VITEEE 2015 :: Mathematics :: General questions

• Mathematics - General questions

 If p; It rains today , q:I go to school , r; I shall meet my friends and s: I shall go for a movie,  then which of the following is the proportion?

 If it does not rain or if i do not go to school , then i shall meet my friend and go for a movie 

Answer:

Explanation:

 Correct  result is as follows

 $(\sim p \vee \sim q)\Rightarrow(r  \wedge s)$

 or   $\sim (p \wedge  q)\Rightarrow r \wedge s$

 If x²+x+1= 0 , then the value of   $\sum_{n=1}^{6}\left(x^{n}+\frac{1}{x^{n}}\right)^{2}$  is

Answer:

Explanation:

 Gtiven equation is x²+x+1=0

 $\Rightarrow x=\omega$ and $x=\omega^{2}$

 Case I ; when x=$\omega$

 Then

$\sum_{n=1}^{6}\left[x^{n}+\frac{1}{x^{n}}\right]^{2}=\sum_{n=1}^{6}\left[\omega^{n}+\omega^{2n}\right]^{2}\left[\because \frac{1}{\omega}=\omega^{2}\right]$

$=(\omega+\omega^{2})^{2}+(\omega^{2}+\omega^{4})^{2}+(\omega^{3}+\omega^{6)^{2}}+(\omega^{4}+\omega^{8)^{2}}+(\omega^{5}+\omega^{10)^{2}}+(\omega^{6}+\omega^{12)^{2}}$

$=(-1)^{2}+(-1)^{2}+(2)^{2}+(-1)^{2}+(-1)^{2}+(2)^{2}=12$

 Case II : when $\omega^{2}$

 Then

  $\sum_{n=1}^{6}\left[x^{n}+\frac{1}{x^{n}}\right]^{2}=\sum_{n=1}^{6}\left[\omega^{2n}+\omega^{n}\right]^{2}\left[\because\frac{1}{\omega^{2}}=\omega\right]$

=12

If the line y=2x+c  is a  normal to the ellipse   $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$   , then 

Answer:

Explanation:

 if the line y=mx+c isa normal to the ellipse   $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , then

 $c^{2}=\frac{m^{2}(a^{2}-b^{2})^{2}}{a^{2}+b^{2}m^{2}}$

 [Here , m=2, a² =9 , and  b² =16]

$=\frac{2^{2}(9-16)^{2}}{9+16\times (2)^{2}}$

 $=\frac{4\times 49}{9+64}=\frac{4\times 49}{73}=\frac{196}{73}$

$\therefore$   $c=\frac{14}{\sqrt{73}}$

If  the complex number z lies on a circle  with centre at the origin and radius 1/4, then the complex number -1+8z lies  on a circle with radius

Answer:

Explanation:

Given : |z|=$\frac{1}{4}$

Let z'=-1+8z

$\Rightarrow  z=\frac{z'+1}{8}\Rightarrow|z|=\frac{|z'+1|}{8}$

 $\Rightarrow  \frac{1}{4}=\frac{|z'+1|}{8}\Rightarrow |z'+1|=2$

 $\therefore$    z' lies ona circle with centre (-1,0) and radius 2.

Find the value of k for which the   simultaneous equations x+y+z=3; x+2y+3z=4  and x+4y +kz=6  will not have a unique solution

Answer:

Explanation:

 The given system of equations will be consistent with unique solution , when

$\begin{bmatrix}1 & 1&1 \\1 & 2&3\\1 &4&k \end{bmatrix}\neq0$

 $\Rightarrow  1(2k-12)+1(3-k)+1(4-2)\neq0$

$\Rightarrow  k-12+3+2\neq0\Rightarrow k-7\neq0\Rightarrow k\neq7$

• Mathematics - General questions