VITEEE 2015 :: Mathematics :: General questions

• Mathematics - General questions

 On the interval [0,1] , the function x²⁵ (1-x)⁷⁵ takes its maximum value at the point

Answer:

Explanation:

 Let  $f(x)=x^{25}(1-x)^{75},x\in[0,1]$

$\Rightarrow f'(x)=25x^{24}(1-x)^{75}-75x^{25}(1-x)^{74}$

$=25x^{24}(1-x)^{74}\left\{(1-x)-3x\right\}$

$=25x^{24}(1-x)^{74}\left\{(1-4x)\right\}$

 We can see that f'(x)  is positive  for $x < \frac{1}{4}$

 and f '(x) is negative for $x > \frac{1}{4}$

 Hence, f(x) attains maximum at x = $\frac{1}{4}$

 If p,q, r are  simple propositions with truth values T,F,T , then  the truth value of  $(\sim p\vee q)\wedge  \sim r)$

$\Rightarrow$ p is 

Answer:

Explanation:

$\sim p\vee q$  means $ F\vee F=F,\sim r$ means F

 $\therefore$    $[(\sim P\vee q)\wedge \sim r]\Rightarrow p $ means  T

If a line segment  OP makes angles of   $\frac{\pi}{4}$ and  $\frac{\pi}{3}$  with X-axis and Y-axis,  respectively . Then, the direction cosines are 

Answer:

Explanation:

Let  $\alpha,\beta$ and $\gamma$ be the angles made by the line segment OP with X-axis, Y-axis and Z-axis , respectively

 Given:  $\alpha=\frac{\pi}{4}$ and $\beta=\frac{\pi}{3} $

We know that  $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1 $

 $\therefore$    $\cos^{2}\frac{\pi}{4}+\cos^{2}\frac{\pi}{3}+\cos^{2}\gamma=1 $

$\Rightarrow \left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\cos^{2}\gamma=1$

$\Rightarrow \frac{1}{2}+\frac{1}{4}+\cos^{2}\gamma=1$

$\Rightarrow \cos^{2}\gamma=\frac{1}{4}$

$\Rightarrow \cos^{}\gamma=\frac{1}{\sqrt{2}}$

$\therefore$      $\gamma=\frac{\pi}{4}$

Hence , direction  cosines are  $\cos\alpha,\cos\beta,\cos\gamma$   

 i.e,  $\frac{1}{\sqrt{2}},\frac{1}{2},\frac{1}{\sqrt{2}}$

 A tetrahedron  has vertices at  O  (0,0,0) , A(1,-2,1) ,B(-2,1,1)  and C  (1,-1,2). Then the angle between the faces OAB and ABC will be

Answer:

Explanation:

 Vector perpendicular  to face OAB=  $\overrightarrow{n_{1}}$

= $\overrightarrow{OA}\times\overrightarrow{OB}$

 =  $(\hat{i}-2\hat{j}+\hat{k})\times (-2\hat{i}+\hat{j}+\hat{k})$

=$(-2-1)\hat{i}-(-2-1)\hat{j}+(1-4)\hat{k})$

= $-3\hat{i}-3\hat{j}-3\hat{k}$

 Vector perpendicular to face ABC=  $\overrightarrow{n_{2}}$.

= $\overrightarrow{AB}\times\overrightarrow{AC}$

= $(-3\hat{i}+3\hat{j})\times  (\hat{j}+\hat{k})$

=   $-3\hat{i}+3\hat{j}-3\hat{k}$

 Since,  angle between faces is equal to angle between their normals.

$\therefore$  $\cos\theta=\frac{\overrightarrow{n_{1}}.\overrightarrow{n_{2}}}{|\overrightarrow{n_{1}}||\overrightarrow{n_{2}}|}$

 $= \frac{(-3)(3)+(-3)(3)+(-3)(-3)}{\sqrt{9+9+9}\sqrt{9+9+9}}$

$= \frac{-9-9+9}{\sqrt{27}\sqrt{27}}=-\frac{1}{3}$

$\Rightarrow\theta=\cos^{-1}(\frac{-1}{3})$

The two liines x=my+n, z=py+q  and x=m'y+n', z=p'y+q' are perpendicular to each other, if

Answer:

Explanation:

Given lines are

 x=my+n,z=py+q   and x=m'y+n'. z=p'y+q'

 Above equations can be rewritten as 

$\frac{x-n}{m}=\frac{y-0}{1}=\frac{z-q}{p}$  and  

 $\frac{x-n'}{m}=\frac{y-0}{1}=\frac{z-q'}{p'}$

Lines will be perpendicular , if

 mm'+1+pp'=0

$\Rightarrow$     mm'+pp'=-1

• Mathematics - General questions