JEE 2013 :: Advanced 2013 :: Mathematics 2013

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

Perpendicular are drawn  from points on the line   $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the  plane x+y+z=3 . The feet of perpendicular lie on the line

Answer:

Explanation:

Concept involved

  To find the foot of perpendicular and find its locus

 Formula used

 Foot  of perpendicular form  $(x_{1},y_{1},z_{1})$   to ax+by+cz+d=0 be    $(x_{2},y_{2},z_{2})$

  then

$\frac{x_{2}-x_{1}}{a}=\frac{y_{2}-y_{1}}{b}=\frac{z_{2}-z_{1}}{a}$

                               = $\frac{-(ax_{1}+by_{1}+cz_{1}+d)}{a^{2}+b^{2}+c^{2}}$

      Any point on,     $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda$

                     $\Rightarrow x=2\lambda-2,y=\lambda-1, z=3\lambda$

Let focus of perpendicular from   $(2\lambda-2,\lambda-1, 3\lambda)$

 to x+y+z=3  be ($(x_{2},y_{2},z_{2})$)

 $\therefore$       $\frac{x_{2}-(2\lambda-2)}{1}=\frac{y_{2}-(-\lambda-1)}{1}=\frac{z_{2}-(3\lambda)}{1}$

   =  $-\frac{(2\lambda-2-\lambda-1+3\lambda-3)}{1+1+1}$

               =  $x_{2}-2\lambda+2=y_{2}+\lambda+1$

   =  $z_{2}-3\lambda=2-\frac{4\lambda}{3}$

 $\therefore$        $x_{2}=\frac{2\lambda}{3}$     ,  $y_{2}=1-\frac{7\lambda}{3}$,  $z_{2}=2+\frac{5\lambda}{3}$

 $\Rightarrow \lambda=\frac{x_{2}-0}{2/3}=\frac{y_{2}-1}{-7/3}=\frac{z_{2}-2}{5/3}$

$\therefore$  foot  of perpendicular lie on

                 $\frac{x}{2/3}=\frac{y_{}-1}{-7/3}=\frac{z_{}-2}{5/3}$

   $\Rightarrow\frac{x}{2}=\frac{y_{}-1}{-7}=\frac{z_{}-2}{5}$

For a>b>c>0, the distance between (1,1)  and the point of intersection of the lines ax+by+c=0  and bx+ay+c=0 is less than 

$2\sqrt{2}$  then 

Answer:

Explanation:

Concept involved Application of inequality sum and difference along with lengths of perpendicular  . For this type of questions involving inequality we should always check all options.

Situation  analysis check all inequalities according to options and use length of perpendicular from the point (x₁,y₁) to ax+by+c=0

 i.e,   $\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}$

   As, a>b>c>0

  a-c>0  and b>0

 $\Rightarrow$ a+b-c >0  ......(i)

  a-b >0 and c>0

 a+c-b>0  .......(ii)

 $\therefore$   (a) and (c) is correct

Also the point of intersection  for  ax+by+c=0 and bx+ay+c=0

 i.e,    $\left(\frac{-c}{a+b},\frac{-c}{a+b}\right)$

  The distance between (1,1) and 

                      $\left(\frac{-c}{a+b},\frac{-c}{a+b}\right)$

 i.e, less than  $2\sqrt{2}$ 

$\Rightarrow\sqrt{\left(1+\frac{c}{a+b}\right)^{2}+\left(1+\frac{c}{a+b}\right)^{2}}<2\sqrt{2}$

    $\Rightarrow\left(\frac{a+b+c}{a+b}\right)\sqrt{2}<2\sqrt{2}$

$\Rightarrow $   a+b+c <2a+2b

   or a+b-c >0

  From Eqs.(i) and (ii) option (c) is correct.

Let complex numbers $\alpha$  and $\frac{1}{\alpha}$ lies on circles   $(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2} $  and 

$(x-x_{0})^{2}+(y-y_{0})^{2}=4r^{2} $ respectively.If z₀=x₀+iy₀ satisfies the equation   2|z₀|²=r²+2 , then |$\alpha$|  is equal to

Answer:

Explanation:

Concept involved intersection of circles , the basic concept is to equations simulataneously and using properties of modules of complex numbers.

 Formula used  |z|²=$z.\overline{z}$

 and   $|z_{1}-z_{2}|^{2}=(z_{1}-z_{2})(\overline{z_{1}}-\overline{z_{2}})$

=  $|z_{1}|^{2}-z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}+|z_{2}|^{2}$

  Here, 

$(x-x_{0})^{2}+(y-y_{0})^{2}=r^{2} $  and 

$(x-x_{0})^{2}+(y-y_{0})^{2}=4r^{2} $    could be written as,

   $|z-z_{0}|^{2}=r^{2}$   and  $ |z-z_{0}|^{2}=4r^{2}$

 since ,$\alpha$ and $\frac{1}{\overline{\alpha}}$ lies on first and second  respectively,

   $\therefore$    $|\alpha-z_{0}|^{2}=r^{2}$   and$ |\frac{1}{\overline{\alpha}}-z_{0}|^{2}=4r^{2}$

$\Rightarrow$            $(\alpha-z_{0})(\overline{\alpha}-\overline{z_{0}})=r^{2}$

$\Rightarrow$         $|\alpha|^{2}-z_{0}\overline{\alpha}-\overline{z_{0}}\alpha+|z_{0}|^{2}=r^{2}$ .....(i)

and     $|\frac{1}{\overline{\alpha}}-z_{0}|^{2}=4r^{2}$

$\Rightarrow$     $(\frac{1}{\overline{\alpha}}-z_{0})(\frac{1}{\alpha}-\overline{z_{0}})=4r^{2}$

      $\Rightarrow$     $\frac{1}{|\alpha|^{2}}-\frac{z_{0}}{\alpha}-\frac{\overline{z_{0}}}{\overline{\alpha}}+|z_{0}|^{2}=4r^{2}$

Since  $|\alpha|^{2}=\alpha.\alpha$

$\Rightarrow$     $\frac{1}{|\alpha|^{2}}-\frac{z_{0}.\overline{\alpha}}{|\alpha|^{2}}-\frac{\overline{z_{0}}}{|{\alpha}|^{2}}\alpha+|z_{0}|^{2}=4r^{2}$

$\Rightarrow$     $1-z_{0}\overline{\alpha}-\overline{z_{0}}\alpha+|\alpha|^{2}|z_{0}|^{2}=4r^{2}|\alpha|^{2}$ ...(ii)

 On substracting Eqs (i) and (ii) , we get

   $(|\alpha|^{2}-1)+|z_{0}|^{2}(1-|\alpha|^{2})$

                                             = $r^{2}(1-4|\alpha|^{2})$

 $\Rightarrow$      $(|\alpha|^{2}-1)(1-|z_{0}|^{2})=r^{2}(1-4|\alpha|^{2})$

    $\Rightarrow$       $(|\alpha|^{2}-1)\left(1-\frac{r^{2}+2}{2}\right)$

                                                            $=r^{2}(1-4|\alpha|^{2})$

Given,     $|z_{0}|^{2}=\frac{r^{2}+2}{2}$

$\Rightarrow$        $(|\alpha|^{2}-1).\left(\frac{-r^{2}}{2}\right)=r^{2}(1-4|\alpha|^{2})$    

$\Rightarrow$   $|\alpha|^{2}-1=-2+8|\alpha|^{2}$

$\Rightarrow$       $7|\alpha|^{2}=1$

 $\therefore$     $|\alpha|=\frac{1}{\sqrt{7}}$

Let   $PR=3\hat{i}+\hat{j}-2\hat{k}$   and  $SQ=\hat{i}-3\hat{j}-4\hat{k}$ determine diagonals of a parallelogram PQRS and

 $PT=\hat{i}+2\hat{j}+3\hat{k}$   be another vector . Then, the volume  of the parallelopiped determined by the vectors PT,PQ and PS is

Answer:

Explanation:

 Concept involved . it involves law of   parallelogram and volume of parallelopiped . i.e,

 a+b=p  and b-a =q

              $a=\frac{p-q}{2}$

 and $b=\frac{p+q}{2}$

i.e, if p and q are diagonals of parallelograms , then its sides are $\frac{p-q}{2}$  and $\frac{p+q}{2}$

Situation analysis

 After finding the sides of parallelogram we should find volume of parallelopiped i.e, [a,b]

 HERE, sides of parallelogram are PQ and PS

 where,  $PQ=\frac{PR+SQ}{2}$

  =$PQ=\frac{(3\hat{i}+\hat{j}-2\hat{k})+(\hat{i}-3\hat{j}-4\hat{k})}{2}$

$PQ=2\hat{i}-\hat{j}-3\hat{k}$

 and PS=$\frac{PR-SQ}{2}$

   = $PS=\frac{(3\hat{i}+\hat{j}-2\hat{k})-(\hat{i}-3\hat{j}-4\hat{k})}{2}$

 PS =$\hat{i}+2\hat{j}+\hat{k}$

 $\therefore$     Volume of parallelopiped

  =[PT PQ PS]

 =  $\begin{bmatrix}1 & 2 &3\\2 & -1&-3\\1&2&1 \end{bmatrix}$

  =1(-1+6)-2(2+3)+3(4+1)

   =5-10+15=10

The value of $cot \left\{\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right\}$  is 

Answer:

Explanation:

Concept involved difference series , when ever , we have summation involving  more than 3 terms we should always covert into diferences

 e.g,  $\sum_{r=1}^{5}\frac{1}{r(r+1)}=\frac{1}{1.(2)}+\frac{1}{2.(3)}+\frac{1}{3.(4)}+\frac{1}{4.5}+\frac{1}{5.6}$

 $=\frac{2-1}{1.(2)}+\frac{3-2}{2.(3)}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}$

 $=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{6})=1-\frac{1}{6}=\frac{5}{6}$

 Situation Analysis

    Convert into

  $\tan^{-1} x-\tan^{-1} y=\tan^{-1} \left(\frac{x-y}{1+xy}\right)$

$\cot\left(\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{n=1}^{n}2k\right)\right)$

$\cot\left(\sum_{n=1}^{23}\cot^{-1}(1+2+4+6+8...+2n)\right)$

$\Rightarrow\cot\left(\sum_{n=1}^{23}\cot^{-1}(1+n(n+1))\right)$

$\Rightarrow\cot\left(\sum_{n=1}^{23}\tan^{-1}\frac{1}{1+n(n+1)}\right)$

$\Rightarrow\cot\left(\sum_{n=1}^{23}\tan^{-1}\frac{(n+1)-n}{1+n(n+1)}\right)$

 $\Rightarrow\cot\left(\sum_{n=1}^{23}\tan^{-1}(n+)-\tan^{-1}ln n\right)$

   $\Rightarrow\cot\left(\tan^{-1}2-\tan^{-1}1\right)+\left(\tan^{-1}3-\tan^{-1}2\right)+\left(\tan^{-1}4-\tan^{-1}3\right)$+   $.............. +\left(\tan^{-1}24-\tan^{-1}23\right)$

$\Rightarrow\cot\left(\tan^{-1}24-\tan^{-1}1\right)$

$\Rightarrow\cot\left(\tan^{-1}\frac{24-1}{1+24-(1)}\right)$

           $\Rightarrow\cot\left(\tan^{-1}\frac{23}{25}\right)$

   = $\cot\left(\cot^{-1}\frac{25}{23}\right)$

    =$\frac{25}{23}$

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