Firstly \(r_1 = \tfrac{1}{8b}(a^2 + 4b^2)\) and \(r_2 = \tfrac{1}{8a}(4a^2 + b^2)\). To see this, for example, suppose that \(P\) has coordinates \((0,\tfrac12a)\), \(Q\) coordinates \((0,-\tfrac12a)\). Then the centre of the first circle will have coordinates \((x,0)\) for some \(x\), and the circle will touch \(RS\) at its midpoint \((b,0)\). Then \(x^2 + \tfrac14a^2 = (b-x)^2\), which gives us \(x\). Then \(r_1=b-x\). While \(x\) could be negative, \(r_1\) cannot.

how to do it? if we can find the number of cubes less than \( 10^8 \) and number of fifth power numbers less than \( 10^8 \) then only we can solve it. Is there another way?

The first question is very easy to solve. First of all, p^6 - 1 is divisible by 7, can be proved by fermat's little theorem. Next, p^6 -1 can be factorised to find whether it is divisible by 3^2, next the factorisation can also help to find whther it is divisble by 2^3. So 504 = 7* (3^2) * (2^3). This means p^6-1 is divisible by 504.

yes, thanks, question 2(a), 5, 7, 8(a) matches with my solution, but can you give some explanation in 4(b).
and I think your 8(b) is wrong. the angle of AC with x-axis is 60 degree.
So, \( \tan{60} = \frac {y - 0}{x - 2} \) Thus equation is \( y= \sqrt{3}x - 2 \sqrt{3} \) which is my answer, and the coordinate also has some error. Am I right?

ok , i went wrong in q.8(b) , it is rotated anti - clockwise , i rotated it clockwise.

4(b) Give end seats to grandchildren in \( 4\times3\) ways.

Now choose 1 seat out of 7 for grandfather( you can't select the seat which second from left or right.).in \(7\) ways.

Then choose 2 seats out of 6 for remaining two grandchildren ( as we can't choose from grandfather's left or right and 3 seats are already occupied) and permutate in \({6 \choose 2}2!\) ways.

Now , six seats for sons and daughters remain , \(6!\) ways for them.

Total no. of ways = \(4\times3\times7\times{6 \choose 2}2!\times6! = 7!4!{6 \choose 2}\)

@Kushagraa Aggarwal
–
am I right in question 8(b), then what is the coordinate of C?
In 4. how did you write- Give end seats to grandchildren in 4×3 ways? as far as I have understood the question,two seats each on the left and right ends of row is occupied by children. Am I wrong or You are mistaken..

@Kumar Ashutosh
–
Left and Right end seats are occupied by grand children .

For left seat it has four options and right has three options ( as one is already taken by left one.). Hence by Fundamental principle of counting , no. of ways = \( 4\times 3\).

@Kushagraa Aggarwal
–
okay, 8(b) is correct and also matches my solution.
In 4, I am still confused.Look and find error in my solution, it goes like this:

The extreme 2 seats on both ends are occupied by grand children. So 4 grand children occupy 4 seats in \( 4! \) ways.
Numbering the seats 1,2,3.....11, we find that Grandfather canno sit on 3rd and 9th seat. So grandfather can sit on any one of 4th,5th,6th,7th,8th seats. This is done in \( 5 \) ways.

Now after this arrangement, we have 6 seats left on which 6 son/daughter is to sit. This can be done in \( 6! \) ways

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## Comments

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TopNewest[6] Let \(\cos \alpha + i\cdot \sin \alpha = e^{i\alpha}\)

\(\cos \beta + i\cdot \sin \beta = e^{i\beta}\)

\(\cos \gamma + i\cdot \sin \gamma = e^{i\gamma}\)

Now Given \(\cos \alpha + \cos \beta + \cos \gamma = 0 .........................(1)\)

and \(\sin \alpha + \sin \beta + \sin \gamma = 0.......................................(2) \times i\)

Add These two equations::

\(e^{i\alpha}+e^{i\beta}+e^{i\gamma} = 0...................................................(3)\)

Now Square both Side

\(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} + 2 \left(e^{i(\alpha+\beta)}+e^{i(\beta+\gamma)}+e^{i(\gamma+\alpha)}\right) = 0\)

Now in a \(\triangle\) Sum of angle i.e \(\alpha+\beta+\gamma = \pi\Leftrightarrow \alpha +\beta = \pi-\gamma\)

So \(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} +e^{i(\pi-\gamma)}+e^{i(\pi-\beta)}+e^{i(\pi-\alpha)} = 0\)

\(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} +e^{i(\pi-\gamma)}-2\left(e^{-i\alpha}+e^{-i\beta}+e^{-i\gamma}\right) = 0\)

Now from \((3)\), Taking Congugate on both side, we get

\(e^{-i\alpha}+e^{-i\beta}+e^{-i\gamma} = 0\)

So \(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} = 0\)

So \(\cos 2\alpha + \cos 2\beta+\cos 2 \gamma+i\left(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma\right) = 0+i\cdot 0\)

we Get \(\cos 2\alpha + \cos 2\beta+\cos 2 \gamma = 0\)

and \(\sin 2\alpha + \sin 2\beta+\sin 2 \gamma = 0\)

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Question 10.

Firstly \(r_1 = \tfrac{1}{8b}(a^2 + 4b^2)\) and \(r_2 = \tfrac{1}{8a}(4a^2 + b^2)\). To see this, for example, suppose that \(P\) has coordinates \((0,\tfrac12a)\), \(Q\) coordinates \((0,-\tfrac12a)\). Then the centre of the first circle will have coordinates \((x,0)\) for some \(x\), and the circle will touch \(RS\) at its midpoint \((b,0)\). Then \(x^2 + \tfrac14a^2 = (b-x)^2\), which gives us \(x\). Then \(r_1=b-x\). While \(x\) could be negative, \(r_1\) cannot.

Then \[ \begin{array}{rcl} r_1+r_2 & = & \frac{a(a^2+4b^2)+b(4a^2+b^2)}{8ab} \; = \; \frac{a^3+b^3 + 4a^2b + 4ab^2}{8ab} \\ & = & \frac{(a+b)(a^2-ab + b^2 + 4ab)}{8ab} \; = \; \frac{(a+b)(a^2 + 3ab + b^2)}{8ab} \\ & = & \tfrac{1}{8}(a+b)\big[3 + \tfrac{a}{b} + \tfrac{b}{a}\big] \; \ge \; \tfrac{1}{8}(a+b)[3+2] \; = \; \tfrac{5}{8}(a+b) \end{array} \]

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2[b]:: Here \(a,b\in \mathbb{R}\) and \(3a^4-4a^3b+b^4\geq 0\)

Now this is a Homogeneous equation, so Let \(a = bx\) and Let \(y = 3b^4x^4-4b^3x^3\cdot b+b^4\)

So \(y = b^4\cdot \left(3x^4-4x^3+1\right) = b^4\cdot (x-1)^2\cdot (3x^2+2x+1)\geq 0\)

bcz \(3x^2+2x+1 = 3\left(x+\frac{1}{3}\right)^2+\frac{8}{3}>0\;\forall x\in \mathbb{R}\)

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There is also a simple answer to this. It goes like this:

We have, \(3a^4-4a^3b+b^4\) \(=a^4-2a^2b^2+b^4+2a^2b^2+2a^4-4a^3b\) \(=(a^2-b^2)^2+2a^2(b^2+a^2-2ab)\) \(= (a^2-b^2)^2+2a^2(a-b)^2 \geq 0 \)

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is the answer of ques 8 34????

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which part, 8(a) or 8(b)

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(A)

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Where and when is this exam held?

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Question 9 is nothing but inclusion-exclusion principle...perhaps!

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how to do it? if we can find the number of cubes less than \( 10^8 \) and number of fifth power numbers less than \( 10^8 \) then only we can solve it. Is there another way?

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The first question is very easy to solve. First of all, p^6 - 1 is divisible by 7, can be proved by fermat's little theorem. Next, p^6 -1 can be factorised to find whether it is divisible by 3^2, next the factorisation can also help to find whther it is divisble by 2^3. So 504 = 7* (3^2) * (2^3). This means p^6-1 is divisible by 504.

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Challenge masters please give some hints!!!!!!!!!!

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Here are the anwers ( not for the proof - type questions.)

2(a) \(\rightarrow 153846\)

4(b) \(\rightarrow \)\(7!4!\binom{6}{2} \)

5 \(\rightarrow\) 90

7 \(\rightarrow \pi x^2\)

8(a) \(\rightarrow 184\)

8(b) \(\rightarrow x - y\sqrt{3} - 2 = 0 \) , \((2 + \sqrt{\frac{3}{2}} , \frac{1}{\sqrt{2}})\)

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yes, thanks, question 2(a), 5, 7, 8(a) matches with my solution, but can you give some explanation in 4(b). and I think your 8(b) is wrong. the angle of AC with x-axis is 60 degree. So, \( \tan{60} = \frac {y - 0}{x - 2} \) Thus equation is \( y= \sqrt{3}x - 2 \sqrt{3} \) which is my answer, and the coordinate also has some error. Am I right?

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ok , i went wrong in q.8(b) , it is rotated anti - clockwise , i rotated it clockwise.

4(b) Give end seats to grandchildren in \( 4\times3\) ways.

Now choose 1 seat out of 7 for grandfather( you can't select the seat which second from left or right.).in \(7\) ways.

Then choose 2 seats out of 6 for remaining two grandchildren ( as we can't choose from grandfather's left or right and 3 seats are already occupied) and permutate in \({6 \choose 2}2!\) ways.

Now , six seats for sons and daughters remain , \(6!\) ways for them.

Total no. of ways = \(4\times3\times7\times{6 \choose 2}2!\times6! = 7!4!{6 \choose 2}\)

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For left seat it has four options and right has three options ( as one is already taken by left one.). Hence by Fundamental principle of counting , no. of ways = \( 4\times 3\).

8(b) . \(( 2 + \frac{1}{\sqrt{2}} , \sqrt{\frac{3}{2}})\)

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In 4, I am still confused.Look and find error in my solution, it goes like this:

The extreme 2 seats on both ends are occupied by grand children. So 4 grand children occupy 4 seats in \( 4! \) ways. Numbering the seats 1,2,3.....11, we find that Grandfather canno sit on 3rd and 9th seat. So grandfather can sit on any one of 4th,5th,6th,7th,8th seats. This is done in \( 5 \) ways.

Now after this arrangement, we have 6 seats left on which 6 son/daughter is to sit. This can be done in \( 6! \) ways

Therefore total possibilities= \( 4!.5.6! \)

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Tell which one you couldn't solve .

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Please give solutions to all the questions you can, I will check the questions done by me and will understand what didn't solved.. Thanks

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Please help me with your solutions..

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Did you try them already? Which did you manage to solve?

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I did q.1,2,4(b), 5,7,8, but try to give solutions of every question, so that I could verify my answers

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There can be subtleties in proof based questions, which would result in a 'similar-looking' answer being marked wrong.

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