VITEEE 2016 :: Mathematics :: General questions

• Mathematics - General questions

  Consider the objective function Z=40x+50y . The minimum number of constraints that are required to maximize Z are

Answer:

Explanation:

 Two constraints are  $x\geq 0,y\geq0$  and the third one will be of the type  $ax+by\leq c$

The value of  $\cos\left( \frac{1}{2}\cos^{-1}\frac{1}{8}\right)$ is equal to 

Answer:

Explanation:

Let  $\cos^{-1}\frac{1}{8}=\theta, where 0<\theta<\frac{\pi}{2}$

 $\Rightarrow \frac{1}{2}\cos^{-1}\frac{1}{8}=\frac{1}{2}\theta, $

$\Rightarrow \cos(\frac{1}{2}\cos^{-1}\frac{1}{8})=\cos\frac{1}{2}\theta, $

Now, $\cos^{-1}\frac{1}{8}=\theta\Rightarrow\cos\theta=\frac{1}{8}$

$\Rightarrow 2\cos^{2}\frac{\theta}{2}-1=\frac{1}{8}$

$\Rightarrow \cos^{2}\frac{\theta}{2}=\frac{9}{16}\Rightarrow\cos\frac{\theta}{2}=\frac{3}{4}$

$[\because 0<\frac{\theta}{2}<\frac{\pi}{4},so \cos\frac{\theta}{2}\neq-\frac{3}{4}]$

The logical expression X, in its simplest form for the truth table

Answer:

Explanation:

X=a.b

The distance moved by the particle in time t is  given by  $s=t^{3}-12t^{2}+6t+8$ . At the instant , when its acceleration is zero the velocity is 

Answer:

Explanation:

$s=t^{3}-12t^{2}+6t+8$

$\frac{ds}{dt}=3t^{2}-24t+6$

$\frac{d^{2}s}{d^{2}t}=6t-24$

 Acceleration =0

 $\Rightarrow$ 6t-24=0

 $\therefore$   t=4

 Required velocity =  $\frac{ds}{dt}]_{t=4}$

$=3\times (4)^{2}-24\times 4+6$

 =48-96+6=42 units

The normal at the point ( $at_{1}^{2}, 2at_{1}$) on the parabola, cuts the parabola again at the point whose parameter is 

Answer:

Explanation:

 Let the normal at 't₁'  cuts the parabola again at the point 't₂' , the equation of the normal at  ( $at_{1}^{2}, 2at_{1}$)  is y+t₁ x =2at₁ +at₁³

Since it passes through the point 't₂' i.e,

 ( $at_{2}^{2}, 2at_{1}$) 

 $\therefore$      $2at_{2}+at_{1}t_2^2=2at_{1}+at_1^3$

 $\Rightarrow2a(t_{1}-t_{2})+at_1(t_1^2-t_2^2)=0$

$\Rightarrow  2+t_{1}(t_{1}+t_{2})=0(\because t_{1}-t_{2}\neq0)$

$\Rightarrow  2+t_{1}^{2}+t_{1}t_{2}=0$

 $\Rightarrow  t_{1}t_{2}=-(t_{1}^{2}+2)\Rightarrow t_{2}=-\left(t_{1}+\frac{2}{t_{1}}\right)$

• Mathematics - General questions