VITEEE 2014 :: Mathematics :: General questions

• Mathematics - General questions

Two line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=z$ intersect at a point , if k is equal to 

Answer:

Explanation:

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}= r$ (say)

$\Rightarrow$   $x=2r+1.y=3r-1,z=4r+1$

Since, the two  lines intersect 

 So, putting above  values  in second lines, we get

$\frac{2r+1-3}{1}=\frac{3r-1-k}{2}=\frac{4r+1}{1}$

$2r-2=4r+1$

$\Rightarrow$   $r=-3/2$

Also, $3r-1-k=8r+2$

 $\Rightarrow$    $k=-5r-3=\frac{15}{2}-3=\frac{9}{2}$

If $\triangle_{r}=\begin{bmatrix}2r-1 & ^{m}C_{r}&1 \\m^{2}-1 &2^{m}&m+1\\\sin^{2}(m^{2})&\sin^{2}(m) &\sin^{2}(m+1) \end{bmatrix}$, then  the value of  $\sum_{r=0}^{m}\triangle_{r}$  is 

Answer:

Explanation:

$\triangle_{r}=\begin{bmatrix}2r-1 & ^{m}C_{r}&1 \\m^{2}-1 &2^{m}&m+1\\\sin^{2}(m^{2})&\sin^{2}(m) &\sin^{2}(m+1) \end{bmatrix}$

$\therefore$   $\sum_{r=0}^{m}\triangle_{r}=$   $\begin{bmatrix}\sum_{r=0}^{m}{(2r-1)} & \sum_{r=0}^{m} {^{m}}C_{r}&\sum_{r=0}^{m} 1 \\m^{2}-1 &2^{m}&m+1\\\sin^{2}(m^{2})&\sin^{2}(m) &\sin^{2}(m+1) \end{bmatrix}$

 $==\begin{bmatrix}m^{2}-1 & 2^{m}&m+1 \\m^{2}-1 &2^{m}&m+1\\\sin^{2}(m^{2})&\sin^{2}(m) &\sin^{2}(m+1) \end{bmatrix}$

=0  ($\because$  two rows are identical)

Let f'(x) , be differentiable $\forall x $ . if f(1)=-2 and $f'(x)\geq 2\forall x \in[1,6]$, then 

Answer:

Explanation:

 f'(x)  is differentiable  $\forall x \in[1,6]$

 By Lagrange's mean value theorm

 $f'(x)= \frac{f(6)-f(1)}{6-1}$

 $f'(x)  \geq 2 \forall x \in [1,6]$ (given)

 $\Rightarrow$   $\frac{f(6)+2}{5} \geq 2$       [$\because$  f(1)=-2]

 $\Rightarrow$      $f(6) \geq 10-2 \Rightarrow f(6) \geq 8$

If $\triangle (x)=\begin{bmatrix}1 & \cos x&1-\cos x\\1+\sin x & \cos x &1+\sin x-\cos x\\ \sin x&\sin x   &1 \end{bmatrix}$

then   $\int_{0}^{\pi/4} \triangle(x) dx$ is equal to 

Answer:

Explanation:

$\triangle (x)=\begin{bmatrix}1 & \cos x&1-\cos x\\1+\sin x & \cos x &1+\sin x-\cos x\\ \sin x&\sin x   &1 \end{bmatrix}$

Applying  $C_{3} \rightarrow C_{3}+C_{2}-C_{1}$

$\triangle (x)=\begin{bmatrix}1 & \cos x&0\\1+\sin x & \cos x &0\\ \sin x&\sin x   &1 \end{bmatrix}$

$= \cos x-\cos x(1+\sin x)$

   [$\because$  expanding along $C_{3}$ =$- \cos x.\sin x=-\frac{1}{2} \sin 2x$

 $\because$    $\int_{0}^{\pi/4}  \triangle(x) dx=-\frac{1}{2}\int_{0}^{\pi/4} \sin 2x dx$

  $= -\frac{1}{2}\left[\frac{-\cos 2x}{2}\right]_{0}^{\pi/4}$

 $= +\frac{1}{2\times2}\left[ \cos\frac{\pi}{2}-\cos 0^{0}\right]$

 $= \frac{1}{4}(0-1)=-\frac{1}{4}$

If the points (1,2,3) and (2,-1,0) lie on the opposite sides of the plane 2x+3y-2z=k. then 

Answer:

Explanation:

The points (1,2,3) and (2,-1,0) lie on the opposite sides of the plane 2x+3y-2z-k=0

so, (2+6-6-k)(4-3-k)<0

$\Rightarrow$      
$(k-1)(k-2)<0$

$\therefore$    $1<k<2$

• Mathematics - General questions