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:

- I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
- 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. - Please make a
**substantial comment**. - 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. - 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. - 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.
- You can use tricks/short methods/specific cases to post the solution of a problem only if they are logically valid.
- In the case of any argument or disagreement among the solvers, the decision/judgement is passed to me.
**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.*- Proof problems are not allowed.
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. (>‿◠)✌

## Comments

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TopNewestHere 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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months ago

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– Shivam Mahajan · 1 year, 3 months ago

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 = 27/32Log in to reply

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 · 1 year, 4 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 · 1 year, 4 months ago

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– Aditya Kumar · 1 year, 4 months ago

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

@Aditya Kumar ,your solution was incorrect. – Saarthak Marathe · 1 year, 4 months ago

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Can you please post that again? – Sandeep Bhardwaj · 1 year, 4 months ago

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

\[\implies \sin C=T(\sin A-\sin B)\] \[\implies a-b=\dfrac{c}{T}=\text{Constant}<c\] Here \(a,b,c\) denote sides opposite to angles \(A,B,C\) respectively.. A hyperbola..... Is it? – Rishabh Cool · 1 year, 4 months ago

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@Ishan Singh

Please reply to Rishabh's comment. – Sandeep Bhardwaj · 1 year, 4 months ago

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– Ishan Singh · 1 year, 4 months ago

Rishabh's solution is correct (I had stated it earlier). Aditya's solution was incorrect.Log in to reply

Problem 45The 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 · 1 year, 3 months agoLog in to reply

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

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

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

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

Therefore x=n , y=6, z= n+1 – Sarvesh Nalawade · 1 year, 3 months ago

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– Kunal Verma · 1 year, 3 months ago

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.Log in to reply

– Vighnesh Shenoy · 1 year, 3 months ago

Don't use the variable x for two purposes in the same question.Log in to reply

Problem 41\(\displaystyle \int tanx. tan2x. tan3x\text{ }\,dx\)

Saarthak Marathe has posted the solution to the problem.Thanks!– Miraj Shah · 1 year, 3 months agoLog in to reply

Now integrating we get,

\(\displaystyle \int tanx.tan2x.tan3x.dx=\frac{ln|sec3x|}{3}-\frac{ln|sec2x|}{2}-ln|secx|+constant \) – Saarthak Marathe · 1 year, 3 months ago

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– Miraj Shah · 1 year, 3 months ago

Correct!Post the next question.Log in to reply

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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months ago

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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 · 1 year, 4 months ago

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– Mayank Chaturvedi · 1 year, 4 months ago

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

– Saarthak Marathe · 1 year, 4 months ago

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 \(d\)Log in to reply

– Mayank Chaturvedi · 1 year, 4 months ago

By solvable, do you mean quadratic type equations with higher degrees?Log in to reply

– Saarthak Marathe · 1 year, 4 months ago

yesLog in to reply

– Mayank Chaturvedi · 1 year, 4 months ago

Great!!!Log in to reply

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 · 1 year, 4 months ago

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– Miraj Shah · 1 year, 4 months ago

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

@Vighnesh Shenoy. @Vighnesh Shenoy and @Mayank Chaturvedi,please make changes in your solutions accordingly. – Saarthak Marathe · 1 year, 4 months ago

I am changing the wording of the question ,as suggested byLog in to reply

– Mayank Chaturvedi · 1 year, 4 months ago

Oh yes.Log in to reply

– Vighnesh Shenoy · 1 year, 4 months ago

I guess he should have asked if the roots were real or complex.Log in to reply

– Vighnesh Shenoy · 1 year, 4 months ago

Mayank should post the new question.Log in to reply

@Mayank Chaturvedi,please post the next question – Saarthak Marathe · 1 year, 4 months ago

OK.Log in to reply

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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months 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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 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 · 1 year, 4 months agoLog in to reply

\( 1^{m} + (-1)^{m} + i^{m} + (-i)^{m} = 4 , m = 4k, \) 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 · 1 year, 4 months ago

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put i, -i,1, -1 in expansion of (1+x)^n and add – Akul Agrawal · 1 year, 4 months 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 · 1 year, 4 months agoLog in to reply

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

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

\( h(x) = (f(x))^{2} + (g(x))^{2} \)

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 · 1 year, 4 months 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 · 1 year, 3 months ago

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

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 · 1 year, 3 months ago

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@Sarvesh Nalawade. I think you should go ahead and post the solution – Miraj Shah · 1 year, 3 months ago

Getting the same thingLog in to reply

– Vighnesh Shenoy · 1 year, 3 months ago

Correct.Log in to reply

Problem 46If \( M = \begin{pmatrix} a & b & c \\ b & c & a \\ c & a & b \end{pmatrix} \) is an orthogonal matrix with real entries , what is the minimum value of abc ? – Sarvesh Nalawade · 1 year, 3 months ago

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\( MM^{T} = I \) 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 · 1 year, 3 months ago

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@Vighnesh Shenoy – Sandeep Bhardwaj · 1 year, 3 months ago

Great! Please post the next problem.Log in to reply

An alternate solution to@Vighnesh Shenoy 'salready 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 pointsof 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 · 1 year, 3 months 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 · 1 year, 3 months ago

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– Vignesh S · 1 year, 3 months ago

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.Log in to reply

\(PROBLEM \quad 29\):

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 · 1 year, 4 months ago

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. 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 · 1 year, 4 months agoLog in to reply

– Sandeep Bhardwaj · 1 year, 4 months ago

Great! Can you please add the graph for the sake of solution? Thanks!Log in to reply

Correct– Mayank Chaturvedi · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months agoLog in to reply

– Mayank Chaturvedi · 1 year, 4 months ago

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.Log in to reply

– Vighnesh Shenoy · 1 year, 4 months ago

Correct.Log in to reply

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 · 1 year, 4 months agoLog in to reply

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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 · 1 year, 4 months ago

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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 · 1 year, 4 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 · 1 year, 4 months agoLog in to reply

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

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

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 · 1 year, 4 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 · 1 year, 4 months agoLog in to reply

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

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 · 1 year, 4 months ago

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this note. @Saakshi Singh – Sandeep Bhardwaj · 1 year, 4 months ago

I've cleaned up your solution for now. But it's very easy to type an equation using latex which you can learn fromLog in to reply

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 · 1 year, 4 months agoLog in to reply

@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 · 1 year, 4 months ago

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– Keshav Tiwari · 1 year, 4 months ago

The equation becomes inconsistent for n=0 . Please check .Log in to reply

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 · 1 year, 4 months agoLog in to reply

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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 · 1 year, 4 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 · 1 year, 4 months agoLog in to reply

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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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months 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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months ago

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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 · 1 year, 4 months ago

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

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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months agoLog in to reply

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

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 · 1 year, 4 months 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 · 1 year, 4 months agoLog in to reply

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

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 · 1 year, 4 months 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 · 1 year, 3 months agoLog in to reply

JEE-Advanced Maths Contest (Continued).

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:Thanks! – Sandeep Bhardwaj · 1 year, 3 months ago

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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 · 1 year, 3 months agoLog in to reply

– Deeparaj Bhat · 1 year, 3 months ago

23.684. I'll post the soln in some time...Log in to reply

– Sandeep Bhardwaj · 1 year, 3 months ago

Waiting for the solution.Log in to reply

– Vignesh S · 1 year, 3 months ago

Please provide the complete solution.Log in to reply

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 · 1 year, 3 months agoLog in to reply

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 · 1 year, 3 months 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 · 1 year, 3 months agoLog in to reply

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 · 1 year, 3 months ago

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– Saarthak Marathe · 1 year, 3 months ago

Correct! But prove your 1st statement in your solution and post the next questionLog in to reply

– Kunal Verma · 1 year, 3 months ago

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.Log in to reply

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 · 1 year, 3 months ago

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Hence the answer is \(\boxed 4\) – Saarthak Marathe · 1 year, 3 months ago

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\( 32^{32^{32}} \equiv (32^{32})^{32} \equiv 9^{32} \equiv 4 \pmod{7} \) – Vighnesh Shenoy · 1 year, 3 months ago

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agreewith 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 · 1 year, 3 months ago

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– Vighnesh Shenoy · 1 year, 3 months ago

Perfect.Log in to reply

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\( a^{b^{c}} \ne (a^{b})^{c} \) – Vighnesh Shenoy · 1 year, 3 months 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 · 1 year, 3 months agoLog in to reply

definite integral as the sum of limits, which is applied as

This problem can be easily solved by using the concept of\( \displaystyle \lim_{n \to \infty} \sum_{k=0}^{n-1} \dfrac{n^{\frac{1}{2}}}{(n+3k)^{\frac{3}{2}}} = \displaystyle \int_{0}^{1} \dfrac{1}{(1+3x)^{\frac{3}{2}}}dx = \left[ \dfrac{2}{3(\sqrt{1+3x})}\right]_{1}^{0} = \dfrac{2}{3} - \dfrac{2}{6} = \dfrac{2}{6} = \dfrac{1}{3} \) – Vighnesh Shenoy · 1 year, 3 months ago

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– Sandeep Bhardwaj · 1 year, 3 months ago

Great! It would be awesome if you can make your solution more detailed. You can post the next problem. Thanks!Log in to reply

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 · 1 year, 3 months ago

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\(\displaystyle \sum_{r=0}^{n} (2r+1)\dbinom{n}{r}\) \( =2\displaystyle\sum_{r=0}^{n}r\dbinom{n}{r} +\displaystyle\sum_{r=0}^{n}\dbinom{n}{r} = n2^n +2^n= \boxed{(n+1)2^n}\) – Miraj Shah · 1 year, 3 months ago

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– Vignesh S · 1 year, 3 months ago

Please post the next questionLog in to reply

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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 3 months ago

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@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 · 1 year, 3 months ago

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PROBLEM 37In 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 · 1 year, 4 months agoLog in to reply

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

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 · 1 year, 4 months ago

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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 · 1 year, 4 months ago

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Crystal clearBut 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 · 1 year, 4 months agoLog in to reply

– Miraj Shah · 1 year, 4 months ago

Can you please tell what is the numerical figure you are getting after solving your expression?Log in to reply

– Saarthak Marathe · 1 year, 4 months ago

16632000Log in to reply

– Miraj Shah · 1 year, 4 months ago

OK! Thanks! Waiting for the next question from you...Log in to reply

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 · 1 year, 4 months ago

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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 · 1 year, 4 months ago

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Thanks! – Miraj Shah · 1 year, 4 months ago

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– Saarthak Marathe · 1 year, 4 months ago

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 \({S}_{j-1}\)Log in to reply

– Miraj Shah · 1 year, 4 months ago

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!Log in to reply

Problem 34:Solve:\(\displaystyle ({x}^{2}+y)\frac{dy}{dx}=6x \) – Saarthak Marathe · 1 year, 4 months ago

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

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 · 1 year, 4 months ago

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– Saarthak Marathe · 1 year, 4 months ago

It is incorrect. You missed a number in your final answerLog in to reply

– Miraj Shah · 1 year, 4 months ago

Got it! it should be \(x^2+y+3\) right?Log in to reply

– Saarthak Marathe · 1 year, 4 months ago

Yes correct!! Post the next questionLog in to reply

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 · 1 year, 4 months ago

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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 eventto thearea of sample space(\(=3600\))gives the answer \(\frac{5}{9}\) – Miraj Shah · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months ago

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– Miraj Shah · 1 year, 4 months ago

Ya! The answer is correct.Log in to reply

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 · 1 year, 4 months ago

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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 · 1 year, 4 months ago

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– Rohit Ner · 1 year, 4 months ago

Correct. :). Waiting for the next problem.Log in to reply

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 · 1 year, 4 months ago

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\(\begin{align} P&=y\left(ax+bz\right)\\&=\dfrac{{R}^2}{2}\sin 2\beta \left(a\sin\alpha+b\cos\alpha\right)\\{P}_{max}&=\dfrac{{R}^2}{2}\sqrt{{a}^2+{b}^2}\\\Rightarrow f(a,b)&=\dfrac{\sqrt{{a}^2+{b}^2}}{2}\\f(3,4)&=\frac{5}{2}\end{align}\) – Rohit Ner · 1 year, 4 months ago

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– Vighnesh Shenoy · 1 year, 4 months ago

Correct.Log in to reply

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 · 1 year, 4 months ago

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– Vighnesh Shenoy · 1 year, 4 months ago

You guys decide who is going to post next.Log in to reply

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 · 1 year, 4 months ago

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\( P(x)(x+1) - 1 = g(x) \)

\( g(x) \) 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 · 1 year, 4 months ago

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– Sarvesh Nalawade · 1 year, 4 months ago

That's correct !!Log in to reply

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 · 1 year, 4 months agoLog in to reply

– Rohit Ner · 1 year, 4 months ago

Sir please extend this page by making a new note as it is becoming slower to load the contents and rendering the latex.Log in to reply

@Sandeep Bhardwaj. The note is lagging too much and taking too much time to load. – Saarthak Marathe · 1 year, 4 months ago

Yes sirLog in to reply

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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 · 1 year, 4 months ago

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@Mayank Chaturvedi – Sandeep Bhardwaj · 1 year, 4 months ago

Great! Can you please post the next problem?Log in to reply

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 · 1 year, 4 months agoLog in to reply

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

\(S = \tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{8} + \tan^{-1}\dfrac{1}{18} + \ldots \)

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

\( \therefore n = 8 \)

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 · 1 year, 4 months ago

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– Mayank Chaturvedi · 1 year, 4 months ago

Correct . And yes i will take care of more terms in the series next time.Log in to reply

– Abhineet Nayyar · 1 year, 4 months ago

I think you mean \(\tan ^{ -1 }{ \frac { 1 }{ 2 } } \) in the third line, and not \(\tan ^{ -1 }{ \frac { 1 }{ 4 } } \)Log in to reply

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 · 1 year, 4 months agoLog in to reply

\( \therefore 15 = A_{1} \cdot h_{1} = A_{2} \cdot h_{2} = A_{3} \cdot h_{3} = A_{4} \cdot h_{4} \)

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 · 1 year, 4 months ago

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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 · 1 year, 4 months 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 · 1 year, 4 months agoLog in to reply

\( e^{-x} = t \rightarrow dx = \dfrac{-dt}{t} \)

\( I = \displaystyle \int_{0}^{1} \left \lfloor nt \right \rfloor \dfrac{dt}{t} \)

\( I = \displaystyle \sum_{k=1}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} \dfrac{k}{t} dt \)

\( I = \displaystyle \sum_{k=1}^{n-1} k\log\left(\dfrac{k+1}{k}\right) \)

\( S = \displaystyle \sum_{k=1}^{n-1} \left[(k+1)\log(k+1)-k\log(k) - \log(k+1)\right]\)

\( S = n\log(n) - log(n!)\) – Vighnesh Shenoy · 1 year, 4 months ago

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PROBLEM 22Let \(a(n)=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1} {2^n-1} \).. 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 · 1 year, 4 months agoLog in to reply

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

AandDare correct. – Aditya Kumar · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months agoLog in to reply

– Saarthak Marathe · 1 year, 4 months ago

The equation is \( {(x-11/6)}^{2}+{(y-25/6)}^{2}=349/9 \). The steps are same as that of Samarth's solutionLog in to reply

l2+(lambda)l3l4=0 lambda can be found by cofficient of xy=0Here 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 · 1 year, 4 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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months ago

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– Rohit Ner · 1 year, 4 months ago

I think it should be -(ac)^2Log in to reply

@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 · 1 year, 4 months ago

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– Prakhar Bindal · 1 year, 4 months ago

No sir saarthak who posted the problem confirmed that the answer is correctLog in to reply

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 · 1 year, 4 months agoLog in to reply

– Saarthak Marathe · 1 year, 4 months ago

Write \( {(1-{x}^{3})}^{n} = {(3x(1-x)+{(1-x)}^{3})}^{n}=\sum _{ r=0 }^{ n }{ { 3 }^{ r } } \binom{ n }{ r }{ (x(1-x)) }^{ r }{ (1-x) }^{ 3n-3r } \)Log in to reply

– Saarthak Marathe · 1 year, 4 months ago

Answer is \({3}^{r}\cdot \dbinom{n}{r} \)Log in to reply

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 · 1 year, 4 months agoLog in to reply

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 · 1 year, 4 months ago

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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 · 1 year, 4 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 · 1 year, 4 months agoLog in to reply

@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 · 1 year, 4 months agoLog in to reply

(8+6+4+2)=220=40 so N/5 =8 this was also my solution – Ashwin Kumar · 1 year, 4 months agoLog in to reply

– Ashu Dablo · 1 year, 4 months ago

We have to SELECT the numbers, not arrange them in order, which is why I said the answer is 4.Log in to reply

– Ashwin Kumar · 1 year, 4 months ago

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 =8Log in to reply

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 · 1 year, 4 months agoLog in to reply

\(\displaystyle = Ax +B\) (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 · 1 year, 4 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 · 1 year, 3 months agoLog in to reply

270Let \[f(x)=9^{a}+2\cdot 9^{b}+3\cdot 9^{c}+4\cdot 9^{d}\] 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 · 1 year, 3 months agoLog in to reply

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 · 1 year, 3 months ago

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– Kunal Verma · 1 year, 3 months ago

Look who's back from the dead.Log in to reply

@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 · 1 year, 3 months ago

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– Vighnesh Shenoy · 1 year, 3 months ago

The answer is correct.Log in to reply