Help!!

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

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

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

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

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.

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