# Help!!

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 $n-7m$

Note by Tanishq Varshney
5 years, 1 month 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 $\displaystyle{\vec { A } =xi+yj+z\overset { . }{ k } \\ \vec { B } =ai+bj+c\overset { . }{ k } \\ \vec { A } .\vec { B } \le \left| \vec { A } \right| \left| \vec { B } \right| \\ \underbrace { ax+by+cz }_{ p } \le \sqrt { { x }^{ 2 }+y^{ 2 }+{ z }^{ 2 } } \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } }$

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

- 5 years, 1 month ago

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

- 5 years, 1 month 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...

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

Plz i want the solution.

- 5 years, 1 month 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... @Ronak Agarwal @Shashwat Shukla

- 5 years, 1 month ago

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

- 5 years, 1 month ago

Alright, thanks a lot!

- 5 years, 1 month ago

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

- 5 years, 1 month ago

ok i'll wait.

- 5 years, 1 month ago

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.

- 5 years, 1 month ago

Raghav do you give fiitjee aits, or aiits

- 5 years, 1 month ago

can u help me with the last two

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

303 is a $very$ good score in my opinion especially as this is AITS.

Do you mean AITS-8?

- 5 years, 1 month ago

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

Surprisingly last AITS was not very tough.

- 5 years, 1 month ago

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 :/

- 5 years, 1 month ago

I also made many careless mistakes and that decreasedy score.

- 5 years, 1 month ago

Okay tommorow's AITS too best of luck.

- 5 years, 1 month ago

Good luck to you as well.

- 5 years, 1 month ago

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.

- 5 years, 1 month ago

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

- 5 years, 1 month ago