JEE-Advanced Maths Contest '16

Hello, guys!

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

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

    • Please write the detailed solutions to the problems.

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :

[Post your solution here]

PROBLEM xxx (number of problem) :

[Post your problem here]

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

Note by Sandeep Bhardwaj
3 years, 4 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

If a^b^b^a=a√2, find a^2/b

Kalyanasundaram Perum - 1 year, 1 month ago

Log in to reply

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)

amrit dubey - 1 year, 2 months ago

Log in to reply

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

A cevian AQAQ of a equilateral Δ\Delta ABCABC is extended to meet circumcircle at PP. If PB=50PB=50 and PC=45PC=45, find PQPQ upto 3 decimal places.

Deeparaj Bhat has provided the answer

Vignesh S - 3 years, 3 months ago

Log in to reply

Solution by Deeparaj Bhat:

Extend CPCP to DD such that ΔBDP\Delta BDP is equilateral. Then, BCD=QCPQPC=ABC( angles in the same segment are equal )=60=BDP    ΔDCBΔPCQ    DBPQ=DCPC=1+PBPC(DC=DP+PC=BP+PC)    1PC+1PB=1PQ(BP=DB) \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.

Sandeep Bhardwaj - 3 years, 3 months ago

Log in to reply

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

Deeparaj Bhat - 3 years, 3 months ago

Log in to reply

Waiting for the solution.

Sandeep Bhardwaj - 3 years, 3 months ago

Log in to reply

@Sandeep Bhardwaj Please provide the complete solution.

Vignesh S - 3 years, 3 months ago

Log in to reply

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 - 3 years, 3 months ago

Log in to reply

Problem 49 :
a,b,c,d a,b,c,d are real numbers such that,
a+2b+3c+4d=15 a + 2b + 3c + 4d = 15
Find the minimum value of
9a+29b+39c+49d 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!

A Brilliant Member - 3 years, 3 months ago

Log in to reply

The answer is 270 Let f(x)=9a+29b+39c+49df(x)=9^{a}+2\cdot 9^{b}+3\cdot 9^{c}+4\cdot 9^{d} and g(x)=a+2b+3c+4dg(x)=a+2b+3c+4d. Using Lagrange Multipliers we have f(a,b,c,d)x=k×g(a,b,c,d)x\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,dx=a,b,c,d (9a+29b+39c+49d)x=k(a+2b+3c+4d)x\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)9a=k×1\ln(9) \cdot 9^{a}=k \times1 Similarly doing for b,c,d we have 2ln(9)9b=k×22\ln(9)\cdot 9^{b}=k \times 2 3ln(9)9c=k×33\ln(9)\cdot9^{c}=k \times 3 4ln(9)9d=k×44\ln(9) \cdot 9^{d}=k \times4     a=b=c=d\implies a=b=c=d. So substituting in gg the above condition we get 10a=15    a=1.510a=15\implies a=1.5 Therefore inff(x)=27×(1+2+3+4)=270\inf{f(x)}=27×(1+2+3+4)=270

Vignesh S - 3 years, 3 months ago

Log in to reply

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 - 3 years, 3 months ago

Log in to reply

@Aayush Patni Look who's back from the dead.

Kunal Verma - 3 years, 3 months ago

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 - 3 years, 3 months ago

Log in to reply

The answer is correct.

A Brilliant Member - 3 years, 3 months ago

Log in to reply

Problem 48 : If 0π2dx(acos2x+bsin2x)2=π(a+b)p(ab)q \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 - 3 years, 3 months ago

Log in to reply

I=0π2dx(acos2(x)+bsin2(x)) I = \displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{(a\cos^{2}(x) + b\sin^{2}(x))}
Put, tan(x)=t \tan(x) = t

I=0dta+bt2=π2abI = \displaystyle \int_{0}^{\infty} \dfrac{dt}{a+bt^{2}} = \dfrac{\pi}{2\sqrt{ab}}
0π2dx(acos2(x)+bsin2(x))=π2(ab)12 \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,
ddyabf(x,y)dx=abyf(x,y)dx\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,

0π2adx(acos2(x)+bsin2(x))=a(π2(ab)12) \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)

0π2cos2dx(acos2(x)+bsin2(x))2=π4b12a32 \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,

0π2bdx(acos2(x)+bsin2(x))=b(π2(ab)12) \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)

0π2sin2dx(acos2(x)+bsin2(x))2=π4b32a12 \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,

0π2dx(acos2(x)+bsin2(x))2=π4ab(1a+1b) \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)

0π2dx(acos2(x)+bsin2(x))2=π(a+b)4(ab)32 \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=32 p = 4 , q = \dfrac{3}{2}
pq=6 pq = 6

A Brilliant Member - 3 years, 3 months ago

Log in to reply

Problem 47 : In a ΔABC \Delta ABC the ratio of side BCBC and AC AC to the circumradius is 2 2 and 32 \dfrac{3}{2} respectively. If the ratio of length of angle bisectors of angle B B to length of angle bisector of angle C C is given by α(α1)βγ \dfrac{\alpha(\sqrt{\alpha}-1)}{\beta\sqrt{\gamma}} find
α+β+γ3 \dfrac{\alpha + \beta + \gamma}{3}

Note :
α,β,γ \alpha, \beta, \gamma are positive integers with gcd(α,β)=gcd(α,γ)=gcd(β,γ=1 gcd(\alpha, \beta ) = gcd(\alpha,\gamma) = gcd( \beta, \gamma = 1

A Brilliant Member - 3 years, 3 months ago

Log in to reply

Firstly Using Sine Rule we get,

asinA=bsinB=2R \frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =2R

Hence, we get A=π2andsinB=34 A = \frac{\pi}{2} and \sin{B} = \frac{3}{4}

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

Let BD and CE be the respective angle bisectors.

ADCD=BABCandAEEB=ACBC \frac { AD }{ CD } =\frac { BA }{ BC } \quad and\quad \frac { AE }{ EB } =\frac { AC }{ BC }

AD=374+7kandAE=37k 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=471+7kandCE=627kBD=\frac { 4\sqrt { 7 } }{ 1+\sqrt { 7 } } k\quad and\quad CE=\frac { 6\sqrt { 2 } }{\sqrt{ 7 }} k\quad

Their ratio will be : 7(71)92 \dfrac{7(\sqrt{7} - 1 )}{9\sqrt2}

Hence, α=7,β=9,γ=2 \alpha =7,\quad \beta =9,\quad \gamma =2

Therefore, final answer is 6

Sarvesh Nalawade - 3 years, 3 months ago

Log in to reply

Correct.

A Brilliant Member - 3 years, 3 months ago

Log in to reply

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

Miraj Shah - 3 years, 3 months ago

Log in to reply

Problem 46

If M=(abcbcacab) 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 - 3 years, 4 months ago

Log in to reply

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

a2+b2+c2=1...(1)ab+bc+ca=0...(2)a^2+b^2+c^2=1...(1) \\ab+bc+ca=0...(2)

Let

a=cosαb=cosβc=cosγa=cos\alpha\\b=cos\beta\\c=cos\gamma

From (1)(1) we can write cos2α+cos2β=sin2γ...(3)cos^2\alpha+cos^2\beta=sin^2\gamma...(3)

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

cosα+cosβ+cosγ=±1cosα+cosβ=±1cosγcos\alpha+cos\beta+cos\gamma=\pm 1\\ cos\alpha+cos\beta=\pm 1-cos\gamma

Squaring both sides we get

cos2α+cos2β+2cosα.cosβ=1+cos2γ2cosγcos^2\alpha+cos^2\beta+2cos\alpha .cos\beta=1+cos^2\gamma \mp 2cos\gamma

From (3)(3) we can write cosα.cosβ=cos2γcosγcos\alpha .cos\beta=cos^2\gamma \mp cos\gamma

Now multiplying cosγcos\gamma on both sides we get:

cosα.cosβ.cosγ=cos3γcos2γcos\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.) {0,23}\{0,\frac{2}{3}\} when cosα.cosβ.cosγ=cos3γcos2γcos\alpha .cos\beta .cos\gamma=cos^3\gamma- cos^2\gamma

(2.) {0,23}\{0,\frac{-2}{3}\} when cosα.cosβ.cosγ=cos3γ+cos2γcos\alpha .cos\beta .cos\gamma=cos^3\gamma+cos^2\gamma

On checking these values, minima occurs at cosγ=23cos\gamma = \frac{2}{3} and the minimum value is 427\frac{-4}{27}, and maximum occur maxima occurs at cosγ=23cos\gamma = \frac{-2}{3} and the maximum value is 427\frac{4}{27}

Miraj Shah - 3 years, 3 months ago

Log in to reply

For an orthogonal matrix M,
MMT=I MM^{T} = I where I is the identity matrix.
Multiplying the matrices on LHS , and comparing with the identity matrix we get,
a2+b2+c2=1 a^{2} + b^{2} + c^{2} = 1
ab+bc+ca=0 ab + bc + ca = 0
From these two equations we get,
a+b+c=±1 a + b + c = \pm 1
f(x)=x3(a+b+c)x2+(ab+bc+ca)xd 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) = 0
f(x)=x3(a+b+c)x2d f(x) = x^{3} - (a+b+c)x^{2} - d
Let a+b+c=p a + b + c = p
f(x)=x3px2d f(x) =x^{3} - px^{2} - d
Differentiating,
f(x)=3x22px f'(x) = 3x^{2} - 2px
The roots of f(x)=0 f'(x) = 0 are,
x=0,2p3 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 p =1
The roots are x=0,23 x = 0 , \dfrac{2}{3}

f(0)f(23)0 \therefore f(0)f\left(\dfrac{2}{3}\right) \le 0
(d)(427d)0 \left(-d\right)\left(\dfrac{-4}{27}-d\right) \le 0
(d)(d+427)0 \left(d\right)\left(d+\dfrac{4}{27}\right) \le 0
d[427,0] d \in \left[\dfrac{-4}{27},0 \right]
When p=1 p = -1
The roots of f(x)=0 f'(x) = 0 are x=0,23 x = 0 , \dfrac{-2}{3}
f(0)f(23)0 f(0)f\left(\dfrac{-2}{3}\right) \le 0
(d)(427d)0 \left(-d\right)\left(\dfrac{4}{27}-d\right)\le 0
d[0,427] d \in \left[0,\dfrac{4}{27}\right]
Range of d [427,427] \left[\dfrac{-4}{27}, \dfrac{4}{27} \right]
The mininum value occurs when a,b,c a,b,c are a permutation of 23,23,13 \dfrac{2}{3}, \dfrac{2}{3}, \dfrac{-1}{3} .
The maximum value occurs when a,b,c a,b,c are a permutation of 23,23,13 \dfrac{-2}{3} , \dfrac{-2}{3} , \dfrac{1}{3}

A Brilliant Member - 3 years, 3 months ago

Log in to reply

Great! Please post the next problem. @Vighnesh Shenoy

Sandeep Bhardwaj - 3 years, 3 months ago

Log in to reply

Problem 45

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

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

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

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

Sarvesh Nalawade has provided a complete solution to the problem.

Kunal Verma - 3 years, 4 months ago

Log in to reply

Let S = n!i=1n(xi(ni1)+(ni)(ni1)) n! \prod_{i=1}^{n} ( \frac{x}{i} - \frac{{n \choose i-1} + {n \choose i}}{{n \choose i-1}})

Therefore,S=n!i=1n(xi(n+1i)(ni1)) Therefore, S= n! \prod_{i=1}^{n} ( \frac{x}{i} - \frac{{n+1 \choose i}}{{n \choose i-1}} )

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

Therefore,S=(xn1)n Therefore, S = (x-n-1)^{n} .

Co-efficient of xn6=(n6)(n+1)6x^{n-6} = {n \choose 6} (n+1)^{6}

Therefore x=n , y=6, z= n+1

Sarvesh Nalawade - 3 years, 4 months ago

Log in to reply

You are required to find the individual values of xx , y y and z z , not the sum ,although it is correct. You may post the next problem after posting the solution.

Kunal Verma - 3 years, 4 months ago

Log in to reply

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

A Brilliant Member - 3 years, 4 months ago

Log in to reply

Problem 44:

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

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

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

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

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

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

Kunal Verma - 3 years, 4 months ago

Log in to reply

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

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

Kunal Verma - 3 years, 4 months ago

Log in to reply

Problem 43 :

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

A Brilliant Member - 3 years, 4 months ago

Log in to reply

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

Hence the answer is 4\boxed 4

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

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

A Brilliant Member - 3 years, 4 months ago

Log in to reply

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

Let me try once.

Reference-euler's theorem

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

Mayank Chaturvedi - 3 years, 4 months ago

Log in to reply

@Mayank Chaturvedi Perfect.

A Brilliant Member - 3 years, 4 months ago

Log in to reply

Problem 42:

Evaluate

limn[n1/2n3/2+n1/2(n+3)3/2+n1/2(n+6)3/2+........+n1/2(n+3(n1))3/2]\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 - 3 years, 4 months ago

Log in to reply

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

limnk=0n1n12(n+3k)32=011(1+3x)32dx=[23(1+3x)]10=2326=26=13 \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}

A Brilliant Member - 3 years, 4 months ago

Log in to reply

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

Sandeep Bhardwaj - 3 years, 4 months ago

Log in to reply

Problem 41

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

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

Miraj Shah - 3 years, 4 months ago

Log in to reply

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

Now integrating we get,

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

Correct!Post the next question.

Miraj Shah - 3 years, 4 months ago

Log in to reply

Problem no. 40.

An easy question:Find the following (n0)+3(n1)+5(n2)+........+(2n+1)(nn)\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 - 3 years, 4 months ago

Log in to reply

The above question can be written in the following manner:

r=0n(2r+1)(nr)\displaystyle \sum_{r=0}^{n} (2r+1)\dbinom{n}{r} =2r=0nr(nr)+r=0n(nr)=n2n+2n=(n+1)2n =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 - 3 years, 4 months ago

Log in to reply

Please post the next question

Vignesh S - 3 years, 4 months ago

Log in to reply

Problem 39:

Solve

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

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

Vignesh S - 3 years, 4 months ago

Log in to reply

Problem 38:

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

Let the person travel dd diagonals in one trip. So the remaining 152d15-2d sides,he travels up or right, which can be done in (152d8d)\dbinom{15-2d}{8-d} ways. The dd diagonals can be placed in (15dd)\dbinom{15-d}{d} ways.

So the total number of ways=d=07(152d8d)(15dd)\displaystyle \sum_{d=0}^{7} \dbinom{15-2d}{8-d}\dbinom{15-d}{d} .

This summation turns out as 108545\boxed {108545}

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

@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 - 3 years, 4 months ago

Log in to reply

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 - 3 years, 4 months ago

Log in to reply

The number of ways distributing nn distinct things into rr groups and arranging them among rr people such that s1,s2,s3,...srs_1, s_2,s_3,...s_r denotes the number of things in the respective groups is :

(n!)(r!)(s1!)(s2!)...(sr!)\large \frac{(n!)( r!)}{(s_1!)( s_2!) ...(s_r!)}

Note in the above formula s1s2s3...srs_1\neq s_2\neq s_3\neq ... s_r

For the given question n=12n=12; r=5r=5; s1=s2=3s_1=s_2=3 and s3=s4=s5=2s_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 s1s_1 and s2s_2 and there is no difference between s3,s4s_3, s_4 and s5s_5. Hence the number of cases will reduce by a factor of 2!2! with respect to s1s_1 and s2s_2 and by 3!3! with respect to s3,s4s_3, s_4 and s5s_5. Hence, in totality the required answer is:

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

Is the above answer correct? @Mayank Chaturvedi

Miraj Shah - 3 years, 4 months ago

Log in to reply

(126)(63)(62)(42)(53)=16632000\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 - 3 years, 4 months ago

Log in to reply

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 - 3 years, 4 months ago

Log in to reply

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

Miraj Shah - 3 years, 4 months ago

Log in to reply

@Miraj Shah 16632000

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

@Saarthak Marathe OK! Thanks! Waiting for the next question from you...

Miraj Shah - 3 years, 4 months ago

Log in to reply

Problem 36:

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

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

A non-calculus approach.

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

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

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

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

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

Substitute these values to get aa.

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

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

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

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

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

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

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

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

Mayank Chaturvedi - 3 years, 4 months ago

Log in to reply

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

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

Mayank Chaturvedi - 3 years, 4 months ago

Log in to reply

@Mayank Chaturvedi yes

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

@Saarthak Marathe Great!!!

Mayank Chaturvedi - 3 years, 4 months ago

Log in to reply

A graphical approach.

We have to see the nature of discriminant

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

Mayank Chaturvedi - 3 years, 4 months ago

Log in to reply

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

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

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

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

A Brilliant Member - 3 years, 4 months ago

Log in to reply

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

Miraj Shah - 3 years, 4 months ago

Log in to reply

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

@Miraj Shah Oh yes.

Mayank Chaturvedi - 3 years, 4 months ago

Log in to reply

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

A Brilliant Member - 3 years, 4 months ago

Log in to reply

@A Brilliant Member Mayank should post the new question.

A Brilliant Member - 3 years, 4 months ago

Log in to reply

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

Saarthak Marathe - 3 years, 4 months ago

Log in to reply

Problem 35:

It can be proved that the areas S0,S1,S2,S3,...S_0,S_1,S_2,S_3,... bounded by the xaxisx-axis