TS EAMCET 2018 :: Mathematics :: General questions

• Mathematics - General questions

In a $\triangle$ ABC, if A=2B and the sides opposite to the angles A, B,C  are $\alpha+1, \alpha-1 $ and $\alpha $ respectively, then $\alpha$= 

Answer:

Explanation:

 According to sine law,

$\frac{ \sin A}{ \alpha +1}= \frac{ \sin B}{\alpha-1}$

 $\Rightarrow$    $\frac{2 \sin B \cos B}{\alpha+1}=\frac{\sin B}{\alpha-1}$     [$\because$   A=2B]

$\Rightarrow$           $ \cos B=\frac{\alpha+1}{2(\alpha-1)}$

$\Rightarrow$     $\frac{(\alpha+1)^{2}+\alpha^{2}-(\alpha-1)^{2}}{2\alpha(\alpha+1)}=\frac{\alpha+1}{2(\alpha-1)}$

$\Rightarrow$              $\frac{4\alpha+\alpha^{2}}{2\alpha(\alpha+1)}=\frac{\alpha+1}{2(\alpha-1)}$

$\Rightarrow$      $(4+\alpha)(\alpha-1)=(\alpha+1)^{2}$               

$\Rightarrow$        $\alpha^{2}+3\alpha-4=\alpha^{2}+2\alpha+1$

$\Rightarrow$    $\alpha$=5

when  $\frac{\sin 9 \theta}{\cos 27 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin  \theta}{\cos 3 \theta}=k$

 $(\tan 27 \theta -\tan \theta)$  is defined , then k=

Answer:

Explanation:

$\frac{\sin 9 \theta}{\cos 27 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin  \theta}{\cos 3 \theta}=k(\tan 27 \theta -\tan \theta)$ 

LHS=$\frac{\sin 9 \theta\cos 9 \theta\sin 3 \theta\cos 27 \theta}{\cos 27 \theta\cos 9 \theta}+\frac{\sin  \theta}{\cos 3 \theta}$

=$\frac{\sin 18 \theta+\sin 30 \theta-\sin 24 \theta}{2\cos 27 \theta\cos 9 \theta}+\frac{\sin  \theta}{\cos 3 \theta}$

=$\frac{\sin 21 \theta+\sin 15 \theta+\sin 33 \theta+\sin 27 \theta-\sin 27 \theta-\sin 21 \theta +4\sin \theta\cos 9 \theta \cos 27 \theta}{4\cos 3 \theta\cos 9 \theta \cos 27 \theta}$

=$\frac{\sin 15 \theta+\sin 33 \theta +4\sin \theta\cos 9 \theta \cos 27 \theta}{4\cos 3 \theta\cos 9 \theta \cos 27 \theta}$

=$\frac{2\sin 24 \theta \cos 9 \theta+4\sin \theta\cos 9 \theta \cos 27 \theta}{4\cos 3 \theta\cos 9 \theta \cos 27 \theta}$

=$\frac{\sin 24 \theta+2\sin \theta\cos 27 \theta}{2\cos 3 \theta\cos 27 \theta}$

=$\frac{\sin (27 \theta-3\theta)}{2\cos 3 \theta\cos 27 \theta}+\frac{\sin \theta}{\cos 3 \theta}$

=$\frac{1}{2}\left[\frac{\sin 27 \theta}{\cos 27 \theta}-\frac{\sin 3 \theta}{\cos 3 \theta}\right]+\frac{\sin \theta}{\cos 3 \theta}$

=$\frac{1}{2}\frac{\sin 27 \theta}{\cos 27 \theta}-\frac{1}{2}\left(\frac{\sin 3 \theta-2\sin \theta}{\cos 3 \theta}\right)$

=$\frac{1}{2}\tan 27 \theta-\frac{1}{2}\frac{3\sin  \theta-4\sin^{3} \theta-2\sin \theta}{\cos  \theta(4\cos^{2}\theta-3)}$

=$\frac{1}{2}\tan 27 \theta-\tan \theta$

So,k=$\frac{1}{2}$

If the coefficient of $x^{5}$  in the expansion  of   $\left( ax^{2}+\frac{1}{bx}\right)^{13}$ is equal to the coefficient of $x^{-5}$  in the expansion of   $\left( ax^{}-\frac{1}{bx^{2}}\right)^{13}$ , then ab=

Answer:

Explanation:

 The general  term in the expansion of   $\left( ax^{2}+\frac{1}{bx}\right)^{13}$

   $T_{ (p+1)}=^{13}C_{p}(ax^{2})^{13-p}\frac{1}{(bx)^{p}}$

    = $^{13}C_{p}\frac{a^{13-p}}{b^{p}}x^{26-3p}$

  For, $x^{5}$, p=7

 So, coefficient of $x^{5}$ in the expansion of

   $\left(ax^{2}+\frac{1}{bx}\right)^{13} $   is   $^{13}C_{7}\frac{a^{6}}{b^{7}}$     [$\because$  p=7]

 Similarly, the general term in the expansion of   $\left( ax^{}-\frac{1}{bx^{2}}\right)^{13}$

     $T_{q+1}=^{13}C_{q}(ax)^{13-q}\left(-\frac{1}{bx^{2}}\right)^{q}$

    $=^{13}C_{q}(-1)^{q}\frac{a^{13-q}}{b^{q}}x^{13-q}$

  for $x^{-5}, q=6$

 So, coefficient  of $x^{-5}$  in the expansion of   $\left( ax^{}-\frac{1}{bx^{2}}\right)^{13}$  is 

$^{13}C_{6}\frac{a^{7}}{b^{6}}$

 According to the question,

$^{13}C_{7}\frac{a^{6}}{b^{7}}=^{13}C_{6}\frac{a^{7}}{b^{6}}$

$\Rightarrow$   ab=1

If all possible numbers are formed by using the digits 1,2,3,5,7 without repetition and they are arranged in descending order, then the rank of the number 327 is 

Answer:

Explanation:

Number of possible 5 digit numbers is 120 

Number of possible 4 digit number is 120

 Number of possible  3 digit numbers start with 7 is 12

 Number of possible 3 digit numbers start with 5 is 12

 Then, according to order 3 digit numbers are 375,372,371,357, 352,351,327

 So , the rank of number  327 is 271

If the roots  of the equation

$\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{5}{2}$ are p and q (p>q)  and the roots of the equation

$(p+q)x^{4}-pqx^{2}+\frac{p}{q}=0$   are $\alpha,\beta , \gamma ,\delta$ then   

$(\sum \alpha)^{2}-\sum \alpha\beta+\alpha\beta\gamma\delta=0$

Answer:

Explanation:

 Let   $\sqrt{\frac{x}{1-x}}=y$

  So,    $y+\frac{1}{y}=\frac{5}{2}$

  $\Rightarrow$      $2y^{2}-5y+2=0$

$\Rightarrow$       $y=2, \frac{1}{2}$

 So,          $x= \frac{4}{5}, \frac{1}{5}$

 $\because$   p>q

 $\therefore$       $p=\frac{4}{5}, q=\frac{1}{5}$

 Now,  for the equation    $(p+q)x^{4}-pqx^{2}+\frac{p}{q}=0$

$\sum \alpha=0,\sum\alpha\beta=-\frac{pq}{p+q} $   and    $ \alpha\beta\gamma\delta=\frac{p}{q(p+q)}$

 So,  $(\sum \alpha)^{2}-\sum \alpha\beta+\alpha\beta\gamma\delta=0+\frac{pq}{p+q}+\frac{p}{q(p+q)}$

    =  $\frac{p}{(p+q)}\left(q+\frac{1}{q}\right)$

    =  $\frac{\frac{4}{5}}{\frac{4}{5}+\frac{1}{5}}\left(\frac{1}{5}+5\right)=\frac{4}{5}\times\frac{26}{5}=\frac{104}{25}$

• Mathematics - General questions