VITEEE 2015 :: Mathematics :: General questions

• Mathematics - General questions

If (2,7,3)  is one end of a diameter of the sphere x²+y²+z²-6x-12y-2z+20=0, then the coordinate of the other end of the diameter  are 

Answer:

Explanation:

Given sphere is  x²+y²+z²-6x-12y-2z+20=0, Centre =(3,6,1)

 Here, one end of diameter is (2,7,3) , let the other end of the diameter be (x,y,z)

Centre of the sphere will be the mid-point of the ends of diameter.

 So, (3,6,1)=   $\left(\frac{2+x}{2},\frac{7+y}{2}, \frac{3+z}{2}\right)$

 $\Rightarrow 2+x=6\Rightarrow x=4$

$\Rightarrow 7+y=12\Rightarrow y=5$   and  $\Rightarrow 3+z=2\Rightarrow z=-1$

 Therefore     (x,y,z) =(4,5,-1)

$\int \frac{dx}{\cos x+\sqrt{3}sin x}$  equals

Answer:

Explanation:

$\int \frac{dx}{\cos x+\sqrt{3}sin x}$

= $\frac{1}{2}\int \frac{dx}{\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}sin x}$

=$\frac{1}{2}\int \frac{dx}{\cos\frac{\pi}{3}\cos x+\sin\frac{\pi}{3}sin x}$

=  $\frac{1}{2}\int \frac{dx}{\cos(x-\frac{\pi}{3})}$

   =$\frac{1}{2}\int \sec (x-\frac{\pi}{3}) dx$

=$\frac{1}{2}\log \tan (\frac{x}{2}-\frac{\pi}{6}+\frac{\pi}{4})+C$

  = $\frac{1}{2}\log \tan (\frac{x}{2}+\frac{\pi}{12})+C$

 If N  is n set of natural numbers, then under binary operation a.b =a+b, (N, ') is

Answer:

Explanation:

 The structure  (N,.) satisfies the closure property, associativity and commutativity but the identity element 0 does not belong to N.

Do, N is a semi-group

 Which of the following options is not the asymptote of the curve

3x³+2x²y-7xy² +2y³-14xy+7y² +4x+5y=0?

Answer:

Explanation:

$\phi_{3}(m)=3+2m-7m^{2}+2m^{3}$

$\phi_{2}(m)=-14m+7m^{2}$

$\phi_{3}'(m)=2-14m+6m^{2}$

 Now, putting $\phi_{3}(m)=0$ , we have

 3+2m-7m²+2m³ =0

$\Rightarrow$      $(1-m)(1+2m)(3-m)=0$

$\Rightarrow$      $ m=-\frac{1}{2},1,3$

 We know that , $c\phi_{n}'(m)+\phi_{n-1}(m)=0$ , which in the given case becomes

 c(2-14m+6m²)+(-14m+7m²)=0

$\Rightarrow c=\frac{14m-7m^{2}}{2-14m+6m^{2}}$

So, when  $m=-\frac{1}{2},c=-\frac{5}{6}$

 when m=1, c= $-\frac{7}{6}$

when m=3, c= $-\frac{3}{2}$

 Asymptotes are y= $-\frac{1}{2}$ x $-\frac{5}{6}$,

 $y=x-\frac{7}{6}$and   $ y=3x-\frac{3}{2}$

If the matrix  $A=\begin{bmatrix}1 & 3&1 \\-1 & 2&-3 \\0 &1 &2 \end{bmatrix}$ then  adj (adj A) is equal to 

Answer:

Explanation:

$A=\begin{bmatrix}1 &3&1 \\-1 & 2&-3 \\0 &1 &2 \end{bmatrix}$

$\Rightarrow |A|=1.(4+3)-3(-2+0)+1(-1-0)=7+6-1=12$

 So, adj (adj A)= |A|^n-2 =A

  =(12)^3-2 A=12A

   $=12\begin{bmatrix}1 &3&1 \\-1 & 2&-3 \\0 &1 &2 \end{bmatrix}$

 $=\begin{bmatrix}12 &36&12 \\-12 & 24&-3 6\\0 &12 &24 \end{bmatrix}$

• Mathematics - General questions