VITEEE 2016 :: Mathematics :: General questions

• Mathematics - General questions

The probability of India  winning a test match  aganist Austraila is 1/2  assuming independence from match to match . The probability than in a match series India 's second win occurs at third test match is 

Answer:

Explanation:

 Required probability

 = P (First win ) x P( First win) x P( second win )

   + P( First Defeat ) x P( First Win) x P ( Second win )

 =  $\frac{1}{2}\times \frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$

If the tangent to the function y=f(x) at (3,4)  makes an angle of  $\frac{3\pi}{4}$ with the positive directions of x-axis in anticlockwise direction then f '(3) is 

Answer:

Explanation:

$f'(3)= \tan \frac{3\pi}{4}=-\tan \frac{\pi}{4}=-1$

The foci of ellipse  $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola  $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$ coincide then value of b² is 

Answer:

Explanation:

Given Ellipse   $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ 

Now $b^{2}=a^{2}(1-e^{2})$

 $\Rightarrow b^{2}=16(1-e^{2}),\Rightarrow\frac{b^{2}}{16}=1-e^{2}$

$\Rightarrow e^{2}=1-\frac{b^{2}}{16}=\frac{16-b^{2}}{16}$

$\Rightarrow e=\frac{\sqrt{16-b^{2}}}{4}$

  Foci =  $(\pm ae,0)=(\pm\sqrt{16-b^{2}},0)$

 Given hyperbola : $\frac{x^{2}}{14}+\frac{y^{2}}{81}=\frac{1}{25}$ 

 $\Rightarrow \frac{x^{2}}{(\frac{12}{5})^{2}}-\frac{y^{2}}{(\frac{9}{5})^{2}}=1$

Now,   $b^{2}=a^{2}(e^{2}-1)$

$\Rightarrow (\frac{9}{5})^{2}=(\frac{12}{5})^{2}(e^{2}-1)$

 $\Rightarrow (\frac{9}{5})^{2}=(e^{2}-1)$

 $\Rightarrow e^{2}=1+\frac{81}{144}=\frac{144+81}{144}$

 $\Rightarrow e=\frac{15}{12}=\frac{5}{4}$

  Foci =  $(\pm ae,0)$= $(\pm 3,0)$

 Since foci of the given ellipse and hyperbola coincide , therefore

 $ \sqrt{16-b^{2}}=3\Rightarrow 16-b^{2}=9$

$\therefore$   b² =7

A tetrahedron has vertices at O(0,0,0) ,  A( 1,2,1)  , B (2,1,3)  and C (-1,1,2) . Then the angle between the faces OAB and ABC will be

Answer:

Explanation:

$\overrightarrow{AO}= \hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{AC}= -2\hat{i}-\hat{j}+\hat{k}$ 

 Angle between faces OAB  and ABC  = Angle between  $\overrightarrow{AO}$ and $\overrightarrow{AC}$ .

 If Q be the angle between $\overrightarrow{AO}$ and $\overrightarrow{AC}$ , then 

$\cos \theta= \frac{\overrightarrow{AO}.\overrightarrow{AC}}{|\overrightarrow{AO}||\overrightarrow{AC}|}$

$=\frac{1\times(-2)+2\times(-1)+1\times1}{\sqrt{1+4+1}\sqrt{4+1+1}}$
$  \frac{-3}{6}=-\frac{1}{2}=\cos 120^{0}$

$\theta=120^{0}$

 The solution of differential equation $(1+y^{2})+(x-e^{tan^{-1}y})\frac{dy}{dx}=0$

Answer:

Explanation:

$1+y^{2}+(x-e^{\tan^{-1}y})\frac{\text{d}y}{\text{d}x}=0$

$\Rightarrow (1+y^{2}) dx=(e^{\tan^{-1}y}-x)dy$

 $\Rightarrow \frac{dy}{dx}=\frac{e^{\tan^{-1}y}-x}{1+y^{2}}$

$\Rightarrow \frac{dy}{dx}+\frac{1}{1+y^{2}}.x=\frac{e^{\tan^{-1}y}}{1+y^{2}}$

 Which is the linear different equation of the `

form  $ \frac{dx}{dy}+Rx=S$ , where R and S are functions of y or constant (s)

$\therefore$ I.F=$e^{\int_{}^{}\frac{1}{1+y^{2}}.dy }=e^{tan^{-1}y}$

Hence required solution is

   x.I.F = $\int_{}^{} \frac{ae^{\tan^{-1}y}}{1+y^{2}}(I.F) dy$

 $\Rightarrow x.e^{\tan^{-1}y}=\int\frac{e^{\tan^{-1}y}}{1+y^{2}} (e^{\tan^{-1}y}) dy$

  $\Rightarrow x.e^{\tan^{-1}y}=\int\frac{e^{2\tan^{-1}y}}{1+y^{2}}  dy$   ......(i)

 Put t= tan^-1 y

$\therefore$    $\frac{dt}{dy}=\frac{1}{1+y^{2}}\Rightarrow dt=\frac{1}{1+y^{2}} dy  $

 $\therefore$      $\int\frac{e^{2\tan^{-1}y}}{1+y^{2}}  dy=\int e^{2t}.dt=\frac{e^{2t}}{2}+K$

 Hence equation(i)  becomes

 $x e^{\tan^{-1}y}=\frac{1}{2}e^{2t}+K$

  $\Rightarrow x e^{\tan^{-1}y}=\frac{1}{2}e^{2tan^{-1}y}+K$

 $\Rightarrow 2x e^{\tan^{-1}y}=e^{2tan^{-1}y}+K$

• Mathematics - General questions