VITEEE 2016 :: Mathematics :: General questions

• Mathematics - General questions

Number of ways of selecting three squares on a chessboard so that all the three be on a diagonal line of the board or parallel to it is 

Answer:

Explanation:

Number of ways 

 $=[(^{3}C_{3}+^{4}C_{3}+^{5}C_{3}+^{6}C_{3}+^{7}C_{3})\times2+^{8}C_{3}]\times2$

  =392

If the mean of a binomial distribution is 25, then its standard deviation lies in the interval

Answer:

Explanation:

Standard deviation ,   $\sigma= \sqrt{npq}\geq 0$

 Now mean =np= 25 and q <1

 So $\sigma$= $\sqrt{npq}$ < $\sqrt{np}=5$

$\therefore$     $0\leq\sigma<5$

If   $a=\cos 2\alpha+i\sin2\alpha,b=\cos 2\beta+i\sin2\beta,c=\cos2\gamma+i\sin2\gamma$   and   $d=\cos2\delta+i\sin2\delta$

 then

$\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=$

Answer:

Explanation:

 we have,

$abcd=\cos(2\alpha+2\beta+2\gamma+2\delta)+i\sin (2\alpha+2\beta+2\gamma+2\delta)$

$\sqrt{abcd}=\cos(\alpha+\beta+\gamma+\delta)+i\sin(\alpha+\beta+\gamma+\delta)$  ..(i)

$\therefore$   $\frac{1}{\sqrt{abcd}}=\cos(\alpha+\beta+\gamma+\delta)-i\sin(\alpha+\beta+\gamma+\delta)$   ....(ii)

Adding (i) and (ii) , we obtain

$\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=2\cos(\alpha+\beta+\gamma+\delta)$

The value of   $\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+\cot^{-1}(3)$  is 

Answer:

Explanation:

Consider  

$\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+\cot^{-1}3$  ..........(i)

We have,   $\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)=\cot^{-1}2$

 $\therefore$ From equation (i)  , we have 

 $\cos^{-1}2+\cot^{-1}3=\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}$

$\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}\right)$

           2

=  $\tan^{-1}\left(\frac{\frac{5/6}{6-1}}{6}\right)=\tan^{-1}1=\frac{\pi}{4}$

 In a culture the bacteria count is 1,00,000. the number is increased by 10% in 2 hours. In how many hours will be the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present

Answer:

Explanation:

 Let y denote the number of bacteria at any instant t - then according to the question

 $\frac{dy}{dt}\alpha y\Rightarrow \frac{dy}{y}=kdt$  .....(i)

  k is the constant of proportionality , taken  to be +ve integrating (i) , we get

log y=kt+c   ....... (ii)

c is a parameter , let y₀ be the initial number of bacteria

 i.e,  at t=0 using thus in (ii) , c = log y₀

$\Rightarrow \log y=kt+\log y_{0}$

$\Rightarrow \log\frac{y}{y_{0}}=kt$  ....(iii)

$y= \left(y_{0}+\frac{10}{100}y_{0}\right)=\frac{11y_{0}}{10},$, when t=2

 so , from (iii)   , we get   $\log \frac{\frac{11y_{0}}{10}}{y_{0}}=k(2)$

 $\Rightarrow k=\frac{1}{2}\log \frac{11}{10}$    .........(iv)

 Using (iv) in (iii)       $\log\frac{y}{y_{0}}=\frac{1}{2}\left(\log \frac{11}{10}\right)t$   .....(v)

 let the number of bacteria become 1,00,000  to 2,00,000 in t ₁  hours .i.e, y=2y₀ when t= t₁   hours , from (v)

$\log\frac{2y_{0}}{y_{0}}=\frac{1}{2}\left(\log \frac{11}{10}\right)t_{1}\Rightarrow t_{1}=\frac{2\log 2}{\log \frac{11}{10}}$

 Hence , the reqd. no of hours =  $\frac{2\log 2}{\log \frac{11}{10}}$

• Mathematics - General questions