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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
1 year, 4 months ago

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

Problem 1:

In a sequence of independent trials, the probability of success in one trial is $$\frac 14$$. 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! · 1 year, 4 months ago

Solution of problem 1

The required probability is

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

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 · 1 year, 4 months ago

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? · 1 year, 4 months ago

I had posted the solution! Some careless mod has deleted it! I can't believe it. · 1 year, 4 months ago

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

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

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

Comment deleted Apr 10, 2016

Comment deleted Apr 10, 2016

$\dfrac{\tan(A/2)+\tan(B/2)}{\tan(A/2)-\tan(B/2)}=\dfrac{\ln \pi+1}{\ln \pi-1}=T~(Let)$

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

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

@Ishan Singh

Rishabh's solution is correct (I had stated it earlier). Aditya's solution was incorrect. · 1 year, 4 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. · 1 year, 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 · 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. · 1 year, 3 months ago

Don't use the variable x for two purposes in the same question. · 1 year, 3 months ago

Problem 41

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

Saarthak Marathe has posted the solution to the problem.Thanks! · 1 year, 3 months ago

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

Now integrating we get,

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

Correct!Post the next question. · 1 year, 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! · 1 year, 4 months ago

A graphical approach.

We have to see the nature of discriminant

$${ a }^{ 2 }-(4{ a }^{ 2 }-24)\\ or\quad 8-{ a }^{ 2 }\\ Now,\quad graph\quad of\quad { a }^{ 3 }-6a-6=f(a)\quad \\ takes\quad two\quad turns\quad at\quad a\quad =\sqrt { 2 } \quad and\quad -\sqrt { 2 } ,\quad \\ which\quad shows\quad a(root\quad of\quad function)\quad is\quad positive\quad and\quad at\quad least\quad \sqrt { 2 } \\remember\quad f(x)\quad is\quad continuous.\\ at\quad a=\sqrt { 8 } \quad the\quad f(a)\quad is\quad negative.\\ So\quad root\quad of\quad f(a)>\sqrt { 8 } .\quad \\ Hence\quad for\quad equation\quad { a }^{ 3 }-6a-6=0,\\ a>\sqrt { 8 } ;{ a }^{ 2 }>8\\ We\quad have\quad 8-{ a }^{ 2 }<0.\quad so\\the \quad roots\quad are\quad non\quad real$$ · 1 year, 4 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. · 1 year, 4 months ago

Wondering, how did you thought of that a=b+(2/b). · 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 $$d$$ · 1 year, 4 months ago

By solvable, do you mean quadratic type equations with higher degrees? · 1 year, 4 months ago

yes · 1 year, 4 months ago

Great!!! · 1 year, 4 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. · 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? · 1 year, 4 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. · 1 year, 4 months ago

Oh yes. · 1 year, 4 months ago

I guess he should have asked if the roots were real or complex. · 1 year, 4 months ago

Mayank should post the new question. · 1 year, 4 months ago

OK. @Mayank Chaturvedi,please post the next question · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

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

The correct options are $$1$$ and $$4$$ · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

Hence locus is

$\frac { { \left( x-a \right) }^{ 2 } }{ { \left( \frac { a }{ 2 } \right) }^{ 2 } } -\frac { { y }^{ 2 } }{ { \left( \sqrt { 3 } \frac { a }{ 2 } \right) }^{ 2 } } =1$ · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

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

$$\frac { { 2 }^{ \frac { n }{ 2 } +1 }\cos { \frac { n\pi }{ 4 } } +{ 2 }^{ n } }{ 4 }$$

put i, -i,1, -1 in expansion of (1+x)^n and add · 1 year, 4 months ago

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! · 1 year, 4 months ago

Comment deleted Apr 06, 2016

Comment deleted Apr 06, 2016

$$f''(x) = -f(x)$$

$$f'(x) = g(x)$$

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

$$h(x) = (f(x))^{2} + (g(x))^{2}$$
Differentiate,
$$h'(x) = 2 \cdot \left( f(x)f'(x) + g(x)g'(x) \right)$$ $$h'(x) = 2 \cdot \left( f(x)g(x) + g(x)(-f(x)) \right) = 0$$
Thus, $$h(x)$$ is a constant.
$$h(x) = h(5) = 11$$ for all x. · 1 year, 4 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 $$gcd(\alpha, \beta ) = gcd(\alpha,\gamma) = gcd( \beta, \gamma = 1$$ · 1 year, 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, $$\alpha =7,\quad \beta =9,\quad \gamma =2$$

Therefore, final answer is 6 · 1 year, 3 months ago

Getting the same thing @Sarvesh Nalawade. I think you should go ahead and post the solution · 1 year, 3 months ago

Correct. · 1 year, 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 ? · 1 year, 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 $$\dfrac{-2}{3} , \dfrac{-2}{3} , \dfrac{1}{3}$$ · 1 year, 3 months ago

Great! Please post the next problem. @Vighnesh Shenoy · 1 year, 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 $$\frac{4}{27}$$ · 1 year, 3 months ago

Problem 39:

Solve

$$\displaystyle {x}^{3}\frac{dy}{dx}={y}^{3}+{y}^{2}\sqrt{{y}^{2}-{x}^{2}}$$ · 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. · 1 year, 3 months ago

$$PROBLEM \quad 29$$:

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. $$y +3x +2 = 0$$
$$3y- 2x- 5=0$$
$$4y+x-14=0$$ · 1 year, 4 months ago

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

$$\therefore \frac{4}{3} \leq \beta \leq \frac{9}{4}$$

· 1 year, 4 months ago

Great! Can you please add the graph for the sake of solution? Thanks! · 1 year, 4 months ago

Correct · 1 year, 4 months ago

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! · 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. · 1 year, 4 months ago

Correct. · 1 year, 4 months ago

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! · 1 year, 4 months ago

Comment deleted Apr 12, 2016

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

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

Cross multiply and cancel like terms . Finally we will obtai

a = z b = conjugate of z

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

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

Hence mod(a) = 1 · 1 year, 4 months ago

Comment deleted Apr 12, 2016

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

Since the conjugate of a real number is equal to itself,
$$z + \dfrac{1}{z} = \overline{z} + \dfrac{1}{\overline{z}}$$
$$z - \overline{z} = \dfrac{1}{\overline{z}} - \dfrac{1}{z}$$
$$z - \overline{z} = \dfrac{z - \overline{z}}{z\overline{z}}$$
Since, $$z$$ is not purely real, $$z \ne \overline{z}$$
$$\therefore z\overline{z} = 1$$
$$|z|^{2} = 1 \rightarrow |z| = 1$$
$$10|z| = 10$$ · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

Differentiating,

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

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

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

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

$$f(x)_{max} = f(5) = 7$$ · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

So we have sum of the cubes of $$n$$ terms

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

$= c^3 \left( n^2(2n^2 - 1) \right) .$ · 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 from this note. @Saakshi Singh · 1 year, 4 months ago

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! · 1 year, 4 months ago

@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! · 1 year, 4 months ago

The equation becomes inconsistent for n=0 . Please check . · 1 year, 4 months ago

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! · 1 year, 4 months ago

Comment deleted Apr 10, 2016

Just to expand.... $$\{\dfrac{\sin x}{x}\}=\dfrac{\sin x}{x}$$ and $$\{\dfrac{\tan x}{x}\}=\dfrac{\tan x}{x}-1$$ as $$\sin x<x<\tan x$$.

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

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

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

Apply Lhospital to get $$e^{-1/2}$$. · 1 year, 4 months ago

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! · 1 year, 4 months ago

The answer is $$0$$.

· 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

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

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

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

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

So the closed form is $$\binom{2n-1}{n}+(n+1)\binom{2n-1}{n-1}$$ · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

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

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

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

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

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

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

Only the second factor gives non - real roots since it's discriminant is negative, therefore sum is $$\boxed{-3}$$ · 1 year, 4 months ago

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

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

Hence the sum is: $$-3$$ · 1 year, 4 months ago

The equation factors out as,

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

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

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

Therefore the sum of the non-real roots are -3 · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

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

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

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

This way, we can generalize to $$n$$ terms also. · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

(Feel free to correct me) · 1 year, 4 months 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 · 1 year, 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! · 1 year, 3 months ago

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. · 1 year, 3 months ago

23.684. I'll post the soln in some time... · 1 year, 3 months ago

Waiting for the solution. · 1 year, 3 months ago

Please provide the complete solution. · 1 year, 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. · 1 year, 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,

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

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

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

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

$$p = 4 , q = \dfrac{3}{2}$$
$$pq = 6$$ · 1 year, 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! · 1 year, 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 $$\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{i-j}{2}$$

$$\frac{x^2}{a^2} \ + \frac{y^2}{b^2} \ = \sec^2 \frac{\alpha }{2}$$ · 1 year, 3 months ago

Correct! But prove your 1st statement in your solution and post the next question · 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. · 1 year, 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. · 1 year, 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 $$\boxed 4$$ · 1 year, 3 months ago

The answer is right, however you used
$$32^{32^{32}} \equiv (32^{32})^{32} \equiv 9^{32} \equiv 4 \pmod{7}$$ · 1 year, 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

$$\varphi (7)=6\quad and\quad gcd(32,7)=1\\ { 32 }^{ 6 }\equiv 1mod(7)............me-1\\ Now\quad we\quad find\quad { 32 }^{ 32 }mod\quad 6\\ { 32 }^{ 32 }\equiv { 2 }^{ 32 }\equiv 4^{ 4 }\equiv 4mod(6)......me-2\\ Using\quad me-1\quad and\quad me-2\\ { 32 }^{ { 32 }^{ 32 } }\equiv { 32 }^{ 6x+4 }\equiv { 32 }^{ 4 }mod(7)\\ { 32 }^{ 4 }\equiv 4^{ 4 }\equiv 256=4mod(7)\\ So\quad 4\quad is\quad the\quad remainder\quad :)$$ · 1 year, 3 months ago

Perfect. · 1 year, 3 months ago

Comment deleted Apr 27, 2016

That is an incorrect way of evaluation of power towers.
$$a^{b^{c}} \ne (a^{b})^{c}$$ · 1 year, 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! · 1 year, 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

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

The above question can be written in the following manner:

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

Please post the next question · 1 year, 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. · 1 year, 4 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 $$\boxed {108545}$$ · 1 year, 3 months ago

@Saarthak Marathe

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

Thanks! · 1 year, 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 · 1 year, 4 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 · 1 year, 4 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 · 1 year, 4 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. · 1 year, 4 months ago

Can you please tell what is the numerical figure you are getting after solving your expression? · 1 year, 4 months ago

16632000 · 1 year, 4 months ago

OK! Thanks! Waiting for the next question from you... · 1 year, 4 months ago

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. · 1 year, 4 months ago

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 · 1 year, 4 months ago

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

Thanks! · 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 $${S}_{j-1}$$ · 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! · 1 year, 4 months ago

Problem 34:

Solve:

$$\displaystyle ({x}^{2}+y)\frac{dy}{dx}=6x$$ · 1 year, 4 months ago

We can write the question as follows:

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

Now let $$x^2=\lambda$$

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

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

Therefore,

$$t+3=3\large \frac{dt}{dy}$$

Now after rearranging the terms and integrating we get:

$$3ln|x^2+y+3| = y+c$$

Is this correct? @Saarthak Marathe · 1 year, 4 months ago

It is incorrect. You missed a number in your final answer · 1 year, 4 months ago

Got it! it should be $$x^2+y+3$$ right? · 1 year, 4 months ago

Yes correct!! Post the next question · 1 year, 4 months ago

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? · 1 year, 4 months ago

Just an alternate solution:

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

Therefore, the required event is denoted graphically be the area enclosed in $$|x-y|\le 20$$. Now the ratio of the area of the event to the area of sample space($$=3600$$) gives the answer $$\frac{5}{9}$$ · 1 year, 4 months ago

Probability of meeting If the first person comes between 0-40 minutes=$$2/3*1/3$$=$$2/9$$

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

Therefore,total probability=$$5/9$$

Is this correct? @Miraj Shah · 1 year, 4 months ago

Ya! The answer is correct. · 1 year, 4 months ago

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 $$m$$. · 1 year, 4 months ago

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

Given:

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

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

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

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

Therefore finally,

$$m=-2$$ or $$m=4$$

Therefore, the required answer is $$\boxed{2}$$

Is this the answer? · 1 year, 4 months ago

Correct. :). Waiting for the next problem. · 1 year, 4 months ago

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 $$f(3,4)$$ · 1 year, 4 months ago

Let $$x=R\sin\alpha\sin\beta,y=R\cos\beta,z=R\cos\alpha\sin\beta$$

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

Correct. · 1 year, 4 months ago

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

$$\sqrt { ({ a }^{ 2 }+b^{ 2 })(x^{ 2 }{ y }^{ 2 }+y^{ 2 }z^{ 2 }) } \ge (axy+byz)\\ \sqrt { ({ a }^{ 2 }+b^{ 2 }){ y }^{ 2 }(x^{ 2 }+z^{ 2 }) } \ge (axy+byz)\\ \sqrt { ({ a }^{ 2 }+b^{ 2 }){ y }^{ 2 }(R^{ 2 }-y^{ 2 }) } \ge (axy+byz)\\ \frac { { R }^{ 2 } }{ 2 } \sqrt { ({ a }^{ 2 }+b^{ 2 }) } \ge (axy+byz)$$ · 1 year, 4 months ago

You guys decide who is going to post next. · 1 year, 4 months ago

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 $$P(12)$$ . · 1 year, 4 months ago

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

$$P(x)(x+1)-1 = ax(x-1)(x-2)\ldots(x-11)$$
Substituting $$x = - 1$$,
$$-1 = a(-1)(-2)\ldots(-12)$$
$$-1 = 12!a$$
$$a = -\dfrac{1}{12!}$$
$$P(12)(13) - 1 = -\dfrac{1}{12!}\times 12!$$
$$P(12)(13) - 1 = -1$$
$$P(12) = 0$$ · 1 year, 4 months ago

That's correct !! · 1 year, 4 months ago

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! · 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. · 1 year, 4 months ago

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

Comment deleted Apr 17, 2016

Comment deleted Apr 17, 2016

Earlier i wrote an answer telling angle between $$\vec c$$ and -$$\vec b$$ so you too take care of it.

Check this:

i have got answer 150 degrees

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

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

All credits to @Vighnesh Shenoy, · 1 year, 4 months ago

Great! Can you please post the next problem? @Mayank Chaturvedi · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

Next time you give a summation, please give more terms. I am still not sure that the series I used is correct. · 1 year, 4 months ago

Correct . And yes i will take care of more terms in the series next time. · 1 year, 4 months ago

I think you mean $$\tan ^{ -1 }{ \frac { 1 }{ 2 } }$$ in the third line, and not $$\tan ^{ -1 }{ \frac { 1 }{ 4 } }$$ · 1 year, 4 months ago

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! · 1 year, 4 months ago

Volume of tetrahedron = $$\dfrac{A \cdot h}{3}$$
$$\therefore 15 = A_{1} \cdot h_{1} = A_{2} \cdot h_{2} = A_{3} \cdot h_{3} = A_{4} \cdot h_{4}$$

We have to minimize :

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

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

Hence,
$$S_{min} = \dfrac{15 \times 16}{5!} = 2$$ · 1 year, 4 months ago

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

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! · 1 year, 4 months ago

$$I = \displaystyle \int_{0}^{\infty} \left \lfloor \dfrac{n}{e^{x}} \right \rfloor dx$$
$$e^{-x} = t \rightarrow dx = \dfrac{-dt}{t}$$
$$I = \displaystyle \int_{0}^{1} \left \lfloor nt \right \rfloor \dfrac{dt}{t}$$
$$I = \displaystyle \sum_{k=1}^{n-1} \int_{\frac{k}{n}}^{\frac{k+1}{n}} \dfrac{k}{t} dt$$
$$I = \displaystyle \sum_{k=1}^{n-1} k\log\left(\dfrac{k+1}{k}\right)$$
$$S = \displaystyle \sum_{k=1}^{n-1} \left[(k+1)\log(k+1)-k\log(k) - \log(k+1)\right]$$
$$S = n\log(n) - log(n!)$$ · 1 year, 4 months ago

PROBLEM 22

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

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

B: $$a(100)>100$$

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

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

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

Aditya Kumar has solved this problem at the first place and provided the solution. But feel free to share new approaches. Thanks! · 1 year, 4 months ago

Solution to problem 22:

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

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

This is an AGP.

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

Therefore options A and D are correct. · 1 year, 4 months ago

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! · 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 solution · 1 year, 4 months ago

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

Here lambda came out to be - 0.5

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

Is this correct.? · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

Let midpoint be (h,k)

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

2k = 2a(t+s)

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

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

We obtain locus as

p = alpha (inclination of line)

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

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

Credits @Samarth Agarwal for alternate way · 1 year, 4 months ago

Comment deleted Apr 13, 2016

I think it should be -(ac)^2 · 1 year, 4 months ago

@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! · 1 year, 4 months ago

No sir saarthak who posted the problem confirmed that the answer is correct · 1 year, 4 months ago

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! · 1 year, 4 months ago

Write $${(1-{x}^{3})}^{n} = {(3x(1-x)+{(1-x)}^{3})}^{n}=\sum _{ r=0 }^{ n }{ { 3 }^{ r } } \binom{ n }{ r }{ (x(1-x)) }^{ r }{ (1-x) }^{ 3n-3r }$$ · 1 year, 4 months ago

Answer is $${3}^{r}\cdot \dbinom{n}{r}$$ · 1 year, 4 months ago

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! · 1 year, 4 months ago

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 $$\boxed{d_{max}=1}$$ · 1 year, 4 months ago

Absolutely Correct.

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

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

Also you could have avoided calculus by using AM Gm in last step by writing sec and cosec in terms of tan and cot · 1 year, 4 months ago

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! · 1 year, 4 months ago

@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 · 1 year, 4 months ago

the number of possibilities in that order is 8+6+4+2 but the the set containing these in reverse order can also be in gp ie(4,7,10) &(10,7,4) similar is the case with all other elements so the So number of possible GP's is 2(8+6+4+2)=220=40 so N/5 =8 this was also my solution · 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. · 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 =8 · 1 year, 4 months ago

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! · 1 year, 4 months ago

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

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

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

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

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

Comparing coefficients, we have,

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

$$\displaystyle \implies \int_{0}^{1} f(t) dt = A - 1 = \boxed{\dfrac{42}{23}}$$ · 1 year, 4 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! · 1 year, 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 $\inf{f(x)}=27×(1+2+3+4)=270$ · 1 year, 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 · 1 year, 3 months ago

Look who's back from the dead. · 1 year, 3 months ago

@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! · 1 year, 3 months ago

The answer is correct. · 1 year, 3 months ago

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