JEE 2016 :: Advanced 2016 :: Mathematics 2016

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• Advanced 2016 - Mathematics 2016

The total number of distincts   $x \epsilon R$ for  which $\begin{bmatrix}x & x^{2} &1+x^{3} \\2x & 4x^{2}& 1+8x^{3} \\3x&9x^{2}&1+27x^{3} \end{bmatrix}=10 $  is

Answer:

Explanation:

Given,    $\begin{bmatrix}x & x^{2} &1+x^{3} \\2x & 4x^{2}& 1+8x^{3} \\3x&9x^{2}&1+27x^{3} \end{bmatrix}=10 $

  =  $\Rightarrow x.x^{2}\begin{bmatrix}1 & 1 &1+x^{3} \\2 & 4& 1+8x^{3} \\3&9&1+27x^{3} \end{bmatrix}=10 $

 Apply  $R_{2}\rightarrow R_{2}-2R_{1}$  and $R_{3}\rightarrow R_{3}-3R_{1}$

we get,

              $x^{3}\begin{bmatrix}1 & 1 &1+x^{3} \\0 & 2& -1+6x^{3} \\0&6&-2+24x^{3} \end{bmatrix}=10$

  $\Rightarrow$     $x^{3}\begin{bmatrix}2 & 6x^{3} &-1\\6 & 24x^{3}&-2 \end{bmatrix}=10$

   $\Rightarrow$    $x^{3} (48x^{3}-4-36x^{3}+6)=10$

  $\Rightarrow$   $12x^{6}+2x^{3}=10$

   $\Rightarrow$    $6x^{6}+x^{3}-5=0$

  $\Rightarrow$  $x^{3}=\frac{5}{6},-1$

$\therefore$               $x=\left(\frac{5}{6}\right)^{1/3},-1$

   Hence, the number of real solutions is 2

Let $z=\frac{-1+\sqrt{3}i}{2}$ , where   $i=\sqrt{-1}$  , and  $ r,s \epsilon (1,2,3)$. Let  $P=\begin{bmatrix}(-z)^{r} & z^{2s} \\z^{2s} & z^{r} \end{bmatrix}$  and  I be the identity matrix of  order2.Then, the total number of ordered pairs (r,s) for which   $p^{2}=-I$ is

Answer:

Explanation:

Here,

$z=\frac{-1+\sqrt{3}i}{2}$ =ω

$\because$     $P= \begin{bmatrix}(-\omega)^{r} & \omega^{2s} \\\omega^{2s} & \omega^{r}\end{bmatrix}$

   $P^{2}=\begin{bmatrix}(-\omega)^{r} & \omega^{2s} \\\omega^{2s} & \omega^{r} \end{bmatrix} \begin{bmatrix}(-\omega)^{r} & \omega^{2s} \\\omega^{2s} & \omega^{r} \end{bmatrix}$

         $=\begin{bmatrix}\omega^{2r}+\omega^{4s} & \omega^{r+2s}[(-1)^{r} +1\\\omega^{r+2s}[(-1)^{r}+1 & \omega^{4s}+\omega^{2r} \end{bmatrix}$

Given,           $P^{2}=-1$

$\therefore$       $\omega^{2r}+\omega^{4s}=-1$

  and    $\omega^{r+2s}[(-1)^{r}+1]=0$

Since,      $r\epsilon (1,2,3)$  and  $(-1)^{r}+1=0$

$\Rightarrow$          r={1,3}

Also,   $\omega^{2r}+\omega^{4r}=-1$

  If r=1  . then  $\omega^{2}+\omega^{4s}=-1$

 which is only possible , when s=1

As,    $\omega^{2}+\omega^{4}=-1$

$\therefore$    r=1, s=1

  Again, if r=3, then

$\omega^{6}+\omega^{4s}=-1$

$\Rightarrow$  $\omega^{4s}=-2$  [ never possible ]

   $\therefore$      $r\neq 3$

 $\Rightarrow$  (r,s) =(1,1) is the only solution.

 Hence , the total  number of ordered pairs is 1.

Let  $\alpha,\beta\epsilon R$   be such that $\lim_{x \rightarrow0}\frac{x^{2}\sin (\beta x)}{\alpha x-\sin x}=1$   Then, 

 $6(\alpha+\beta)$  equals 

Answer:

Explanation:

Here,   $\lim_{x \rightarrow0}\frac{x^{2}\sin (\beta x)}{\alpha x-\sin x}=1$

$\Rightarrow$    $\lim_{x \rightarrow 0}\frac{x^{2}\left[ \beta x- \frac{(\beta x)^{3}}{3!}+\frac{(\beta x)^{5}}{5!}-.....\right]}{\alpha x-\left[ x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-......\right]}$ =1

  $\Rightarrow$   $\lim_{x \rightarrow 0}\frac{x^{3}\left[\beta-\frac{\beta^{3}x^{2}}{3!}+\frac{\beta^{5}x^{4}}{5!}-.....\right]}{(\alpha-1)x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....}=1$

Limits exist only,

   when  , $\alpha -1=0$

 $\Rightarrow$    $\alpha =1$             ............(i)

$\therefore$    $\lim_{x \rightarrow 0}\frac{x^{3}\left[ \beta-\frac{\beta^{3}x^{2}}{3!}+\frac{\beta^{5}x^{4}}{5!}- .....\right]}{x^{3\left[\frac{1}{3!}-\frac{x^{2}}{5!}-  ....\right]}}=1$

 $\Rightarrow$    $6\beta=1$          ........(ii)

  From Eq .(i) and (ii),  we get   $6(\alpha+\beta)=6\alpha+6\beta$

                      =6+1= 7

The total  number of distincts  $x \epsilon  [0,1]$ for which   $\int_{0}^{x} \frac{t^{2}}{1+t^{4}}dt=2x-1$  is

Answer:

Explanation:

Let  $f(x)=\int_{0}^{x} \frac{t^{2}}{1+t^{4}}dt$

   $\Rightarrow f'(x)=\frac{x^{2}}{1+x^{4}}>0$  , for all $x \epsilon  [0,1]$

  $\therefore$    f(x) is increasing
  At   x=0, f(0)=0    and x=1,

  $f(1)=\int_{0}^{1} \frac{t^{2}}{1+t^{4}}dt$

Because,   $0<\frac{t^{2}}{1+t^{4}}<\frac{1}{2}$

$\Rightarrow$      $\int_{0}^{1} 0.dt <\int_{0}^{1} \frac{t^{2}}{1+t^{4}}dt<\int_{0}^{1} \frac{1}{2}.dt$

$\Rightarrow$    $0<f(1)<\frac{1}{2}$

 Thus, f(x) can be plotted as,

$\therefore$   y= f(x) and  y=2x-1 can be shown as

From the graph , the total number of distinct solutions for  $x \epsilon  [0,1]$ = 1

                               [ as they intersect only at one point]

Let m be the smallesr positive integer such that the coefficient of  $x^{2}$  in the  expansion of $(1+x)^{2}+(1+x)^{3}+.....+(1+x)^{49}+(1+mx)^{50}$   is

   $\left(3n+1\right)^{51}C_{3}$  for some positive integer  n, Then the value of n is

Answer:

Explanation:

Coefficient of $x^{2}$  in the expansion of

  {$(1+x)^{2}+(1+x)^{3}+.....+(1+x)^{49}+(1+mx)^{50}$} 

$\Rightarrow$    $^{2}C_{2}+^{3}C_{2}+^{4}C_{2}+......+^{49}C_{2}+^{50}C_{2}.m^{2}$

$= (3n+1).^{51}C_{3}$

$\Rightarrow$     $^{50}C_{3}+^{50}C_{2}m^{2}=(3n+1).^{51}C_{3}$

[$\because $   $^{r}C_{r}+^{r+1}C_{r}+.....+^{n}C_{r}=^{n+1}C_{r+1}$  ]

$\Rightarrow$     $\frac{50\times 49\times48}{3\times2\times1}+\frac{50\times49}{2}\times m^{2}$

$=(3n+1)\frac{51\times 50\times 49}{3\times2\times1}$

$m^{2}=51n+1$

$\therefore$    Minimum  value of  $m^{2}$   for which (51n+1)  is integer (perfect square) for n=5.

$\therefore$    $m^{2}=51\times5+1$

$\Rightarrow$     $m^{2}=256$

$\therefore$         m=16  and n=5    

Hence, the value of n is 5

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