1) The minimum value of x2+y2+z2x^{2}+y^{2}+z^{2} if ax+by+cz=pax+by+cz=p is

2) The general solution of the equation tan2(x+y)+cot2(x+y)=12xx2tan^{2}(x+y)+cot^{2}(x+y)=1-2x-x^{2} lie on the line is

3) A fair coin is tossed 10 times, if the probability that heads never occur on consecutive tosses be mn\frac{m}{n} (where m,n are coprime natural numbers) then find the value of n7mn-7m

Note by Tanishq Varshney
4 years, 5 months ago

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All of are Very very Elementry Questions..... As Solution of 3rd is already posted by shashwat... So i give rest of 2

1) Let A=xi+yj+zk.B=ai+bj+ck.A.BABax+by+czpx2+y2+z2a2+b2+c2\displaystyle{\vec { A } =xi+yj+z\overset { . }{ k } \\ \vec { B } =ai+bj+c\overset { . }{ k } \\ \vec { A } .\vec { B } \le \left| \vec { A } \right| \left| \vec { B } \right| \\ \underbrace { ax+by+cz }_{ p } \le \sqrt { { x }^{ 2 }+y^{ 2 }+{ z }^{ 2 } } \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } }

2)- Clearly Such Question is tackle by using boundedness of function... LHS=tan2(x+y)+cot2(x+y)RHS=2(x+1)2LHS2&RHS2LHS=RHS=2x=1&tan2(x+y)=1y+1=nπ±π4y=(n±14)π1\displaystyle{LHS=\tan ^{ 2 }{ (x+y) } +\cot ^{ 2 }{ (x+y) } \\ RHS=2-{ (x+1) }^{ 2 }\\ LHS\ge 2\quad \& \quad RHS\le 2\\ \Rightarrow LHS=RHS=2\\ \therefore \boxed { x=-1 } \quad \& \quad \tan ^{ 2 }{ (x+y) } =1\\ \\ y+1=n\pi \pm \cfrac { \pi }{ 4 } \\ \boxed { y=(n\pm \cfrac { 1 }{ 4 } )\pi -1 } }

Karan Shekhawat - 4 years, 5 months ago

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3) Let the number of heads that appear in a trial be nn.

Then there will be n+1n+1 'gaps' created.

Thus for a particular number of heads(=n), the number of ways of this happening will be equal to the number of solutions of the equation: x1+x2+x3...xn+1=10nx_{1}+x_{2}+x_{3}...x_{n+1}=10-n

(as after n heads occur, 10-n positions are left)

The two gaps at the 'end' of the sequence can have zero elements in them while the other gaps must contain at least one element as the heads are not consecutive.

Thus, we have to find the solutions of
x1+x2+x3...xn+1=2+10n=12nx_{1}+x_{2}+x_{3}...x_{n+1}=2+10-n=12-n

in the naturalnatural numbers.

The number of solutions is (11nn)\binom{11-n}{n}.

Now n can vary from 0 to 5 (as 11-n has to be greater than or equal to n)and thus we sum up all the corresponding number of solutions. This divided by 2102^{10} is the required probability

@Tanishq Varshney

Shashwat Shukla - 4 years, 5 months ago

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I did the same in the exam... I did not get the right answer :/ I guess I made some careless mistake. Also, this was a previous JEE problem right? It is there in archive...

Raghav Vaidyanathan - 4 years, 5 months ago

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Don't remember if this was in the archive...But I guess you must have miscalculated during the exam.

Shashwat Shukla - 4 years, 5 months ago

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The answers are 1) p2a2+b2+c2\frac{p^2}{a^2+b^2+c^2} 2) x=1x=-1 3) 11.

Plz i want the solution.

@Raghav Vaidyanathan @Brian Charlesworth sir, @Shashwat Shukla

Tanishq Varshney - 4 years, 5 months ago

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Not sure about this, but couldn't you use Lagrange Multipliers for the first question? I thought of this since it is an optimization question with constraints. Haven't had time to try it though... @Ronak Agarwal @Shashwat Shukla

Ishan Dasgupta Samarendra - 4 years, 5 months ago

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Sure, that would also work...But that would be tad bit lengthy. More importantly, the geometric interpretation is lost in the process.

@Ishan Dasgupta Samarendra

Shashwat Shukla - 4 years, 5 months ago

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@Shashwat Shukla Alright, thanks a lot!

Ishan Dasgupta Samarendra - 4 years, 5 months ago

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First question is very easy, the second equation can be considered as an equation of a plane in 3-D space, with x,y,z being the co-ordinate axes. You just have to find square of shortest distance of said plane from origin.

these problems are from some aiits... i will answer rest after lunch

Raghav Vaidyanathan - 4 years, 5 months ago

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ok i'll wait.

Tanishq Varshney - 4 years, 5 months ago

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Sorry Tanishq actually I am mobile hence unable to post long comments or solutions

My laptop is broken and I will keep it damaged till Jee Advanced.

Ronak Agarwal - 4 years, 5 months ago

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Raghav do you give fiitjee aits, or aiits

Ronak Agarwal - 4 years, 5 months ago

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can u help me with the last two

Tanishq Varshney - 4 years, 5 months ago

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Did you write AITS 8 (last Sunday)? How did it go for you?

Shashwat Shukla - 4 years, 5 months ago

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@Shashwat Shukla Got 303/360 not good enough first paper went bad could get more in first paper surprisingly got 3 rd rank.

Did you write for AITS -7 last sunday

@Shashwat Shukla

Ronak Agarwal - 4 years, 5 months ago

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@Ronak Agarwal 303 is a veryvery good score in my opinion especially as this is AITS.

Do you mean AITS-8?

Shashwat Shukla - 4 years, 5 months ago

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@Shashwat Shukla I belive last paper was Fiitjee AITS Full Test -7 Advanced.

Surprisingly last AITS was not very tough.

Ronak Agarwal - 4 years, 5 months ago

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@Ronak Agarwal Ahh...it was 7 after all...And yes, the paper this time was easier than the previous ones....But I made too many careless mistakes and am getting only about 250 :/

@Ronak Agarwal

Shashwat Shukla - 4 years, 5 months ago

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@Shashwat Shukla I also made many careless mistakes and that decreasedy score.

Ronak Agarwal - 4 years, 5 months ago

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@Shashwat Shukla Okay tommorow's AITS too best of luck.

Ronak Agarwal - 4 years, 5 months ago

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@Ronak Agarwal Good luck to you as well.

Shashwat Shukla - 4 years, 5 months ago

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1) we can solve the problem by simply finding the distance of origin from the given line as given line is tangent to the equation of sphere.

shivam sharma - 4 years, 5 months ago

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3) Not a full solution But the number of cases are given by the Fibonacci recurrence relation. Let P(n) be the number of possibilities . We can have either a head or a tail at the end. If we have a tail then possibilities are P(n-1) and if we have a head then we certainly have a tail before it. so possibilities are P(n-2). Therefore we have the recuurence relation P(n) = P(n-1) + P(n-2) @Raghav Vaidyanathan

Rohit Shah - 4 years, 5 months ago

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