JEE 2016 :: Advanced 2016 :: Mathematics 2016

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016

Let   $-\frac{\pi}{6}<\theta <-\frac{\pi}{12}$ . Suppose  $\alpha_{1}$ and $\beta_{1}$ are the roots of the equation $x^{2}-2x\sec\theta+1=0,$  and $\alpha_{2}$  and $\beta_{2}$ are roots of the equation $x^{2}+2x\tan\theta-1=0$. If  $\alpha_{1}>\beta_{1}$  and $\alpha_{2}>\beta_{2}$  , $\alpha_{1}+\beta_{2}$ equals

Answer:

Explanation:

Here,  $x^{2}-2x\sec\theta+1=0$ has roots  $\alpha_{1}$ and $\beta_{1}$.

 $\therefore$     $\alpha_{1},\beta_{1}=\frac{2\sec \theta\pm \sqrt{4\sec^{2}\theta-4}}{2\times 1}$

$=\frac{2\sec \theta\pm 2|\tan\theta|}{2}$

Since, $\theta \epsilon(-\frac{\pi}{6},-\frac{\pi}{12}),$

i.e,    Θ ε IV   quadrant = $\frac{2\sec \theta\pm2\tan \theta}{2}$

$\therefore$          $\alpha_{1}=\sec \theta-\tan\theta$      and    $\beta_{1}=\sec \theta+\tan\theta$

 (as $\alpha_{1}$ > $\beta_{1}$ )

 and     $x^{2}+2x\tan \theta-1=0$ has roots  $\alpha_{2}$  and $\beta_{2}$.

 i.e. $\alpha_{2} , \beta_{2}= \frac{-2\tan \theta\pm\sqrt{4\tan^{2}\theta+4}}{2}$

 $\therefore$      $\alpha_{2} =-\tan\theta+\sec \theta$

  and   $\beta_{2} =-\tan\theta-\sec \theta $  [as   $\alpha_{2}>\beta_{2} $ ]

 Thus,  $\alpha_{1}+\beta_{2}=-2\tan\theta$

The  least value of  $\alpha   \epsilon R$ for which  $4\alpha x^{2}+\frac{1}{x}\geq 1,$ , for all x >0, is 

Answer:

Explanation:

 Here, to find the least value of   $\alpha   \epsilon R$ . for which  

                           $4\alpha x^{2}+\frac{1}{x}\geq 1,$ , for all x >0

 i.e. to find the minimum value of $\alpha$ when

                  $y=4\alpha x^{2}+\frac{1}{x};x>0$  attains minimum value of $\alpha$

                 $\therefore$           $\frac{\text{d}y}{\text{d}x}=8\alpha x-\frac{1}{x^{2}}$    .....(i)

           Now,      $\frac{d^{2}y}{dx^{2}}=8\alpha +\frac{2}{x^{3}}$       ........(ii)

                        when       $\frac{\text{d}y}{\text{d}x}=0,$

         then                $8x^{3}\alpha =1$

           $\frac{d^{2}y}{dx^{2}}=8\alpha+16\alpha=24\alpha ,$ Thus, y attains minimum when

                            $x=(\frac{1}{8\alpha})^{\frac{1}{3}}: \alpha >0.$

     $\therefore$ y attains minimum when   $x=(\frac{1}{8\alpha})^{\frac{1}{3}}$

 i.e,   $4\alpha(\frac{1}{8\alpha})^{\frac{2}{3}}+(8\alpha)^{\frac{1}{3}}\geq1$

    $\Rightarrow \alpha^{\frac{1}{3}}+2\alpha^{\frac{1}{3}}\geq1\Rightarrow 3\alpha^{\frac{1}{3}}\geq1$

$\Rightarrow $                     $\alpha \geq\frac{1}{27} $

     Hence, the least value of $\alpha$   is  $\frac{1}{27}$

A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include almost one boy, the number of ways of selecting the team is

Answer:

Explanation:

We have 6 girls and 4 boys. To select 4 members (at most one boy)

  i.e. (1 boy and 3 girls ) or (4 girls)

                   =  $^{6}C_{3}.^{4}C_{1}+^{6}C_{4}$            ............(i)

         Now, selection of captain from 4 members= $^{4}C_{1}$                  ...........(ii)

     $\therefore$     Number of ways to select 4 members ( including the selection of a captain , from these 4 members)

           =     $ (^{6}C_{3}.^{4}C_{1}+^{6}C_{4} )^{4}C_{1} $

           =   $(20\times 4  +15)\times 4  =380$

Let  $F_{1}(x_{1},0)$ and $F_{2}(x_{2},0)$  , for  $x_{1}<0$ and $x_{2}>0$ , be the foci of the ellipise $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$ . Suppose a parabola  having vertex at the origin and focus at $F_{2}$ intersects the  ellipse  at the point M in the first quadrant and at point N  in the fourth quadrant

If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the X-axis at Q, then the ratio of area of $\triangle MQR$ to area of the quadrilateral   $MF_{1}NF_{2}$ is

Answer:

Explanation:

Equation of tangent at $M(3/2,\sqrt{6})$ to     $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$ is

    $\frac{3}{2}.\frac{x}{9}+\sqrt{6}.\frac{y}{8}=1$                 .........(i)

 which intersect X-axis at (6,0)

   Also, equation of tangent at $N(3/2. -\sqrt{6})$ is

  $\frac{3}{2}.\frac{x}{9}-\sqrt{6}.\frac{y}{8}=1$          ...........(ii)

Eqs. (i) and (ii) intersect on X-axis at R (6.0)                        ..........(iii)

Also normal at  $M(3/2. \sqrt{6})$ is

$Y-\sqrt{6}=-\frac{\sqrt{6}}{2}(x-\frac{3}{2})$

 On solving with y=0,

   we get $Q(\frac{7}{2},0$    ............(iv)

$\therefore$     Area of  $\triangle MQR=\frac{1}{2}(6-\frac{7}{2})\sqrt{6}=\frac{5\sqrt{6}}{4}$ sq units

and area of quadrilateral  $MF_{1}NF_{2}$

              $=2\times\frac{1}{2}[1-(-1)]\sqrt{6}=2\sqrt{6}$ sq units

$\therefore$  Area of  $\triangle MQR$ / Area of quadrilateral  $MF_{1}NF_{2}$    = $\frac{5}{8}$

Let  $F_{1}(x_{1},0)$ and $F_{2}(x_{2},0)$  , for  $x_{1}<0$ and $x_{2}>0$ , be the foci of the ellipise $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$ . Suppose a parabola  having vertex at the origin and focus at $F_{2}$ intersects the  ellipse  at the point M in the first quadrant and at point N  in the fourth quadrant

The orthocentre of  $\triangle F_{1}MN$ is 

Answer:

Explanation:

Here,  $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$             ..........(i)

 has foci (± ae,0)

  where ,   $a^{2}e^{2}=a^{2}-b^{2}\Rightarrow a^{2}e^{2}=9-8$

   $\Rightarrow$  ae=± 1, i.e.   $F_{1}.F_{2}=(\pm1,0)$

Equation of parabola having vertex  O(0,0) and   $F_{2}(1,0)$    (as $x_{2}>0$)

                                             $y^{2}=4x$             ........(ii)

On solving   $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$ and  $y^{2}=4x$ , we get

$x=\frac{3}{2}$ and  $y=\pm\sqrt{6}$

   Equation of altitude through M on NF₁ is

 $\frac{y-\sqrt{6}}{x-3/2}=\frac{5}{2\sqrt{6}}$

$\Rightarrow (y-\sqrt{6})=\frac{5}{2\sqrt{6}}(x-3/2)$         ............(iii)

and equation of altitude through

                          F₁   is y=0               ........(iv)

 On solving Eqs. (iii) and (iv), we get    $(-\frac{9}{10},0)$  as orthocentre.

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016