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The number of values of k, for which the system of equations (k+1)x+8y=4k

    kx+(k+3)y=3k-1   has no solution , is 

Answer:

Explanation:

Given equations can be written in matrix form

 AX=B

 Where, A= $\begin{bmatrix}k+1 & 8 \\k& k+3 \end{bmatrix}=\begin{bmatrix}x   \\y  \end{bmatrix} and \begin{bmatrix}4k  \\3k-1  \end{bmatrix}$

 For no solution , |A|= 0  and (adj A)B≠ 0

nOW, |A|=   $\begin{bmatrix}k+1 & 8 \\k& k+3 \end{bmatrix}=0$

  $\Rightarrow$     (k+1)(k+3)-8K=0

$\Rightarrow$     k²+4k+3-8k=0

$\Rightarrow$    k²-4k+3=0

$\Rightarrow$     (k-1) (k-3)=0

$\Rightarrow$    k=1,k=3

   Now, adj A=   $\begin{bmatrix}k+3 & -8 \\-k & k+1 \end{bmatrix}$

  Now, (adj A)B=  $\begin{bmatrix}k+3 & -8 \\-k & k+1 \end{bmatrix}$   $\begin{bmatrix}4k  \\3k-1  \end{bmatrix}$

     =   $\begin{bmatrix}(k+3)(4k)-8(3k-1)  \\-4k^{2}+(k+1)(3k-1)  \end{bmatrix}$

   =  $\begin{bmatrix}4k^{2}-12k+8 \\-k^{2}+2k-1  \end{bmatrix}$

  Put k=1

(adj A)B=   $\begin{bmatrix}4-12-8\\-1+2-1 \end{bmatrix}=\begin{bmatrix}0  \\0  \end{bmatrix}$   not true

 Put k=3

 (adj A) B=   $\begin{bmatrix}36-36+8  \\-9+6-1  \end{bmatrix}=\begin{bmatrix}8  \\-4  \end{bmatrix}\neq0$   true

 Hence , required value of k is 3

    only onne value of k exist

A ray of light along  $x+\sqrt{3}y=\sqrt{3}$  gets reflected upon reaching x-axis , the equation of the reflected ray is

Answer:

Explanation:

Take any point B(0,1) on gives line, equation of AB'

 $y-0= \frac{-1-0}{0-\sqrt{3}}(x-\sqrt{3})$

$\Rightarrow$     $ -\sqrt{3}y=-x+\sqrt{3}$

 $\Rightarrow$     $ x-\sqrt{3}y=\sqrt{3}$

$\Rightarrow $     $\sqrt{3}y=x-\sqrt{3}$

 The sum of first 20 terms of the sequence 0.7 ,0.77,0.777, .... is 

Answer:

Explanation:

 Let S=0.7+0.77+0.777+.....=  $\frac{7}{10}+\frac{77}{10^{2}}+\frac{777}{10^{3}}+....$  + up to 20 terms

$=7\left[\frac{1}{10}+\frac{11}{10^{2}}+\frac{111}{10^{3}}+....upto 20 terms \right]$

   $=\frac{7}{9}\left[\frac{9}{10}+\frac{99}{10^{2}}+\frac{999}{10^{3}}+....upto 20 terms \right]$

  $=\frac{7}{9}\left[(1-\frac{1}{10})+(1-\frac{1}{10^{2}})+(1-\frac{1}{10^{3}})+....upto 20 terms \right]$

$=\frac{7}{9}\left[(1+1+......+upto 20 terms)-(\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+....upto 20 terms) \right]$

$=\frac{7}{9}\left[20-\frac{\frac{1}{10}\left\{{1-(\frac{1}{10})^{20}}\right\}}{1-\frac{1}{10}}\right]$

[   $\therefore$ $\sum_{i=1}^{20}=20$  and sum of n terms of GPS   $s_{n}=\frac{a(1-r^{n})}{1-r}$ when (r<1)]

   $=\frac{7}{9}\left[20-\frac{1}{9}\left\{1-(\frac{1}{10})^{20}\right\}\right]$

 $=\frac{7}{9}\left[\frac{179}{9}+\frac{1}{9}(\frac{1}{10})^{20}\right]$

 $=\frac{7}{81}[179+(10)^{-20}]$

The real number k for which the equation,   $2x^{3}+3x+k=0$   has two  distinct real roots in [0,1]

Answer:

Explanation:

Let f(x)= $2x^{3}+3x+k$

 On differentiating  w.r,t x , we get

$f'(x)= 6x^{2}+3> 0,\forall x\epsilon R$

   $\Rightarrow$    f(x) is strictly increasing function.

$\Rightarrow$    f(x)= 0 has only one real root, so two roots are not possible 

 If the vectors AB= $3\hat{i}+4\hat{k} $  and $AC= 5\hat{i}-2\hat{j}+4\hat{k}$  are the sides of $\triangle$ABC  , then the length of the median through A is

Answer:

Explanation:

 We know that, the sum of three  vectors of a triangle is zero.

$\therefore$   AB+BC+CA=0

  $\Rightarrow$   BC=AC-AB

 $\Rightarrow$      $BM=\frac{AC-AB}{2}$

   ($\because$  M is a mid point of BC )

Also, AB+BM+MA=0

                  (By properties of a triangle)

   $\Rightarrow $     $  AB+\frac{AC-AB}{2}=AM$

$\Rightarrow$     $   AM=\frac{AB+AC}{2}$

      = $\frac{3\hat{i}+4\hat{k}+5\hat{i}-2\hat{k}+4\hat{k}}{2}$

        =  $4\hat{i}-\hat{j}+4\hat{k}$

 $\Rightarrow$|AM|=   $\sqrt{4^{2}+1^{2}+4^{2}}=\sqrt{33}$

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