VITEEE 2016 :: Mathematics :: General questions

• Mathematics - General questions

 The moment about the point  $\hat{i}+2\hat{j}+3\hat{k}$ of a force represented by  $\hat{i}+\hat{j}+\hat{k}$ acting through the point  $2i+3j+k$  is 

Answer:

Explanation:

 Here,   $r= (2\hat{i}+3\hat{j}+\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})$

 $\Rightarrow$    $ r=\hat{ i}+\hat{j}-2\hat{k}$   and

$ F=\hat{ i}+\hat{j}+\hat{k}$

 Then, the required moment is given by

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

$( \hat{i}+\hat{j}+\hat{k})$

 $\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\1 & 1 &-2\\1&1&1 \end{bmatrix}=3\hat{i}-3\hat{j}$

$\therefore$ moment about given point = $3\hat{i}-3\hat{j}$

The value of   $(1+\omega-\omega^{2})^{7}$

Answer:

Explanation:

$(1+\omega-\omega^{2})^{7}=(-\omega^{2}-\omega^{2})^{7}=(-2\omega^{2})^{7}$

 $=-128(\omega^{4})^{3}\omega^{2}=-128\omega^{2}$

 If A and B are matrices and B= ABA^-1 then the value of (A+B) (A-B) is 

Answer:

Explanation:

 B= ABA^-1(Given)

 But B= BAA^-A

$\therefore$   $ABA^{-1}=BAA^{-1}\Rightarrow AB=BA$

 Now   $(A+B)(A-B)=A^{2}-AB+BA-B^{2}$

$A^{2}-AB+BA-B^{2}[\because AB=BA]$

   = $A^{2}-B^{2}$

If f(x) = x², g(x) = 2x,  $0\leq x \leq2$ then the value of  $I(x)= \int_{0}^{2} max(f(x),g(x) ) $ is 

Answer:

Explanation:

Let r(x) = f(x).g(x)

 = x^2..2x= 2 x³

 r'(x) =6x²

 Put 6 x²  =0, $\therefore$   x=0

 Max r(x) =2 (2)³ =16

 or Max (f(x) ,g(x) )=16

 $I(x)=\int_{0}^{2} 16 dx$

$I(x)=[ 16x ]_0^2=32-0=32$

If  $|\overrightarrow{a}|=3 |\overrightarrow{b}|=2, |\overrightarrow{c}|=1$ then the value of $|\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}|$ is (given that  $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ )

Answer:

Explanation:

$\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$

 $\Rightarrow(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})^{2}=0$

 $\Rightarrow|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}+|\overrightarrow{c}|^{2}+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$

$\Rightarrow9+4+1+2(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a})=0$

 $\Rightarrow\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=7$

• Mathematics - General questions