alpha in N के लिए, N पर एक संबंध R, R ={( x , | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

For $R$ to be reflexive $\Rightarrow xRx$

$\Rightarrow 3 x +\alpha x =7 x \Rightarrow(3+\alpha) x =7 K$

$\Rightarrow 3+\alpha=7 \lambda \Rightar

Vedclass · Accepted Answer

$\alpha$ को $7$ से विभाजित करने पर शेषफल $4$ है

Vedclass · Answer

For $R$ to be reflexive $\Rightarrow xRx$

$\Rightarrow 3 x +\alpha x =7 x \Rightarrow(3+\alpha) x =7 K$

$\Rightarrow 3+\alpha=7 \lambda \Rightarrow \alpha=7 \lambda-3=7 N +4, K , \lambda, N \in I$

$	herefore$ when $\alpha$ divided by $7$, remainder is $4$.

$R$ to be symmetric $x R y \Rightarrow y R x$

$3 x +\alpha y =7 N _{1}, 3 y +\alpha x =7 N _{2}$

$\Rightarrow(3+\alpha)( x + y )=7\left( N _{1}+ N _{2}ight)=7 N _{3}$

Which holds when $3+\alpha$ is multiple of $7$

$	herefore \alpha=7 N +4 	ext { (as did earlier) }$

$R$ to be transitive

$xRy \& yRz \Rightarrow xRz$.

$3 x +\alpha y =7 N _{1} \& 3 y +\alpha z =7 N _{2}$

$3 x +\alpha z =7 N _{3}$

$	herefore 3 x +7 N _{2}-3 y =7 N _{3}$

$	herefore 7 N _{1}-\alpha y +7 N _{2}-3 y =7 N _{3}$

$	herefore 7\left( N _{1}+ N _{2}ight)-(3+\alpha) y =7 N _{3}$

$	herefore(3+\alpha) y =7\,N$

Which is true again when $3+\alpha$ divisible by $7$ when $\alpha$ divided by $7$, remainder is $4$ .

Similar Questions

$A $ के घात समुच्चय $P(A) $ पर संबंध “का उपसमुच्चय है” है

माना $ R$ समुच्चय $A$ पर संबंध इस प्रकार है कि $R = {R^{ - 1}}$ तब $R $ है