TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

If A+B+C =270°  , then

$\cos 2A+\cos 2B+\cos 2C+4\sin A \sin B \sin C$=

Answer:

Explanation:

Given , A+B+C =270° 

and   $\cos 2A+\cos 2B+\cos 2C+4\sin A \sin B \sin C$

 = $2 \cos (A+B) \cos (A-B)+1-2 \sin ^{2} C + 4 \sin A\sin B \sin C$

  =$2 \cos (270 ^{0}-C) \cos (A-B)+1-2 \sin^{2} C+ 4 \sin A \sin B \sin C$

 = $1-2 \sin C [ \cos (A-B)+\sin C]+ 4 \sin A \sin B \sin C$

 = $1-2 \sin C[ \cos (A-B)+\sin C] +4 \sin A  \sin B \sin C$

 =$1-2 \sin C[ \cos (A-B)- \cos  (A+B) ]+4 \sin A \sin B \sin C$

 =$1-4 \sin A  \sin B \sin C+4 \sin A \sin B \sin C=1$

If  $\frac{2x+7}{(x^{2}+4)(x^{2}+9)(x^{2}+16)}= \frac{Ax+1}{x^{2}+4}+\frac{Bx+m}{x^{2}+9}+\frac{Cx+n}{x^{2}+16}$  , then 

$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$=........

Answer:

Explanation:

We have,

 $\frac{2x+7}{(x^{2}+4)(x^{2}+9)(x^{2}+16)}= \frac{Ax+1}{x^{2}+4}+\frac{Bx+m}{x^{2}+9}+\frac{Cx+n}{x^{2}+16}$ 

$2x+7= (Ax+l)(x^{2}+9)(x^{2}+16) +(Bx+m)(x^{2}+4)$

$(x^{2}+16)+(Cx+n)(x^{2}+4)(x^{2}+9)$

put x=2i

      4i+7=120Ai+60l  $\Rightarrow \frac{1}{A} =\frac{60}{2}$

 put x=3i

     6(i) +7 =-105 Bi-35m $\Rightarrow \frac{1}{B}=\frac{-35}{2}$

 pit x=4i

 8i+7=336 Ci+84n

$\Rightarrow$     $\frac{1}{C} = \frac{84}{2}$

 $\therefore$     $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$= $\frac{60-35+84}{2}= \frac{109}{2}$

If $\alpha$  and $\beta$  are the roots of the equation

$x^{2}-2x+4=0$  , then $\alpha^{12}+\beta^{12}$=

Answer:

Explanation:

 We have,

  $\alpha, \beta$  are the  roots of equation  $x^{2}-2x+4$=0

  $x^{2}-2x+4=0$

$x^{2}-2x+1= -3 \Rightarrow  (x-1)^{2}=-3$

 $x-1= \pm\sqrt{3}i\Rightarrow x=1\pm\sqrt{3}i$

 $x=-2\omega,-2\omega^{2}$

   $\left[\because \frac{-1+\sqrt{3}i}{2}=\omega , \frac{-1-\sqrt{3}i}{2}=\omega^{2}\right]$

$\therefore$   $\alpha =-2 \omega$ , $\beta = -2 \omega^{2}$

 $\alpha^{12}+\beta ^{12}= (-2 \omega) ^{12} +(-2 \omega^{2})^{12}$

 = $2^{12} (\omega^{12}+ \omega^{24})= 2^{12} (1+1)$   [ $\because \omega^{3}=1$]

 =$2^{13}$

If the point $\left(\frac{k-1}{k},\frac{k-2}{k}\right)$ lies on the  locus of z satisfying  the inequality  $|\frac{z+3i}{3z+i}|$  <1, then the interval in which k lies is 

Answer:

Explanation:

Given ,

  $|\frac{z+3i}{3z+i}|$  <1

 $|\frac{x+(y+3)i}{3(x+iy)+i}|  <1$  [Let z=x+iy]

  | x+(y+3)i|  < | 3x+(3y+1)i|

 $\Rightarrow$   $x^{2} +(y+3)^{2}   <   9 x^{2}+(3y+1)^{2}$

$\Rightarrow$      $x^{2}+y^{2}+6y+9  < 9 x^{2}+9 y^{2}+6y+1$

$\Rightarrow$      $8x^{2}+8 y^{2}-8 >0$

 $\Rightarrow$       $x^{2}+y^{2}-1 >0$

 $\frac{k-1}{k}, \frac{k-2}{k}$ lie on locus of z

 $\therefore$       $\frac{(k-1)^{2}}{k^{2}}+\frac{(k-2)^{2}}{k^{2}}-1 >0$

 $\Rightarrow$    $k^{2}-2k+1+k^{2}-4k+4-k^{2} >0$

 $\Rightarrow$ $k^{2}-6k+5 >0 \Rightarrow  (k-5)(k-1) > 0$

 $\Rightarrow$    k ε  $(-\infty ,1) \cup (5, \infty)$

Consider  the following system of equations in matrix form 

$\begin{bmatrix}1  \\2 \\\lambda \end{bmatrix}$  (1    2    $\lambda$) $\begin{bmatrix}x  \\y \\z\end{bmatrix}  =0$

Then which one of the following statements  is ture?

Answer:

Explanation:

We have ,

$\begin{bmatrix}1  \\2 \\\lambda \end{bmatrix}$  (1    2    $\lambda$) $\begin{bmatrix}x  \\y \\z\end{bmatrix}  =0$

   $\begin{bmatrix}1 & 2 &\lambda \\2 & 4 & 2\lambda \\\lambda &2\lambda &\lambda^{2} \end{bmatrix}\begin{bmatrix}x  \\y \\z \end{bmatrix}=0$

 Let        A=   $\begin{bmatrix}1 & 2 &\lambda \\2 & 4 & 2\lambda \\\lambda &2\lambda &\lambda^{2} \end{bmatrix}$

 |A|  = $1(4 \lambda^{2}-4 \lambda^{2})-2(2 \lambda^{2}-2 \lambda^{2})+\lambda (4 \lambda-4 \lambda)$

|A|= 0, $\forall \lambda \epsilon R$

$\therefore$ the given syste m has non trivial solution. 

• Mathematics - General questions