Relations and Functions

$\frac{1}{7}{\displaystyle \sum _{i=1}^7{\left(x_i-62\right)}^2}=20$ $\overline{x}=62$      

${\displaystyle \sum _{i=1}^7{\left(x_i-62\right)}^2}$ = 140……………… (i)

 If any one student get less 50 marks then   ${\left(x_i-62\right)}^2\ge 144$

but    ${\displaystyle \sum _{i=1}^7{\left(x_i-62\right)}^2}=140$

$\therefore$ both condition cannot satisfy together hence no students fails.

Total number of possible relation = $2^{n^2}=2^4=16$

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability = $\frac{5}{16}$

Boys (10)            Girls (5)

(3)                        (3)

B₁ & B₂ should not be selected together

Total number of ways

           

= (56 + 56) * 10 = 1120

  $?p+q=3$ $$ ….(i)

and $p^4+q^4$ ${}^{}{}^{}$ = 369 ….(ii)

${\left\{{\left(p+q\right)}^2-2pq\right\}}^2-2p^2{q}^2=369$

or ${\left(9-2pq\right)}^2-2{\left(pq\right)}^2=369$ ${}^{}{}^{}$

or ${\left(pq\right)}^2-18pq-144=0$ ${}^{}$

$\therefore pq=-6or24$

But pq 24 is not possible

$\therefore pq=-6$

$Hence,{\left(\frac{1}{p}+\frac{1}{q}\right)}^{-2}={\left(\frac{pq}{p+q}\right)}^2={\left(-2\right)}^2=4$

Use aRb = a is related to b, belongs to A iff a belongs to A.

In simple terms, aRb is true if both a & b belongs to the same set.

For reflexive

aRa, a $\in$ A, so it is true.

For symmetric

Let aRb be true

Þ a & b belongs to the same set.

Þ b & a also belongs to the same set

Þ bRa will be true

For transitive

Let aRb and bRc be true.

aRb Þ a, b belongs to the same set

bRc Þ b, c belongs to the same set

Þ (a, c) belongs to the same set

Þ so aRc will be true.

So R is an equivalence relation.

(D)

$\sim \left(p\iff \sim q\right)\wedge q=\left(p\iff \sim q\right)\wedge qis:$

$\therefore \left(\sim \left(P\iff \sim Q\right)\right)\wedge$ q is equivalent to $\left(p\Rightarrow q\right)\wedge$ p

$p\iff \left(q\Rightarrow p\right)\sim \left(p\iff \left(q\iff p\right)=p\iff \sim \left(q\Rightarrow p\right)\right)$

$=\sim p\wedge \left(\sim q\vee p\right)$

$=\left(\sim p\wedge \sim q\right)$

 $f\left(x\right)=\{\begin{array}{cc}\hfill x+a,\hfill & \hfill x\le 0\hfill \\ \hfill \left|x-4\right|,\hfill & \hfill x>0\hfill \end{array}andg\left(x\right)=\{\begin{array}{cc}\hfill x+1,\hfill & \hfill x<0\hfill \\ \hfill {\left(x-4\right)}^2+b,\hfill & \hfill x\ge 0\hfill \end{array}$

$?$ f (x) and g (x) are continuous on R $\therefore$ a = 4 and b = 1 – 16 = 15

then (gof) (2) + (fog) (2) = g (2) + f (-1) = -11 + 3 = -8

Consider the following image

For x < 0 0 < e^x < 1 [e^x] = 0

$0\le x<1a{e}^x+\left[x-1\right]$               

$=a{e}^x+\left[x\right]-1$               

= a e^x – 1             b + [sin px]

$\therefore f\left(x\right)=\left[\begin{array}{l}0x<0\\ a{e}^x-10\le x<1\\ b-11\le x<2\\ -cx\ge 2\end{array}\right]$               

For f to be continuous at x = 0

a – 1 = 0 ⇒ a = 1

$\begin{array}{l}\therefore a+b+c=1+e+1-e=2\\ a+b+c\ne 1\end{array}$