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 $n-7m$

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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... $\displaystyle{LHS=\tan ^{ 2 }{ (x+y) } +\cot ^{ 2 }{ (x+y) } \\ RHS=2-{ (x+1) }^{ 2 }\\ LHS\ge 2\quad \& \quad RHS\le 2\\ \Rightarrow LHS=RHS=2\\ \therefore \boxed { x=-1 } \quad \& \quad \tan ^{ 2 }{ (x+y) } =1\\ \\ y+1=n\pi \pm \cfrac { \pi }{ 4 } \\ \boxed { y=(n\pm \cfrac { 1 }{ 4 } )\pi -1 } }$

Log in to reply

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

Log in to reply

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

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

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

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

Log in to reply

ok i'll wait.

Log in to reply

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.

Log in to reply

Raghav do you give fiitjee aits, or aiits

Log in to reply

can u help me with the last two

Log in to reply

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

Log in to reply

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

303 is aDo you mean AITS-8?

Log in to reply

Surprisingly last AITS was not very tough.

Log in to reply

@Ronak Agarwal

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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.

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

Log in to reply