JEE 2013 :: Advanced 2013 :: Mathematics 2013

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013

A box B₁ contains 1 white ball, 3 red balls and 2 black balls. Another box B₂ contains 2 white balls, 3 red balls and 4 black balls. A third box B₃ contains 3 white balls, 4 red balls and 5 black balls.

If 1 ball is drawn from each of the  boxes B₁,B₂ and B₃  the probability that all 3 drawn  balls are of the same colour is 

Answer:

Explanation:

P (All 3 drawn balls are of the same colour) 

=  P(www)+P(RRR)+P(BBB)

    $=(\frac{1}{6}\times\frac{2}{9}\times\frac{3}{12})+(\frac{3}{6}\times\frac{3}{9}\times\frac{4}{12})$

    $+(\frac{2}{6}\times\frac{4}{9}\times\frac{5}{12})=\frac{82}{648}$

Let  $S:S_{1}\cap S_{2} \cap S_{3}$ , where   

$S_{1}=\left\{ z\in C:|z|<4\right\}, S_{2}$ $=\left\{ z\in C :In\left[\frac{z-1+\sqrt{3}i}{1-\sqrt{3}i}\right]>0\right\}$

and   $S_{3}=\left\{ z\in C :Re z>0\right\}$

 $min_{z\in s}|1-3i-z|$ is equal to 

Answer:

Explanation:

 Here , $S=S_{1}\cap S_{2} \cap S_{3}$

 $S_{1}=\left\{ z\in C:|z|<4\right\}$

$\therefore$    S₁ :x²+y²   < 4   ...........(i)

 and   $S_{2}$ $=\left\{ z\in C :In\left[\frac{z-1+\sqrt{3}i}{1-\sqrt{3}i}\right]>0\right\}$

 where   $\frac{z-1+i\sqrt{3}}{1-i\sqrt{3}}= \frac{(x-1) +i(y+\sqrt{3})}{1-i\sqrt{3}}$

 =  $\frac{(x-1)(1+i\sqrt{3})+i(y+\sqrt{3}) (1+i\sqrt{3})}{1+3}$

 =$\frac{(x-1)+i\sqrt{3}(x-1)+i(y+\sqrt{3}) -\sqrt{3}(y+\sqrt{3})}{4}$

=$\frac{(x-1-\sqrt{3}y-3)}{4}+\frac{i(\sqrt{3}x+y)}{4}$

  $\therefore $     $s_{2}:\sqrt{3}x+y>0$  .......(ii)

  $S_{3}=\left\{ z\in C :Re z>0\right\}$

  $s_{3}:x>0$     ......(iii)

  $min_{z\in s}|1-3i-z|$ = perpendicular distance of point (1,-3) from the line  $\sqrt{3}x+y=0$

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

Let  $S:S_{1}\cap S_{2} \cap S_{3}$ , where   

$S_{1}=\left\{ z\in C:|z|<4\right\}, S_{2}$ $=\left\{ z\in C :In\left[\frac{z-1+\sqrt{3}i}{1-\sqrt{3}i}\right]>0\right\}$

and   $S_{3}=\left\{ z\in C :Re z>0\right\}$

Area of S is equal to 

Answer:

Explanation:

 Here , $S=S_{1}\cap S_{2} \cap S_{3}$

 $S_{1}=\left\{ z\in C:|z|<4\right\}$

$\therefore$    S₁ :x²+y²   < 4   ...........(i)

 and   $S_{2}$ $=\left\{ z\in C :In\left[\frac{z-1+\sqrt{3}i}{1-\sqrt{3}i}\right]>0\right\}$

 where   $\frac{z-1+i\sqrt{3}}{1-i\sqrt{3}}= \frac{(x-1) +i(y+\sqrt{3})}{1-i\sqrt{3}}$

 =  $\frac{(x-1)(1+i\sqrt{3})+i(y+\sqrt{3}) (1+i\sqrt{3})}{1+3}$

 =$\frac{(x-1)+i\sqrt{3}(x-1)+i(y+\sqrt{3}) -\sqrt{3}(y+\sqrt{3})}{4}$

=$\frac{(x-1-\sqrt{3}y-3)}{4}+\frac{i(\sqrt{3}x+y)}{4}$

  $\therefore $     $s_{2}:\sqrt{3}x+y>0$  .......(ii)

  $S_{3}=\left\{ z\in C :Re z>0\right\}$

  $s_{3}:x>0$     ......(iii)

  $s_{3}:x>0$     ......(iii)

Since , 

                $S=S_{1}\cap S_{2} \cap S_{3}$

 Clearly , the shaded region represents the area of sector

 $\therefore$     $S= \frac{1}{2}r^{2}\theta=\frac{1}{2}\times 4^{2}\times\frac{5\pi}{6}=\frac{20\pi}{3}$

Let PQ be a focal chord of the parabola y²=4ax.The tangents to the parabola at P and Q meet at a point lying on the line y=2x+a, a >0

 If chord PQ subtends an angle $\theta$ at the vertex of y² =4ax, then tan $\theta$  is equal to 

Answer:

Explanation:

Concept involved

 Intersection point of tangents at  $(at_1^2,2at_{1})$ and $(at_2^2,2at_{2})$ is ( at₁t₂ , a/t₁ +t₂) ) , also tangents drawn  at end point of focal chord are perpendicular and intersect on directrix

$m_{op}=\frac{2at-0}{at^{2}-0}=\frac{2}{t}$

$m_{oQ}=\frac{-2a/t}{a/t^{2}}=-2t$

$\therefore$   $\tan\theta =\frac{\frac{2}{t}+2t}{1-\frac{2}{t}.2t}=\frac{2(t+\frac{1}{t})}{1-4}$

 where   ,   $t+\frac{1}{t}=\sqrt{5}=\frac{2\sqrt{5}}{-3}$

Let PQ be a focal chord of the parabola y²=4ax.The tangents to the parabola at P and Q meet at point lying on the line y=2x+a, a >0

 Length  of chord PQ is 

Answer:

Explanation:

Concept involved

 Intersection point of tangents at  $(at_1^2,2at_{1})$ and $(at_2^2,2at_{2})$ is ( at₁t₂ , a/t₁ +t₂) ) , also tangents drawn  at end point of focal chord are perpendicular and intersect on directrix

 Since  $R\left(-a,a(t-\frac{1}{t})\right)$  lies on y =2x+a

$\Rightarrow$    $ a.(t-\frac{1}{t})=-2a+a$

 $\Rightarrow$     $t-\frac{1}{t}=-1$

 Thus, length of focal chord

 $a( t+\frac{1}{t})^{2}=a\left\{\left(t-\frac{1}{t}\right)^{2}+4\right\}=5a$

• Advanced 2013 - Physics 2013
• Advanced 2013 - Chemistry 2013
• Advanced 2013 - Mathematics 2013