + 9^c + 9^c + 9^c + 9^d + 9^d + 9^d + 9^d

Now applying AM - GM inequality

S/10 >= ( 9 ^(a+2b+3c+4d) )^1/10

S/10 >= 9^(15/10)

S/10 >= 9^3/2 = 27

Therefore S >= 270

and hence minimum value of S is 270 Aayush Patni · 6 months, 4 weeks ago

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@Aayush Patni Look who's back from the dead. Kunal Verma · 6 months, 4 weeks ago

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@Vignesh S @Vignesh S

Please post the next problem(Problem 50) of the contest. And after this problem on this note, the contest will be continued on JEE-Advanced Maths Contest(Continued) note.

Thanks! Sandeep Bhardwaj · 7 months, 2 weeks ago

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@Vignesh S The answer is correct. Vighnesh Shenoy · 7 months, 2 weeks ago

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