VITEEE 2016 :: Mathematics :: General questions

• Mathematics - General questions

If  $f(a+b-x) =f(x)$ then   $\int_{a}^{b}  x f(x) dx$ is  equal to 

Answer:

Explanation:

Let I=  $\int_{a}^{b}  x f(x) dx$

 Let a+b-x= z $\Rightarrow$ -dx=dz

 when x=a,z=b and when x=b,z=a

$\therefore$    $I= -\int_{b}^{a} (a+b-z)f(z) dz$

 $I= (a+b)\int_{b}^{a} f(x) dx-\int_{a}^{b} x f(x) dx$

 $I= (a+b)\int_{b}^{a} f(x) dx-I;$

$2I= (a+b)\int_{a}^{b} f(x) dx$

 Hence , I=  $\left(\frac{a+b}{2}\right)\int_{a}^{b} f(x) dx$

The number of surjective  functions from A to B where A= {1,2,3,4} and B= {a,b} is 

Answer:

Explanation:

 If  A and B are two sets having m and n elements such that

 $1\leq n \leq m=\sum_{r=1}^n (-1)^{n-r}$  $^{n}C_{r}r^{m}$

 Number of surjection from A to B

 $\sum_{r=1}^n(-1)^{2-r}$   $^{2}a_{r}(r)^{4}$

 =  $(-1)^{2-1} .{^{2}C_{1}}(1)^{4}+(-1)^{2-2}. {^{2}C_{2}}(2)^{4}=-2+16$

   =14

The domain and range of the function f given by f(x)= 2-|x-5| is

Answer:

Explanation:

Given  f(x)=2-|x-5|

 Domain of f(x)  is defined for all real values  of x

 since ,|x-5|  $\geq 0\Rightarrow-|x-5|\leq0$

$\Rightarrow2-|x-5|\leq2\Rightarrow f(x)\leq 2$

 Hence , range of f(x)  is (-$\infty$,2]

The shortest distance between  the lines x=y+2=6z-6 and x+1=2y=-12z is 

Answer:

Explanation:

The line are   $\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}$

 and   $\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}$

 Here,   $\overrightarrow{a_{1}}=-2\hat{j}+\hat{k},\overrightarrow{b_{1}}=6\hat{i}+6\hat{j}+\hat{k},\overrightarrow{a_{2}}=-\hat{i}$

   $\overrightarrow{b_{2}}=12\hat{i}+6\hat{j}-\hat{k}$

 $\overrightarrow{b_{1}}\times \overrightarrow{b_{2}}=\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\6& 6&1\\12&6&-1 \end{bmatrix}=-12\hat{i}+18\hat{j}-36\hat{k}$.

 Shortest distance =   $\frac{|(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}).(\overrightarrow{b_{1}}-\overrightarrow{b_{2}})|}{|\overrightarrow{b_{1}}\times\overrightarrow{b_{2}}|}$

=$\frac{|(-\hat{i}+2\hat{j}-\hat{k}).(-12\hat{i}+18\hat{j}-36\hat{k})|}{\sqrt{(-12)^{2}+(18)^{2}+(-36)^{2}}}$

 =  $\frac{|12+36+36|}{\sqrt{1764}}=\frac{84}{42}=2$

If the letters of the word KRISNA are arranged in all possible  ways  and these words are written out as in a dictionary, then the rank of the work 'KRISNA' is 

Answer:

Explanation:

 The number of words starting from A are  5!=120

The number of words starting from I are  5!=120

The number of words starting from KA are  4!=24

The number of words starting from KI are  4!=24

The number of words starting from KN are  4!=24

The number of words starting from KRA are  3!=6

The number of words starting from KRIA are  2!=2

The number of words starting from KRIN are  2!=2

The number of words starting from KRISA are  1!=1

The number of words starting from KRISNA are  1!=1

Hence , rank of word KRISNA= 2(120)+3(24)+6+2(2)+2(1)=324

• Mathematics - General questions