Question

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

Answer 1

We are given two relations

(a)

Relation (R)

$R=[((k-8.3+2.4k),-5),(-\frac{3}{4}k,4)]$

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

$k-8.3+2.4k=-\frac{3}{4} k$

$3.4k-8.3=-\frac{3}{4}k$

$3.4k\cdot \:10-8.3\cdot \:10=-\frac{3}{4}k\cdot \:10$

$4k-83=-\frac{15}{2}k$

$34k=-\frac{15}{2}k+83$

$\frac{83}{2}k=83$

$83k=166$

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

(b)

S = {(2−|k+1| , 4), (−6, 7)}

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