VITEEE 2017 :: Mathematics :: General questions

• Mathematics - General questions

The rank of the matrix $\begin{bmatrix}-1 & 2 & 5\\2 & -4 & a-4\\ 1 & -2 & a+1 \end{bmatrix}$ is

Answer:

Explanation:

Let

$A=\begin{bmatrix}-1 & 2 & 5\\2 & -4 & a-4\\ 1 & -2 & a+1 \end{bmatrix}\sim\begin{bmatrix}-1 & 2 & 5\\0 & 0 & a+6\\ 0 & 0 & a+6 \end{bmatrix}$

[R₂ → R₂ + 2R₁, R₃ → R₃ +R₁]

Clearly rank of A is 1 if a = -6

Also, for a= 1, | A | = $\begin{vmatrix}-1 & 2 & 5\\2 & -4 & -3\\ 1 & -2 & 2 \end{vmatrix}=0$

and $\begin{vmatrix}2 & 5 \\-4 & -3 \end{vmatrix}= -6+20=14\neq0$

.'. rank of A is 2 if a = 1

The parabola having its focus at (3, 2) and directrix along the y-axis has its vertex at

Answer:

Explanation:

Vertex of the parabola is a point which lies on the axis of the parabola, which is a line ⊥ to the directrix through the focus, i.e., y=2 and equidistant from the focus and directrix x = 0, so that the vertex is $(\frac{3}{2},2)$

The value of $\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{7}{8}$ is

Answer:

Explanation:

$\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{7}{8}$

$=\left[\frac{\frac{1}{2}+\frac{1}{3}+\frac{7}{8}-\frac{1}{2}*\frac{1}{3}*\frac{7}{8}}{1-\frac{1}{2}*\frac{1}{3}-\frac{1}{3}*\frac{7}{8}-\frac{7}{8}*\frac{1}{2}}\right]$

$[\therefore\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}(\frac{x+y+z-xyz}{1-xy-yz-zx})]$

$=\tan^{-1}[\frac{\frac{41}{24}-\frac{7}{48}}{1-\frac{1}{6}-\frac{7}{24}-\frac{7}{16}}]$

$=\tan^{-1}[\frac{\frac{75}{48}}{1-\frac{43}{48}}]=\tan^{-1}(\frac{75}{48-43})$

$=\tan^{-1}[\frac{75}{5}]=\tan^{-1}15$

A random variable X has the probability distribution

For the events E = {X is a prime number} and F={X<4}, then p(E ∪ F) is

Answer:

Explanation:

P(E) = P(2 or 3 or 5 or 7)
= 0.23 + 0.12 + 0.20 + 0.07 = 0.62
P(F)= P(1 or 2 or 3)
=0.15 +0.23+ 0.12 = 0.50
P(E ∩ F) = P(2 or 3)
  =0.23+0.12=0.35
.'. P(EUF) = P(E)+ P(F)- P(E ∩ F)
=0.62+ 0.50-0.35 = 0.77

The condition that the line $\frac{x}{p}+\frac{y}{q}$ = 1 be a normal to the parabola y² = 4ax is

Answer:

Explanation:

The line  $\frac{x}{p}+\frac{y}{q}$ = 1 be a normal to the parabola y² = 4ax if, for some value of m, it is identical with

y = mx-2am - am³ i.e.  mx-y = (2am+ am³)

Comparing coefficients, we get 

$\frac{m}{1/p}=\frac{-1}{1/q}=\frac{2am +am^{3}}{1}$ → mp = -q

.'. m= -q/p and mp = m(2a+am²)

or p = $2a+\frac{aq^{2}}{p^{2}}$ or $p^{3} = 2ap^{2}+aq^{2}$

• Mathematics - General questions