VITEEE 2015 :: Mathematics :: General questions

• Mathematics - General questions

The area of the region bounded by the curves x²+y²=9  and x+y=3 is

Answer:

Explanation:

 Area of required region

   =$\frac{1}{4}$  x Area of circle - Area of $\triangle$ OAB

=  $\frac{1}{4}\times\pi\times(3)^{2}-\frac{1}{2}\times 3\times3$

= $9\left[\frac{\pi}{4}-\frac{1}{2}\right]$

$\lim_{x \rightarrow {\pi/4}}\frac{\tan x-1}{\cos 2x}$ is equal to 

Answer:

Explanation:

$\lim_{x \rightarrow {\pi/4}}\frac{\tan x-1}{\cos 2x}=\lim_{h \rightarrow 0}\frac{\tan(\frac{\pi}{4}+h)-1}{\cos2(\frac{\pi}{4}+h)}$

[$\because$   $x=\frac{\pi}{4}+h$]

=$\lim_{h \rightarrow 0}\frac{(\frac{1+\tan h}{1-\tan h})-1}{cos (\frac{\pi}{2}+2h)}$

=$\lim_{h \rightarrow 0}\frac{1+\tanh-1+\tan h}{-\sin 2h(1-\tan h)}$

= $\lim_{h \rightarrow 0}\frac{-2\tanh}{2\sin h\cos h(1-\tan h)}$

= $\lim_{h \rightarrow 0}\frac{-1}{\cos^{2}h(1-\tan h)}=-1$

The equation  of tangent  to the hyperbola 3x²-2y²=6 , which is perpendicular  to the line x-3y=3, are 

Answer:

Explanation:

Equation of hyperbola is 

 3x²-2y²=6

 $\Rightarrow $     $\frac{x^{2}}{2}-\frac{y^{2}}{3}=1$

So, a² =2   and b²=3

 Given,  equation of line is x-3y=3

$\therefore$  Slope of given line = $\frac{1}{3}$

$\therefore$   Slope of line  perpendicular to given line, m=-3

 The equation of tangents are

$y=mx\pm\sqrt{a^{2}m^{2}-b^{2}}$

 $=-3x\pm\sqrt{2\times9-3}$

$=-3x\pm\sqrt{18-3}$

 $=-3x\pm\sqrt{15}$

 If the rectangular hyperbola x²-y² =64  , Then which of the following is not correct?

Answer:

Explanation:

Given equation of rectangular hyperbola is x²-y²=8²

Length of latusrectum= 2 x(8)=16

 and eccentricity = $\sqrt{2}$

 The asymptotes are perpendicular lines.

 x $\pm$ y=0

 Now, directrices are  $x=\pm\frac{8}{\sqrt{2}}=\pm4\sqrt{2}$

If  there is  an error of m%  in measuring the edge  of cube, then the percent error is estimating  its surface area is 

Answer:

Explanation:

 Surface area A of a cube of side x is given by A=6x².

 on differentiating w.r.t x , we get  $\frac{dA}{dx}=12x$

 Let the change in x be  $\triangle x$ = m% of x = $\frac{mx}{100}$

 Change in surface area,

$\triangle A= \left(\frac{dA}{dx}\right)\triangle x=(12x)\triangle x$

= $12 x \left(\frac{mx}{100}\right)=\frac{12x^{2}m}{100}$

 The approximate change in surface area =   $\frac{2m}{100}\times6x^{2}$

 =2m%  of original surface area

• Mathematics - General questions