JEE 2013 :: Advanced 2013 :: Mathematics 2013

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

Consider the lines

$L_{1}=\frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1},$
$L_{2}=\frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2}$
 and the planes p₁:7x+y+2z=3,

 p₂:3x+5y-6z=4, Let  ax+by+cz=d the equation of the line passing through the point of intersection of lines L₁ and L₂ and perpendicular to plane p₁  and p₂.

 Match List I with List II and select the correct answer using the code given below the lists.

Answer:

Explanation:

$L_{1}:\frac{x-1}{2}=\frac{y-0}{-1}=\frac{z-(-3)}{1}$

  Normal of plane p:n=  $\begin{bmatrix}\hat{i} & \hat{j} & \hat{k}\\7 & 1&2\\3&5&-6 \end{bmatrix}$

 = $\hat{i} (-16)-\hat{j}(-42-6)+\hat{k}(32)$

   $= -16\hat{i}+48\hat{j}+32\hat{k}$

 DR's  of normal n=  $\hat{i}-3\hat{j}-2\hat{k}$

 Point of intersection of L₁  and L₂ 

$\Rightarrow $    $ 2K_{1}+1=K_{2}+4 and -k_{1}=k_{2}-3$

 $\therefore$   k₁= 2 and k₂ =1

 $\therefore$  point of intersection (5,-2,-1)

 $\therefore$ Equation of plane ,

  1.(x-5)-3 (y+2)-2(z+1)=0

 $\Rightarrow  $  x-3y-2z-13=0

 $\Rightarrow $   x-3y-2z=13

$\therefore$  a=1, b=-3, c=-2, d=13

 Match List I with List II and select the correct answer using the code given below the lists

Answer:

Explanation:

 (P) Given ,[a b c]=2

 $\therefore $ Volume of 

  2 (a x b),3(b x c), (c x a)

 $\Rightarrow$      6[axb  bxc  cxa]

$\Rightarrow$   6[a b c]²

 $\Rightarrow$   6 x 4=24

 (Q)    [a b c] =5 , given

 $\therefore$ Volume of 

                           3(a+b),(b+c),2(c+a)

$\Rightarrow$    6[a+b  b+c  c+a]

$\Rightarrow$  6 x2 [a b c]

 $\Rightarrow$ 12 x 5=60

  (R)      $\frac{1}{2}|a\times b|=20$  , given

 $\therefore$     $\triangle_{1}=\frac{1}{2}|(2a+3b)\times(a-b)|$

 $=\frac{1}{2}|2a\times a-2a \times b+3b \times a-3b\times b|$

 $=\frac{1}{2}|2b\times a+3b \times a|=\frac{5}{2}|a\times b|$

     =5 x20=100

  (S)    Given, |a x b|=30

 $\therefore$  |(a+b)x a|=|axb + bxa|

  = |b x a|

 = |a x b|=30

A line L:y=mx+3 meets y-axis at E (0,3) and the arc of the parabola y²  =16 x ,  $0\leq y\leq6$ at the point F (x₀,y₀ ) . The tangent to the parabola at F (x₀,y₀) intersects the y axis at G(0,y₁). The slope m of the line L is chosen such that the area of the $\triangle$ EFG  has a local maximum

 Match List I with List II and select the correct answer using the code given below the lists

Answer:

Explanation:

Here,   $y^{2}=16x,0\leq y\leq6$

Tangent at F,

                   yt=x+at²

 at     x= 0, y= at=4t

 Also , (4t²,8t ) satisfy

   y=mx+c

 $\Rightarrow$    $ 8t=4mt^{2}+3$

$\Rightarrow $    $4mt^{2}-8t+3=0$

  $\therefore$   Area of $\triangle$  = $\frac{1}{2}\begin{bmatrix}0 & 3&1 \\0 &4t&1\\ 4t^{2}&8t&1 \end{bmatrix}=\frac{1}{2}4t^{2}(3-4t)$

A= 2[3t²-4t³]

 $\therefore$       $\frac{dA}{dt}=2[6t-12t^{2}]=-12t(2t-1)$

$\therefore$ Maximum at t=1/2

 and   $4mt^{2}-8t+3=0$

 $\Rightarrow$    m-4+3=0

 $\Rightarrow$   m=1

 G(0,4t)$\Rightarrow$  G(0,2)

y₁=2

 (x₀,y₀)  =(4 t²,8t)=(1,4)

 y₀=4

 Area   $=2\left(\frac{3}{4}-\frac{1}{2}\right)=\frac{1}{2}$

 Match List i with List II and select the correct answer using the code given below the lists

Answer:

Explanation:

Here, innermost function is inverse

$\therefore$   put,   $\tan^{-1}y=\theta\Rightarrow\tan\theta=y$

$\Rightarrow \left[\frac{1}{y^{2}}\left(\frac{\cos(\tan^{-1}y) +y \sin(\tan ^{-1}y)}{\cot^{-1}(\tan^{-1}y) +\tan(\sin ^{-1}y)}\right)^{2}+y^{4}\right]^{1/2}$

$\Rightarrow \left[\frac{1}{y^{2}}\left(\frac{\frac{1}{\sqrt{1+y^{2}}}+\frac{y^{2}}{\sqrt{1+y^{2}}}}{\frac{\sqrt{1-y^{2}}}{y}+\frac{y}{\sqrt{1-y^{2}}}}\right)^{2}+y^{4}\right]^{1/2}$

$\Rightarrow\left[\frac{1}{y^{2}}.y^{2}(1-y^{4})+y^{4}\right]^{1/2}=1$

 cos x +cos y= -cos z

 sinx +sin y =-sinz

 on squaring and adding , we get

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 2 balls are drawn (without replacement)   from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B₂ is 

Answer:

Explanation:

 P(Ball drawn from box 2/one is white and one is red)

$=\frac{p(A\cap B)}{p(B)}$

  $=\frac{\frac{1}{3}\times\frac{2\times3}{^{9}C_{2}}}{\frac{1}{3}\left[\frac{1\times3}{^{6}C_{2}}+\frac{2\times3}{^{9}C_{2}}+\frac{3\times4}{^{12}C_{2}}\right]}$

 =   $\frac{\frac{1}{6}}{\frac{1}{5}+\frac{1}{6}+\frac{2}{11}}=\frac{\frac{1}{6}}{\frac{181}{55\times60}}=\frac{55}{181}$

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