TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

$\int\frac{{dx}}{x(x^{4}+1)} =$

Answer:

Explanation:

$\int\frac{{dx}}{x(x^{4}+1)}=\int\frac{x^{4}+1-x^{4}}{x(x^{4}+1)}dx$

           $=\int\frac{x^{4}+1}{x(x^{4}+1)}dx-\int\frac{x^{4}}{x(x^{4}+1)}dx$

     $=\int\frac{1}{x}dx-\int\frac{x^{3}}{x^{4}+1}dx$

          $=\log|x| dx-\int\frac{x^{3}}{x^{4}+1}dx+C$

 Let $x^{4}+1=t \Rightarrow x^{3}dx=dt$

$\Rightarrow$    $x^{3} dx=\frac{1}{4}$

$\therefore\int \frac{dx}{4(x^{4}+1)}=\log |x|-\frac{1}{4}\int \frac{dt}{t}+C$

 = $\log |x|-\frac{1}{4} \log |t|+C$

=$\log|x|- \frac{1}{4} \log |x^{4}+1|+C$

   =$\frac{1}{4} \log |x^{4}|-\frac{1}{4} \log |x^{4}+1|+C$

$=\frac{1}{4}[ \log |x^{4}|-\log |x^{4}+1|+C$

       =$\frac{1}{4}\log\left(\frac{x^{4}}{x^{4}+1}\right)+C$

An integer is chosen from $\left\{2k/-9\leq k\leq10\right\}$   .The probability that it is divisible  by both 4 and 6 is 

Answer:

Explanation:

Let S=$\left\{2k/-9\leq k\leq10\right\}$ =

 {-18,-16,-14,.......,0,2,4,6,....20}

 Total number of possible outcomes =20

  Favourable  outcomes are -12,0 and 12

 So, number of favourable outcomes =3

  $\therefore$   Required probability= $\frac{3}{20}$

The equation  of the straight  line passing  through the point of intersection of $5x-6y-1$ , $3x+2y+5=0$, and perpendicular to the line $3x-5y+11=0$ is 

Answer:

Explanation:

We know that point of intersection of two lines 

i.e,   $ax_{1}+by_{1}+c_{1}=0$ and $ ax_{2}+by_{2}+c_{2}=0$  is 

    $\left(\frac{b_{1}c-b_{2}c}{a_{1}b_{2}-a_{2}b_{1}},\frac{c_{1}a_{2}-c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}\right)$

$\therefore$ point of intersection of two lines  i.e,  

5x-6y-1=0  and 3x+2y+5=0 is 

     $\left(\frac{-6\times5-2\times-1}{5\times2-3\times-6},\frac{-1\times3-5\times5}{5\times2-3\times-6}\right)$

     =  $\left(\frac{-30+2}{10+18},\frac{-3-25}{10+18}\right)$

  = $\left(\frac{-28}{28},\frac{-28}{28}\right)=\left(-1,-1\right)$

 and slope of line 3x-5+11=0 is m= $\frac{3}{5}$

Equation of the line passing through (-1,-1)  and perpendicular to 3x-5y+11=0 is 

  (y+1)= $-\frac{5}{3} (x+1)$

 $\Rightarrow $    $3(y+1)=-\frac{5}{3} (x+1)$

 $\Rightarrow $    $3y+3=-5x-5$

 $\Rightarrow $   $5x+3y+3+3+5=0$

 $\Rightarrow $    $5x+3y+8=0$

A village has players. A team  of 6 players is to be formed .5 members are chosen out of these 10 players and then the captain is chosen from the remaining players. Then choose from the remaining players. The total number of ways  of choosing such teams is 

Answer:

Explanation:

Total number  of ways of choosing  such terms is $^{10}C_{5}$.$^{5} C_{1}$

    $\frac{ 10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2\times 1} \times 5$

      = $10 \times 3 \times 7 \times 6=1260$

If $\triangle=\begin{bmatrix}1 & 5&6 \\0 & 1&7\\0&0&1 \end{bmatrix}$

and $\triangle ^{'}=\begin{bmatrix}1 & 0&1 \\3 & 0&3\\4&6&100 \end{bmatrix}$

Answer:

Explanation:

Given, 

$\triangle=\begin{bmatrix}1 & 5&6 \\0 & 1&7\\0&0&1 \end{bmatrix}$

     = $1(1-0)-5(0-0)+6(0-0)$

   =$1-0+0$

$\triangle =1$

 If $\triangle=1$ and $\triangle ^{'}=\begin{bmatrix}1 & 0&1 \\3 & 0&3\\4&6&100 \end{bmatrix}$

             =$1(0-18)-0(300-12)+1(18-0)$

=$-18-0+18$

    $\triangle '=0$

  Now, $(\triangle +\triangle')^{2}-3(\triangle+\triangle')+2$

 =$(1+0)^{2}-3(1+0)+2$

  =-2+2=0

• Mathematics - General questions