1) The minimum value of \(x^{2}+y^{2}+z^{2}\) if \(ax+by+cz=p\) is

2) The general solution of the equation \(tan^{2}(x+y)+cot^{2}(x+y)=1-2x-x^{2}\) lie on the line is

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

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

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

2)- Clearly Such Question is tackle by using boundedness of function... \(\displaystyle{LHS=\tan ^{ 2 }{ (x+y) } +\cot ^{ 2 }{ (x+y) } \\ RHS=2-{ (x+1) }^{ 2 }\\ LHS\ge 2\quad \& \quad RHS\le 2\\ \Rightarrow LHS=RHS=2\\ \therefore \boxed { x=-1 } \quad \& \quad \tan ^{ 2 }{ (x+y) } =1\\ \\ y+1=n\pi \pm \cfrac { \pi }{ 4 } \\ \boxed { y=(n\pm \cfrac { 1 }{ 4 } )\pi -1 } }\) – Karan Shekhawat · 2 years, 5 months ago

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3) Let the number of heads that appear in a trial be \(n\).

Then there will be \(n+1\) '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: \[x_{1}+x_{2}+x_{3}...x_{n+1}=10-n\]

(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

\[x_{1}+x_{2}+x_{3}...x_{n+1}=2+10-n=12-n\]

in the \(natural\) numbers.

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

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 \(2^{10}\) is the required probability

@Tanishq Varshney – Shashwat Shukla · 2 years, 5 months ago

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– Raghav Vaidyanathan · 2 years, 5 months ago

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...Log in to reply

– Shashwat Shukla · 2 years, 5 months ago

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

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 · 2 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 · 2 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 · 2 years, 5 months ago

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– Ronak Agarwal · 2 years, 5 months ago

Raghav do you give fiitjee aits, or aiitsLog in to reply

– Shashwat Shukla · 2 years, 5 months ago

Did you write AITS 8 (last Sunday)? How did it go for you?Log in to reply

Did you write for AITS -7 last sunday

@Shashwat Shukla – Ronak Agarwal · 2 years, 5 months ago

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Do you mean AITS-8? – Shashwat Shukla · 2 years, 5 months ago

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Surprisingly last AITS was not very tough. – Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal – Shashwat Shukla · 2 years, 5 months ago

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– Ronak Agarwal · 2 years, 5 months ago

I also made many careless mistakes and that decreasedy score.Log in to reply

– Ronak Agarwal · 2 years, 5 months ago

Okay tommorow's AITS too best of luck.Log in to reply

– Shashwat Shukla · 2 years, 5 months ago

Good luck to you as well.Log in to reply

– Tanishq Varshney · 2 years, 5 months ago

can u help me with the last twoLog in to reply

– Tanishq Varshney · 2 years, 5 months ago

ok i'll wait.Log in to reply

My laptop is broken and I will keep it damaged till Jee Advanced. – Ronak Agarwal · 2 years, 5 months ago

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

Plz i want the solution.

@Raghav Vaidyanathan @Brian Charlesworth sir, @Shashwat Shukla – Tanishq Varshney · 2 years, 5 months ago

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@Ronak Agarwal @Shashwat Shukla – Ishan Dasgupta Samarendra · 2 years, 5 months ago

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...Log in to reply

@Ishan Dasgupta Samarendra – Shashwat Shukla · 2 years, 5 months ago

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– Ishan Dasgupta Samarendra · 2 years, 5 months ago

Alright, thanks a lot!Log in to reply