TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

If the range of the function f(x)=-3x-3 is (3,-6,-9,-18) , them which of the following elements is not in the domain of f?

Answer:

Explanation:

 We have ,f(x)=-3x-3

(i)  let f(x)=3 , then we get

         $ 3=-3x-3 \Rightarrow 6=-3x \Rightarrow x=-2$

 (ii)    Let f(x)=-6 , then we get

              $ -6=-3x-3 \Rightarrow -3=-3x \Rightarrow x=1$

(iii)  let f(x)=-9 , then we get

$ -9=-3x-3 \Rightarrow -6=-3x \Rightarrow x=2$

(iv) Let f(x)=-18, then we get

$ -18=-3x-3 \Rightarrow -15=-3x \Rightarrow x=5$

Thus, domain of f={-2,1,2,5}

Hence,-1 cannot  be in the domain of f.

$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i\sin \left(\frac{\pi}{12}\right)}{1+ \cos\left(\frac{\pi}{12}\right)-i\sin\left(\frac{\pi}{12}\right)}\right]^{72}=$

Answer:

Explanation:

Consider ,$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i\sin \left(\frac{\pi}{12}\right)}{1+ \cos\left(\frac{\pi}{12}\right)-i\sin\left(\frac{\pi}{12}\right)}\right]^{72}$

=$\left[\frac{2\cos^{2} \frac{\pi}{24}+i2\sin \frac{\pi}{24}\cos \frac{\pi}{24}}{2\cos^{2} \frac{\pi}{24}-i2\sin \frac{\pi}{24}\cos \frac{\pi}{24}}\right]^{72}$

 =$\left[\frac{2\cos^{} \frac{\pi}{24}\left(\cos \frac{\pi}{24}+i\sin \frac{\pi}{24}\right)}{2\cos^{} \frac{\pi}{24}\left(\cos \frac{\pi}{24}-i\sin \frac{\pi}{24}\right)}\right]^{72}$

       =$\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\right]^{72}$

=   $\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\times \frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\right]^{72}$

     =   $\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{2}}{\left(\cos^{2} \frac{\pi}{24}-i^{2} \sin^{2} \frac{\pi}{24}\right)}\right]^{72}$

=$\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{2}}{\cos^{2} \frac{\pi}{24}+ \sin^{2} \frac{\pi}{24}}\right]^{72}=\left( \cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{144}$

       = $\cos \left(\frac{\pi}{24}\times144\right)+i\sin\left( \frac{\pi}{24}\times144\right)$ 

                                                  [ By Dermoivre's theorm]

=$\cos 6\pi+i \sin 6 \pi$

=1                  $\left[ \because \sin n\pi=0 \forall n\in Z\right]$   and  

                                [ $\cos n \pi= \begin{cases}1, & if n is even\\-1, & if n is odd\end{cases}$]

The foci of the ellipse

$25x^{2}+4y^{2}+100x-4y+100=0$ are

Answer:

Explanation:

 Given equation of ellipse can be rewritten as 

$((5x)^{2}+2(5)(10)x+10^{2})+$

                   $((2y)^{2}-2(2)(1)y+1^{2})-10^{2}-1^{2}+100=0$

$\Rightarrow$    (5x+10)^{2}+(2y-1)^{2}=1$

$\Rightarrow$  $25(x+2)^{2}+4(y-\frac{1}{2})^{2}=1$

$\Rightarrow$    $\frac{(x+2)^{2}}{(1/5)^{2}}+\frac{(y-\frac{1}{2})^{2}}{(1/2)^{2}}=1$ , which is of the form

                                                       $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$  , where a< b

 Here, $a=\frac{1}{5},b=\frac{1}{2}$ and major axis of ellipse

x+2=0 , i.e, x=-2

 Now,  $e=\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{4}{25}}=\sqrt{\frac{21}{25}}=\frac{\sqrt{21}}{5}$

 $therefore$     foci are   $(-2,\frac{1}{2}\pm be)=\left(-2,\frac{1}{2}\pm \frac{\sqrt{21}}{10}\right)$

    $=\left(-2,\frac{5\pm\sqrt{21}}{10}\right)$

For an integer  $n\geq1,\sum_{k=1}^{n}K(K+2)=$

Answer:

Explanation:

Consider,

$\sum_{k=1}^{n}k(k+2)=\sum_{k=1}^{n}(k^{2}+2k)=\sum_{k=1}^{n}k^{2}+2\sum_{k=1}^{n}k$

    =$\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}$

      $=n(n+1)\left[\frac{(2n+1)}{6}+1\right]=n(n+1)\left[\frac{2n+7}{6}\right]$

  =$\frac{n(n+1)(2n+7)}{6}$

If $\cosh^{-1} x=2 \log_{e} (\sqrt{2}+1),$ then x= 

Answer:

Explanation:

We have,   $\cosh^{-1} x=2 \log_{e} (\sqrt{2}+1),$

$\Rightarrow$   $\log_{e}(x+\sqrt{x^{2}-1})= \log_{e}(\sqrt{2}+1)^{2}$

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

$\Rightarrow$     $x+\sqrt{x^{2}-1}= 3+2\sqrt{2}=3+\sqrt{8}$

 On comparing rational and irrational part, we get

   x=3

• Mathematics - General questions