VITEEE 2014 :: Mathematics :: General questions

• Mathematics - General questions

If $r= \alpha b \times c+\beta c  \times a+ \gamma a \times b$ and [a b c]=2 , then $\alpha+\beta+\gamma$ is equal to 

Answer:

Explanation:

$r.a= \alpha( a.b \times c)+\beta(a.c \times a)+\gamma(a.a \times b)$

$=\alpha[abc]+0+0$ Similarly 

$r.b=\beta [a bc]$ and $r.c=\gamma [abc]$

 $\therefore$  $\frac{1}{2} r.(a+b+c)=\frac{1}{2} (r.a+r.b+r.c)$

$\frac{1}{2} (\alpha+\beta+\gamma)[abc]$

 =$\frac{1}{2}(\alpha+\beta+\gamma)\times 2=\alpha+\beta+\gamma$

The length of longer diagonal of the parallelogram constructed on 5a+2b and a-3b , if it is given that $|a|= 2\sqrt{2}$.|b|=3 and the angle between a and b is $\frac{\pi}{4}$ , is 

Answer:

Explanation:

Given : |a|=$2 \sqrt{2}$,|b|=3

One diagonal is 5a+2b+a-3b=6a-b

 Length of one diagonal

=|6a-b|

=$\sqrt{36a^{2}+b^{2}-2 \times 6|a|.|b| \cos 45^{0}}$

 =$\sqrt{36\times 8+9-12\times2\sqrt{2}\times3\times\frac{1}{\sqrt{2}}}$

=$\sqrt{288+9-12\times6}=\sqrt{225}=15$

other diagonal is (5a+2b)-(a-3b)=4a+5b

Its length is 

=$\sqrt{(4a)^{2}+(5b)^{2}+2\times |4a||5b| \cos 45^{0}}$

=$\sqrt{16\times8+25\times9+40\times6}=\sqrt{593}$

The differential equation of the rectangular hyperbola, where axes are the asymptotes of the hyperbola, is 

Answer:

Explanation:

 The differential  equation of the rectangular hyperbola $xy=c^{2}$  is 

$y+x \frac{dy}{dx}=0$

 $\Rightarrow$   $x \frac{dy}{dx}=-y$

If $x \frac{dy}{dx}+y=x.\frac{f(xy)}{f'(xy)}$, then f(xy) is equal to 

Answer:

Explanation:

$x \frac{dy}{dx}+y=x.\frac{f(xy)}{f'(xy)}$,

i.e, $\frac{d}{dx}(xy)=x \frac{f(x,y)}{f'(x,y)}$

$\Rightarrow$        $\frac{f'(xy)}{f(xy)} d(xy)=x dx$

$\Rightarrow$        $\int \frac{f'(xy)}{f(xy)} d(xy)=\int xdx$

$\Rightarrow$   $\log [f(xy)]= \frac{x^{2}}{2}+C$

$\Rightarrow$      $f(xy)= e^{(x^{2}/2+C)}$

$\Rightarrow$    $e^{x^{2}/2} .e^{C}$= $k.e^{\frac{x^{2}}{2}}$

The solution of $\frac{dy}{dx}=\frac{x^{2}+y^{2}+1}{2xy}$ , satisfying y(1)=0 is given by 

Answer:

Explanation:

 $\frac{dy}{dx}=\frac{x^{2}+y^{2}+1}{2xy}$ ,

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

 $\Rightarrow$    $\frac{xd(y^{2})-y^{2}dx}{x^{2}}=\left(\frac{x^{2}+1}{x^{2}}\right)dx$

  $\Rightarrow$   $\int  d\left(\frac{y^{2}}{x}\right)=\int \left(1+\frac{1}{x^{2}}\right)dx$

$\Rightarrow$   $\frac{y^{2}}{x}=x-\frac{1}{x}+C$

$\Rightarrow$   $y^{2}=(x^{2}-1+Cx)$

where x=1;y=0

$\Rightarrow$     0=1-1+C

$\Rightarrow$  C=0

$\therefore$  The solution is $x^{2}-y^{2}=1$ i.e, hyperbola

• Mathematics - General questions