JEE 2015 :: Advanced 2015 :: Mathematics 2015

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015

If the normals of the parabola  $y^{2}=4x$    drawn at the endpoints of the latus rectum  are tangents to the circle $(x-3)^{2}+(y+2)^{2}=r^{2}$, then the value of r² is 

Answer:

Explanation:

 End points of latusrectum are  $(a,\pm2a)$  i.e,  $(1,\pm2)$

 Equation  of normal at ( x₁, y₁) is   $\frac{y-y_{1}}{x-x_{1}}=-\frac{y_{1}}{2a}$

  i.e,                  $\frac{y-2}{x-1}=-\frac{2}{2}$

  and                $\frac{y+2}{x-1}=\frac{2}{2}$

  $\Rightarrow$             x+y=3

                and                 x-y=3

 which is tangent to 

           $(x-3)^{2}+(y+2)^{2}=r^{2}$

$\therefore$   Length of perpendicular from centre = Radius

 $\Rightarrow$   $\frac{|3-2-3|}{\sqrt{1^{2}+1^{2}}}=r$

   $r^{2}=2$

 Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then, the value of  $\frac{m}{n}$  is

Answer:

Explanation:

Here,   $-B_{1}-B_{2}-B_{3}-B_{4}-B_{5}-$

  out of 5 girls , 4 girls are together and 1 girl is seperate  . Now , to select 2 positions  out of 6 positions between boys = $^{6}C_{2}$        .............(i)

 4 girls are to be selected out of 5=  $^{5}C_{4}$      ...........(ii)

 Now, 2 groups of girls can be arranged in  2! ways...........(iii)

 Also, the group of 4 girls and  5 boys  is arranged in 4! X 5!  ways .....(iv)

Now, total number of ways  $=^{6}C_{2}\times ^{5}C_{4}\times 2!\times4!\times5!$

        [ from Eqs. (i),(ii) ,(iii) and (iv)]

$\therefore$       m$=^{6}C_{2}\times ^{5}C_{4}\times 2!\times4!\times5!$

  and                 n= 5! X6!

 $\Rightarrow$  $\frac{m}{n}=\frac{^{6}C_{2}\times ^{5}C_{4}\times 2!\times4!\times5!}{6!\times5!}$

                       = $\frac{15\times5\times2\times4!}{6\times5\times4!}=5$

 The minimum number of times a fair coin needs to be tossed, so that the probability of getting atleast two heads is atleast 0.96, is 

Answer:

Explanation:

 Using Binomial  distribution,

    $P(X\geq2)=1-P(X=0)-P(X=1)$

    =   $1-\left(\frac{1}{2}\right)^{n}-\left[^{n}C_{1}.\left(\frac{1}{2}\right).\left(\frac{1}{2}\right)^{n-1}\right]$

   =  $1-\frac{1}{2^{n}}-^{n}C_{1}.\frac{1}{2^{n}}=1-\left(\frac{1+n}{2^{n}}\right)$

   Given,   $P(X\geq2)\geq0.96$

 $\therefore$   $1-\frac{(n+1)}{2^{n}}\geq\frac{24}{25}\Rightarrow\frac{n+1}{2^{n}}\leq\frac{1}{25}$

  $\therefore$     n=8

Let the curve C be the mirror image of the parabola  $y^{2}=4x$  with respect to the line  x+y+4=0, If A and B are the points  of intersection  of C with the line y=-5  , then the distance between A and B are

Answer:

Explanation:

Let  P( t² ,2t)  be a point on the curve  $y^{2}=4x$ , whose image is Q(x,y)  on x+y+4=0, then

$\frac{x-t^{2}}{1}=\frac{y-2t}{1}$

$=\frac{-2(t^{2}+2t+4)}{1^{2}+1^{2}}$

$\Rightarrow$   x=-2t-4

   and            y$=-t ^{2}-4$

 Now, the straight line y=-5 meets the mirror image

  $\therefore$                  $-t ^{2}-4=-5$

  $\Rightarrow$     $t ^{2}=1$

$\Rightarrow$           $t=\pm1$

Thus, points of intersection of A and B are (-6,-5) and (-2,-5)

 $\therefore$ Distance , AB= $\sqrt{(-2+6)^{2}+(-5+5)^{2}}=4$

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015