Question

if f={(1,2),(2,-3),(3,-1)}then find (i)2f(ii)2+f(iii) f square (iv)√f​

Answer 1

Answer:

Given that

f = {(1, 2), (2, -3), (3, -1)}

According to it , we have

f(1) = 2 , f(2) = - 3 , f(3) = - 1

i. 2f

Solution: 2f(1) = 2(2) = 4 , 2f(2) = 2(- 3) = -6 , 2f(3) = 2(- 1) = -2

ii. 2 + f

Solution: 2 + f(1) = 2 + 2 = 4 , 2 + f(2) = 2 + ( - 3 ) = 2 - 3 = - 1 ,

2 + f(3) = 2 + (- 1) = 2 - 1 = 1

iii. f²

Solution: f^2(1) = 2^2 = 4 , f^2(2) = (- 3)^2 = 9 , f^2(3) = (- 1)^2 = 1

iv. √f

Solution:

\begin{gathered}\sqrt{f(1)} = \sqrt{2} \\\sqrt{f(2)} = \sqrt{-3} \\\sqrt{f(3)} =\sqrt{-1}\end{gathered}

f(1)

=

2

f(2)

=

−3

f(3)

=

−1

Answer 2

Given:

The $f={(1,2),(2,-3),(3,-1)}$.

To find:

Find the $(1).2f (2).2+f (3). f^{2}(4).\sqrt{f}$.

Step-by-step explanation:

(1).$2f$

$f(1)=2\\f(2)=-3\\f(3)=-1$

$2f(1)=2*2=4\\2f(2)=2*-3=-6\\2f(3)=2*-1=-2$

(2).$2+f$

$2+2=4\\2-3=-1\\2-1=1$

(3).$f^{2}$

$2^{2}=4\\(-3)^{2}=9\\(-1)^{2}=1$

(4)$\sqrt{f}$

$(1,\sqrt{2}).$

Answer:

Therefore, ${(1,4),(2,-6),(3,-2)},{(1,4),(2.-1),(3,1)},{(1,4),(2,9),(3,1)},({1,\sqrt{2})}$.