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) :

PROBLEM xxx (number of problem) :

Note by Sandeep Bhardwaj
5 years, 3 months ago

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is this contest still on ???

- 1 year, 9 months ago

If a^b^b^a=a√2, find a^2/b

If the equation x^4-Px^2+9=0 has 4 real roots then P lies in the interval:- A} (0,infinity) B} (6,infinity) C} (-infinty,-6)

- 3 years, 1 month ago

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

- 5 years, 3 months ago

Solution by Deeparaj Bhat:

Extend $CP$ to $D$ such that $\Delta BDP$ is equilateral. Then, \begin{aligned}\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{aligned}

Substituting the given values the answer comes out to be 23.684.

- 5 years, 3 months ago

23.684. I'll post the soln in some time...

- 5 years, 3 months ago

Waiting for the solution.

- 5 years, 3 months ago

- 5 years, 3 months ago

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!

- 5 years, 3 months ago

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!

- 5 years, 3 months ago

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 $\inf{f(x)}=27×(1+2+3+4)=270$

- 5 years, 3 months ago

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

- 5 years, 2 months ago

Look who's back from the dead.

- 5 years, 2 months ago

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!

- 5 years, 3 months ago

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

$I = \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))}$
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,

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

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

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

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

$p = 4 , q = \dfrac{3}{2}$
$pq = 6$

- 5 years, 3 months ago

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 $gcd(\alpha, \beta ) = gcd(\alpha,\gamma) = gcd( \beta, \gamma = 1$

- 5 years, 3 months ago

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, $\alpha =7,\quad \beta =9,\quad \gamma =2$

- 5 years, 3 months ago

Correct.

- 5 years, 3 months ago

Getting the same thing @Sarvesh Nalawade. I think you should go ahead and post the solution

- 5 years, 3 months ago

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 ?

- 5 years, 3 months ago

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 $\frac{4}{27}$

- 5 years, 3 months ago

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 $\dfrac{-2}{3} , \dfrac{-2}{3} , \dfrac{1}{3}$

- 5 years, 3 months ago

Great! Please post the next problem. @Vighnesh Shenoy

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

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

- 5 years, 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.

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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!

- 5 years, 3 months ago

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

$\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{\alpha }{2}$

- 5 years, 3 months ago

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

- 5 years, 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.

- 5 years, 3 months ago

Problem 43 :

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

- 5 years, 3 months ago

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

Hence the answer is $\boxed 4$

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

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

- 5 years, 3 months ago

Perfect.

- 5 years, 3 months ago

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!

- 5 years, 3 months ago

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

$\displaystyle \lim_{n \to \infty} \sum_{k=0}^{n-1} \dfrac{n^{\frac{1}{2}}}{(n+3k)^{\frac{3}{2}}} = \displaystyle \int_{0}^{1} \dfrac{1}{(1+3x)^{\frac{3}{2}}}dx = \left[ \dfrac{2}{3(\sqrt{1+3x})}\right]_{1}^{0} = \dfrac{2}{3} - \dfrac{2}{6} = \dfrac{2}{6} = \dfrac{1}{3}$

- 5 years, 3 months ago

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

- 5 years, 3 months ago

Problem 41

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

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

- 5 years, 3 months ago

$\displaystyle tan3x=\frac{tan2x+tanx}{1-tanx.tan2x}\\ tanx.tan2x.tan3x=tan3x-tanx-tan2x$

Now integrating we get,

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

- 5 years, 3 months ago

Correct!Post the next question.

- 5 years, 3 months ago

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!

- 5 years, 3 months ago

The above question can be written in the following manner:

$\displaystyle \sum_{r=0}^{n} (2r+1)\dbinom{n}{r}$ $=2\displaystyle\sum_{r=0}^{n}r\dbinom{n}{r} +\displaystyle\sum_{r=0}^{n}\dbinom{n}{r} = n2^n +2^n= \boxed{(n+1)2^n}$

- 5 years, 3 months ago

- 5 years, 3 months ago

Problem 39:

Solve

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

- 5 years, 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.

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

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 $\boxed {108545}$

- 5 years, 3 months ago

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

Thanks!

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

$\dbinom{12}{6}\dbinom{6}{3}\dbinom{6}{2}\dbinom{4}{2}\dbinom{5}{3}=16632000$

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

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

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

- 5 years, 3 months ago

16632000

- 5 years, 3 months ago

OK! Thanks! Waiting for the next question from you...

- 5 years, 3 months ago

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!

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

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

- 5 years, 3 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 $d$

- 5 years, 3 months ago

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

- 5 years, 3 months ago

yes

- 5 years, 3 months ago

Great!!!

- 5 years, 3 months ago

A graphical approach.

We have to see the nature of discriminant

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

- 5 years, 3 months ago

$\Delta = (a)^{2} -4(a^{2}-6) = 24-3a^{2} = 3(8-a^{2})$
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.

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

Oh yes.

- 5 years, 3 months ago

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

- 5 years, 3 months ago

Mayank should post the new question.

- 5 years, 3 months ago

OK. @Mayank Chaturvedi,please post the next question

- 5 years, 3 months ago

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.
Answer: ${e}^{-\frac{\alpha\pi}{\beta}}$