TS EAMCET 2017 :: Mathematics :: General questions

• Mathematics - General questions

A parallelogram has vertices A(4,4,-1), B(5,6,-1),C(6,5,1) and D(x,y,z) . Then the vertex D is 

Answer:

Explanation:

Given , ABCD is a parallelogram with vertices  A(4,4,-1), B(5,6,-1), C(6,5,1) and D(x,y,z)

 We know that digonals  of parallelogram ABCD bisects each other. 

  $\therefore$  Mid poinr of Ac= Mid point of BD

 $\Rightarrow$          $\left(\frac{4+6}{2},\frac{4+5}{2},\frac{-1+1}{2}\right)=\left(\frac{x+5}{2},\frac{y+6}{2},\frac{z-1}{2}\right)$

$\Rightarrow$         $\left(\frac{10}{2},\frac{9}{2},0\right)=\left(\frac{x+5}{2},\frac{y+6}{2},\frac{z-1}{2}\right)$

On comparing both sides, we get

      $\frac{x+5}{2}=\frac{10}{2},\frac{y+6}{2}=\frac{9}{2}$ and $\frac{z-1}{2}=0$

$\Rightarrow$   $x+5=10,y+6=9$ and $z-1=0$

$\Rightarrow$   $x=10-5,y=9-6 $ and z=1

$\Rightarrow$   x=5 ,y=3 and z=1

Thus ,        D(x,y,z)=D(5,3,1)

If a and b are unit vectors and $\alpha$  is the angle between  them, then a+b is a unit vector when $\cos \alpha$=$

Answer:

Explanation:

We have, |a|=|b|=1 and $\alpha$ is the angle between a and b

 Clearly , $\cos \alpha=\frac{a.b}{|a|.|b|}$

$\Rightarrow$  $\cos \alpha=a.b$   .........(i)

 Now, let a+b is a unit vector , then

       |a+b|=1

$\Rightarrow$     $|a+b|^{2}=1$

$\Rightarrow$   $(a+b).(a+b)=1$

$\Rightarrow$   $a.a+a.b+b.a+b.b=1$

$\Rightarrow$   $|a|^{2}-2a.b+|b|^{2}=1$     $[ \because a.b=b.a]$

$\Rightarrow$    $1+2\cos \alpha+1=1$   [using Eq.(i)]

$\Rightarrow$     $1+2\cos \alpha=0$

$\Rightarrow$  $\cos \alpha= -\frac{1}{2}$

The differential equation of the simple harmonic  motion given by $x=A \cos (nt+\alpha)$ is 

Answer:

Explanation:

Given ,$x=A \cos (nt+\alpha)$   ...........(i)

On differentiating x w.r.t .'t' we get

  $\frac{dx}{dt}=A\frac{d}{d} \cos (nt+\alpha)$

=$-A \sin (nt+\alpha) \frac{d}{dt}(nt+\alpha)$

=$-A \sin(nt+\alpha)(n)$

$\Rightarrow$    $\frac{dx}{dt}=-An\sin(nt+\alpha)$

$\Rightarrow$  $\frac{dx}{dt}=-\frac{x n \sin (nt+\alpha)}{(\cos (nt+\alpha)}$      [From Eq.(i)]

$\Rightarrow$     $\frac{dx}{dt}=-nx \tan (nt+\alpha)$     ........(ii)

Again differentiating w.r.t 't'  , we get

$\frac{d^{2}x}{dt^{2}}=-\left[nx \frac{d}{dt} \tan (nt+\alpha)+\tan (nt+\alpha) \frac{d}{dx} nx\right]$

 $=-\left[nx \sec^{2}(nt+\alpha)\frac{d}{dt}(nt+\alpha)+\tan(nt+\alpha)\frac{dx}{dt} \times n\right]$

    $=-\left[nx \sec^{2}(nt+\alpha)n+n \tan(nt+\alpha)x-nx \tan(nt+\alpha)\right]$

                                                                                                   [From Eq.(ii)]

$=-\left[n^{2}x \sec^{2}(nt+\alpha)-n^{2}x \tan^{2}(nt+\alpha)\right]$

$=-n^{2}x[\sec^{2}(nt+\alpha)-\tan^{2}(nt+\alpha)]$

 $= -n^{2}x \times 1$       $[ \because \sec^{2} \theta-\tan^{2} \theta=1]$

$\Rightarrow$     $\frac{d^{2}x}{dt^{2}}=-n^{2}x$

$\Rightarrow$    $\frac{d^{2}x}{dt^{2}}+n^{2}x=0$

If a, b and c are unit vectors  such that  a+b+c=0  and (a,b)=$\frac{\pi}{3}$, then

$|a \times b|+|b \times c|+|c \times a|=$

Answer:

Explanation:

 We have,

a,b,c are unit vector

$\therefore$   |a|=|b|=|c|=1  and a+b+c=0 , angle between a and b is $\frac{\pi}{3}$

Now,      a+b+c=0

$\Rightarrow$    $a \times (a+b+c)=0$

$\Rightarrow$  $a \times a+a \times b+a \times c=0$

$\Rightarrow$      $a \times b= c \times a$

$\Rightarrow$   $|a \times b|=|c \times a|$

Similarly , $|a \times b|=|c \times a|$

$\therefore$  $|a \times b|+|b \times c|+|c \times a|$

                                      =$3 |a \times b|$

                                    =$3|a||b| \sin (a,b)$

                                      =$ 3 \times 1 \times 1 \times \sin \frac{\pi}{3}= \frac{3 \sqrt{3}}{2}$

$f:(-\infty,0] \rightarrow[0,\infty)$ is defined as f(x)=x² . The domain  and range of its inverse is 

Answer:

Explanation:

We have a function $f:(-\infty,0] \rightarrow [0, \infty)$ defined as f(x)=x²

The graph  of above function is 

Since, each line parallel to x-axis cuts the above curve at a maximum of one point, therefore f is one-one . Also from the graph it is clear that range $f= (0,\infty)$

$\therefore$     f is onto

 Thus, f is invertible

Hence, $f^{-1}:[0, \infty) \rightarrow (-\infty,0]$

$\Rightarrow$    Domain$(f^{-1})=[0,\infty)$,  and Range of $(f^{-1})=(-\infty,0]$

• Mathematics - General questions