JEE 2014 :: Main 2014 :: Mathematics 2014

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If   $\alpha,\beta\neq0$   and f(n)=  $\alpha^{n}+\beta^{n}$   and $\begin{bmatrix}3 & 1+f(1)&1+f(2) \\1+f(1) & 1+f(2)&1+f(3)\\ 1+f(2)&1+f(3)&1+f(4) \end{bmatrix}$  = $K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$  , then K is equal to

Answer:

Explanation:

 Use the property that, two determinants can be multipied row-to row or row -to-column to 

write the given determinant as the product of two determinants and then expand.

Given,  f(n)=  $\alpha^{n}+\beta^{n}$,  f(1)=  $\alpha^{1}+\beta^{1}$, f(2)=  $\alpha^{2}+\beta^{2}$, 

 f(3)=  $\alpha^{3}+\beta^{3}$, f(4)=  $\alpha^{4}+\beta^{4}$

Let     $\triangle=\begin{bmatrix}3 & 1+f(1)&1+f(2) \\1+f(1) & 1+f(2)&1+f(3)\\ 1+f(2)&1+f(3)&1+f(4) \end{bmatrix}$

       $\Rightarrow\triangle=\begin{bmatrix}3 & 1+\alpha+\beta&1+\alpha^{2}+\beta^{2} \\1+\alpha+\beta & 1+\alpha^{2}+\beta^{2}&1+\alpha^{3}+\beta^{3}\\ 1+\alpha^{2}+\beta^{2}&1+\alpha^{3}+\beta^{3}&1+\alpha^{4}+\beta^{4} \end{bmatrix}$

$\Rightarrow\triangle=\begin{bmatrix}1.1+1.1+1.1 & 1.1+1.\alpha+1.\beta&1.1+1.\alpha^{2}+1.\beta^{2} \\1.1+1.\alpha+1.\beta & 1.1+\alpha.\alpha^{}+\beta.\beta^{}&1.1+\alpha.\alpha^{2}+\beta.\beta^{2}\\ 1.1+1.\alpha^{2}+1.\beta^{2}&1+\alpha^{2}.\alpha+\beta^{2}.\beta&1.1+\alpha^{2}.\alpha^{2}+\beta^{2}.\beta^{2} \end{bmatrix}$

=$\begin{bmatrix}1 & 1 &1\\1 & \alpha &\beta\\1&\alpha^{2}&\beta^{2}\end{bmatrix}\begin{bmatrix}1 & 1 &1\\1 & \alpha &\beta\\1&\alpha^{2}&\beta^{2}\end{bmatrix}=\begin{bmatrix}1 & 1 &1\\1 & \alpha &\beta\\1&\alpha^{2}&\beta^{2}\end{bmatrix}^{2}$

On expanding , we get 

$\triangle=(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$

 But given     $\triangle=K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$

 hence,   $K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}=(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$

                          K=1

Let   $\alpha $   and $\beta$ be the roots of equation $px^{2}+qx+r=0$ , $p\neq0$ . If p,q and r in AP and  $\frac{1}{\alpha}+\frac{1}{\beta}=4$  , then the value of   $|\alpha-\beta|$ is 

Answer:

Explanation:

If    $ax^{2}+bx+c=0$ ,  has roots  $\alpha$ and $\beta$   , then 

  $\alpha+\beta=\frac{-b}{a}$    and    $\alpha\beta=\frac{c}{a}$ . Find the valuesw of   $\alpha+\beta$  and $\alpha \beta$ and then put in     $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta$    to get required value.

  Given   $\alpha $   and $\beta$    are roots of   

                          $px^{2}+qx+r=0$ , $p\neq0$ 

  $\therefore$    $\alpha+\beta$  = $\frac{-q}{p}$  ,   $\alpha \beta$=   $\frac{r}{p}$    ......(i)

Since p,q and r are in A.P

$\therefore$      2q=p+r        ................(ii)

Also,    $\frac{1}{\alpha}+\frac{1}{\beta}=4$

$\Rightarrow$     $\frac{\alpha+\beta}{\alpha\beta}=4$

$\Rightarrow$       $\alpha+\beta=4\alpha\beta$

$\Rightarrow$   $\frac{-q}{p}=\frac{4r}{p}$  [ from Eq.(i)]

$\Rightarrow$     q=-4r

 On putting the value of q in Eq(ii) , we get  $\Rightarrow$  2(-4r)=p+r  $\Rightarrow$  p=-9r

Now  $\alpha+\beta=\frac{-q}{p}=\frac{4r}{p}=\frac{4r}{-9r}=-\frac{4}{9}$    and $\alpha\beta=\frac{r}{p}=\frac{r}{-9r}=\frac{1}{-9}$

$\therefore$    $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta=\frac{16}{81}+\frac{4}{9}=\frac{16+36}{81}$

$\Rightarrow$    $(\alpha-\beta)^{2}=\frac{52}{81}\Rightarrow|\alpha-\beta|=\frac{2}{9}\sqrt{31}$

If  a ε R  and equation -3(x-[x])²+2(x-[x])+a²=0    (where,[x] denotes the greatest integer   $ \leq$ x)  has no integral solution, then all possible values of  a lie  in the interval

Answer:

Explanation:

Put  t= x -[x]={X}, which  is a fractional part function and lie between    $0\leq \left\{x\right\}$  <1 and then solve it,

  Given    a ε R and equation is     

   -3(x-[x])²+2(x-[x])+a²=0   

  t=x-[x] , then equation is

$-3t^{2}+2t+a^{2}=0$

$\Rightarrow $      $  t=\frac{1\pm\sqrt{1+3a^{2}}}{3}$

$\therefore$    t= x-[x]= {X}

                          { fractional part}

$\therefore$         $0\leq t\leq1$

                   $ 0\leq\frac{1\pm\sqrt{1+3a^{2}}}{3}\leq1$

Taking positive sign , we get 

     $   0\leq\frac{1+\sqrt{1+3a^{2}}}{3}<1$                   [ $\therefore$ {x}>0}

$\Rightarrow$                $ \sqrt{1+3a^{2}}<2\Rightarrow1+3a^{2}<4$

$\Rightarrow$                    a²-1<0

$\Rightarrow$            (a+1)(a-1)<0

   for no integral  solution of a we consider the interval (-1,0) $ \cup$  (0,1) .

Note.  here, we figure out the integral solution , we get a=0, This implies any interval excluding zero should be correct answer as it gives either  no solution or no integral.

If z is complex number such that   $|z|\geq 2$  then the minimum value of   $|z+\frac{1}{2}|$  

Answer:

Explanation:

$|z|\geq 2$  is the region on or outside circle whose centre is (0,0) and radius is 2 .

 Minimum   $|z+\frac{1}{2}|$   is distance of z, which

lie on circle |z|=2 from ( -$\frac{1}{2},0$)

   $\therefore$  Minimum  $|z+\frac{1}{2}|$ = Distance of ($-\frac{1}{2},0$) from (-2,0)

                       =$\sqrt{(-2+\frac{1}{2})^{2}+0}=\frac{3}{2}$

            = $\sqrt{(-\frac{1}{2}+2)^{2}+0}=\frac{3}{2}$

Geometrically min   $|z+\frac{1}{2}|$ = AD

Here minimum value of   $(z+\frac{1}{2}) $   lies in the intervel (1,2)

If   $X= \left\{  4^{n}-3n-1:n\epsilon N\right\}$  and   $Y= \left\{  9(n-1):n\epsilon N\right\}$  when N is the set of natural numbers, then   $X\cup Y$ is equal to

Answer:

Explanation:

We have,

 $X= \left\{  4^{n}-3n-1:n\epsilon N\right\}$

     X= { 0,9,54,243.....}  [put n= 1,2,3 ......]

 $Y= \left\{  9(n-1):n\epsilon N\right\}$

           Y= {0,9,18,27....}

              [Put    n= 1,2,3....]

  It is clear that    $X\subset Y$

$\therefore $      $X\cup Y=Y$

• Main 2014 - Physics 2014
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• Main 2014 - Mathematics 2014