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For what value(s) of k will the relation not be a function

R={(k-8.3+2.4k,-5),(-(3)/(4)k,4)}
S = (2−
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1 Answer

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We are given two relations

(a)

Relation (R)


R=[((k-8.3+2.4k),-5),(-(3)/(4)k,4)]

We know that

any relation can not be function when their inputs are same

so, we can set both x-values equal

and then we can solve for k


k-8.3+2.4k=-(3)/(4) k


3.4k-8.3=-(3)/(4)k


3.4k\cdot \:10-8.3\cdot \:10=-(3)/(4)k\cdot \:10


4k-83=-(15)/(2)k


34k=-(15)/(2)k+83


(83)/(2)k=83


83k=166


k=2............Answer

(b)

S = (2−

We know that

any relation can not be function when their inputs are same

so, we can set both x-values equal

and then we can solve for k


2-|k+1|=-6


2-\left|k+1\right|-2=-6-2


-\left|k+1\right|=-8


\left|k+1\right|=8

Since, this is absolute function

so, we can break it into two parts


|f\left(k\right)|=a\quad \Rightarrow \:f\left(k\right)=-a\quad \mathrm{or}\quad \:f\left(k\right)=a


k+1=-8\quad \quad \mathrm{or}\quad \:\quad \:k+1=8

we get


k+1=-8\quad


k=-9


k+1=8\quad


k=7

so,


k=-9\quad \mathrm{or}\quad \:k=7...............Answer

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