Question

If f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\} and g = \left\\{ {(4, - 4),(6,5),(8,5)} \right\\} then find
a.f + g
b.f – g
c.2f + 4g
d.f + 4
e. fg
f.$f/g$
g.$\left| f \right|$
h.$\sqrt f$
i.${f^2}$
j.${f^3}$

Answer 1

We are given two two sets with ordered pairs and if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f + g is given as f + g = \left\\{ {(a + c,b + d)} \right\\}, if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f - g is given as f - g = \left\\{ {(a - c,b - d)} \right\\}, if we have function f = \left\\{ {(a,b)} \right\\} then kf is given as kf = \left\\{ {(ka,kb)} \right\\} , if we have function f = \left\\{ {(a,b)} \right\\} then f + k is given as f + k = \left\\{ {(a + k,b + k)} \right\\} , if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f g is given as fg = \left\\{ {(ac,bd)} \right\\}, if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f g is given as f/g = \left\\{ {(a/c,b/d)} \right\\}, if we have function f = \left\\{ {(a,b)} \right\\} then $\left| f \right|$ is given as \left| f \right| = \left\\{ {(\left| a \right|,\left| b \right|)} \right\\}, if we have function f = \left\\{ {(a,b)} \right\\} then $\sqrt f$ is given as \sqrt f = \left\\{ {(\sqrt a ,\sqrt b )} \right\\}, if we have function f = \left\\{ {(a,b)} \right\\} then ${f^3}$ is given as {f^3} = \left\\{ {({a^3},{b^3})} \right\\} Using these we have find the given values.

Complete step-by-step answer:
We are given two two sets with ordered pairs
a.f + g
if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f + g is given as f + g = \left\\{ {(a + c,b + d)} \right\\}
therefore we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\} and g = \left\\{ {(4, - 4),(6,5),(8,5)} \right\\}
so
\Rightarrow f + g = \left\\{ {(4 + 4,5 - 4),(5 + 6,6 + 5),(6 + 8, - 4 + 5)} \right\\} \\\ \Rightarrow f + g = \left\\{ {(8,1),(11,11),(14,1)} \right\\} \\\ \\\
b.f – g
if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f - g is given as f - g = \left\\{ {(a - c,b - d)} \right\\}
therefore we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\} and g = \left\\{ {(4, - 4),(6,5),(8,5)} \right\\}
so
\Rightarrow f - g = \left\\{ {(4 -4,5 + 5),(5 - 6,6 - 5),(6 - 8, - 4 - 5)} \right\\} \\\ \Rightarrow f - g = \left\\{ {(0,10),( - 1,1),(-2, - 9)} \right\\} \\\
c.2f + 4g
if we have function f = \left\\{ {(a,b)} \right\\} then kf is given as kf = \left\\{ {(ka,kb)} \right\\}
therefore we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\} and g = \left\\{ {(4, - 4),(6,5),(8,5)} \right\\}
so
\Rightarrow 2f = \left\\{ {(2(4),2(5)),(2(5),2(6)),(2(6),2( - 4))} \right\\} \\\ \Rightarrow 2f = \left\\{ {(8,10),(10,12),(12, - 8)} \right\\} \\\
Same way
\Rightarrow 4g = \left\\{ {(4(4),4( - 4)),(4(6),4(5)),(4(8),4(5))} \right\\} \\\ \Rightarrow 4g = \left\\{ {(16, - 16),(24,20),(32,20)} \right\\} \\\
And we know that if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f + g is given as f + g = \left\\{ {(a + c,b + d)} \right\\}
\Rightarrow 2f + 4g = \left\\{ {(8 + 16,10 - 16),(10 + 24,12 + 20),(12 + 32, - 8 + 20)} \right\\} \\\ \Rightarrow 2f + 4g = \left\\{ {(24, - 6),(34,32),(44,12)} \right\\} \\\ \\\
d.f + 4
if we have function f = \left\\{ {(a,b)} \right\\} then f + k is given as f + k = \left\\{ {(a + k,b + k)} \right\\}
we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\}
hence,
\Rightarrow f + 4 = \left\\{ {(4 + 4,5 + 4),(5 + 4,6 + 4),(6 + 4, - 4 + 4)} \right\\} \\\ \Rightarrow f + 4 = \left\\{ {(8,9),(9,10),(10,0)} \right\\} \\\ \\\

e.fg
if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f g is given as fg = \left\\{ {(ac,bd)} \right\\}
therefore we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\} and g = \left\\{ {(4, - 4),(6,5),(8,5)} \right\\}
so
\Rightarrow fg = \left\\{ {((4*4),(5* - 4)),((5*6),(6*5)),((6*8),( - 4*5))} \right\\} \\\ \Rightarrow fg = \left\\{ {(16, - 20),(30,30),(48, - 20)} \right\\} \\\
f.$f/g$
if we have functions f = \left\\{ {(a,b)} \right\\} and g = \left\\{ {(c,d)} \right\\}then f g is given as f/g = \left\\{ {(a/c,b/d)} \right\\}
therefore we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\} and g = \left\\{ {(4, - 4),(6,5),(8,5)} \right\\}
so
\Rightarrow f/g = \left\\{ {((4/4),(5/ - 4)),((5/6),(6/5)),((6/8),( - 4/5))} \right\\} \\\ \Rightarrow f/g = \left\\{ {(1, - \dfrac{5}{4}),(\dfrac{5}{6},\dfrac{6}{5}),(\dfrac{3}{4}, - \dfrac{4}{5})} \right\\} \\\
if we have function f = \left\\{ {(a,b)} \right\\} then $\left| f \right|$ is given as \left| f \right| = \left\\{ {(\left| a \right|,\left| b \right|)} \right\\}
we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\}
hence,

\Rightarrow \left| f \right| = \left\\{ {(\left| 4 \right|,\left| 5 \right|),(\left| 5 \right|,\left| 6 \right|),(\left| 6 \right|,\left| { - 4} \right|)} \right\\} \\\ \Rightarrow \left| f \right| = \left\\{ {(4,5),(5,6),(6,4)} \right\\} \\\ \\\

$\sqrt f$
if we have function f = \left\\{ {(a,b)} \right\\} then $\sqrt f$ is given as \sqrt f = \left\\{ {(\sqrt a ,\sqrt b )} \right\\}
we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\}
hence,

\Rightarrow \sqrt f = \left\\{ {(\sqrt 4 ,\sqrt 5 ),(\sqrt 5 ,\sqrt 6 ),(\sqrt 6 ,\sqrt { - 4} )} \right\\} \\\ \Rightarrow \sqrt f = \left\\{ {(2,\sqrt 5 ),(\sqrt 5 ,\sqrt 6 ),(\sqrt 6 ,\sqrt { - 4} )} \right\\} \\\ \\\

g.${f^2}$
if we have function f = \left\\{ {(a,b)} \right\\} then ${f^2}$ is given as {f^2} = \left\\{ {({a^2},{b^2})} \right\\}
we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\}
hence,

\Rightarrow {f^2} = \left\\{ {({4^2},{5^2}),({5^2},{6^2}),({6^2}, - {4^2})} \right\\} \\\ \Rightarrow {f^2} = \left\\{ {(16,25),(25,36),(36,16)} \right\\} \\\ \\\

h.${f^3}$
if we have function f = \left\\{ {(a,b)} \right\\} then ${f^3}$ is given as {f^3} = \left\\{ {({a^3},{b^3})} \right\\}
we are given f = \left\\{ {(4,5),(5,6),(6, - 4)} \right\\}
hence,

\Rightarrow {f^3} = \left\\{ {({4^3},{5^3}),({5^3},{6^3}),({6^3}, - {4^3})} \right\\} \\\ \Rightarrow {f^3} = \left\\{ {(64,125),(125,216),(216, - 64)} \right\\} \\\ \\\

Note: The operations (addition, subtraction, division, multiplication, etc.) can be generalised as a binary operation is performed on two elements (say a and b) from set X. The result of the operation on a and b is another element from the same set X.
Thus, the binary operation can be defined as an operation * which is performed on a set A. The function is given by *: A * A → A. So the operation * performed on operands a and b is denoted by a * b.