JEE 2013 :: Advanced 2013 :: Mathematics 2013

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Let   $w=\frac{\sqrt{3}+i}{2}$  and   $P= \left\{ W^{n}:n=1,2,3...\right\}$   Further

  $H_{1}= \left\{ z \epsilon C;Rez>\frac{1}{2}\right\}$     and

$H_{2}= \left\{ z \epsilon C;Rez<-\frac{1}{2}\right\} $

where C is the set of all complex numbers, if   $z_{1}\in P\cap H_{1},z_{2}\in P\cap H_{2}$    and O represents the origin, then   $\angle z_{1}Oz_{2}$ is equal to 

Answer:

Explanation:

Concept involved

 It is simple representation of points on argand plane and to find the angle between the points

 here,  $P=W^{n}=\left( \cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)^{n}$

  $= \cos\frac{n\pi}{6}+i\sin\frac{n\pi}{6}$

 $H_{1}=\left\{Z\in C: Re (z)>\frac{1}{2}\right\}$

 $\therefore$   $P\cap H_{1}$ represents those points for which $\cos\frac{n \pi}{6}$ is +ve

$\therefore$ it belongs to I to IV quadrant

$\Rightarrow$    $ z_{1}=P\cap H_{1}=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}$

 or    $\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}$

 $\therefore$        $z_{1}=\frac{\sqrt{3}}{2}+\frac{i}{2} or\frac{\sqrt{3}}{2}-\frac{i}{2}$    .....(i)

 Similarly ,   $ z_{2}=P\cap H_{2}$ i,e.   those points for which 

 $\cos \frac{n\pi}{6}<0$

 $\therefore$     $z_{2}=\cos\pi+i\sin\pi,\cos\frac{5\pi}{6}$

                                     $+i\sin\frac{5\pi}{6},\frac{\cos 7\pi}{6}+i\sin\frac{7\pi}{6}$

 $\Rightarrow z_{2}=-1, \frac{-\sqrt{3}}{2}+\frac{i}{2},\frac{-\sqrt{3}}{2}-\frac{i}{2}$

 Thus,  $\angle z{1}Oz_{2}=\frac{2\pi}{3},\frac{5\pi}{6},\pi$

 In a $\triangle$ PQR, P is the largest angle and  $\cos p=\frac{1}{3}$ . Further in circle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of Pn, QL, and RM are consecutive even integers. Then  possible length(s) of the side(s)  of the triangle is (are)

Answer:

Explanation:

Concept involved

 When ever cosine of angle and sides are given or to find out , we should always use cosine law

 i.e,    $\cos A= \frac{b^{2}+c^{2}-a^{2}}{2bc}$

     $\cos B= \frac{a^{2}+c^{2}-b^{2}}{2ac}$

$\cos C= \frac{a^{2}+b^{2}-c^{2}}{2ba}$

$\therefore$   $\cos P= \frac{b^{2}+c^{2}-a^{2}}{2bc}$

 $\Rightarrow \frac{1}{3}=\frac{(2n+4)^{2}+(2n+2)^{2}-(2n+6)^{2}}{2(2n+4)(2n+2)}$

    ($\because$    cos P= $ \frac{1}{3}$)

 $\Rightarrow \frac{4n^{2}-16}{8(n+1)(n+2)}=\frac{1}{3}$

  $\Rightarrow \frac{n^{2}-4}{2(n+1)(n+2)}=\frac{1}{3}$

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

$\Rightarrow$     3n-6=2n+2 $\Rightarrow$ n=8

$\therefore$   sides are 2n+2, 2n+4,2n+6

$\Rightarrow$  18,20,22

 Two lines L₁ :x=5 , $\frac{y}{3-\alpha}=\frac{z}{-2}$    and   $L_{2}:x=\alpha,\frac{y}{-1}=\frac{z}{2-\alpha}$ are coplannar , Then , $\alpha$ can take value(s)

Answer:

Explanation:

Concept involved

 If two straight lines are coplannar

i.e,   $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$  and 

$\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$ are coplannar

  Then ,(x₂-x₁ , y₂-y₁,z₂-z₁), (a₁,b₁,c₁)  and (a₂,b₂,c₂) are coplanar

 i,e,

   $\begin{bmatrix}x_{2}-x_{1} & y_{2}-y_{1}&z_{2}-z_{1} \\a_{1} & b_{1}&c_{1}\\a_{2}&b_{2}&c_{2} \end{bmatrix}=0$

Here  x=5, $\frac{y}{3-\alpha}=\frac{z}{-2}$

 $\Rightarrow \frac{x-5}{0}=\frac{y-0}{-(\alpha-3)}=\frac{z-0}{-2}$......(i)

 and   $x=\alpha,\frac{y}{-1}=\frac{z}{2-\alpha}$

  $\Rightarrow \frac{x-\alpha}{0}=\frac{y-0}{-1}=\frac{z-0}{2-\alpha}$   .....(ii)

$\Rightarrow    \begin{bmatrix}5-\alpha & 0&0\\0 & 3-\alpha&-2\\0&-1&2-\alpha \end{bmatrix}=0$

  $\Rightarrow$     $ (5-\alpha)[(3-\alpha)(2-\alpha)-2]=0$

 $\Rightarrow$    $(5-\alpha)[\alpha^{2}-5\alpha+4]=0$

 $\Rightarrow $    $ (5-\alpha)(\alpha-1)(\alpha-4)=0$

                     $\alpha$  =1,4,5

Circle (s) touching x-axis at a distance 3 from the origin and having an intercept of length  $2\sqrt{7}$ on the y-axis is (are)

Answer:

Explanation:

Concept involved

 Here, thr length of intercept on y-axis 

  $\Rightarrow$           $  2\sqrt{f^{2}-c}$

 and if circle touches x-axis

 $\Rightarrow$     $  g^{2}=c$

 for     $x^{2}+y^{2}+2gx+2fy+c=0$

 here, $x^{2}+y^{2}+2gx+2fy+c=0$

 Passes through (3,0)

 $\Rightarrow$        9+6g+c=0  ........(i)

    $g^{2}=c$    ..........(ii)

 and  $2\sqrt{f^{2}-c}=2\sqrt{7}$

          $f^{2}-c=7$     .........(iii)

 From Eqs .(i) and (ii), we get

 $g^{2}+6g+9=0$

    (g+3)²  =0

 g=-3

 and c=9  $\therefore$   f²=16

 $f=\pm 4$

 $\therefore$     $x^{2}+y^{2}-6x \pm 8y+9=0$

 For a ε R ( the set of all real numbers ), a ≠ -1, 

$\lim_{n \rightarrow \infty}\frac{(1^{a}+2^{a}+....+n^{a})}{(n+1)^{a-1}[(na+1)+(na+2)+....+(na+n)]}$

=  $\frac{1}{60}$  . Then ,a is equal to

Answer:

Explanation:

Concept involved

 Converting infinite series into definite intergral

 ie     $\lim_{n\rightarrow \infty}\frac{h(n)}{n}$

 $\lim_{n \rightarrow \infty}\frac{1}{n} \sum_{r=g(n)}^{h(n)}f(\frac{r}{n})=\int_{}^{} f(x) dx$

$\lim_{n \rightarrow \infty}\frac{g(n)}{n}$

 where,r/n  is replaced with x

$\sum$  is replaced with integral

 Here  

$\lim_{n \rightarrow \infty}\frac{(1^{a}+2^{a}+....+n^{a})}{(n+1)^{a-1}[(na+1)+(na+2)+....+(na+n)]}$

=  $\frac{1}{60}$

 $\Rightarrow \lim_{n \rightarrow \infty}\frac{\sum_{r=1}^{n}r^{a}}{(n+1)^{a-1}\left[n^{2}a+\frac{n(n+1)}{2}\right]}=\frac{1}{60}$

$\Rightarrow \lim_{n \rightarrow \infty}\frac{2\sum_{r=1}^{n}(\frac{r}{n})^{a}}{(1+\frac{1}{n})^{a-1}.(2na+n+1)}$

$\Rightarrow \lim_{n \rightarrow \infty}\frac{1}{n}\left(2\sum_{r=1}^{n}(\frac{r}{n})^{a}\right)$

                $\times  \lim_{n \rightarrow \infty}\frac{1}{(1+\frac{1}{n})^{a-1}(2a+1+\frac{1}{n})}$

 $\Rightarrow  2\int_{0}^{1} (x^{a})dx.\frac{1}{1.(2a+1)}$

  $\Rightarrow \frac{2.(x^{a+1})_0^1}{(2a+1).)(a+1)}=\frac{2}{(2a+1)(a+1)}$

 $\therefore$     $\frac{2}{(2a+1)(a+1)}=\frac{1}{60}$

  $\Rightarrow(2a+1)(a+1)=120$

 $\Rightarrow2a^{2}+3a+1-120=0$

$\Rightarrow2a^{2}+3a-119=0$

 $\Rightarrow $     $ (2a+17) (a-7)=0$

 $\Rightarrow $    a=7, $\frac{-17}{2}$

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