Maths

It is not possible to pursue PhD in Mathematics from IIIT Guwahati with 50% marks because minimum percentage requirement is 60%. Even if you are a reserved category candidate, you get a relaxation of 5%. For improving your chances of admission, try to score 60% or more in post-graduation.

For calculating the percentage of marks in which multiple subjects are involved sum up all marks of all subjects. Then sum up all total marks in respective subjects. In next step divide total obtained marks in all subjects by total maximum marks in all subjects then multiply the result by 100. Here's the formula- (Total marks in all subjects/Total maximum marks for these subjects)x100

The process to calculating the percentage of marks of 6 subjects is pretty simple and the formula is-

(Total marks in 6 subjects/Total maximum marks for these 6 subjects)x100

For calculating percentage of marks on calculator divide total marks that you scored by total maximum marks (M.M.) then multiply result by 100. For eg. if you scored 362 marks out of 500 - divide 362 by 500 on calculator and multiply result by 100. This gives 72.4% marks. 

  $\overrightarrow{a}+5\overrightarrow{b}$  is collinear with $\overrightarrow{c}$  

⇒   $\overrightarrow{a}+5\overrightarrow{b}$ = $\overrightarrow{c}$           …(1)

$\overrightarrow{b}+6\overrightarrow{c}$ is collinear with $\overrightarrow{a}$  

⇒   $\overrightarrow{b}+6\overrightarrow{c}=\mu \overrightarrow{a}$               …(2)

From (1) and (2)

  $\overrightarrow{b}+6\overrightarrow{c}=\mu \left(\lambda \overrightarrow{c}-5\overrightarrow{b}\right)$          

-> $\left(1+5\mu \right)\overrightarrow{b}+\left(6-\lambda \mu \right)\overrightarrow{c}=0$

$?\overrightarrow{b}$ and $\overrightarrow{c}$  are non-collinear

-> 1 + 5m = 0 $\mu =\frac{-1}{5}$  and 6 – lm = 0 Þ lm = 6

-> l = – 30

Now,

$\overrightarrow{b}=6\overrightarrow{c}=\frac{-1}{5}\overrightarrow{a}$

$5\overrightarrow{b}+30\overrightarrow{c}=-\overrightarrow{a}$

$\begin{array}{cc}\hfill \overrightarrow{a}+5\overrightarrow{b}+30\overrightarrow{c}=0\hfill & \hfill \overrightarrow{a}+\alpha \overrightarrow{b}+\beta \overrightarrow{c}=0\hfill \end{array}]$

On comparing

α = 5, β = 30 ⇒ α + β = 35

  $\left|\frac{1}{2}-2i+1\right|=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$  

$\sqrt[]{\frac{9}{4}+4}=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$

$\frac{5}{2}=\alpha \left(\frac{1}{2}\right)+\beta +i\left(-2\alpha +\beta \right)$             

$\frac{\alpha }{2}+\beta =\frac{5}{2}$      .(1)

 –2α + β = 0                    …(2)

Solving (1) and (2)

$\frac{\alpha }{2}+2\alpha =\frac{5}{2}$

$\frac{5}{2}\alpha =\frac{5}{2}$            

a = 1

b = 2

-> a + b = 3

Variance = $\frac{\sum x^2}{n}-{\left(\overline{x}\right)}^2$  

$\frac{60^2+60^2+44^2+58^2+68^2+{\alpha }^2+{\beta }^2+56^2}{8}={\left(58\right)}^2=66.2$            

$\frac{7200+1936+3364+4624+3136+{\alpha }^2+{\beta }^2}{8}=3364=66.2$             

$2532.5+\frac{{\alpha }^2+{\beta }^2}{8}-3364=66.2$            

α² + β² = 897.7 * 8

= 7181.6

y (x) = 2^x – x²

y? (x) = 2x log 2 – 2x

M = 3

N = 2

M + N = 5